<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/' xmlns:georss='http://www.georss.org/georss' xmlns:thr='http://purl.org/syndication/thread/1.0' xmlns:gd='http://schemas.google.com/g/2005' gd:etag='W/&quot;DkICQ3k5cSp7ImA9WxFbFkw.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171</id><updated>2010-07-08T11:42:42.729-07:00</updated><title>Adriano G. S.</title><subtitle type='html'></subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default?redirect=false&amp;v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><link rel='next' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default?start-index=26&amp;max-results=25&amp;redirect=false&amp;v=2'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>53</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>25</openSearch:itemsPerPage><entry gd:etag='W/&quot;DkICQ3Y9eip7ImA9WxFbFkw.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-6548215294916625247</id><published>2010-07-08T11:38:00.000-07:00</published><updated>2010-07-08T11:42:42.862-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T11:42:42.862-07:00</app:edited><title>Simulado de Matemática do Ensino Médio (Parte 2)</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 11:&lt;/span&gt; O valor de um determinado automóvel diminui com o passar do tempo, como mostra o gráfico. Decorridos quantos anos esse carro passará a não ter valor algum?&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s1600/ques+11.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 392px; height: 386px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s400/ques+11.jpg" alt="" id="BLOGGER_PHOTO_ID_5487034773099396386" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-11.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 12:&lt;/span&gt; Analise a alternativa que apresenta uma função crescente.&lt;table&gt;&lt;br /&gt;&lt;tbody&gt;&lt;tr&gt;&lt;br /&gt;&lt;th style="text-align: left;"&gt;(A)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXijdrefdI/AAAAAAAAANk/St72zqPzSMs/s1600/ques12-1.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 259px; height: 312px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXijdrefdI/AAAAAAAAANk/St72zqPzSMs/s320/ques12-1.jpg" alt="" id="BLOGGER_PHOTO_ID_5487040819911097810" border="0" /&gt;&lt;/a&gt;&lt;/th&gt;&lt;br /&gt;&lt;th style="text-align: left;"&gt;(B)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXiqwfz7TI/AAAAAAAAANs/qCidU2Y4eqs/s1600/ques12-2.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 320px; height: 314px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXiqwfz7TI/AAAAAAAAANs/qCidU2Y4eqs/s320/ques12-2.jpg" alt="" id="BLOGGER_PHOTO_ID_5487040945221528882" border="0" /&gt;&lt;/a&gt;&lt;/th&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;tr style="font-weight: bold;"&gt;&lt;br /&gt;&lt;td&gt;(C)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXizEisd9I/AAAAAAAAAN0/rsjnYneTWtQ/s1600/ques12-3.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 286px; height: 233px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXizEisd9I/AAAAAAAAAN0/rsjnYneTWtQ/s320/ques12-3.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041088041285586" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;td&gt;(D)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXi5SOLssI/AAAAAAAAAN8/YX2MHYmyv6w/s1600/ques12-4.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 316px; height: 316px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXi5SOLssI/AAAAAAAAAN8/YX2MHYmyv6w/s320/ques12-4.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041194792563394" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;tr&gt;&lt;br /&gt;&lt;td&gt;&lt;span style="font-weight: bold;"&gt;(E)&lt;/span&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXjAGaCU3I/AAAAAAAAAOE/fBvXcF49jr8/s1600/ques12-5.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 320px; height: 215px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXjAGaCU3I/AAAAAAAAAOE/fBvXcF49jr8/s320/ques12-5.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041311880139634" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;td&gt;&lt;br /&gt;&lt;/td&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;/tbody&gt;&lt;/table&gt;&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-12.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 13:&lt;/span&gt; A discussão sobre número de raízes distintas de uma equação do 2º grau é comumente feita por meio do dicriminante da equação. Para o caso da equação &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x%5E2-px+q%5E2=0,%20%28p%3E0,%20q%3E0%29" title="" 2="0," /&gt;, isso pode ser feito geometricamente, como mostra a figura.&lt;br /&gt;Nele, o arco é uma semicircunferência de diâmetro &lt;span style="font-weight: bold;"&gt;AB&lt;/span&gt;, com &lt;span style="font-weight: bold;"&gt;AB=p&lt;/span&gt; e&lt;span style="font-weight: bold;"&gt; CD=EF=q&lt;/span&gt;.&lt;br /&gt;As raízes r e s da equação são representadas pelos seguimentos &lt;span style="font-weight: bold;"&gt;AF&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;BF&lt;/span&gt;, respectivamente.&lt;br /&gt;De fato, &lt;span style="font-weight: bold;"&gt;r+s=p&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;rs=q²&lt;/span&gt;, uma vez que o triângulo &lt;span style="font-weight: bold;"&gt;AEB&lt;/span&gt; é retângulo e &lt;span style="font-weight: bold;"&gt;EF&lt;/span&gt; a altura relativa à hipotenusa.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXn4xndpOI/AAAAAAAAAOU/OQmhQzPq8hw/s1600/quest13.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 400px; height: 136px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXn4xndpOI/AAAAAAAAAOU/OQmhQzPq8hw/s400/quest13.jpg" alt="" id="BLOGGER_PHOTO_ID_5487046683598365922" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;a) a partir da construção acima, conclua qual é a relação entre &lt;span style="font-weight: bold;"&gt;r&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;s&lt;/span&gt;, caso em que &lt;span style="font-weight: bold;"&gt;q=p/2&lt;/span&gt;.&lt;br /&gt;b) Calcule o valor do discriminante da equação para &lt;span style="font-weight: bold;"&gt;p=q/2&lt;/span&gt; e compare o que você concluiu com o observado em a).&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/questao-13-discussao-sobre-numero-de.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 14:&lt;/span&gt; O brasil, em 1997, com cerca de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?160%5Ctimes10%5E6" title="160\times10^6" alt="160\times10^6" style="vertical-align: -2pt;" /&gt; habitantes, apresentou um consumo de energia da ordem de 250.000 TEP (tonelada equivalente de petróleo), proveniente de diversas fontes primárias.&lt;br /&gt;O grupo com renda familiar de mais de vinte salários mínimos representa 5% da população brasileira e utiliza cerca de 10% da energia total consumida no país.&lt;br /&gt;O grupo com renda familiar de até três salárioas mínimos representa 50% da população e consome 30% do total de energia.&lt;br /&gt;&lt;br /&gt;Com base nessas informações, pode-se concluir que o consumo médio de energia para um indivíduo do grupo de renda é &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt; vezes maior do que para um indivíduo do grupo de renda inferior. Qual o valor aproximado de &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;?&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-14.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 15:&lt;/span&gt; Um lojista comprou de seu fornecedor um artigo por p reais (preço de custo) e revende com lucro de 40%. A seguir, ao fazer uma liquidação, ele dá aos compradores um desconto de 30% sobre o preço de venda desse artigo. Pode-se afirmar que esse comerciante tem, sobre p,&lt;br /&gt;a) prejuízo de 2%&lt;br /&gt;b) prejuízo de 10%&lt;br /&gt;c) lucro de 10%&lt;br /&gt;d) lucro de 12%&lt;br /&gt;e) lucro de 82%&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-15.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;div style="text-align: center;"&gt;&lt;span style="font-weight: bold;"&gt;Use o texto abaixo para responder as questões 16, 17 e 18&lt;/span&gt;&lt;br /&gt;&lt;/div&gt;&lt;br /&gt;No quadro abaixo estão as contas de luz e água de uma mesma residência. Além do valor a pagar, cada conta mostra como calculá-lo, em função do consumo de água (em m³) e de eletricidade (em kwh). Observe que, na conta de luz, o valor a pagar é igual ao consumo multiplicado por um certo fator. Já na conta de água, existe uma tarifa mínima e diferentes faixas de tarifação.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCdVMs_eS2I/AAAAAAAAAOk/0bUSAwpUqtE/s1600/tarifas.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 356px; height: 290px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCdVMs_eS2I/AAAAAAAAAOk/0bUSAwpUqtE/s400/tarifas.jpg" alt="" id="BLOGGER_PHOTO_ID_5487448347698416482" border="0" /&gt;&lt;/a&gt;&lt;hr /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 16:&lt;/span&gt; Suponha que, no próximo mês, dobre o consumo de energia elétrica dessa residência. Qual o novo valor da conta?&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questoes-16-17-e-18.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 17:&lt;/span&gt; Suponha que dobre o consumo d'água. Qual o novo valor da conta?&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questoes-16-17-e-18.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 18:&lt;/span&gt; Dos gráficos abaixo, o que melhor representa o valor da conta de água, de acordo com o consumo é:&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_Z_uiHIsRFf0/TCdVHgN9UNI/AAAAAAAAAOc/rNE3zFOEhfE/s1600/quest+18.JPG"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 320px; height: 223px;" src="http://1.bp.blogspot.com/_Z_uiHIsRFf0/TCdVHgN9UNI/AAAAAAAAAOc/rNE3zFOEhfE/s400/quest+18.JPG" alt="" id="BLOGGER_PHOTO_ID_5487448258370162898" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questoes-16-17-e-18.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 19:&lt;/span&gt; Se um ângulo de 30° é aumentado em 15°, a medida do seno:&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;a)&lt;/span&gt; e do cosseno desse ângulo aumentam.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;b)&lt;/span&gt; e do cosseno desse ângulo não se alteram.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;c) &lt;/span&gt;e do cosseno desse ângulo diminuem.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;d)&lt;/span&gt; diminui e a do cosseno aumenta.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;e)&lt;/span&gt; aumenta e a do cosseno diminui.&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-19.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 20:&lt;/span&gt; O gráfico abaixo representa o sistema:&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCdmkcSUI9I/AAAAAAAAAO0/_SE0rUQrV4Q/s1600/quest+12.jpg"&gt;&lt;img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 400px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCdmkcSUI9I/AAAAAAAAAO0/_SE0rUQrV4Q/s400/quest+12.jpg" alt="" id="BLOGGER_PHOTO_ID_5487467447228572626" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;a) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;6%5C%5Cx-y&amp;amp;=&amp;amp;4%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;b) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;4%5C%5Cx-y&amp;amp;=&amp;amp;6%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;4\\x-y&amp;amp;=&amp;amp;6\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;4\\x-y&amp;amp;=&amp;amp;6\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;c) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;-4%5C%5Cx+y&amp;amp;=&amp;amp;6%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;-4\\x+y&amp;amp;=&amp;amp;6\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;-4\\x+y&amp;amp;=&amp;amp;6\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;d) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx-4y&amp;amp;=&amp;amp;1%5C%5Cx+y&amp;amp;=&amp;amp;5%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x-4y&amp;amp;=&amp;amp;1\\x+y&amp;amp;=&amp;amp;5\end{array}\right." alt="\left\{\begin{array}{rcl}x-4y&amp;amp;=&amp;amp;1\\x+y&amp;amp;=&amp;amp;5\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;e) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx&amp;amp;=&amp;amp;y%5C%5C6x-4y&amp;amp;=&amp;amp;0%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;y\\6x-4y&amp;amp;=&amp;amp;0\end{array}\right." alt="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;y\\6x-4y&amp;amp;=&amp;amp;0\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-20.html"&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/a&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-6548215294916625247?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/6548215294916625247/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/simulado-de-matematica-do-ensino-medio.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6548215294916625247?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6548215294916625247?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/simulado-de-matematica-do-ensino-medio.html' title='Simulado de Matemática do Ensino Médio (Parte 2)'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s72-c/ques+11.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;Ak8DQ387cCp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-5566510532345939553</id><published>2010-07-08T04:32:00.000-07:00</published><updated>2010-07-08T04:34:32.108-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T04:34:32.108-07:00</app:edited><title>SMEM 1 - Questão 35</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 35:&lt;/span&gt; Com o objetivo de discutir a resolução de equações quadráticas com enfase em procedimentos historicamente construídos, por meio de complementação de quadrados, um professor propôs a seguinte tarefa a seus alunos:&lt;br /&gt;&lt;br /&gt;"Determine o comprimento x de um retângulo sabendo que a sua largura é 10 unidades e que, se juntarmos esse retângulo a um quadrado cujo lado tem a mesma medida do comprimento desse retângulo, obtemos uma área de 39 unidades quadradas."&lt;br /&gt;&lt;br /&gt;Para a resolução deste problema, o professor sugeriu que os alunos utilizassem métodos algébricos e geométricos. Após as tentativas de alguns alunos, o professor apresenta as seguintes figuras:&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDW3eH-Q7jI/AAAAAAAAAQk/Ih-W7FfA30Q/s1600/screen+-+5.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 591px; height: 198px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDW3eH-Q7jI/AAAAAAAAAQk/Ih-W7FfA30Q/s400/screen+-+5.jpg" alt="" id="BLOGGER_PHOTO_ID_5491497048812351026" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;Com base nessas informações, considere as afirmativas a seguir.&lt;br /&gt;I - A figura 3 é uma representação geométrica do problema proposto, e a representação algébrica do mesmo é x²+10x=39&lt;br /&gt;II - A área da figura 4 é igual à área da figura 3 (39 unidades quadradas) mais as áreas dos quadrados de lado &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{10}{4}" title="\dfrac{10}{4}" alt="\dfrac{10}{4}" style="vertical-align:-13pt;" /&gt; unidades, totalizando 49 unidades quadradas.&lt;br /&gt;III - O método de completar quadrados em sua versão algebrizada transforma uma equação do 2° grau em uma equação do 1° grau, nesse caso, basta encontrar o quadrado perfeito (x+5)²=64, obtendo x=3.&lt;br /&gt;IV- A equação geral correspondente a este problema é x²+px=q e o método de completar quadrados, sugerido pelo professor, nos leva a seguinte resposta &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x=\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}" title="x=\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}" alt="x=\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Estão corretas as afirmativas:&lt;br /&gt;a) I e III&lt;br /&gt;b) II e III&lt;br /&gt;c) II e IV&lt;br /&gt;d) I, II e IV&lt;br /&gt;e) I, III e IV&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Analisando as afirmações&lt;br /&gt;&lt;br /&gt;I verdadeira: a figura 3 representa o primeiro retângulo cortado em 4 outros retângulos de comprimento x justapostos aos lados do quadrado de lado x. A área do retângulo é 10x, a do quadrado é x², como a soma destas áreas é 39, a expressão que descreve esta situação é &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x^2+10x=39" title="x^2+10x=39" alt="x^2+10x=39" style="vertical-align:-2pt;" /&gt;&lt;br /&gt;&lt;br /&gt;II falsa: a área da figura 4 não totaliza 49 unidades quadradas e sim 64 unidades quadradas.&lt;br /&gt;&lt;br /&gt;III verdadeira: Na versão agebrizada procura-se somar a ambos os membros da equação um número de tal forma a encontrar uma expressão que seja um "trinômio quadrado perfeito", no caso deste equação somamos 25 a ambos os lados.&lt;br /&gt;&lt;br /&gt;IV verdadeira: somando &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{p^2}{4}" title="\dfrac{p^2}{4}" alt="\dfrac{p^2}{4}" style="vertical-align:-13pt;" /&gt; a ambos os membros da equação &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x^2+px=q" title="x^2+px=q" alt="x^2+px=q" style="vertical-align:-2pt;" /&gt; encontramos um trinômio quadrado perfeito no lado esquedo da equação, a saber&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x^2+px+\dfrac{p^2}{4}=\left(x+\dfrac{p}{2}\right)^2" title="x^2+px+\dfrac{p^2}{4}=\left(x+\dfrac{p}{2}\right)^2" alt="x^2+px+\dfrac{p^2}{4}=\left(x+\dfrac{p}{2}\right)^2" style="vertical-align:-13pt;" /&gt;. Resolvendo a equação desta forma temos:&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\begin{array}{rcl}\left(x+\dfrac{p}{2}\right)^2&amp;amp;=&amp;amp;q+\dfrac{p^2}{4}\\x+\dfrac{p}{2}&amp;amp;=&amp;amp; \sqrt{q+\dfrac{p^2}{4}}\\ x&amp;amp;=&amp;amp;\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}\end{array}" title="\begin{array}{rcl}\left(x+\dfrac{p}{2}\right)^2&amp;amp;=&amp;amp;q+\dfrac{p^2}{4}\\x+\dfrac{p}{2}&amp;amp;=&amp;amp; \sqrt{q+\dfrac{p^2}{4}}\\ x&amp;amp;=&amp;amp;\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}\end{array}" alt="\begin{array}{rcl}\left(x+\dfrac{p}{2}\right)^2&amp;amp;=&amp;amp;q+\dfrac{p^2}{4}\\x+\dfrac{p}{2}&amp;amp;=&amp;amp; \sqrt{q+\dfrac{p^2}{4}}\\ x&amp;amp;=&amp;amp;\sqrt{q+\dfrac{p^2}{4}}-\dfrac{p}{2}\end{array}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Logo a alternativa correta é a (E)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-5566510532345939553?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/5566510532345939553/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-35.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5566510532345939553?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5566510532345939553?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-35.html' title='SMEM 1 - Questão 35'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDW3eH-Q7jI/AAAAAAAAAQk/Ih-W7FfA30Q/s72-c/screen+-+5.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;AkQESXs5eyp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-8306159309930000458</id><published>2010-07-08T04:19:00.001-07:00</published><updated>2010-07-08T04:25:08.523-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T04:25:08.523-07:00</app:edited><title>SMEM 1 - Questão 34</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 34:&lt;/span&gt; Os esquemas I e II representam arestas planificadas de poliedros, cujos esqueletos foram construidos com canudos.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDW1FkygyBI/AAAAAAAAAQc/RkBk0lk5v4I/s1600/quest34.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 231px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDW1FkygyBI/AAAAAAAAAQc/RkBk0lk5v4I/s400/quest34.jpg" alt="" id="BLOGGER_PHOTO_ID_5491494428027701266" border="0" /&gt;&lt;/a&gt;Com base nas figuras considere as afirmações:&lt;br /&gt;I - Ambos os esquemas podem representar as arestas de um cubo.&lt;br /&gt;II - Ambos os esquemas podem representar as arestas de um octaedro.&lt;br /&gt;III - O esquema II pode representar as arestas tanto do cubo como do octaedro.&lt;br /&gt;IV - O esquema I pode representar as arestas de uma pirâmide de base quadrangular.&lt;br /&gt;Estão corretas apenas as afirmações contidas nos itens:&lt;br /&gt;a) I e II&lt;br /&gt;b) II e III&lt;br /&gt;c) III e IV&lt;br /&gt;d) I, II, e IV&lt;br /&gt;e) I, III e IV&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Analisando as afirmações&lt;br /&gt;&lt;br /&gt;I falsa: no esquena I há vértices formados pelo encontro de 4 arestas, o que não ocorre no cubo.&lt;br /&gt;II verdadeira: (As linhas mais finas representam como os vértices devem ser ligados)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDW0uuowwcI/AAAAAAAAAQU/muiNvrxk8FU/s1600/screen+-+1.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 221px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDW0uuowwcI/AAAAAAAAAQU/muiNvrxk8FU/s400/screen+-+1.jpg" alt="" id="BLOGGER_PHOTO_ID_5491494035534168514" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDW0h4zBo3I/AAAAAAAAAQM/Qn_nuGx7mwg/s1600/screen+-+2.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 186px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDW0h4zBo3I/AAAAAAAAAQM/Qn_nuGx7mwg/s400/screen+-+2.jpg" alt="" id="BLOGGER_PHOTO_ID_5491493814923273074" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;III verdadeira: Na afirmação II vemos como este esquema repesenta um octaedro, aqui vemos como ele pode representar um cubo&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TDW0aCrHI6I/AAAAAAAAAQE/cwTBhpk2qFY/s1600/screen+-+3.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 292px; height: 272px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TDW0aCrHI6I/AAAAAAAAAQE/cwTBhpk2qFY/s400/screen+-+3.jpg" alt="" id="BLOGGER_PHOTO_ID_5491493680135480226" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TDW0UFHrlTI/AAAAAAAAAP8/BMule95ul0o/s1600/screen+-+4.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 201px; height: 221px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TDW0UFHrlTI/AAAAAAAAAP8/BMule95ul0o/s400/screen+-+4.jpg" alt="" id="BLOGGER_PHOTO_ID_5491493577712964914" border="0" /&gt;&lt;/a&gt;IV falsa: A pirâmide de base quadrada possui 8 vétices no total enquanto o esquema I possui 12.&lt;br /&gt;&lt;br /&gt;Alternativa correta (B).&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-8306159309930000458?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/8306159309930000458/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-34.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8306159309930000458?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8306159309930000458?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-34.html' title='SMEM 1 - Questão 34'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDW1FkygyBI/AAAAAAAAAQc/RkBk0lk5v4I/s72-c/quest34.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DEMNQ387eCp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-8382660874355670328</id><published>2010-07-08T03:51:00.000-07:00</published><updated>2010-07-08T03:54:52.100-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:54:52.100-07:00</app:edited><title>SMEM 1 - Questão 33</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 33:&lt;/span&gt; A equação &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;³-6&lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;²+&lt;span style="font-weight: bold; font-style: italic;"&gt;ax&lt;/span&gt;+&lt;span style="font-weight: bold; font-style: italic;"&gt;b&lt;/span&gt;=0, na qual a e b são números reais, tem uma rais igual a 2+3&lt;span style="font-weight: bold; font-style: italic;"&gt;i&lt;/span&gt;. Assinale a(s) alternativa(s) correta(s).&lt;br /&gt;a) -2+3&lt;span style="font-weight: bold; font-style: italic;"&gt;i&lt;/span&gt; também é raiz da euquação.&lt;br /&gt;b) A equação não possui raízes reais.&lt;br /&gt;c) A soma de todas as raízes da equação é 6.&lt;br /&gt;d) o valor de &lt;span style="font-weight: bold; font-style: italic;"&gt;b&lt;/span&gt; é -26&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Se &lt;span style="font-weight: bold; font-style: italic;"&gt;p&lt;/span&gt;, &lt;span style="font-weight: bold; font-style: italic;"&gt;q&lt;/span&gt; e&lt;span style="font-style: italic; font-weight: bold;"&gt; r&lt;/span&gt; são as raízes de uma equação cúbica podemos escrever esta equação por:&lt;br /&gt;&lt;br /&gt;&lt;div style="text-align: center;"&gt;(&lt;span style="font-style: italic; font-weight: bold;"&gt;x&lt;/span&gt;-&lt;span style="font-style: italic; font-weight: bold;"&gt;p&lt;/span&gt;)(&lt;span style="font-style: italic; font-weight: bold;"&gt;x&lt;/span&gt;-&lt;span style="font-style: italic; font-weight: bold;"&gt;q&lt;/span&gt;)(&lt;span style="font-style: italic; font-weight: bold;"&gt;x&lt;/span&gt;-&lt;span style="font-style: italic; font-weight: bold;"&gt;r&lt;/span&gt;)=0&lt;br /&gt;&lt;/div&gt;&lt;br /&gt;desenvolvendo os produtos temos&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;³-(&lt;span style="font-weight: bold; font-style: italic;"&gt;p&lt;/span&gt;+&lt;span style="font-style: italic; font-weight: bold;"&gt;q&lt;/span&gt;+&lt;span style="font-weight: bold; font-style: italic;"&gt;r&lt;/span&gt;)&lt;span style="font-style: italic; font-weight: bold;"&gt;x&lt;/span&gt;²+&lt;span style="font-weight: bold; font-style: italic;"&gt;pq&lt;/span&gt;+&lt;span style="font-weight: bold; font-style: italic;"&gt;pr&lt;/span&gt;+&lt;span style="font-weight: bold; font-style: italic;"&gt;qr&lt;/span&gt;)&lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;-&lt;span style="font-weight: bold; font-style: italic;"&gt;pqr&lt;/span&gt;=0&lt;br /&gt;&lt;br /&gt;Repare que o coeficiente de &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;² é o oposto da soma das raízes desta equação, isto é sempre válido quando o coeficiente do &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;³ é 1. A alternativa correta é a (C)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-8382660874355670328?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/8382660874355670328/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-33.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8382660874355670328?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8382660874355670328?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-33.html' title='SMEM 1 - Questão 33'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DEUDQH85eCp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-5773118081196239726</id><published>2010-07-08T03:45:00.001-07:00</published><updated>2010-07-08T03:51:11.120-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:51:11.120-07:00</app:edited><title>SMEM 1 - Questão 32</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 32:&lt;/span&gt; Asinale a afirmação verdadeira sobre logaritmos:&lt;br /&gt;a) Se log a&gt;log b, então a&gt;b.&lt;br /&gt;b) log&lt;sub&gt;2&lt;/sub&gt;x=6 então x=12&lt;br /&gt;c) log&lt;sub&gt;a&lt;/sub&gt;a=10, com a&gt;0&lt;br /&gt;d) log(a.c)=log a. log c&lt;br /&gt;e) log a - log c=log(a-c)&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Uma forma que podemos definir é "&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_ax=b" title="\hbox{log}_ax=b" alt="\hbox{log}_ax=b" style="vertical-align: -2pt;" /&gt; se, e somente se, &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%5Eb=x" title="a^b=x" alt="a^b=x" style="vertical-align: -2pt;" /&gt;". 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	mso-header-margin:36.0pt; 	mso-footer-margin:36.0pt; 	mso-paper-source:0;} div.Section1 	{page:Section1;} --&gt; &lt;/style&gt;&lt;!--[if gte mso 10]&gt; &lt;style&gt;  /* Style Definitions */  table.MsoNormalTable 	{mso-style-name:"Tabela normal"; 	mso-tstyle-rowband-size:0; 	mso-tstyle-colband-size:0; 	mso-style-noshow:yes; 	mso-style-priority:99; 	mso-style-qformat:yes; 	mso-style-parent:""; 	mso-padding-alt:0cm 5.4pt 0cm 5.4pt; 	mso-para-margin:0cm; 	mso-para-margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:11.0pt; 	font-family:"Calibri","sans-serif"; 	mso-ascii-font-family:Calibri; 	mso-ascii-theme-font:minor-latin; 	mso-fareast-font-family:"Times New Roman"; 	mso-fareast-theme-font:minor-fareast; 	mso-hansi-font-family:Calibri; 	mso-hansi-theme-font:minor-latin; 	mso-bidi-font-family:"Times New Roman"; 	mso-bidi-theme-font:minor-bidi;} &lt;/style&gt; &lt;![endif]--&gt;&lt;span style="font-size: 12pt; font-family: &amp;quot;Times New Roman&amp;quot;,&amp;quot;serif&amp;quot;;"&gt;&amp;gt;&lt;/span&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\hbox{log }b" title="\hbox{log }b" alt="\hbox{log }b" style="vertical-align:-2pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;(B) é incorreta: se log&lt;sub&gt;2&lt;/sub&gt;x=6 então x=2&lt;sup&gt;6&lt;/sup&gt;=64&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cneq" title="\neq" alt="\neq" style="vertical-align: -2pt;" /&gt;12.&lt;br /&gt;&lt;br /&gt;(C) é incorreta:  se &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_aa=x" title="\hbox{log}_aa=x" alt="\hbox{log}_aa=x" style="vertical-align: -2pt;" /&gt;, então &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%5Ex=a" title="a^x=a" alt="a^x=a" style="vertical-align: -2pt;" /&gt; e x=1 é o único valor que satisfaz esta equação, portanto &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_aa%5Cneq10" title="\hbox{log}_aa\neq10" alt="\hbox{log}_aa\neq10" style="vertical-align: -2pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;(D) é incorreta: tome a=100 e c=1000 &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_%7B10%7Da=2" title="\hbox{log}_{10}a=2" alt="\hbox{log}_{10}a=2" style="vertical-align: -2pt;" /&gt; e &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_%7B10%7Dc=3" title="\hbox{log}_{10}c=3" alt="\hbox{log}_{10}c=3" style="vertical-align: -2pt;" /&gt;, entretanto &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_%7B10%7D%28a%5Ccdot%20c%29=5%5Cneq%206=%5Chbox%7Blog%7D_%7B10%7D%28a%29%5Ccdot%20%5Chbox%7Blog%7D_%7B10%7D%28c%29" title="\hbox{log}_{10}(a\cdot c)=5\neq 6=\hbox{log}_{10}(a)\cdot \hbox{log}_{10}(c)" alt="\hbox{log}_{10}(a\cdot c)=5\neq 6=\hbox{log}_{10}(a)\cdot \hbox{log}_{10}(c)" style="vertical-align: -2pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;(E) é incorreta: tome os mesmos a e c, veja que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Blog%7D_%7B10%7D%28c%29-%5Chbox%7Blog%7D_%7B10%7D%28a%29=%201=%5Chbox%7Blog%7D_%7B10%7D10%5Cneq%5Chbox%7Blog%7D_%7B10%7D900" title="\hbox{log}_{10}(c)-\hbox{log}_{10}(a)= 1=\hbox{log}_{10}10\neq\hbox{log}_{10}900" alt="\hbox{log}_{10}(c)-\hbox{log}_{10}(a)= 1=\hbox{log}_{10}10\neq\hbox{log}_{10}900" style="vertical-align: -2pt;" /&gt;.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-5773118081196239726?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/5773118081196239726/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-32.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5773118081196239726?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5773118081196239726?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-32.html' title='SMEM 1 - Questão 32'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;D08HRn49eCp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-3685130214443419871</id><published>2010-07-08T03:42:00.000-07:00</published><updated>2010-07-08T03:43:57.060-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:43:57.060-07:00</app:edited><title>SMEM 1 - Questão 31</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 31:&lt;/span&gt; Um restaurante do tipo self-service oferece 3 opções de entrada, 5 pratos principais e 4 sobremesas. Um cliente desse restaurante deseja compor sua refeição com exatamente 1 entrada, 2 pratos principais e 2 sobremesas. De quantas maneiras diferentes esse cliente poderá compor a sua refeição?&lt;br /&gt;a) 4.&lt;br /&gt;b) 5.&lt;br /&gt;c) 12.&lt;br /&gt;d) 60.&lt;br /&gt;e) 180.&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Temos 3 opções para a entrada. Os dois pratos principais podem ser escolhidos de 10 modos. Note que este número equivale a combinação de 5 tomados 2 a 2. Por sua vez temos 6 modos de escolher as duas sobremesas que é igual a combinação de 4 tomados 2 a 2.&lt;br /&gt;&lt;br /&gt;Pelo princípio Multiplicativo o número  de modos que o cliente pode compor sua refeição é&lt;br /&gt;&lt;br /&gt;3x10x6=180&lt;br /&gt;&lt;br /&gt;Alternativa (E)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-3685130214443419871?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/3685130214443419871/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-31.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/3685130214443419871?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/3685130214443419871?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-31.html' title='SMEM 1 - Questão 31'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;D0ENRHozeyp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-2272579243919366886</id><published>2010-07-08T03:22:00.000-07:00</published><updated>2010-07-08T03:41:35.483-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:41:35.483-07:00</app:edited><title>SMEM 1 - Questão 30</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 30:&lt;/span&gt; Assinale a relação trigonométrica verdadeira:&lt;br /&gt;a) para todo ângulo a, sen a&lt;meta equiv="Content-Type" content="text/html; charset=utf-8"&gt;&lt;meta name="ProgId" content="Word.Document"&gt;&lt;meta name="Generator" content="Microsoft Word 12"&gt;&lt;meta name="Originator" content="Microsoft Word 12"&gt;&lt;link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CAdriano%5CCONFIG%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml"&gt;&lt;link rel="themeData" href="file:///C:%5CDOCUME%7E1%5CAdriano%5CCONFIG%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx"&gt;&lt;link rel="colorSchemeMapping" href="file:///C:%5CDOCUME%7E1%5CAdriano%5CCONFIG%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml"&gt;&lt;!--[if gte mso 9]&gt;&lt;xml&gt;  &lt;w:worddocument&gt;   &lt;w:view&gt;Normal&lt;/w:View&gt;   &lt;w:zoom&gt;0&lt;/w:Zoom&gt;   &lt;w:trackmoves/&gt;   &lt;w:trackformatting/&gt;   &lt;w:hyphenationzone&gt;21&lt;/w:HyphenationZone&gt;   &lt;w:punctuationkerning/&gt; 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 &lt;/w:LatentStyles&gt; &lt;/xml&gt;&lt;![endif]--&gt;&lt;style&gt; &lt;!--  /* Font Definitions */  @font-face 	{font-family:"Cambria Math"; 	panose-1:2 4 5 3 5 4 6 3 2 4; 	mso-font-charset:1; 	mso-generic-font-family:roman; 	mso-font-format:other; 	mso-font-pitch:variable; 	mso-font-signature:0 0 0 0 0 0;}  /* Style Definitions */  p.MsoNormal, li.MsoNormal, div.MsoNormal 	{mso-style-unhide:no; 	mso-style-qformat:yes; 	mso-style-parent:""; 	margin:0cm; 	margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:12.0pt; 	font-family:"Times New Roman","serif"; 	mso-fareast-font-family:"Times New Roman";} .MsoChpDefault 	{mso-style-type:export-only; 	mso-default-props:yes; 	font-size:10.0pt; 	mso-ansi-font-size:10.0pt; 	mso-bidi-font-size:10.0pt;} @page Section1 	{size:612.0pt 792.0pt; 	margin:70.85pt 3.0cm 70.85pt 3.0cm; 	mso-header-margin:36.0pt; 	mso-footer-margin:36.0pt; 	mso-paper-source:0;} div.Section1 	{page:Section1;} --&gt; &lt;/style&gt;&lt;!--[if gte mso 10]&gt; &lt;style&gt;  /* Style Definitions */  table.MsoNormalTable 	{mso-style-name:"Tabela normal"; 	mso-tstyle-rowband-size:0; 	mso-tstyle-colband-size:0; 	mso-style-noshow:yes; 	mso-style-priority:99; 	mso-style-qformat:yes; 	mso-style-parent:""; 	mso-padding-alt:0cm 5.4pt 0cm 5.4pt; 	mso-para-margin:0cm; 	mso-para-margin-bottom:.0001pt; 	mso-pagination:widow-orphan; 	font-size:11.0pt; 	font-family:"Calibri","sans-serif"; 	mso-ascii-font-family:Calibri; 	mso-ascii-theme-font:minor-latin; 	mso-fareast-font-family:"Times New Roman"; 	mso-fareast-theme-font:minor-fareast; 	mso-hansi-font-family:Calibri; 	mso-hansi-theme-font:minor-latin; 	mso-bidi-font-family:"Times New Roman"; 	mso-bidi-theme-font:minor-bidi;} &lt;/style&gt; &lt;![endif]--&gt;&lt;span style=";font-family:&amp;quot;;font-size:12pt;"  &gt;&amp;lt;&lt;/span&gt;1.&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Analisando alternativa por alternativa:&lt;br /&gt;&lt;br /&gt;(A) Incorreta: temos que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Chbox%7Bcossec%20%7Da=%5Cdfrac%7B1%7D%7B%5Csin%20a%7D" title="\hbox{cossec }a=\dfrac{1}{\sin a}" alt="\hbox{cossec }a=\dfrac{1}{\sin a}" style="vertical-align: -13pt;" /&gt;, para a=90 temos ambas as funçãos (sin e cossec) valem 1.&lt;br /&gt;&lt;br /&gt;(B) Correta:  no intervalo (0°, 90°) as funções &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\cos a" title="\cos a" alt="\cos a" style="vertical-align:-2pt;" /&gt; e &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\sin a" title="\sin a" alt="\sin a" style="vertical-align:-2pt;" /&gt; são respectivamente decrescente e crescente. Como &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\cos 45=\sin 45" title="\cos 45=\sin 45" alt="\cos 45=\sin 45" style="vertical-align:-2pt;" /&gt; qualquer valor &lt;span style="font-style: italic; font-weight: bold;"&gt;a&lt;/span&gt;&amp;gt;45, &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\cos a&lt;\sin a" title="\cos a&lt;\sin a" alt="\cos a&lt;\sin a" style="vertical-align:-2pt;" /&gt; e qualquer valor &lt;span style="font-style: italic; font-weight: bold;"&gt;a&lt;/span&gt;&amp;lt;45, &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\cos a" title="\cos a" alt="\cos a" style="vertical-align:-2pt;" /&gt;&amp;gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\sin a" title="\sin a" alt="\sin a" style="vertical-align:-2pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;(C) Incorreta: explicado na (B)&lt;br /&gt;&lt;br /&gt;(D) Incorreta: &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Csin%2045=%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3C1=%5Ctan%2045" title="\sin 45=\dfrac{\sqrt{2}}{2}&lt;1=\tan 45" alt="\sin 45=\dfrac{\sqrt{2}}{2}&lt;1=\tan 45" style="vertical-align: -13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;(E) Correta: Uma vez que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Ccos%20a%3C%5Csin%20a,%5Cquad%20a%3C45" title="\cos a&lt;\sin a,\quad a&lt;45" alt="\cos a&lt;\sin a,\quad a&lt;45" style="vertical-align: -2pt;" /&gt;, teremos que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Ctan%20a=%5Cdfrac%7B%5Csin%20a%7D%7B%5Ccos%20a%7D%3C1" title="\tan a=\dfrac{\sin a}{\cos a}&lt;1" alt="\tan a=\dfrac{\sin a}{\cos a}&lt;1" style="vertical-align: -13pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;É, de fato esta questão tem duas alternativas corretas. A questão é que muitas vezes provas de classificação como Vestibulares e o ENEM eliminam algumas questões por causa de erros na formulação das mesmas. Querendo ser fiel a este teste não modifiquei a questão, uma vez que tal erro pode levar ao aluno um tempo maior para responder a questão e leve a não confiar no que encontrou.&lt;a&gt;&lt;br /&gt;&lt;/a&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-2272579243919366886?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/2272579243919366886/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-30.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/2272579243919366886?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/2272579243919366886?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-30.html' title='SMEM 1 - Questão 30'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DkMCQ3ozeSp7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-5093914762155655239</id><published>2010-07-08T03:20:00.000-07:00</published><updated>2010-07-08T03:21:02.481-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:21:02.481-07:00</app:edited><title>SMEM 1 - Questão 29</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 29:&lt;/span&gt; De acordo com a tabela, a fórmula que expressa a relação entre x e y é:&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\begin{array}{|c|c|c|c|c|c|}\hline x&amp;amp;0&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;4 \\ \hline y&amp;amp;100&amp;amp;90&amp;amp;70&amp;amp;40&amp;amp;0\\ \hline\end{array}" title="\begin{array}{|c|c|c|c|c|c|}\hline x&amp;amp;0&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;4 \\ \hline y&amp;amp;100&amp;amp;90&amp;amp;70&amp;amp;40&amp;amp;0\\ \hline\end{array}" alt="\begin{array}{|c|c|c|c|c|c|}\hline x&amp;amp;0&amp;amp;1&amp;amp;2&amp;amp;3&amp;amp;4 \\ \hline y&amp;amp;100&amp;amp;90&amp;amp;70&amp;amp;40&amp;amp;0\\ \hline\end{array}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Repare que podemos escrever esta relação por uma recorrencia como&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\begin{array}{rcl} f(0)&amp;amp;=&amp;amp;100\\ f(1)&amp;amp;=&amp;amp;100-(10)\\ f(2)&amp;amp;=&amp;amp;100-(10+20)\\ f(3)&amp;amp;=&amp;amp;100-(10+20+30)\\ f(4)&amp;amp;=&amp;amp;100-(10+20+30+40)\end{array}" title="\begin{array}{rcl} f(0)&amp;amp;=&amp;amp;100\\ f(1)&amp;amp;=&amp;amp;100-(10)\\ f(2)&amp;amp;=&amp;amp;100-(10+20)\\ f(3)&amp;amp;=&amp;amp;100-(10+20+30)\\ f(4)&amp;amp;=&amp;amp;100-(10+20+30+40)\end{array}" alt="\begin{array}{rcl} f(0)&amp;amp;=&amp;amp;100\\ f(1)&amp;amp;=&amp;amp;100-(10)\\ f(2)&amp;amp;=&amp;amp;100-(10+20)\\ f(3)&amp;amp;=&amp;amp;100-(10+20+30)\\ f(4)&amp;amp;=&amp;amp;100-(10+20+30+40)\end{array}" style="vertical-align:-2pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Se este padrão continuar teremos que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f(n)=100-(10+20+30+...+10n)" title="f(n)=100-(10+20+30+...+10n)" alt="f(n)=100-(10+20+30+...+10n)" style="vertical-align:-2pt;" /&gt;. A parte entre parenteses desta função é uma PA onde o primeiro termo é 10 e a razão é 10, neste caso podemos escrever esta soma por:&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?10+20+30+...+10n=\dfrac{(10+10n)n}{2}=5n^2+5n" title="10+20+30+...+10n=\dfrac{(10+10n)n}{2}=5n^2+5n" alt="10+20+30+...+10n=\dfrac{(10+10n)n}{2}=5n^2+5n" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;A fórmula que queremos encontrar portanto é:&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f(x)=100-5x^2-5x" title="f(x)=100-5x^2-5x" alt="f(x)=100-5x^2-5x" style="vertical-align:-2pt;" /&gt;.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-5093914762155655239?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/5093914762155655239/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-29.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5093914762155655239?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5093914762155655239?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-29.html' title='SMEM 1 - Questão 29'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DkQMQXY9fip7ImA9WxFbFUU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-4149043444369469274</id><published>2010-07-08T03:17:00.000-07:00</published><updated>2010-07-08T03:19:40.866-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-08T03:19:40.866-07:00</app:edited><title>SMEM 1 - Questão 28</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 28:&lt;/span&gt; Assinale a alternativa correta:&lt;br /&gt;a) Duplicando-se a altura de um triângulo sua área fica duplicada.&lt;br /&gt;b) Duplicando-se os lados de um quadrado sua área fica duplicada.&lt;br /&gt;c) Duplicando-se o raio de um círculo sua área fica duplicada.&lt;br /&gt;d) Duplicando-se o perímetro de um retângulo sua área fica duplicada.&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;A única alternativa correta desta questão é a (A), pois se &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?h" title="h" alt="h" style="vertical-align:-2pt;" /&gt; e &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?b" title="b" alt="b" style="vertical-align:-2pt;" /&gt; representam a altura e a base de um triângulo,  a área deste é portanto &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?A=\dfrac{b\cdot h}{2}" title="A=\dfrac{b\cdot h}{2}" alt="A=\dfrac{b\cdot h}{2}" style="vertical-align:-13pt;" /&gt;. Se a altura passa a ser &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?2h" title="2h" alt="2h" style="vertical-align:-2pt;" /&gt; a área passará a ser &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?A=\dfrac{b\cdot 2h}{2}=2\cdot\dfrac{b\cdot h}{2}" title="A=\dfrac{b\cdot 2h}{2}=2\cdot\dfrac{b\cdot h}{2}" alt="A=\dfrac{b\cdot 2h}{2}=2\cdot\dfrac{b\cdot h}{2}" style="vertical-align:-13pt;" /&gt;, portanto o dobro.&lt;br /&gt;&lt;br /&gt;Nas outras alternativas quando fazemos a operação de duplicação a área da figura em questão ficará quadriplicada e não duplicada.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-4149043444369469274?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/4149043444369469274/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-28.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4149043444369469274?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4149043444369469274?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-28.html' title='SMEM 1 - Questão 28'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;D08FR30yfCp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-5260124151885067777</id><published>2010-07-07T19:21:00.001-07:00</published><updated>2010-07-07T19:23:36.394-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T19:23:36.394-07:00</app:edited><title>SMEM 1 - Questão 27</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 27:&lt;/span&gt; Para executar a cobertura de uma casa, optou-se pela telha francesa, que exige que o ângulo formado pelo telhado com a horizontal seja de, no mínimo, 30°. Qual deve ser o comprimento do &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_0"&gt;pontalete&lt;/span&gt;, sabendo-se que a largura da casa é 10 metros?&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDU2VZUPUoI/AAAAAAAAAPk/5UILdYA9kPw/s1600/quest27.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 335px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDU2VZUPUoI/AAAAAAAAAPk/5UILdYA9kPw/s400/quest27.jpg" alt="" id="BLOGGER_PHOTO_ID_5491355061849117314" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;a) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{5\sqrt{3}}{3}" title="\dfrac{5\sqrt{3}}{3}" alt="\dfrac{5\sqrt{3}}{3}" style="vertical-align:-13pt;" /&gt;m&lt;br /&gt;b) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{5}{2}" title="\dfrac{5}{2}" alt="\dfrac{5}{2}" style="vertical-align:-13pt;" /&gt;m&lt;br /&gt;c) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{5\sqrt{3}}{2}" title="\dfrac{5\sqrt{3}}{2}" alt="\dfrac{5\sqrt{3}}{2}" style="vertical-align:-13pt;" /&gt;m&lt;br /&gt;d) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{\sqrt{3}}{3}" title="\dfrac{\sqrt{3}}{3}" alt="\dfrac{\sqrt{3}}{3}" style="vertical-align:-13pt;" /&gt;m&lt;br /&gt;e) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?3\sqrt{5}" title="3\sqrt{5}" alt="3\sqrt{5}" style="vertical-align:-13pt;" /&gt;m&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Observe que o telhado, a horizontal e o &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_1"&gt;pontalete&lt;/span&gt; formam um triângulo &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_2"&gt;retângulo&lt;/span&gt; onde o &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_3"&gt;pontalete&lt;/span&gt; é o cateto oposto do ângulo de 30°.&lt;br /&gt;&lt;br /&gt;Além disso, neste triângulo o valor do cateto adjacente, relativo ao ângulo que conhecemos, é a  metade da largura da casa, portanto 5m&lt;br /&gt;&lt;br /&gt;Representando a medida do &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_4"&gt;pontalete&lt;/span&gt; por h, temos a tangente de 30° é dada por:&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\begin{array}{rcl}\tan(30^o)&amp;amp;=&amp;amp;\dfrac{h}{5}\\\dfrac{\sqrt{3}}{3}&amp;amp;=&amp;amp;\dfrac{h}{5}\\h&amp;amp;=&amp;amp;\dfrac{5\sqrt{3}}{3}\end{array}" title="\begin{array}{rcl}\tan(30^o)&amp;amp;=&amp;amp;\dfrac{h}{5}\\\dfrac{\sqrt{3}}{3}&amp;amp;=&amp;amp;\dfrac{h}{5}\\h&amp;amp;=&amp;amp;\dfrac{5\sqrt{3}}{3}\end{array}" alt="\begin{array}{rcl}\tan(30^o)&amp;amp;=&amp;amp;\dfrac{h}{5}\\\dfrac{\sqrt{3}}{3}&amp;amp;=&amp;amp;\dfrac{h}{5}\\h&amp;amp;=&amp;amp;\dfrac{5\sqrt{3}}{3}\end{array}" style="vertical-align:-2pt;" /&gt;&lt;br /&gt;&lt;br /&gt;A alternativa correta é a (A).&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-5260124151885067777?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/5260124151885067777/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-27.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5260124151885067777?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5260124151885067777?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-27.html' title='SMEM 1 - Questão 27'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDU2VZUPUoI/AAAAAAAAAPk/5UILdYA9kPw/s72-c/quest27.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DkYDSHw6fyp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-8985831246002826837</id><published>2010-07-07T18:42:00.000-07:00</published><updated>2010-07-07T18:56:19.217-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T18:56:19.217-07:00</app:edited><title>SMEM 1 - Questão 26</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 26:&lt;/span&gt; A única afirmação correta é&lt;br /&gt;a)  &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%5Ex%3Ca%5E2" title="a^x&lt;a^2" alt="a^x&lt;a^2" style="vertical-align: -2pt;" /&gt;, então x&lt;2 para qualquer valor de a.&lt;br /&gt;b) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%28x%29=a%5Ex" title="f(x)=a^x" alt="f(x)=a^x" style="vertical-align: -2pt;" /&gt; é definida para qualquer valor de a.&lt;br /&gt;c) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%28x%29=a%5Ex" title="f(x)=a^x" alt="f(x)=a^x" style="vertical-align: -2pt;" /&gt; é decrescente se x&lt;0.&lt;br /&gt;d) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%28x%29=a%5Ex" title="f(x)=a^x" alt="f(x)=a^x" style="vertical-align: -2pt;" /&gt; é crescente se x&gt;0.&lt;br /&gt;e) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%5Ex%3Ea%5E2" title="" /&gt;, então x&gt;2 se a&gt;1.&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Analisando as alternativas:&lt;br /&gt;&lt;br /&gt;(A) incorreta: tome, por exemplo &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a=%5Cdfrac%7B1%7D%7B2%7D" title="a=\dfrac{1}{2}" alt="a=\dfrac{1}{2}" style="vertical-align: -13pt;" /&gt; e &lt;span style="font-style: italic; font-weight: bold;"&gt;x&lt;/span&gt;=1, neste caso &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%5Ex=%5Cdfrac%7B1%7D%7B2%7D%3C%5Cdfrac%7B1%7D%7B4%7D=a%5E2" title="a^x=\dfrac{1}{2}&lt;\dfrac{1}{4}=a^2" alt="a^x=\dfrac{1}{2}&lt;\dfrac{1}{4}=a^2" style="vertical-align: -13pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;(B) incorreta: A função exponencial é definida somente para &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?a%3E0" title="" /&gt; e diferente de 1.&lt;br /&gt;&lt;br /&gt;(C) e (D) incorretas: A função exponencial é estritamente crescente ou decrescente quando &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?1%3Ca" title="1&lt;a" alt="1&lt;a" style="vertical-align: -2pt;" /&gt; ou &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?0%3Ca%3C1" title="0&lt;a&lt;1" alt="0&lt;a&lt;1" style="vertical-align: -2pt;" /&gt; respectivemente.&lt;br /&gt;&lt;br /&gt;(E) Correta: A função é crescente nesta condição para &lt;span style="font-weight: bold; font-style: italic;"&gt;a&lt;/span&gt;.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-8985831246002826837?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/8985831246002826837/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-26.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8985831246002826837?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8985831246002826837?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-26.html' title='SMEM 1 - Questão 26'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;CUUMQXw_eip7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-4748847455662911410</id><published>2010-07-07T18:30:00.000-07:00</published><updated>2010-07-07T18:41:20.242-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T18:41:20.242-07:00</app:edited><title>SMEM 1 - Questão 25</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 25:&lt;/span&gt; Em uma empresa, existe um &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_0"&gt;galpão&lt;/span&gt; que precisa ser dividido em três depósitos e um &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_1"&gt;hall&lt;/span&gt; de entrada de 20m², conforme a figura abaixo. Os depósitos I, II e III serão &lt;span class="blsp-spelling-corrected" id="SPELLING_ERROR_2"&gt;construídos&lt;/span&gt; para o armazenamento de, respectivamente, 90, 60 e 120 fardos de igual volume, e suas áreas devem ser &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_3"&gt;proporcionais&lt;/span&gt; a essas capacidades.&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDUsR8VhHOI/AAAAAAAAAPc/reta934hybU/s1600/quest25.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 316px; height: 269px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDUsR8VhHOI/AAAAAAAAAPc/reta934hybU/s400/quest25.jpg" alt="" id="BLOGGER_PHOTO_ID_5491344007413964002" border="0" /&gt;&lt;/a&gt;Qual a largura do depósito III em metros?&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Quando dizemos que as áreas dos &lt;span class="blsp-spelling-corrected" id="SPELLING_ERROR_4"&gt;depósitos&lt;/span&gt; I, II, e II são &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_5"&gt;proporcionais&lt;/span&gt; a 90, 60 e 120 respectivamente, estamos dizendo que existe uma constante K de tal forma que a área destes depósitos são 90K, 60K e 120K. Assim, como sabemos a área total do &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_6"&gt;galpão&lt;/span&gt; (10x11) e que esta área é a soma das áreas dos depósitos e do &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_7"&gt;hall&lt;/span&gt; de entrada chegamos a equação:&lt;br /&gt;&lt;br /&gt;10x11=20+90K+60K+120K&lt;br /&gt;110-20=270K&lt;br /&gt;K=&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{1}{3}" title="\dfrac{1}{3}" alt="\dfrac{1}{3}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;A á&lt;span class="blsp-spelling-error" id="SPELLING_ERROR_8"&gt;rea&lt;/span&gt; do &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_9"&gt;galpão&lt;/span&gt; III portanto é 120x&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{1}{3}" title="\dfrac{1}{3}" alt="\dfrac{1}{3}" style="vertical-align:-13pt;" /&gt;=40m². Como o comprimento deste &lt;span class="blsp-spelling-error" id="SPELLING_ERROR_10"&gt;galpão&lt;/span&gt; é 10, sua largura é 40/10=4m.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-4748847455662911410?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/4748847455662911410/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-25.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4748847455662911410?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4748847455662911410?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-25.html' title='SMEM 1 - Questão 25'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_Z_uiHIsRFf0/TDUsR8VhHOI/AAAAAAAAAPc/reta934hybU/s72-c/quest25.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;CEIHR3wyeSp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-1984663033591254919</id><published>2010-07-07T18:22:00.000-07:00</published><updated>2010-07-07T18:28:56.291-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T18:28:56.291-07:00</app:edited><title>SMEM 1 - Questão 23 e 24</title><content type='html'>Leia o texto a seguir e responda às questões 23 e 24.&lt;br /&gt;&lt;br /&gt;Um apostador tem três opções para participar de certa modalidade de jogo. que consiste no sorteio aleatório de um número dentre dez.&lt;br /&gt;1ª opção: comprar três números para um único sorteio.&lt;br /&gt;2ª opção: comprar dois números para um sorteio e um para o segundo sorteio.&lt;br /&gt;3ª opção: comprar um número para cada sorteio, num total de três sorteios.&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 23:&lt;/span&gt; Se X, Y e Z representam as probabilidades de o apostador ganhar algum prêmio, escolhendo, respectivamente, a 1ª, a 2ª ou a 3ª opção, é correto afirmar que:&lt;br /&gt;a) X&amp;lt;Y&amp;lt;Z&lt;br /&gt;b) X=Y=Z&lt;br /&gt;c) X&gt;Y=Z&lt;br /&gt;d) X=Y&gt;Z&lt;br /&gt;e) X&gt;Y&gt;Z&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Antes de responder estas duas questões vamos calcular as probabilidades.&lt;br /&gt;&lt;br /&gt;Na primeira opção as chances do apostador é de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{3}{10}" title="\dfrac{3}{10}" alt="\dfrac{3}{10}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Vale aqui ressaltar que a soma das probabilidades (chances de ganhar pelo menos um prêmio + chances de não ganhar prêmio algum) é sempre 1. Assim, as chances do apostador não ganhar nenhum prêmio na segunda opção é de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" title="\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" alt="\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" style="vertical-align:-13pt;" /&gt;. Note que as frações representam os 8 números não escolhidos dentre os 10 do primeiro sorteio, e os 9 não escolhidos dentre os 10 do segundo.&lt;br /&gt;&lt;br /&gt;Como a soma das probabilidade é sempre 1, as chances do apostador ganhar alguma coisa nesta 2º opção é &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?1-\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{28}{100}" title="1-\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{28}{100}" alt="1-\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{28}{100}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Da mesma forma, a probabilidade do apostador ganhar alguma coisa na 3º opção é &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?1-\dfrac{9}{10}\cdot\dfrac{9}{10}\cdot\dfrac{9}{10}=\dfrac{271}{1000}" title="1-\dfrac{9}{10}\cdot\dfrac{9}{10}\cdot\dfrac{9}{10}=\dfrac{271}{1000}" alt="1-\dfrac{9}{10}\cdot\dfrac{9}{10}\cdot\dfrac{9}{10}=\dfrac{271}{1000}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Para responder esta questão repare que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{3}{10}" title="\dfrac{3}{10}" alt="\dfrac{3}{10}" style="vertical-align:-13pt;" /&gt; é maior que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{28}{100}" title="\dfrac{28}{100}" alt="\dfrac{28}{100}" style="vertical-align:-13pt;" /&gt; que é maior que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{271}{1000}" title="\dfrac{271}{1000}" alt="\dfrac{271}{1000}" style="vertical-align:-13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;A alternativa correta é a (E)&lt;br /&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 24:&lt;/span&gt; Escolhendo a 2ª opção, a probabilidade de o apostador não ganhar em qualquer dos sorteios é igual a:&lt;br /&gt;a) 90%&lt;br /&gt;b) 81%&lt;br /&gt;c) 72%&lt;br /&gt;d) 70%&lt;br /&gt;e) 65%&lt;br /&gt;&lt;span style="color: rgb(204, 0, 0); font-weight: bold;"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Acima, na resposta da questão 23, calculamos qual é a probabilidade do apostador não ganhar nada na 2º opção que é &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" title="\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" alt="\dfrac{8}{10}\cdot\dfrac{9}{10}=\dfrac{72}{100}" style="vertical-align:-13pt;" /&gt;, ou seja 72%. Alternativa (C).&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-1984663033591254919?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/1984663033591254919/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-23-e-24.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/1984663033591254919?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/1984663033591254919?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-23-e-24.html' title='SMEM 1 - Questão 23 e 24'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;CEcNQ3Y7fCp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-792918204360459435</id><published>2010-07-07T18:20:00.000-07:00</published><updated>2010-07-07T18:21:32.804-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T18:21:32.804-07:00</app:edited><title>SMEM 1 - Questão 22</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 22:&lt;/span&gt; Comprando-se 5 números de uma rifa com 50 números, qual a chance de ganhar o prêmio?&lt;br /&gt;&lt;span style="color: rgb(204, 0, 0); font-weight: bold;"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Como os cinco números são diferentes e cada um tem a chance de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{1}{50}" title="\dfrac{1}{50}" alt="\dfrac{1}{50}" style="vertical-align:-13pt;" /&gt; de ser sorteado, a chance de pelo menos 1 dentre os 5 de ser sorteado é de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?5\cdot\dfrac{1}{50}=\dfrac{1}{10}" title="5\cdot\dfrac{1}{50}=\dfrac{1}{10}" alt="5\cdot\dfrac{1}{50}=\dfrac{1}{10}" style="vertical-align:-13pt;" /&gt;, ou seja 10%.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-792918204360459435?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/792918204360459435/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-22.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/792918204360459435?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/792918204360459435?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-22.html' title='SMEM 1 - Questão 22'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;C04DSXo9cCp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-8310301728742036339</id><published>2010-07-07T18:17:00.000-07:00</published><updated>2010-07-07T18:19:38.468-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T18:19:38.468-07:00</app:edited><title>SMEM 1 - Questão 21</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 21:&lt;/span&gt; O gráfico abaixo mostra o crescimento de uma planta pelo tempo em dias.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDUnZkegYQI/AAAAAAAAAPU/hf5hcGVpnTE/s1600/quest21.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 267px; height: 162px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDUnZkegYQI/AAAAAAAAAPU/hf5hcGVpnTE/s400/quest21.jpg" alt="" id="BLOGGER_PHOTO_ID_5491338640890028290" border="0" /&gt;&lt;/a&gt;Mantido o crescimento, no 35° dia, qual será a altura da planta?&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Observando o gráfico podemos verificara que a planta cresce 1 cm a cada 5 dias, portanto crescendo &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\dfrac{1}{5}" title="\dfrac{1}{5}" alt="\dfrac{1}{5}" style="vertical-align:-13pt;" /&gt; cm por dia. No 35º dia a planta terá crescido &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?35\cdot\dfrac{1}{5}=7" title="35\cdot\dfrac{1}{5}=7" alt="35\cdot\dfrac{1}{5}=7" style="vertical-align:-13pt;" /&gt; cm.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-8310301728742036339?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/8310301728742036339/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-21.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8310301728742036339?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/8310301728742036339?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/07/smem-1-questao-21.html' title='SMEM 1 - Questão 21'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://2.bp.blogspot.com/_Z_uiHIsRFf0/TDUnZkegYQI/AAAAAAAAAPU/hf5hcGVpnTE/s72-c/quest21.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;CkEDRXk9fip7ImA9WxFUFkg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-4941743166669019372</id><published>2010-06-27T07:55:00.001-07:00</published><updated>2010-06-27T07:57:54.766-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-27T07:57:54.766-07:00</app:edited><title>SMEM 1 - Questão 20</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 20:&lt;/span&gt; O gráfico abaixo representa o sistema:&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCdmkcSUI9I/AAAAAAAAAO0/_SE0rUQrV4Q/s1600/quest+12.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 360px; height: 400px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCdmkcSUI9I/AAAAAAAAAO0/_SE0rUQrV4Q/s400/quest+12.jpg" alt="" id="BLOGGER_PHOTO_ID_5487467447228572626" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;a) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;6%5C%5Cx-y&amp;amp;=&amp;amp;4%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;b) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;4%5C%5Cx-y&amp;amp;=&amp;amp;6%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;4\\x-y&amp;amp;=&amp;amp;6\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;4\\x-y&amp;amp;=&amp;amp;6\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;c) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;-4%5C%5Cx+y&amp;amp;=&amp;amp;6%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;-4\\x+y&amp;amp;=&amp;amp;6\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;-4\\x+y&amp;amp;=&amp;amp;6\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;d) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx-4y&amp;amp;=&amp;amp;1%5C%5Cx+y&amp;amp;=&amp;amp;5%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x-4y&amp;amp;=&amp;amp;1\\x+y&amp;amp;=&amp;amp;5\end{array}\right." alt="\left\{\begin{array}{rcl}x-4y&amp;amp;=&amp;amp;1\\x+y&amp;amp;=&amp;amp;5\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;e) &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx&amp;amp;=&amp;amp;y%5C%5C6x-4y&amp;amp;=&amp;amp;0%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;y\\6x-4y&amp;amp;=&amp;amp;0\end{array}\right." alt="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;y\\6x-4y&amp;amp;=&amp;amp;0\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;No grágico temos duas retas, uma decrescente e outra crescente. Chamaremos estas retas de &lt;span style="font-style: italic; font-weight: bold;"&gt;r&lt;/span&gt; e &lt;span style="font-style: italic; font-weight: bold;"&gt;s&lt;/span&gt; respectivamente.&lt;br /&gt;&lt;br /&gt;Dois pontos da reta &lt;span style="font-weight: bold; font-style: italic;"&gt;r&lt;/span&gt; são (0,6) e (6,0), este dois pontos nos permite calcular a equação da reta, entretanto podemos apenas verificar que estee pontos satisfazem a relação &lt;span style="font-style: italic; font-weight: bold;"&gt;x+y&lt;/span&gt;=6.&lt;br /&gt;&lt;br /&gt;Dois pontos da reta &lt;span style="font-weight: bold; font-style: italic;"&gt;s&lt;/span&gt; são (0,-4) e (4,0), se fizermos&lt;span style="font-weight: bold; font-style: italic;"&gt; y-x&lt;/span&gt; veremos que o resuldade é 4 para os dois pontos&lt;br /&gt;&lt;br /&gt;Assim o gráfico representa o sistema&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx+y&amp;amp;=&amp;amp;6%5C%5Cx-y&amp;amp;=&amp;amp;4%5Cend%7Barray%7D%5Cright." title="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." alt="\left\{\begin{array}{rcl}x+y&amp;amp;=&amp;amp;6\\x-y&amp;amp;=&amp;amp;4\end{array}\right." style="vertical-align: -13pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Vale resaltar que, dados dois pontos &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28x_1,y_1%29" title="(x_1,y_1)" alt="(x_1,y_1)" style="vertical-align: -2pt;" /&gt; e &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28x_2,y_2%29" title="(x_2,y_2)" alt="(x_2,y_2)" style="vertical-align: -2pt;" /&gt; de uma reta podemos encontrar a equação da mesma fazendo &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?y-y_1=%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%28x-x_1%29" title="y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)" alt="y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)" style="vertical-align: -13pt;" /&gt;, o que também poderia resolver o problema.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-4941743166669019372?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/4941743166669019372/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-20.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4941743166669019372?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4941743166669019372?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-20.html' title='SMEM 1 - Questão 20'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCdmkcSUI9I/AAAAAAAAAO0/_SE0rUQrV4Q/s72-c/quest+12.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;D0UCRHgzfCp7ImA9WxFbFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-4278588432090162733</id><published>2010-06-27T07:38:00.001-07:00</published><updated>2010-07-07T19:14:25.684-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-07-07T19:14:25.684-07:00</app:edited><title>SMEM 1 - Questão 19</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 19:&lt;/span&gt; Se um ângulo de 30° é aumentado em 15°, a medida do seno:&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;a)&lt;/span&gt; e do cosseno desse ângulo aumentam.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;b)&lt;/span&gt; e do cosseno desse ângulo não se alteram.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;c) &lt;/span&gt;e do cosseno desse ângulo diminuem.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;d)&lt;/span&gt; diminui e a do cosseno aumenta.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;e)&lt;/span&gt; aumenta e a do cosseno diminui.&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Dado um triângulo retângulo con um dos ângulos &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\alpha" title="\alpha" alt="\alfa" style="vertical-align: -2pt;" /&gt; diferente do ângulo de 90°, por definição o seno deste ângulo é a razão entre o cateto oposto ao ângulo pela hipotenusa. Analogamente cos &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?\alpha" title="\alfa" alt="\alfa" style="vertical-align: -2pt;" /&gt; é a razão do cateto adjacente ao ângulo pela hipotenusa.&lt;br /&gt;&lt;br /&gt;Considere uma circunferência de raio 1 como na figura abaixo:&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCdigyjfwkI/AAAAAAAAAOs/k8_6sgL37kY/s1600/quest+19.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 357px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCdigyjfwkI/AAAAAAAAAOs/k8_6sgL37kY/s400/quest+19.jpg" alt="" id="BLOGGER_PHOTO_ID_5487462986440229442" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;Aqui os pontos ACF formam um triângulo retângulo com um dos ângulos igual a 30°, o seno e o coseno de 30 são respectivamente CF/AC e AF/AC...como o raio da circunferência é 1 podemos simplificar o seno e o coseno respectivamente para CF e AF.&lt;br /&gt;&lt;br /&gt;O triângulo ADE é retângulo com um dos ângulo igual a 30°+15°, e o seno e cosseno respectivos deste ângulo são DE e AE.&lt;br /&gt;&lt;br /&gt;Temos aqui que DE é maior que CF, portanto o seno de 30° aumenta quando este ângulo é acrescido de 15°. Da mesma forma AE é meor que AF mostrando que o cosseno de 30° diminui quando este ângulo é acrescido de 15°.&lt;br /&gt;&lt;br /&gt;A alternativa correta é a &lt;span style="font-weight: bold;"&gt;(E)&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-4278588432090162733?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/4278588432090162733/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-19.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4278588432090162733?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/4278588432090162733?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-19.html' title='SMEM 1 - Questão 19'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCdigyjfwkI/AAAAAAAAAOs/k8_6sgL37kY/s72-c/quest+19.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DEQDSH0_eCp7ImA9WxFUFkk.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-6986871522638629636</id><published>2010-06-27T06:40:00.001-07:00</published><updated>2010-06-27T06:46:19.340-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-27T06:46:19.340-07:00</app:edited><title>SMEM 1 - Questões 16, 17 e 18</title><content type='html'>&lt;div style="text-align: center;"&gt;&lt;span style="font-weight: bold;"&gt;Use o texto abaixo para responder as questões 16, 17 e 18&lt;/span&gt;&lt;br /&gt;&lt;/div&gt;&lt;br /&gt;No quadro abaixo estão as contas de luz e água de uma mesma residência. Além do valor a pagar, cada conta mostra como calculá-lo, em função do consumo de água (em m³) e de eletricidade (em kwh). Observe que, na conta de luz, o valor a pagar é igual ao consumo multiplicado por um certo fator. Já na conta de água, existe uma tarifa mínima e diferentes faixas de tarifação.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCdVMs_eS2I/AAAAAAAAAOk/0bUSAwpUqtE/s1600/tarifas.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 356px; height: 290px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCdVMs_eS2I/AAAAAAAAAOk/0bUSAwpUqtE/s400/tarifas.jpg" alt="" id="BLOGGER_PHOTO_ID_5487448347698416482" border="0" /&gt;&lt;/a&gt;&lt;hr /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 16:&lt;/span&gt; Suponha que, no próximo mês, dobre o consumo de energia elétrica dessa residência. Qual o novo valor da conta?&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Nesta situação o consumo passaria de 401 KWH para 802 KWH e o valor da conta seria 802 X 0,13276=R$ 106,47. Praticamente o dobro do valor da conta.&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 17:&lt;/span&gt; Suponha que dobre o consumo d'água. Qual o novo valor da conta?&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Pela tarifa notamos que o residente pagou R$ 5,50 por 10m³ gastos e R$ 5,95 pelos 7m³ que excederam 10m³, ou seja, foram gastos 17m³.&lt;br /&gt;&lt;br /&gt;Se o consumo fosse o dobro, isto é 34m³, o valor da conta seria R$ 5,50 pelos primeiros 10m³, 0,85x10=R$ 8,50 pelos 10m³ entre 11 e 20m³ e 2,13x14=R$ 29,82 pelos 14m³ que excedem 20m³ (note que o valor da conta onde o consumo está entre 21 a 30 e 31 a 50 é o mesmo). Portanto o valor da conta seria 5,50+8,50+29,82=R$ 43,82.&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Questão 18:&lt;/span&gt; Dos gráficos abaixo, o que melhor representa o valor da conta de água, de acordo com o consumo é:&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_Z_uiHIsRFf0/TCdVHgN9UNI/AAAAAAAAAOc/rNE3zFOEhfE/s1600/quest+18.JPG"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 320px; height: 223px;" src="http://1.bp.blogspot.com/_Z_uiHIsRFf0/TCdVHgN9UNI/AAAAAAAAAOc/rNE3zFOEhfE/s400/quest+18.JPG" alt="" id="BLOGGER_PHOTO_ID_5487448258370162898" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Não é difícil responder esta questão excluindo as alternativas erradas. Veja que de 0m³ a 10m³ o valor da tarifa é constante, só isto já exclui as alternativas (B), (C) e (D). Por outro lado a tarifa não permanece constante após 10m³ o que exclui a alternativa (E)&lt;br /&gt;A alternativa (A) realmente é a que melhor representa o problema, pois a conta é uma função contínua em relação ao consumo e ainda, quanto mais se gasta maior é o valor por metro cúbico fazendo com que a função tenha coeficientes angulares diferentes em diferentes intervalos.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-6986871522638629636?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/6986871522638629636/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questoes-16-17-e-18.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6986871522638629636?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6986871522638629636?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questoes-16-17-e-18.html' title='SMEM 1 - Questões 16, 17 e 18'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCdVMs_eS2I/AAAAAAAAAOk/0bUSAwpUqtE/s72-c/tarifas.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;Dk8HRXw6fip7ImA9WxFUFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-812124382517276777</id><published>2010-06-26T05:19:00.000-07:00</published><updated>2010-06-26T05:20:34.216-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-26T05:20:34.216-07:00</app:edited><title>SMEM 1 - Questão 15</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 15:&lt;/span&gt; Um lojista comprou de seu fornecedor um artigo por p reais (preço de custo) e revende com lucro de 40%. A seguir, ao fazer uma liquidação, ele dá aos compradores um desconto de 30% sobre o preço de venda desse artigo. Pode-se afirmar que esse comerciante tem, sobre p,&lt;br /&gt;a) prejuízo de 2%&lt;br /&gt;b) prejuízo de 10%&lt;br /&gt;c) lucro de 10%&lt;br /&gt;d) lucro de 12%&lt;br /&gt;e) lucro de 82%&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Preço de venda com 40% de lucro é:&lt;br /&gt;&lt;br /&gt;p+40% de p=p+p(0,4)=p(1,4).&lt;br /&gt;&lt;br /&gt;Calculando o preço do desconto de 30% sobre o preço de venda temos:&lt;br /&gt;&lt;br /&gt;p(1,4)-30% de p=p(1,4)-p(1,4)(0,3)=p(1,4)(1-0,3)=p(1,4)(0,7)=p(0,98).&lt;br /&gt;&lt;br /&gt;Ou seja, o lojista venderá o artigo por 98% do preço de custo, tendo um prejuíso de 2%. alternativa a)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-812124382517276777?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/812124382517276777/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-15.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/812124382517276777?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/812124382517276777?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-15.html' title='SMEM 1 - Questão 15'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DkQGSXgzcCp7ImA9WxFUFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-6335510079113388694</id><published>2010-06-26T05:10:00.001-07:00</published><updated>2010-06-26T05:12:08.688-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-26T05:12:08.688-07:00</app:edited><title>SMEM 1 - Questão 14</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 14:&lt;/span&gt; O brasil, em 1997, com cerca de &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?160%5Ctimes10%5E6" title="160\times10^6" alt="160\times10^6" style="vertical-align: -2pt;" /&gt; habitantes, apresentou um consumo de energia da ordem de 250.000 TEP (tonelada equivalente de petróleo), proveniente de diversas fontes primárias.&lt;br /&gt;O grupo com renda familiar de mais de vinte salários mínimos representa 5% da população brasileira e utiliza cerca de 10% da energia total consumida no país.&lt;br /&gt;O grupo com renda familiar de até três salárioas mínimos representa 50% da população e consome 30% do total de energia.&lt;br /&gt;&lt;br /&gt;Com base nessas informações, pode-se concluir que o consumo médio de energia para um indivíduo do grupo de renda é &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt; vezes maior do que para um indivíduo do grupo de renda inferior. Qual o valor aproximado de &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;?&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Apesar da quantidade de informações do problema, a maior dificuldade que ele impõe é a quantidade de contas que devemos fazer, desta forma apresentamos as contas comentadas linha por linha.&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(1)&lt;/span&gt; Número de pessoas do grupo que tem renda maior que 20 salários&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28160%5Ctimes%2010%5E6%29%5Ctimes%280,05%29=8%5Ctimes10%5E6" title="(160\times 10^6)\times(0,05)=8\times10^6" alt="(160\times 10^6)\times(0,05)=8\times10^6" style="vertical-align: -2pt;" /&gt; habitantes&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(2)&lt;/span&gt; Número de pessoas do grupo que tem renda inferior a 3 salários&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28160%5Ctimes%2010%5E6%29%5Ctimes%280,5%29=80%5Ctimes10%5E6" title="(160\times 10^6)\times(0,5)=80\times10^6" alt="(160\times 10^6)\times(0,5)=80\times10^6" style="vertical-align: -2pt;" /&gt; habitantes&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(3)&lt;/span&gt; Energia consumida pelo grupo (1)&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28250.000%29%5Ctimes%280,1%29=25.000" title="(250.000)\times(0,1)=25.000" alt="(250.000)\times(0,1)=25.000" style="vertical-align: -2pt;" /&gt; TEP&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(4)&lt;/span&gt; Energia consumida pelo grupo (2)&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%28250.000%29%5Ctimes%280,3%29=75.000" title="(250.000)\times(0,3)=75.000" alt="(250.000)\times(0,3)=75.000" style="vertical-align: -2pt;" /&gt; TEP&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(5)&lt;/span&gt; Consumo por pessoa no grupo (1)&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%2825.000%29/%288%5Ctimes10%5E6%29=3.125%5Ctimes10%5E%7B-6%7D" title="(25.000)/(8\times10^6)=3.125\times10^{-6}" alt="(25.000)/(8\times10^6)=3.125\times10^{-6}" style="vertical-align: -2pt;" /&gt; TEP / habitante&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(6)&lt;/span&gt; Consumo por pessoa no grupo (2)&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%2875.000%29/%2880%5Ctimes10%5E6%29=937,5%5Ctimes10%5E%7B-6%7D" title="(75.000)/(80\times10^6)=937,5\times10^{-6}" alt="(75.000)/(80\times10^6)=937,5\times10^{-6}" style="vertical-align: -2pt;" /&gt; TEP / habitante&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;(7)&lt;/span&gt; Encontrando &lt;span style="font-weight: bold; font-style: italic;"&gt;x&lt;/span&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x=%5Cdfrac%7B3.125%5Ctimes10%5E%7B-6%7D%7D%7B937,5%5Ctimes10%5E%7B-6%7D%7D=%5Cdfrac%7B3.125%7D%7B937,5%7D=3,33" title="x=\dfrac{3.125\times10^{-6}}{937,5\times10^{-6}}=\dfrac{3.125}{937,5}=3,33" alt="x=\dfrac{3.125\times10^{-6}}{937,5\times10^{-6}}=\dfrac{3.125}{937,5}=3,33" style="vertical-align: -13pt;" /&gt;&lt;br /&gt;O que resolve o problema.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-6335510079113388694?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/6335510079113388694/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-14.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6335510079113388694?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6335510079113388694?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-14.html' title='SMEM 1 - Questão 14'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;CUMHQHs9fCp7ImA9WxFUFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-7790107693995844642</id><published>2010-06-26T04:42:00.001-07:00</published><updated>2010-06-26T04:57:11.564-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-26T04:57:11.564-07:00</app:edited><title>SMEM 1 - Questão 13</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 13:&lt;/span&gt; A discussão sobre número de raízes distintas de uma equação do 2º grau é comumente feita por meio do dicriminante da equação. Para o caso da equação &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x%5E2-px+q%5E2=0,%20%28p%3E0,%20q%3E0%29" title="" 2="0," /&gt;, isso pode ser feito geometricamente, como mostra a figura.&lt;br /&gt;Nele, o arco é uma semicircunferência de diâmetro &lt;span style="font-weight: bold;"&gt;AB&lt;/span&gt;, com &lt;span style="font-weight: bold;"&gt;AB=p&lt;/span&gt; e&lt;span style="font-weight: bold;"&gt; CD=EF=q&lt;/span&gt;.&lt;br /&gt;As raízes r e s da equação são representadas pelos seguimentos &lt;span style="font-weight: bold;"&gt;AF&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;BF&lt;/span&gt;, respectivamente.&lt;br /&gt;De fato, &lt;span style="font-weight: bold;"&gt;r+s=p&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;rs=q²&lt;/span&gt;, uma vez que o triângulo &lt;span style="font-weight: bold;"&gt;AEB&lt;/span&gt; é retângulo e &lt;span style="font-weight: bold;"&gt;EF&lt;/span&gt; a altura relativa à hipotenusa.&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXn4xndpOI/AAAAAAAAAOU/OQmhQzPq8hw/s1600/quest13.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 136px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXn4xndpOI/AAAAAAAAAOU/OQmhQzPq8hw/s400/quest13.jpg" alt="" id="BLOGGER_PHOTO_ID_5487046683598365922" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;a) a partir da construção acima, conclua qual é a relação entre &lt;span style="font-weight: bold;"&gt;r&lt;/span&gt; e &lt;span style="font-weight: bold;"&gt;s&lt;/span&gt;, caso em que &lt;span style="font-weight: bold;"&gt;q=p/2&lt;/span&gt;.&lt;br /&gt;b) Calcule o valor do discriminante da equação para &lt;span style="font-weight: bold;"&gt;p=q/2&lt;/span&gt; e compare o que você concluiu com o observado em a).&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;a) Neste caso q é iguala ao raio da circunferência, desta forma o ponto E será o ponto mais alto do arco de forma que F divide o seguimento AB no meio, Temos dai que &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?r=AF=BF=s" title="r=AF=BF=s" alt="r=AF=BF=s" style="vertical-align: -2pt;" /&gt;.&lt;br /&gt;&lt;br /&gt;b) Temos &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x%5E2-px+%5Cleft%28%5Cdfrac%7Bp%7D%7B2%7D%5Cright%29%5E2=0" title="x^2-px+\left(\dfrac{p}{2}\right)^2=0" alt="x^2-px+\left(\dfrac{p}{2}\right)^2=0" style="vertical-align: -13pt;" /&gt;. o discriminante desta equação é &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5CDelta=%28-p%29%5E2-4%5Ccdot1%5Ccdot%5Cleft%28%5Cdfrac%7Bp%7D%7B2%7D%5Cright%29%5E2=p%5E2-4%5Ccdot%5Cdfrac%7Bp%5E2%7D%7B4%7D=0" title="\Delta=(-p)^2-4\cdot1\cdot\left(\dfrac{p}{2}\right)^2=p^2-4\cdot\dfrac{p^2}{4}=0" alt="\Delta=(-p)^2-4\cdot1\cdot\left(\dfrac{p}{2}\right)^2=p^2-4\cdot\dfrac{p^2}{4}=0" style="vertical-align: -13pt;" /&gt;. Portanto a equação tem duas raízes iguais, como vimos no caso geométrico.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-7790107693995844642?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/7790107693995844642/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/questao-13-discussao-sobre-numero-de.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7790107693995844642?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7790107693995844642?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/questao-13-discussao-sobre-numero-de.html' title='SMEM 1 - Questão 13'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXn4xndpOI/AAAAAAAAAOU/OQmhQzPq8hw/s72-c/quest13.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;C04MRHszfSp7ImA9WxFUFUg.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-5271087358521054773</id><published>2010-06-26T04:18:00.000-07:00</published><updated>2010-06-26T04:33:05.585-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-26T04:33:05.585-07:00</app:edited><title>SMEM 1 - Questão 12</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 12:&lt;/span&gt; Analise a alternativa que apresenta uma função crescente.&lt;table&gt;&lt;br /&gt;&lt;tbody&gt;&lt;tr&gt;&lt;br /&gt;&lt;th style="text-align: left;"&gt;(A)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXijdrefdI/AAAAAAAAANk/St72zqPzSMs/s1600/ques12-1.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 259px; height: 312px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXijdrefdI/AAAAAAAAANk/St72zqPzSMs/s320/ques12-1.jpg" alt="" id="BLOGGER_PHOTO_ID_5487040819911097810" border="0" /&gt;&lt;/a&gt;&lt;/th&gt;&lt;br /&gt;&lt;th style="text-align: left;"&gt;(B)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXiqwfz7TI/AAAAAAAAANs/qCidU2Y4eqs/s1600/ques12-2.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 320px; height: 314px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXiqwfz7TI/AAAAAAAAANs/qCidU2Y4eqs/s320/ques12-2.jpg" alt="" id="BLOGGER_PHOTO_ID_5487040945221528882" border="0" /&gt;&lt;/a&gt;&lt;/th&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;tr style="font-weight: bold;"&gt;&lt;br /&gt;&lt;td&gt;(C)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXizEisd9I/AAAAAAAAAN0/rsjnYneTWtQ/s1600/ques12-3.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 286px; height: 233px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXizEisd9I/AAAAAAAAAN0/rsjnYneTWtQ/s320/ques12-3.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041088041285586" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;td&gt;(D)&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXi5SOLssI/AAAAAAAAAN8/YX2MHYmyv6w/s1600/ques12-4.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 316px; height: 316px;" src="http://4.bp.blogspot.com/_Z_uiHIsRFf0/TCXi5SOLssI/AAAAAAAAAN8/YX2MHYmyv6w/s320/ques12-4.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041194792563394" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;tr&gt;&lt;br /&gt;&lt;td&gt;&lt;span style="font-weight: bold;"&gt;(E)&lt;/span&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXjAGaCU3I/AAAAAAAAAOE/fBvXcF49jr8/s1600/ques12-5.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 320px; height: 215px;" src="http://2.bp.blogspot.com/_Z_uiHIsRFf0/TCXjAGaCU3I/AAAAAAAAAOE/fBvXcF49jr8/s320/ques12-5.jpg" alt="" id="BLOGGER_PHOTO_ID_5487041311880139634" border="0" /&gt;&lt;/a&gt;&lt;/td&gt;&lt;br /&gt;&lt;td&gt;&lt;br /&gt;&lt;/td&gt;&lt;br /&gt;&lt;/tr&gt;&lt;br /&gt;&lt;/tbody&gt;&lt;/table&gt;&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Note que nos gráficos &lt;span style="font-weight: bold;"&gt;(A)&lt;/span&gt;, &lt;span style="font-weight: bold;"&gt;(B) &lt;/span&gt;e &lt;span style="font-weight: bold;"&gt;(C)&lt;/span&gt; as funções são decrescentes quando &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x&lt;0" title="x&lt;0" alt="x&lt;0" style="vertical-align:-2pt;" /&gt;, no gráfico &lt;span style="font-weight: bold;"&gt;(E)&lt;/span&gt; a função é crescente em alguns pontos, mas decrescente en outros. A única figura que apresenta um gráfico crescente em todos os pontos é a &lt;span style="font-weight: bold;"&gt;(D)&lt;/span&gt;.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-5271087358521054773?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/5271087358521054773/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-12.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5271087358521054773?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/5271087358521054773?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-12.html' title='SMEM 1 - Questão 12'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXijdrefdI/AAAAAAAAANk/St72zqPzSMs/s72-c/ques12-1.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;A04CRHoyfSp7ImA9WxFUFUk.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-7212698525643647527</id><published>2010-06-26T03:47:00.000-07:00</published><updated>2010-06-26T03:59:25.495-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-26T03:59:25.495-07:00</app:edited><title>SMEM 1 - Questão 11</title><content type='html'>&lt;span style="font-weight: bold;"&gt;Questão 11:&lt;/span&gt; O valor de um determinado automóvel diminui com o passar do tempo, como mostra o gráfico. Decorridos quantos anos esse carro passará a não ter valor algum?&lt;br /&gt;&lt;br /&gt;&lt;a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s1600/ques+11.jpg"&gt;&lt;img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 392px; height: 386px;" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s400/ques+11.jpg" alt="" id="BLOGGER_PHOTO_ID_5487034773099396386" border="0" /&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold; color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;Neste problema o valor do automóvel é descrico como uma função afim que tem a forma &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%28t%29=At+B" title="f(t)=At+B" alt="f(t)=At+B" style="vertical-align: -2pt;" /&gt;. Conhecemos dois pontos deta função, (0;25,5) e (8;13,5), assim temos.&lt;br /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%280%29=A%5Ccdot%200+B=25,5%5CRightarrow%20B=25,5" title="f(0)=A\cdot 0+B=25,5\Rightarrow B=25,5" alt="f(0)=A\cdot 0+B=25,5\Rightarrow B=25,5" style="vertical-align: -2pt;" /&gt;&lt;br /&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%288%29=A%5Ccdot%208+25,5=13,5%5CRightarrow%20A=%2813,5-25,5%29/8=-1,5" title="f(8)=A\cdot 8+25,5=13,5\Rightarrow A=(13,5-25,5)/8=-1,5" alt="f(8)=A\cdot 8+25,5=13,5\Rightarrow A=(13,5-25,5)/8=-1,5" style="vertical-align: -2pt;" /&gt;&lt;br /&gt;&lt;br /&gt;Encontramos portanto &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?f%28t%29=-1,5t+25,5" title="f(t)=-1,5t+25,5" alt="f(t)=-1,5t+25,5" style="vertical-align: -2pt;" /&gt;. O tempo (em anos) que o automóvel não terá valor será &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?t=%28-25,5%29/1,5" title="t=(-25,5)/1,5" alt="t=(-25,5)/1,5" style="vertical-align: -2pt;" /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-7212698525643647527?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/7212698525643647527/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-11.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7212698525643647527?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7212698525643647527?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/smem-1-questao-11.html' title='SMEM 1 - Questão 11'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_Z_uiHIsRFf0/TCXdDfjOPSI/AAAAAAAAANc/TGHIUpG2DOI/s72-c/ques+11.jpg' height='72' width='72'/><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;DkIGRHo_fip7ImA9WxFUE0w.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-6335912375900277512</id><published>2010-06-23T10:14:00.000-07:00</published><updated>2010-06-23T10:35:25.446-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-23T10:35:25.446-07:00</app:edited><title>7x13=28?</title><content type='html'>Não compreender bem a estrutura de formação dos números na representação decimal pode acarretar a erros de cálculo que normalmente passam desapercebido para quem está resolvendo um exercício.&lt;br /&gt;&lt;br /&gt;Todo sabemos (ou espero que saibamos) que o resultado de 7x13 é 91 e não 28. No vídeo abaixo a dupla de humorista dos anos 40, Abbott e Costello, mostram uma manipulação  na conta 7x13 onde o resultado é 28.&lt;br /&gt;&lt;br /&gt;O vídeo mostra o que acontece quando desrespeitamos a ordem das dezenas e das unidades dos números.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;center&gt;&lt;object height="385" width="480"&gt;&lt;param name="movie" value="http://www.youtube.com/v/yHZzObQUgE8&amp;amp;hl=pt_PT&amp;amp;fs=1&amp;amp;"&gt;&lt;param name="allowFullScreen" value="true"&gt;&lt;param name="allowscriptaccess" value="always"&gt;&lt;embed src="http://www.youtube.com/v/yHZzObQUgE8&amp;amp;hl=pt_PT&amp;amp;fs=1&amp;amp;" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" height="385" width="480"&gt;&lt;/object&gt;&lt;/center&gt;&lt;br /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-6335912375900277512?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/6335912375900277512/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/7x1328.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6335912375900277512?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/6335912375900277512?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/7x1328.html' title='7x13=28?'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><thr:total>0</thr:total></entry><entry gd:etag='W/&quot;D0EERX8yfSp7ImA9WxFVGEU.&quot;'><id>tag:blogger.com,1999:blog-939914245193593171.post-7630642406440625765</id><published>2010-06-18T11:20:00.000-07:00</published><updated>2010-06-18T11:26:44.195-07:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-06-18T11:26:44.195-07:00</app:edited><title>Simulado de Matemática do Ensino Médio (Parte 1)</title><content type='html'>&lt;div&gt;Esta avaliação foi aplicada em 2010 no primeiro dia de aula da disciplina de estágio dos alunos do 4º ano de licenciatura em matemática. Esta prova teve como objetivo avaliar o que os alunos conheciam dos assuntos de matemática do Ensino Médio para a preparação das aulas de estágio.&lt;br /&gt;&lt;br /&gt;Aqui apresentamos os problemas, as resoluções podem ser verificadas no link abaixo de cada problema.&lt;br /&gt;&lt;br /&gt;&lt;div align="center"&gt;&lt;strong&gt;&lt;big&gt;Simulado de Matemática do Ensino Médio&lt;/big&gt;&lt;/strong&gt;&lt;/div&gt;&lt;br /&gt;&lt;br /&gt;&lt;div align="left"&gt;&lt;strong&gt;Questão 1:&lt;/strong&gt; A medida do menor ângulo inteiro de um polígono convexo é 139° e as medidas dos outros ângulos formam com a medida deste uma progressão aritmética cuja razão é 2°. Nestas condições, o polígono pode ter quantos lados?&lt;/div&gt;&lt;br /&gt;&lt;div align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-1.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;br /&gt;&lt;br /&gt;&lt;hr /&gt;&lt;br /&gt;&lt;strong&gt;Questão 2:&lt;/strong&gt; De um grupo de 6 homens e 4 mulheres, deseja-se escolher 5 pessoas, incluindo, pelo menos, 2 mulheres. Qual o número de escolhas distintas que se pode fazer?&lt;/div&gt;&lt;div align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-2.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;strong&gt;Questão 3&lt;/strong&gt;: A figura a seguir monstra uma sequência de triângulos de Sierpinski.&lt;/div&gt;&lt;br /&gt;&lt;img style="margin: 0px auto 10px; text-align: center; width: 400px; display: block; height: 141px;" id="BLOGGER_PHOTO_ID_5481235760172867106" alt="" src="http://3.bp.blogspot.com/_Z_uiHIsRFf0/TBFC4g9FniI/AAAAAAAAANI/zeGo_k_ONNs/s400/triangulo-de-sierpinski.jpg" border="0" /&gt;&lt;br /&gt;&lt;div align="left"&gt;O processo começa no nível zero, com um triângulo equilátero de área 1. Em cada passo a seguir, cada triângulo equilátero é dividido através dos seguimentos que ligam os pontos médios dos seus lados e é eliminado o triângulo central assim formado. É correto afirmar que a área que resta no nível &lt;span style="font-weight: bold; font-style: italic;"&gt;n&lt;/span&gt; (indicada nas figuras pela parte branca) é dada por:&lt;/div&gt;&lt;br /&gt;&lt;div align="center"&gt;a) &lt;img style="vertical-align: -13pt;" title="1-\left(\dfrac{1}{4}\right)^n" alt="1-\left(\dfrac{1}{4}\right)^n" src="http://www.forkosh.dreamhost.com/mathtex.cgi?1-%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%5En" /&gt; b) &lt;img style="vertical-align: -13pt;" title="\left(\dfrac{3}{4}\right)^n" alt="\left(\dfrac{3}{4}\right)^n" src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%28%5Cdfrac%7B3%7D%7B4%7D%5Cright%29%5En" /&gt; c) &lt;img style="vertical-align: -13pt;" title="\left(\dfrac{1}{4}\right)^n" alt="\left(\dfrac{1}{4}\right)^n" src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%5En" /&gt; d) &lt;img style="vertical-align: -13pt;" title="1-\left(\dfrac{3}{4}\right)^n" alt="1-\left(\dfrac{3}{4}\right)^n" src="http://www.forkosh.dreamhost.com/mathtex.cgi?1-%5Cleft%28%5Cdfrac%7B3%7D%7B4%7D%5Cright%29%5En" /&gt; e) &lt;img style="vertical-align: -13pt;" title="\left(\dfrac{1}{2}\right)^n" alt="\left(\dfrac{1}{2}\right)^n" src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5En" /&gt;&lt;/div&gt;&lt;div align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-3.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;strong&gt;Questão 4:&lt;/strong&gt; Qual o domínio e a imagem da função &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?x\to \sec x" title="x\to \sec x" alt="x\to \sec x" style="vertical-align:-2pt;" /&gt;?&lt;br /&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-4.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;br /&gt;&lt;hr /&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;&lt;strong&gt;Questão 5&lt;/strong&gt;: A transposição do rio São Francisco é um assunto que desperta grande interesse. Questiona-se, entre outros aspectos, os efeitos no meio ambiente, o elevado custo do empreendimento relativamente a população benificiada e à quantidade de água a ser retirada - o que poderia prejudicar a vazão do riop, que hoje é de 1.850 m³/s. Visãodo promover em sala de aula um debate acerca desse assunto, um professor de matemática propôs a seus alunos o problema seguinte, baseando-se em dados obtidos do Ministério da Integração Nacional. Considere que o projeto prevê a retirada de x m³/s de água. Denote por y o custo total estimado da obra, em bilhões de reais, e por z o número, me milhões, de habitantes que serão beneficiados pelo projeto. Relacionando-se essas quantidades, obtém-se o sistema de equações lineares &lt;img style="vertical-align: -2pt;" title="AX=" alt="AX=" src="http://www.forkosh.dreamhost.com/mathtex.cgi?AX=B" /&gt; em que&lt;/p&gt;&lt;p&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?A=\left[\begin{array}{ccc}1&amp;amp;2&amp;amp;-2\\0&amp;amp;4&amp;amp;-1\\1&amp;amp;0&amp;amp;-2\end{array}\right], B=\left[\begin{array}{c}11\\4\\2\end{array}\right]" title="A=\left[\begin{array}{ccc}1&amp;amp;2&amp;amp;-2\\0&amp;amp;4&amp;amp;-1\\1&amp;amp;0&amp;amp;-2\end{array}\right], B=\left[\begin{array}{c}11\\4\\2\end{array}\right]" alt="A=\left[\begin{array}{ccc}1&amp;amp;2&amp;amp;-2\\0&amp;amp;4&amp;amp;-1\\1&amp;amp;0&amp;amp;-2\end{array}\right], B=\left[\begin{array}{c}11\\4\\2\end{array}\right]" style="vertical-align:-2pt;" /&gt; e &lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?X=\left[\begin{array}{c}x\\y\\z\end{array}\right]" title="X=\left[\begin{array}{c}x\\y\\z\end{array}\right]" alt="X=\left[\begin{array}{c}x\\y\\z\end{array}\right]" style="vertical-align:-2pt;" /&gt;&lt;br /&gt;&lt;/p&gt;&lt;p&gt;Com base nessas informações, é correto afirmar:&lt;/p&gt;&lt;p&gt;a) O sistema linear proposto pelo professor é indeterminado, uma vez que det(A)=0.&lt;/p&gt;&lt;p&gt;b) A transposição proposta vai benificiar menos de 11 milhões de habitantes.&lt;/p&gt;&lt;p&gt;c) Mais de 2% da vazão do rio São Francisco serão retirados com a transposições, o que pode provocar sérios danos ambientais.&lt;/p&gt;&lt;p&gt;d) O cusato total estimado da obra é superior a 4 bilhões de reais.&lt;/p&gt;&lt;p&gt;e) A matriz linha reduzida à forma escalonada, que é linha equivalente à matris A, possui uma coluna nula.&lt;/p&gt;&lt;p&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-5.html"&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;&lt;strong&gt;&lt;span&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/span&gt;&lt;/a&gt;&lt;/p&gt;&lt;p&gt;&lt;span&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;&lt;hr /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;&lt;strong&gt;&lt;span&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;Questão&lt;/span&gt; &lt;/span&gt;&lt;/strong&gt;&lt;/span&gt;&lt;strong&gt;6&lt;/strong&gt;: Dez pessoas fundaram, no início do ano, um clube. Um dos regulamentos de seu regime interno prevê que cada sócio pode apresentar, no máximo, 2 novos sócios ao final de cada ano. A expressão que permite calcular o número máximo de sócios após decorrerem x anos é:&lt;/p&gt;&lt;p align="center"&gt;a) &lt;img style="vertical-align: -2pt;" title="3.10^x+10" alt="3.10^x+10" src="http://www.forkosh.dreamhost.com/mathtex.cgi?3.10%5Ex+10" /&gt; b) &lt;img style="vertical-align: -2pt;" title="2.10^x" alt="2.10^x" src="http://www.forkosh.dreamhost.com/mathtex.cgi?2.10%5Ex" /&gt; c) &lt;img style="vertical-align: -2pt;" title="10+2^x" alt="10+2^x" src="http://www.forkosh.dreamhost.com/mathtex.cgi?10+2%5Ex" /&gt; d) &lt;img style="vertical-align: -2pt;" title="10.2^x" alt="10.2^x" src="http://www.forkosh.dreamhost.com/mathtex.cgi?10.2%5Ex" /&gt; e) &lt;img style="vertical-align: -2pt;" title="10.3^x" alt="10.3^x" src="http://www.forkosh.dreamhost.com/mathtex.cgi?10.3%5Ex" /&gt;&lt;/p&gt;&lt;p&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-6.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;div&gt;&lt;hr /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;&lt;strong&gt;Questão 7: &lt;/strong&gt;&lt;/span&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;A população de uma cidade cresce a uma taxa de 10% ao ano. Se no início de 1995, sua população era de 20.000 habitantes, no decorrer de que ano ela ultrapassará a marca de 40.000 habitantes? (Use log 2=0,3010 e log1,1-0,0414)&lt;/span&gt;&lt;br /&gt;&lt;/div&gt;&lt;p align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-7.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;hr /&gt;&lt;p align="left"&gt;&lt;strong&gt;Questão 8&lt;/strong&gt;: Qual é o valor de k para o qual a reta kx-y-3=0 seja perpendicular à reta&lt;/p&gt;&lt;p align="left"&gt;&lt;img src="http://www.forkosh.dreamhost.com/mathtex.cgi?%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Brcl%7Dx&amp;amp;=&amp;amp;1+2t%5C%5Cy&amp;amp;=&amp;amp;2+3t%5Cend%7Barray%7D%5Cright" title="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;1+2t\\y&amp;amp;=&amp;amp;2+3t\end{array}\right." alt="\left\{\begin{array}{rcl}x&amp;amp;=&amp;amp;1+2t\\y&amp;amp;=&amp;amp;2+3t\end{array}\right." style="vertical-align: -2pt;" /&gt;?&lt;/p&gt;&lt;p align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-8.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;p align="left"&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;&lt;hr /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p align="left"&gt;&lt;strong&gt;Questão 9&lt;/strong&gt;: Um determinado processo diário de produção é descrito por função de custo C(x)=100x+10500 e de remuneração R(x)=600x-5x². Considerando a função lucro, L(x)=R(x)-C(x), determine o número x de bens que fornece o lucro máximo diário.&lt;/p&gt;&lt;p align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-9.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;hr /&gt;&lt;p align="left"&gt;&lt;strong&gt;Questão 10&lt;/strong&gt;: Uma mercadoria foi vendida por R$ 300,00 com um lucro de 20% sobre o preço de custo. Determine o preço de custo da mercadoria.&lt;/p&gt;&lt;p align="left"&gt;&lt;a href="http://adrianogsant.blogspot.com/2010/06/smem-1-questao-10.html"&gt;&lt;strong&gt;&lt;span style="color: rgb(204, 0, 0);"&gt;Resposta.&lt;/span&gt;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/939914245193593171-7630642406440625765?l=adrianogsant.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://adrianogsant.blogspot.com/feeds/7630642406440625765/comments/default' title='Postar comentários'/><link rel='replies' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/simulado-de-matematica-do-ensino-medio_18.html#comment-form' title='0 Comentários'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7630642406440625765?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/939914245193593171/posts/default/7630642406440625765?v=2'/><link rel='alternate' type='text/html' href='http://adrianogsant.blogspot.com/2010/06/simulado-de-matematica-do-ensino-medio_18.html' title='Simulado de Matemática do Ensino Médio (Parte 1)'/><author><name>Adriano G.S.</name><uri>http://www.blogger.com/profile/06032563415124970126</uri><email>noreply@blogger.com</email><gd:extendedProperty name='OpenSocialUserId' value='04325694168470608379'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_Z_uiHIsRFf0/TBFC4g9FniI/AAAAAAAAANI/zeGo_k_ONNs/s72-c/triangulo-de-sierpinski.jpg' height='72' width='72'/><thr:total>0</thr:total></entry></feed>