physics and mathematics

2012-07-30T04:12:00.000+08:00

Question 1

Which of the following points lie on the line

Question 3

What is the equation of the line between the two points and

a)				b)
c)				d)

Question 4

What is the gradient, and -intercept, of the line

a)		and
b)		and
c)		and
d)		and

Question 5

What is the equation of the line below?

a)				b)
c)				d)

Question 6

Which of the following points lie on the line

a)				b)
c)				d)

Question 7

What is the gradient, of the line between the points and

Question 8

What is the equation of the line between the two points and

Question 9

What is the gradient, and -intercept, of the line

Question 10

Which of the following points lie on the line

Question 11

What is the gradient, of the line between the points and

Question 12

What is the equation of the line between the two points and

Question 13

What is the gradient, and -intercept, of the line

Question 14

What is the equation of the line below?

2012-07-21T22:36:00.001+08:00

Question 1

Which of the following points lie on the line

Question 3

What is the equation of the line between the two points and

a)				b)
c)				d)

Question 4

What is the gradient, and -intercept, of the line

a)		and
b)		and
c)		and
d)		and

Question 5

What is the equation of the line below?

a)				b)
c)				d)

Question 6

Which of the following points lie on the line

a)				b)
c)				d)

Question 7

What is the gradient, of the line between the points and

Question 8

What is the equation of the line between the two points and

Question 9

What is the gradient, and -intercept, of the line

Question 10

Which of the following points lie on the line

Question 11

What is the gradient, of the line between the points and

Question 12

What is the equation of the line between the two points and

Question 13

What is the gradient, and -intercept, of the line

Question 14

What is the equation of the line below?

2012-07-21T22:21:00.001+08:00

The Straight Line

Gradient of a Straight Line

Gradient of a Straight Line in Cartesian Coordinates

Intercepts

Equation of a Straight Line

Parallel Lines

2012-07-20T23:55:00.003+08:00

Question 1

A painter mixes gallons of paint in a large cylindrical bucket so that there will be no difference in colour among individual gallons.

If one gallon of paint has a volume of approximately 8000 cm³, what is the maximum number of whole gallons of paint that can be poured into the bucket?

Answer: 3

The Math formulas page will be helpful for this question. You must use the appropriate formula for the volume of a cylinder and recognize that its radius is half of the diameter shown. Once the volume is computed, that answer is divided by 8000. Since only whole gallons are to be poured into the bucket, the quotient is rounded DOWN. The concept of rounding up or down (as appropriate to a particular situation) to produce a whole-number answer is an important one for you to understand.

Volume of bucket:
(3.14) x (20)² x (60) = 75,360 cm³
75,360 / 8000 = 9.42 gallons

Answer is rounded DOWN because 10 whole gallons would not fit. Final answer: 9 gallons

Question 2

A surveyor made the measurements shown in the diagram below.

What is the measure, in feet, of AB, the straight-line distance across the stream?

Answer: 4

You must recognize similar triangles in the diagram and set up the correct proportion, which could be expressed as:

AC	=	AB
EC	=	ED

Letting AB be represented by x and substituting the other measures from the diagram, the proportion is then solved for the unknown measure.

AC	=	AB	becomes	40	=	x
EC	=	ED	becomes	10	=	25

Cross-multiplying: 10x = 1000
Dividing by 10: x = 100
The final answer is 100 feet.

Question 3

Last month, the balance in Tisha's checkbook was $1219.17. Since then she has deposited her latest paycheck of $2425.66 and written checks for $850.00 (rent), $235.89 (car payment), and $418.37 (credit card payment).

What is the current balance in Tisha's checking account?

$ 921.40
$2140.57
$3215.27
$3929.92
$5149.09

Answer: 2

The paycheck would be added to the existing balance, and the checks written would be subtracted. The realistic amounts in the question would require considerable time for the arithmetic to be done by hand, but the values can be entered into the calculator in one continuous string as they are presented in the expression.

New balance = Old balance + Paycheck deposited - Checks written

New balance = 1219.17 + 2425.66 - 850.00 - 235.89 - 418.37

New balance = $2140.57

Question 4

Byron purchased a $5000 certificate of deposit (CD) at his local bank. The CD will pay him 7% simple interest at the end of 2 years.

How much INTEREST, in dollars, will Byron have earned from his CD at the end of the 2-year period?

Mark your answer in the circles in the grid on your answer sheet.

Answer: 700

The formula for simple interest is found on the formulas page in the front of the Mathematics Test.

simple interest = principal x rate x time
simple interest = $5000 x 0.07 x 2
simple interest = $700

Alternate-format responses can be right-, left-, or center-justified, as shown in the examples below.

Question 5

One ingredient in the sauce Kyle is preparing for tonight's dinner is a 1/2 teaspoon of red pepper. If the recipe he is using is designed to make enough sauce for 8 servings, but Kyle is making only 4 servings, what fraction of a teaspoon of red pepper should he use?

Mark your answer in the circles in the grid on your answer sheet.

Answer: 1/4, 0.25, or .25

Since Kyle is serving four from a recipe designed for eight, he needs 4/8, or 1/2, the amount of each ingredient. One half of 1/2 teaspoon is found by multiplying:

(1/2) x (1/2) = 1/4

This answer could also be gridded using the decimal equivalent of 1/4, namely 0.25, or .25 but, in this example, the fraction is gridded to illustrate the use of the fraction bar.

Note: In the last two examples, a decimal answer and a fraction were gridded. Mixed numbers, such as 3 1/2, CANNOT be gridded. They must be changed to decimal or fraction form. In the case of 3 1/2 ,the answer could be gridded as 3.5 or 7/2.

Answers for alternate-format questions using the standard grid shown in the previous two examples will NEVER be a negative number, such as -8, since the grid does not accommodate negative numbers.

Both of the points mentioned above are included in the directions for alternate-format questions found on the actual test.

Question 6

Show the location of the point whose coordinates are (3, -4).

Mark your answer on the coordinate plane grid on your answer sheet.

Answer: See coordinate plane grid

The coordinates in the ordered pair are listed with the x-coordinate (horizontal) first and the y-coordinate (vertical) second. The point (3 -4) is located in the lower-right, or fourth, quadrant of the graph.

2012-07-20T23:41:00.002+08:00

1. Find the slope of the line that passes through the points (-1 , 0) and (3 , 8).

2. Find the slope of the line that passes through the points (2 , 0) and (2 , 4).

3. Find the slope of the line that passes through the points (7 , 4) and (-9 , 4).

4. Are the three points A(2 , 3) , B(5 , 6) and C(0 , -2) collinear?

5. What is the slope of the line parallel to the line whose equation is given by -y = - 2 x + 4?

6. What is the slope of the line perpendicular to the line whose equation is given
by - 2y = - 8 x + 9?

7. Is the triangle whose vertices are the points A(0 , -1) , B(2 , 1) and C(-4 , 3) a right triangle?

8. What is the slope of the line -7y + 8x = 9

9. What is the slope of the line y = 9?

10. What is the slope of the line x = -5?

2012-07-20T23:33:00.001+08:00

1. The product of 2 and 98 can be determined by viewing the problem as 2 × (100 – 2) and using which

of the following number properties?

A. distributive

B. associative

C. commutative

D. transitive

2. A building contractor has n days to complete a house project. If he uses one-third of the time to repair the roof and five days to paint the house, how many days will he have left to complete the project?

A. ⅓n + 5

B. ⅓n – 5

B. ⅔n + 5

D. ⅔n – 5

3. The equation x = ⅞y + 18 can be solved for y to yield which of the following?

A. y = 8/7x – 18

B. y = ⅞x + 18

C. y = 8/7(x – 18)

D. y = ⅞(18 – x)

4. True or False. The two functions f and g defined by

f(x) = 3x + 3 for x real and g(t) = 3t + 3 for t real and positive
are equal?
5. If functions f and g have domains Df and Dg respectively, then the domain of f / g is given by

(A) the union of Df and Dg

(B) the intersection of Df and Dg

(C) the intersection of Df and Dg without the zeros of function g

(D) None of the above
6. True or False. The graph of f(x) and that of f(x + 2) are the same

7. Let the closed interval [a , b] be the domain of function f. The domain of f(x - 3) is given by

(A) the open interval (a , b)

(B) the closed interval [a , b]

(C) the closed interval [a - 3 , b - 3]

(D) the closed interval [a + 3 , b + 3]

8. Let the interval (a , +infinity) be the range of function f. The range of f(x) - 4 is given by

(A) the interval (a - 4 , +infinity)

(B) the interval (a + 4, +infinity)

(C) the interval (a, +infinity)

(C) None of the above

9. True or False. The equation y = | x | , with y >= 0, represents y as a function of x.

10. True or False. The equation x = | y | , with x >= 0, represents y as a function of x.

---------------------------------------------------------------------------------------------------------

Answer:

4. False. Two functions are equal if their rules are equal and their domains are the same.

5. (C). Division by zero is not allowed in mathematics. Students tend to forget this point.

6. False. The graph of f(x + 2) is that of f(x) shifted 2 units to the left.

7. (D). The graph of f(x - 3) is that of f(x) shifted 3 units to the right. To shift the closed interval [a , b] to the right you need to add 3 units to the endpoints a and b of the interval.

8. (A). If the range of f is given by the interval (a , +infinity), we can write the following inequality

f(x) > a

add - 4 to both sides on the inequality to obtain

f(x) - 4 > a - 4

The last inequality suggests that the range of f(x) - 4 is (a - 4, +infinity)

9. True.

10. False. Solve for y to find that y = | x | or y = -| x |; for one value of the independent variable x we have two values of the dependent variable y.

EXAMPLE OF QUESTIONS

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1. Calculate the exact value of

i) 4 - 1.67

ii) 0.4 x 3.5

2. (a) Express 486mm in cm, giving your answer correct to the nearest cm.

(b) Express 5.126 correct to two significant figures.

3. (a) Calculate the total time required for 5 journeys each of 1 hour 34 minutes.

(b) (i) A marathon race started at 11.55 and one of he runners reached the finish at 15.37. Calculate the time

taken by this runner.

(ii) Another runner took 25min for the first 8km. If he continued to run at the same steady speed, calculate

the time taken for the next 12km.

4. Solve the simultaneous equations:

3x - 4y = 25
4x - 5y = 32

x =

y =

5. All of the pupils in a class have four choices in their study of languages.

They may study French only, or French and German, or French and Latin, or all three of these languages.

Draw a clearly labelled Venn diagram to illustrate this information.

6. (a) A rectangular courtyard 17m long by 9¼m wide is to be covered by square paving slabs each of side ½m.

Calculate the number of paving slabs which will be required.

(b) A different area to be paved needs 390 slabs. Given that one-third of these cost £1.20 each and the

remainder cost £1.40 each, calculate the total cost of the slabs.

7. i) On a packet of lettuce seeds it is stated that 85% should produce seedlings. Find the number of seedlings

which would be expected from a packet containing 800 seeds.

ii) (a) A market gardener sows 25000 tomato seeds. Given that 21500 produce seedlings, calculate the

percentage which did not produce seedlings.

(b) In one season he sold 130 tonnes of tomatoes, which was 4% more than he had expected. Calculate

the number of tonnes he had expected to sell.

2012-07-20T02:19:00.001+08:00

Ratios

A ratio is a comparison between two or more like quantities in the same units.

This suggests that a ratio can be simplified by dividing (or multiplying) its terms by the same number.

Note:

The ratio 1 : 2 is read as '1 to 2' or '1 is to 2'.

Simplification of Ratios

Example 1

Express the ratio 6 : 9 in its simplest form.

Solution:

Note:

The ratio 2 : 3 is in its simplest form since the numbers 2 and 3 have no common factor.

Scale Factor

If the ratio is expressed in the form 1 : n, then n is called the scale factor.

E.g. 10 : 100 = 1 : 10

So, 10 is the scale factor.

Equivalent Ratios

Clearly, 5 : 20 = 1 : 4

We say that 5 : 20 and 1 : 4 are equivalent.

In general:

Multiplying both terms of the ratio a : b by the same number, c, results in the equivalent ratio ac : bc.

Finding the Ratio of Two Quantities

We must express both quantities in the same unit of measurement to find the ratio of two quantities in its simplest form.

Example 2

Find the ratio of 40 centimetres to 2 metres in its simplest form.

Solution:

Note:

We can use ratios to compare more than two quantities conveniently. Fractions are not usually suitable for this.

Dividing a Quantity in a Given Ratio

Consider a line segment AB of length 9 cm, which is divided into two line segments of length 3 cm and 6 cm, as shown in the diagram.

That is, AP = 3 cm and BP = 6 cm.

Conversely, if we know that the point P divides the line segment AB of length 9 cm in the ratio
1 : 2, we can determine the lengths of AP and BP as follows:

So, the lengths of AP and BP are 3 cm and 6 cm, respectively.

Check:

3 cm + 6 cm = 9 cm

Example 3

Anne and Allen worked together on a project and received $250 for their completed work. Allen worked for 2 days and Anne worked for 3 days, and they agree to divide the money between them in the ratio 2 : 3. How much should each receive?

Solution:

We picture the $250 divided into equal parts.

Now, there are 5 parts and the smaller amount is 2 of them.

So, the smaller amount (Allen's share) is $100, and so Anne's share is $250 – $100 = $150.

Example 4

An urn contains red and black marbles in the ratio 2 : 3. If there are 40 red marbles, find the total number of marbles in the urn.

Solution:

Ratio of red marbles to black marbles = 2 : 3

Number of red marbles (2 parts out of 5) = 40

Let x be the number of marbles in the urn.

Hence the number of marbles in the urn is 100.

Proportion

The statement of equality of two ratios is called a proportion.

E.g. 2 : 3 = 8 : 12 is a proportion.

The proportion 2 : 3 = 8 : 12 is read as '2 is to 3' as '8 is to 12'.

Example 5

If 2 : 3 = x : 6, find x.

Solution:

Example 6

If 6 : x = 18 : 27, find x.

Solution:

LOGARITHMS

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Log Rules:

1) log_b(mn) = log_b(m) + log_b(n)

2) log_b(^m/_n) = log_b(m) – log_b(n)

3) log_b(mⁿ) = n · log_b(m)

In less formal terms, the log rules might be expressed as:

1) Multiplication inside the log can be turned into addition outside the log, and vice versa.

2) Division inside the log can be turned into subtraction outside the log, and vice versa.

3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.

Warning: Just as when you're dealing with exponents, the above rules work only if the bases are the same. For instance, the expression "log_d(m) + log_b(n)" cannot be simplified, because the bases (the "d" and the "b") are not the same, just as x² × y³ cannot be simplified (because the bases x and y are not the same).

Expanding logarithms

Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values.

Expand log₃(2x).

When they say to "expand", they mean that they've given you one log expression with lots of stuff inside it, and they want you to use the log rules to take the log apart into lots of separate logs, each with only one thing inside. That is, they've given you one log with a complicated argument, and they want you to convert this to many logs, each with a simple argument.

I have a "2x" inside the log. Since "2x" is multiplication, I can take this expression apart and turn it into an addition outside the log:

log₃(2x) = log₃(2) + log₃(x)

The answer they are looking for is:

log₃(2) + log₃(x)

Do not try to evaluate "log₃(2)" in your calculator. While you would be correct in saying that "log₃(2)" is just a number, they're actually looking here for the "exact" form of the log, as shown above, and not a decimal approximation from your calculator.

Expand log₄(¹⁶/_x ).

I have division inside the log, which can be split apart as subtraction outside the log, so:

log₄(¹⁶/_x ) = log₄(16) – log₄(x)

The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is:

log₄(16) = 2

Then the original expression expands fully as:

log₄(¹⁶/_x ) = 2 – log₄(x)

Always remember to take the time to check to see if any of the terms in your expansion (such as thelog₄(16) above) can be simplified.

Expand log₅(x³). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved'

The exponent inside the log can be taken out front as a multiplier:

log₅(x³) = 3 · log₅(x) =

3log₅(x)

Expand the following:

The

is divided into the

8x⁴

, so split the numerator and denominator by using subtraction:

Don't take the exponent out front yet; it is only on the

, not the

, and you can only take the exponent out front if it is "on" everything inside the log. The

is multiplied onto the

x⁴

, so split the factors by using addition:

log₂(

8x⁴

) – log₂(5) = log₂(

) + log₂(

x⁴

) – log₂(5)

The

has an exponent (which is now "on" everything inside its log), so move the exponent out front as a multiplier:

log₂(8) + log₂(x

⁴

) – log₂(5)

= log₂(8) +

log₂(x) – log₂(5)

Since

is a power of

, I can simplify the first log to an exact value:

log₂(8)

+ 4log₂(x) – log₂(5)

Each log contains only one thing, so this is fully simplified. The answer is:

3 + 4log₂(x) – log₂(5)

Expand the following:

Use the log rules, and don't try to do too much in one step:

Then the final answer is:

log₃(4) + 2log₃(x – 5) – 4log₃(x) – 3log₃(x – 1)

Simplify log₂(x) + log₂(y).

Since these logs have the same base, the addition outside can be turned into multiplication inside:

log₂(x) + log₂(y) = log₂(xy)

The answer is log₂(xy).

Simplify log₃(4) – log₃(5).

Since these logs have the same base, the subtraction outside can be turned into division inside:

log₃(4) – log₃(5) = log₃(⁴/₅)

The answer is log₃(⁴/₅).

Simplify 2log₃(x). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

The multiplier out front can be taken inside as an exponent:

2log₃(x) = log₃(x²)

Simplify 3log₂(x) – 4log₂(x + 3) + log₂(y).

I will get rid of the multipliers by moving them inside as powers:

3log₂(x) – 4log₂(x + 3) + log₂(y)

= log₂(x³) – log₂((x + 3)⁴) + log₂(y)

Then I'll put the added terms together, and convert the addition to multiplication:

log₂(x³) – log₂((x + 3)⁴) + log₂(y)

= log₂(x³) + log₂(y) – log₂((x + 3)⁴)

= log₂(x³y) – log₂((x + 3)⁴)

Then I'll account for the subtracted term by combining it inside with division:

Evaluate log₃(6).

The argument is 6 and the base is 3. I'll plug them into change-of-base, using the natural log as my new log:

Then the answer, rounded to three decimal places, is:

log₃(6) = 1.631

You would get the same answer if you used the common log, though the numerator and denominator of the intermediate fraction would be different from what I did above:

While I showed the numerator and denominator values in the above calculations, it is actually best to do the calculations entirely within your calculator. You don't need to bother writing out the intermediate step. To minimize on round-off errors, try to do all the steps for the division and evaluation in your calculator, all in one go. In the above computation, rather than writing down the first eight or so decimal places in the values of ln(6) and ln(3) and then dividing, you would just do "ln(6) ÷ ln(3)" in your calculator.

You may also get some simple (but useless) exercises on this topic, such as:

Convert log₃(6) to base 5.

I can't think of any particular reason why a base-5 log might be useful, so I think the only point of these problems is to give you practice using change-of-base:

Convert ln(4) to an expression written in terms of the common log.

Again, why would you do this (in "real life"), since you can already evaluate the natural log in your calculator? You wouldn't; this exercise is just for practice:

Since getting an actual decimal value is not the point in exercises of this sort (the converting using change-of-base is the point), just leave the answer as a logarithmic fraction.

On the other hand, using change-of-base is handy for finding plot-points when graphing non-standard logs, especially when you are supposed to be using a graphing calculator.

Use your graphing utility to graph y = log₂(x).

If you were working by hand, you would use the definition of logs to note that:

since 2^–1 = ¹/₂, then log₂(1/2) = –1
since 2⁰ = 1, then log₂(1) = 0
since 2¹ = 2, then log₂(2) = 1
since 2² = 4, then log₂(4) = 2
since 2³ = 8, then log₂(8) = 3
since 2⁴ = 16, then log₂(16) = 4

LINEAR GRAPHS

2012-07-13T20:27:00.002+08:00

Relations

In mathematics, we use symbols to denote relations and we build mathematical sentences using numbers, pronumerals and relations.

Consider the following sentence:

The cost (in dollars) of buying pens is equal to ten times the number of pens bought.
If c represents the cost in dollars and p represents the number of pens bought, then this sentence can be expressed mathematically as

This suggests the following definition:

A relation is a set of ordered pairs, and is usually defined by a rule.

Domain

The domain of a relation is the set of all first elements (usually x values) of its ordered pairs.

Range

The range of a relation is the set of all second elements (usually y values) of its ordered pairs.

Note:

The graph of c against p is discrete because p is an element of the set of natural numbers. The values of c depend upon p. So, we say that p is anindependent variable and c is a dependent variable.

Example 1

State the domain and range of the following relations:

Solution:

Functions

A relation is said to be a function if each element of the domain determines exactly one element of the range.

Domain of a Function

The domain of a function is the set of all first elements (usually x values) of its ordered pairs.

Range of a Function

The range of a function is the set of all second elements (usually y values) of its ordered pairs.

Example 2

State the domain and range of the following functions:

Solution:

Gradient of a Straight Line

The gradient of a straight line is the rate at which the line rises (or falls) vertically for every unit across to the right.
That is:

Note:

The gradient of a straight line is denoted by m where:

Example 3

Find the gradient of the straight line joining the points P(– 4, 5) and Q(4, 17).

Solution:

So, the gradient of the line PQ is 1.5.

Note:

If the gradient of a line is positive, then the line slopes upward as the value of x increases.

Example 4

Find the gradient of the straight line joining the points A(6, 0) and B(0, 3).

Solution:

Note:

If the gradient of a line is negative, then the line slopes downward as the value of x increases.

Applications of Gradients

Gradients are an important part of life. The roof of a house is built with a gradient to enable rain water to run down the roof. An aeroplane ascends at a particular gradient after take off, flies at a different gradient and descends at another gradient to safely land. Tennis courts, roads, football and cricket grounds are made with a gradient to assist drainage.

Example 5

A horse gallops for 20 minutes and covers a distance of 15 km, as shown in the diagram.
Find the gradient of the line and describe its meaning.

Solution:

In the above example, we notice that the gradient of the distance-time graph gives the speed (in kilometres per minute); and the distance covered by the horse can be represented by the equation:

Example 6

The cost of transporting documents by courier is given by the line segment drawn in the diagram. Find the gradient of the line segment; and describe its meaning.

Solution:

So, the gradient of the line is 3. This means that the cost of transporting documents is $3 per km plus a fixed charge of $5, i.e. it costs $5 for the courier to arrive and $3 for every kilometre travelled to deliver the documents.

Equation of a Straight Line

To establish a rule for the equation of a straight line, consider the previous example.An increase in distance by 1 km results in an increase in cost of $3. We say that the rate of change of cost with respect to distance is $3 per kilometre.
The information given in the graph can be represented by the equation c = 5 + 3d. That is:

In general:

A line with equation y = mx + c has gradient m and y-intercept c.

The gradient of a straight line is the coefficient of x.

Particular Case

If a straight line passes through the origin, then its y-intercept is 0. So, the equation of a straight line passing through the origin is

y = mx

where m is the gradient of the line.

Example 7

Solution:

Example 8

Write down the equation of the straight line that has m = 5 and c = 3.

Solution:

Example 9

Calculate the gradient of the straight line given in the following diagram; and find its equation.

Solution:

Example 10

Find the equation of the line joining the points (2, 3) and (4,7).

Solution:

Sketch Graphs

Often we need to know the general shape and location of a graph. In such cases, a sketch graph is drawn instead of plotting a number of points to obtain the graph.Two points are needed to obtain a straight line graph. It is simpler to find the points of intersection of the graph with the axes. These points are called thex- and y- intercepts.

x-intercept:

The y-coordinate of any point on the x-axis is 0. Therefore to find the x-intercept we put y = 0 in the equation and solve it for x.

y-intercept:

The x-coordinate of any point on the y-axis is 0. Therefore to find the y-intercept we put x = 0 in the equation and solve it for y.

Example 11

Sketch the graph of y = 3x + 6.

Solution:

y = 3x + 6x-intercept:

y-intercept:

Note:

We often represent the gradient and the y-intercept of the straight line by m and c respectively.

In the previous example:

From the ongoing discussion we can infer that y = 3x + 6 is a straight line with a gradient of 3 and y-intercept of 6.
In the example under consideration, the gradient of the straight line is positive. So, the straight line slopes upward as the value of x increases.

Example 12

Sketch the graph of y = –2x + 4.

Solution:

y = –2x + 4

x-intercept:

y-intercept:

Note:

From the ongoing discussion we find that the linear function y = –2x + 4 represents the equation of a straight line with a gradient of –2 and y-intercept of 4.

In the example under consideration, the gradient of the straight line is negative. So, the straight line slopes downward as the value of x increases.

Example 13

Sketch the graph of y = 2x.

Solution:

y = 2xx-intercept:

y-intercept:

When x = 0, y = 0.

As both the x- and y- intercepts are (0, 0), another point is needed.
We find when x = 5, y = 10. So, (5, 10) is an example of another point that can be used to form the straight line graph.

Alternative technique:

Use the gradient-intercept method:

So, the straight line passes through (0, 0). Use this point to draw a line of slope 2 (i.e. go across 3 units and up 6 units).

Note:

It is simpler to find the run and rise if we start from the y-intercept.

Example 14

Sketch the graph of 7y – 5x = 35.

Solution:

7y – 5x = 35
x-intercept:

y-intercept:

Horizontal Lines

A horizontal line is parallel to the x-axis, as shown in the following diagram.

Note:
The value of the y-coordinate on a horizontal line is always equal to c, the y-intercept.

Example 15

Sketch the graph of y = 8.

Solution:

Vertical Lines

A vertical line is parallel to the y-axis, as shown in the following diagram.

Note:

The value of the x-coordinate on a vertical line is always equal to k, the x-intercept.

Example 16

Sketch the graph of x = 8.

Solution:

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Quadratic Equations

If the highest power of the pronumeral in an equation is two, then it is said to be a quadratic equation.

Solving Quadratic Equations

We will consider the following methods to solve a quadratic equation:

Factorising and using the Null Factor Law
Quadratic formula

The Null Factor Law

From this we can infer that:

If the product of any two numbers is zero, then one or both of the numbers is zero.
That is, if ab = 0, then a = 0 or b = 0 (which includes the possibility that a = b = 0).
This is called the Null Factor Law; and we use it often to solve quadratic equations.

Example 1

Solution:

Equations Involving Factorisation

To solve a quadratic equation, make the right-hand side of the equation zero by transposition. Then factorise the left-hand side, if possible; and then use theNull Factor Law.

Example 2

Solution:

Note:

Example 3

Solution:

Example 4

Solution:

Use of Substitution

Sometimes the solution of an equation can be simplified by the use of substitution, as illustrated in the following example.

Example 5

Solution:

Quadratic Equations Containing Fractions

The solution of an equation containing fractions is obtained by multiplying all terms by the lowest common denominator, making the RHS equal to zero, factorising the LHS and using the Null Factor Law.

Example 6

Solution:

Use of a Square Root

We write the solution as follows:

Example 7

Solution:

Use of the Difference of Two Squares

Recall that:

A solution of a quadratic equation may be obtained by using DOTS, as illustrated below.

Example 8

Solution:

The Quadratic Formula

Recall that:

A quadratic equation can be obtained by using the square completion method, as illustrated below.
Consider the quadratic equation:

This is called the quadratic formula.

Example 9

Solution:

Alternative way:

Note:

If the LHS of an equation can not be factorised, then we use the quadratic formula.
The quadratic formula can be used to solve any quadratic equation.

Example 10

Solution:

FACTORISATION

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Factors of a Number

If a number can be expressed as a product of two whole numbers, then the whole numbers are called factors of that number.

So, the factors of 6 are 1, 2, 3 and 6.

Example 1

Find all factors of 45.

Solution:

So, the factors of 45 are 1, 3, 5, 9, 15 and 45.

Common Factors

10 = 2 × 5 = 1 × 10

Thus, the factors of 10 are 1, 2, 5 and 10.

15 = 1 × 15 = 3 × 5

Thus, the factors of 15 are 1, 3, 5 and 15.

Clearly, 5 is a factor of both 10 and 15. It is said that 5 is a common factor of 10 and 15.

Example 2

Find a common factor of:

a. 6 and 8
b. 14 and 21

Solution:

Prime Numbers

If a number has only two different factors, 1 and itself, then the number is said to be a prime number.

For example, 31 = 1 × 31

31 is a prime number since it has only two different factors.

Note:

But 1 is not a prime number since it does not have two different factors.

Example 3

Express 150 as a product of prime numbers, i.e. find its prime factor decomposition.

Solution:

Note:

We try the prime numbers in order of their magnitude.

We observe that:

The prime factor decomposition of a number is unique.
This is called the Fundamental Theorem of Arithmetic. It provides us with a good reason for defining prime numbers so as to exclude 1. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. This is because we could multiply by 1 as many times as we like in the decomposition.

Highest Common Factor

The highest common factor (HCF) of two numbers (or expressions) is the largest number (or expression) that is a factor of both.

Consider the highest common factor of 16 and 24.

The common factors are 2, 4 and 8. So, the highest common factor is 8.

Note:

The highest common factor is the product of the common prime factors.

In general:

To find the highest common factor of two (or more) numbers, make prime factors of the numbers and identify the common prime factors. Then the highest common factor is the product of the common prime factors.

Example 4

Find the highest common factor of 60 and 150.

Solution:

The prime factorisation of 60 is:

The prime factorisation of 150 is:

Note:

The highest common factor can also be obtained by a trial and error method.
For example, the highest common factor of 40 and 45 is 5 because 5 is the largest number which divides into both 40 and 45 exactly.

Likewise, the highest common factor of 27 and 36 is 9 because 9 is the largest number which divides into both 27 and 36 exactly.

Highest Common Factor of Algebraic Expressions

The highest common factor (HCF) of algebraic expressions is obtained in the same way as that of numbers.

Example 5

Solution:

Factorisation using the Common Factor

We know that:

a(b + c) = ab + ac

The reverse process, ab + ac = a(b + c), is called taking out the common factor.
Consider the factorisation of the expression 5x + 15.

Note that the common factor 5 has been taken out and placed in front of the brackets. The expression inside the brackets is obtained by dividing each term by 5.

In general:

To factorise an algebraic expression, take out the highest common factor and place it in front of the brackets. Then the expression inside the brackets is obtained by dividing each term by the highest common factor.

Example 6

Solution:

Note:

The process of taking out a common factor is of great importance in algebra. With practice you will be able to find the highest common factor (HCF) readily and hence factorise the given expression.

Example 7

Solution:

Factors by Grouping 'Two and Two'

Now, consider the expression 7x + 14y + bx + 2by. Clearly, there is no factor common to every
term. However, it is clear that 7 is a common factor of the first two terms and b is a common factor
of the last two terms. So, the expression can be grouped into two pairs of two terms as shown.

This factorisation technique is called grouping 'Two and Two'; and it is used to factorise an
expression consisting of four terms.

Example 8

Solution:

Factorisation of a Difference of Two Squares

Example 9

Solution:

Taking out a Common Factor

To factorise an expression, take out any common factor; and then use the formula:

Example 10

Solution:

A Calculation Short Cut

Some arithmetic expressions can be evaluated easily using the factorisation of the difference of two squares.

Example 11

Solution:

Factorisation of Quadratic Trinomials

The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below.

Consider the expansion of (x + 2)(x + 3).

We notice that:

5, the coefficient of x, is the sum of 2 and 3.
6, the independent term, is the product of 2 and 3.

Note:

The product of two linear factors yields a quadratic trinomial; and the factors of a quadratic trinomial are linear factors.

Now consider the expansion of (x + a)(x + b).

Coefficient of x = a + b = Sum of a and b.

Independent term = ab = Product of a and b.

In general:

To factorise a quadratic trinomial, find two numbers whose sum is equal to the coefficient of x, and whose product is equal to the independent term.

Example 12

Solution:

Check:

Cross-Multiplication Method

Place the linear factors one above the other as shown below.

Multiply the numbers along the arms of the cross, and then add the products. That is,
x×10 + x×1 = 10x+ x = 11x. This is not the middle term.

x×5 + x×2 = 5x+ 2x = 7x. This is the middle term.

Note:

The solution is read across.
10 = 1 × 10 = 2 × 5 = –1 × –10 = –2 × –5
We did not try –1 × –10 and –2 × –5 because the middle term is positive.
The answer can be checked using the Distributive Law. That is:

Example 13

Factorise the following:

Solution:

     Multiply the numbers along the arms of the cross, and then add the products.

     x×15 + x×1 = 15x+ x = 16x. This is not the middle term.

     We reject this pair as the middle term is 8x. Now, try the next pair.

     x×5 + x×3 = 5x+ 3x = 8x. We accept this pair as the middle term is 8x.

     x× – 4 + x× –3 = – 4x – 3x = –7x. We accept this pair as the middle term is –7x.

Note:

We did not try 1 × 12, 2 × 6 and 3 × 4 because the middle term is negative. Also, we did not try
–2 × –6 and –1 × –12 because we have already obtained the middle term by using –3 × – 4.

x × 7 + x × –1 = 7x – x = 6x. We accept this pair as the middle term is 6x.

Note:

We did not try 1 × –7 because we have already obtained the middle term by using –1 × 7.

Further Quadratic Trinomials

7x×2 + x×1 = 14x+ x = 15x. We reject this pair as the middle term is 9x.

Note:
The solution is read across.

Use of a Common Factor

If a quadratic trinomial has a common factor, take it out and place it in front of the brackets. Then use the cross-multiplication method to factorise the quadratic trinomial.

Example 14

Solution:

Check:

Use of Perfect Squares

Recall that:

Example 15

Solution:

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Simultaneous Equations

Two linear equations in two variables are said to be simultaneous if they are considered at the same time.

Solution of Linear Simultaneous Equations

Simultaneous equations are solved exactly either by the substitution method or the elimination method. An approximate solution can be found by using the graphical method.

Substitution Method

To solve simultaneous equations, find the value of y in terms of x (or vice versa) for one of the two equations and then substitute this value into the other equation.

Example 1

Solve the following simultaneous equations by using the substitution method:

Solution:

Label the equations as follows:

From (1) we have:

y = 3x ...(3)

Substituting y = 3x in (2) gives:

So, the solution is (2, 6).

The Graphical Method

The graphical solution of the simultaneous equations

is given by the point of intersection of the graphs.
Consider the graph of

It passes through the origin (0, 0) and the point (1, 3).

Consider the graph of x + y = 8.

x-intercept: When y = 0, x = 8.

y-intercept: When x = 0, y = 8.

The lines intersect at (2, 6). So, the solution is (2, 6) as shown in the diagram.

Elimination Method

To solve the simultaneous equations, make the coefficients of one of the variables the same value in both equations. Then either add the equations or subtract one equation from the other (whichever is appropriate) to form a new equation that contains only one variable. This is referred to as eliminating the variable.Solve the equation thus obtained. Then substitute the value found for the variable in one of the given equations and solve it for the other variable. Write the solution as an ordered pair.

Example 2

Solve the following simultaneous equations by using the elimination method:

Solution:

Label the equations as follows:

Notice that 3y appears in the left-hand side of both equations. Adding the left-hand side of (1) and (2), and then the right-hand sides, gives:

Note:

We have added equals to equals, and addition eliminates y.
Substituting x = 3 in (1) gives:

So, the solution is (3, 3).

Example 3

Solve the following simultaneous equations by using the elimination method:

Solution:

Label the equations as follows:

Subtracting (2) from (1) gives:

Substituting y = 4 in (1) gives:

So, the solution is (5, 4).

Example 4

Solve the following simultaneous equations by using the elimination method:

Solution:

Label the equations as follows:

Multiplying (1) by 2 and (2) by 3 gives:

Subtracting (3) from (4) gives:

Substituting x = 2 in (1) gives:

So, the solution is (2, 3).

DIFFERENTATION

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Example 1 Differentiate each of the following functions.

(a)

(b)

(c)

(d)

(e)

Solution

(a)

In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the first property from above and then put them back together with the proper sign. Also, for each term with a multiplicative constant remember that all we need to do is “factor” the constant out (using the second property) and then do the derivative.

Notice that in the third term the exponent was a one and so upon subtracting 1 from the original exponent we get a new exponent of zero. Now recall that . Don’t forget to do any basic arithmetic that needs to be done such as any multiplication and/or division in the coefficients.

]

(b)

The point of this problem is to make sure that you deal with negative exponents correctly. Here is the derivative.

Make sure that you correctly deal with the exponents in these cases, especially the negative exponents. It is an easy mistake to “go the other way” when subtracting one off from a negative exponent and get instead of the correct .

(c)

Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator. So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.

Note that we left the 3 in the denominator and only moved the variable up to the numerator. Remember that the only thing that gets an exponent is the term that is immediately to the left of the exponent. If we’d wanted the three to come up as well we’d have written,

so be careful with this! It’s a very common mistake to bring the 3 up into the numerator as well at this stage.

Now that we’ve gotten the function rewritten into a proper form that allows us to use the Power Rule we can differentiate the function. Here is the derivative for this part.

(d)

All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents. Again, remember that the Power Rule requires us to have a variable to a number and that it must be in the numerator of the term. Here is the function written in “proper” form.

In the last two terms we combined the exponents. You should always do this with this kind of term. In a later section we will learn of a technique that would allow us to differentiate this term without combining exponents, however it will take significantly more work to do. Also don’t forget to move the term in the denominator of the third term up to the numerator. We can now differentiate the function.

Make sure that you can deal with fractional exponents. You will see a lot of them in this class.

(e)

In all of the previous examples the exponents have been nice integers or fractions. That is usually what we’ll see in this class. However, the exponent only needs to be a number so don’t get excited about problems like this one. They work exactly the same.

The answer is a little messy and we won’t reduce the exponents down to decimals. However, this problem is not terribly difficult it just looks that way initially.

Example 2 Differentiate each of the following functions.

(a)

(b)

Solution

(a)

In this function we can’t just differentiate the first term, differentiate the second term and then multiply the two back together. That just won’t work. We will discuss this in detail in the next section so if you’re not sure you believe that hold on for a bit and we’ll be looking at that soon as well as showing you an example of why it won’t work.

It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents (as we should anyway) and then multiply this through the parenthesis.

Now we can differentiate the function.

(b)

As with the first part we can’t just differentiate the numerator and the denominator and the put it back together as a fraction. Again, if you’re not sure you believe this hold on until the next section and we’ll take a more detailed look at this.

We can simplify this rational expression however as follows.

This is a function that we can differentiate.

Example 3 Is increasing, decreasing or not changing at ?

Solution

We know that the rate of change of a function is given by the functions derivative so all we need to do is it rewrite the function (to deal with the second term) and then take the derivative.

Note that we rewrote the last term in the derivative back as a fraction. This is not something we’ve done to this point and is only being done here to help with the evaluation in the next step. It’s often easier to do the evaluation with positive exponents.

So, upon evaluating the derivative we get

So, at the derivative is negative and so the function is decreasing at

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Naming the Sides of a Right-Angled Triangle

The side opposite the right angle is called the hypotenuse. It is the largest side of a right-angled triangle.
If you stand at A in the triangle ABC, the side BC is opposite to you and the side AB is next to you. We therefore say that BC is the opposite side to angleA and AB is the adjacent side to angle A.

Notation

Trigonometric Ratios

If you have trouble remembering the definitions, just remember SOH CAH TOA.

Recall that:

This is abbreviated as follows:

Finding Side-Lengths

We can use the trigonometric ratios sine, cosine and tangent to find any side-length of a right-angled triangle if we know another angle and one side-length.

Using the Sine Ratio

Example 1

Find the value of the pronumeral, correct to 2 decimal places, in the given diagram.

Solution:

Example 2

Find the length of the hypotenuse, correct to 2 decimal places, in the given diagram.

Solution:

Let the length of the hypotenuse be x cm.

So, the length of the hypotenuse is 55.01 cm.

Using the Cosine Ratio

Example 3

Find the value of the pronumeral in the following diagram, correct to 2 decimal places.

Solution:

Example 4

Find the length of the hypotenuse in the following diagram, correct to 2 decimal places.

Solution:

Let the length of the hypotenuse be x cm.

So, the length of the hypotenuse is 109.45 cm.

Using the Tangent Ratio

Example 5

Find the value of the pronumeral in the following diagram, correct to 2 decimal places.

Solution:

Example 6

Find the value of the pronumeral in the following diagram, correct to 2 decimal places.

Solution:

Finding Angles

We can use inverse trigonometric functions to find the angle(s) of a right-angled triangle if we know the length of any two sides.

Using Inverse Sine

Example 7

Solution:

Example 8

Solution:

Using Inverse Cosine

Example 9

Solution:

Using Inverse Tangent

Example 10

Solution:

Composite Figures

A diagram consisting of more than one triangle is said to be a composite figure.For trigonometric problems involving a composite figure, first decide whether to use sine, cosine or tangent, and then calculate the required length or angle.

Example 11

In the given diagram, find:
a. xb. y

Solution:

Using a Construction Line

To solve some trigonometric problems, we need to convert the given triangle into two right-angled triangles by drawing a perpendicular construction line from the vertex to the opposite side.

Example 12

Find BC in the given diagram, rounded to 2 decimal places.

Solution:

Draw BD perpendicular to AC. Let BD = x cm, BC = y cm.

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Data

We collect information in order to draw conclusions, to make business decisions or to predict trends. Information that has been gathered in some way is called data. We may collect data ourselves or obtain data from an existing source.

Surveys

The collection of information for a specific purpose is called a survey. Data is needed to help solve a problem. If it is not already available, we collect it by conducting a survey. The main methods of collecting data are by observation, interview and questionnaire. Some surveys use a combination of these.

The following terms are often used when we collect data:

Population

The population is the whole set of items under consideration. E.g. If the information on the best selling book in Australia is required, the population is all books on sale in bookshops in Australia.

Sample

A sample is drawn from a population. So a sample is only a part of the population that has been selected to find an information about the population.

For example, a store manager wants to know the favourite washing powder of customers in his area, so that he can stock the more popular items (i.e. different brands of washing powders). He decides to conduct a survey and choose the first 200 customers as a sample group, asking the question 'what is your favourite washing powder?'

When selecting a sample, two things are taken into consideration:

The sample should be representative of the population (i.e. it should have the same traits as that of the population).
All items of the population should have the same chance of being selected for the sample.

If a sample has these two features, it should be a fair representation of the population and unbiased towards particular sub-groups in the population.

Mean, Median and Mode

We use statistics such as the mean, median and mode to obtain information about a population from our sample set of observed values.

Mean

The mean (or average) of a set of data values is the sum of all of the data values divided by the number of data values. That is:

Example 1

The marks of seven students in a mathematics test with a maximum possible mark of 20 are given below:

15 13 18 16 14 17 12

Find the mean of this set of data values.

Solution:

So, the mean mark is 15.

Symbolically, we can set out the solution as follows:

So, the mean mark is 15.

Median

The median of a set of data values is the middle value of the data set when it has been arranged in ascending order. That is, from the smallest value to the highest value.

Example 2

The marks of nine students in a geography test that had a maximum possible mark of 50 are given below:

47 35 37 32 38 39 36 34 35

Find the median of this set of data values.

Solution:

Arrange the data values in order from the lowest value to the highest value:

32 34 35 35 36 37 38 39 47

The fifth data value, 36, is the middle value in this arrangement.

Note:

In general:

If the number of values in the data set is even, then the median is the average of the two middle values.

Example 3

Find the median of the following data set:

12 18 16 21 10 13 17 19

Solution:

Arrange the data values in order from the lowest value to the highest value:

10 12 13 16 17 18 19 21

The number of values in the data set is 8, which is even. So, the median is the average of the two middle values.

Alternative way:

There are 8 values in the data set.

The fourth and fifth scores, 16 and 17, are in the middle. That is, there is no one middle value.

Note:

Half of the values in the data set lie below the median and half lie above the median.
The median is the most commonly quoted figure used to measure property prices. The use of the median avoids the problem of the mean property price which is affected by a few expensive properties that are not representative of the general property market.

Mode

The mode of a set of data values is the value(s) that occurs most often.The mode has applications in printing. For example, it is important to print more of the most popular books; because printing different books in equal numbers would cause a shortage of some books and an oversupply of others.
Likewise, the mode has applications in manufacturing. For example, it is important to manufacture more of the most popular shoes; because manufacturing different shoes in equal numbers would cause a shortage of some shoes and an oversupply of others.

Example 4

Find the mode of the following data set:

48 44 48 45 42 49 48

Solution:

The mode is 48 since it occurs most often.

Note:

It is possible for a set of data values to have more than one mode.
If there are two data values that occur most frequently, we say that the set of data values is bimodal.
If there is no data value or data values that occur most frequently, we say that the set of data values has no mode.

Frequency and Frequency Tables

The frequency of a particular data value is the number of times the data value occurs.

For example, if four students have a score of 80 in mathematics, and then the score of 80 is said to have a frequency of 4. The frequency of a data value is often represented by f.

A frequency table is constructed by arranging collected data values in ascending order of magnitude with their corresponding frequencies.

Example 5

The marks awarded for an assignment set for a Year 8 class of 20 students were as follows:

6 7 5 7 7 8 7 6 9 7
4 10 6 8 8 9 5 6 4 8

Present this information in a frequency table.

Solution:

To construct a frequency table, we proceed as follows:

Step 1:

Construct a table with three columns. The first column shows what is being arranged in ascending order (i.e. the marks). The lowest mark is 4. So, start from 4 in the first column as shown below.

Step 2:

Go through the list of marks. The first mark in the list is 6, so put a tally mark against 6 in the second column. The second mark in the list is 7, so put a tally mark against 7 in the second column. The third mark in the list is 5, so put a tally mark against 5 in the third column as shown below.

We continue this process until all marks in the list are tallied.

Step 3:

Count the number of tally marks for each mark and write it in third column. The finished frequency table is as follows:

In general:

We use the following steps to construct a frequency table:

Step 1:

Construct a table with three columns. Then in the first column, write down all of the data values in ascending order of magnitude.

Step 2:

To complete the second column, go through the list of data values and place one tally mark at the appropriate place in the second column for every data value. When the fifth tally is reached for a mark, draw a horizontal line through the first four tally marks as shown for 7 in the above frequency table. We continue this process until all data values in the list are tallied.

Step 3:

Count the number of tally marks for each data value and write it in the third column.

Class Intervals (or Groups)

When the set of data values are spread out, it is difficult to set up a frequency table for every data value as there will be too many rows in the table. So we group the data into class intervals (or groups) to help us organise, interpret and analyse the data.

Ideally, we should have between five and ten rows in a frequency table. Bear this in mind when deciding the size of the class interval (or group).

Each group starts at a data value that is a multiple of that group. For example, if the size of the group is 5, then the groups should start at 5, 10, 15, 20 etc. Likewise, if the size of the group is 10, then the groups should start at 10, 20, 30, 40 etc.

The frequency of a group (or class interval) is the number of data values that fall in the range specified by that group (or class interval).

Example 6

The number of calls from motorists per day for roadside service was recorded for the month of December 2003. The results were as follows:

Set up a frequency table for this set of data values.

Solution:

To construct a frequency table, we proceed as follows:

Step 1: Construct a table with three columns, and then write the data groups or class intervals in the first column. The size of each group is 40. So, the groups will start at 0, 40, 80, 120, 160 and 200 to include all of the data. Note that in fact we need 6 groups (1 more than we first thought).

Step 2: Go through the list of data values. For the first data value in the list, 28, place a tally mark against the group 0-39 in the second column. For the second data value in the list, 122, place a tally mark against the group 120-159 in the second column. For the third data value in the list, 217, place a tally mark against the group 200-239 in the second column.

We continue this process until all of the data values in the set are tallied.

Step 3: Count the number of tally marks for each group and write it in the third column. The finished frequency table is as follows:

Frequency Tables and the Mean

To find the mean of a large set of data values, we can use a frequency table. Add an extra column to the frequency table and label it Frequency × Data Value. Then calculate the sum of the values in this fourth column and use it to find the mean.

Example 7

A computer repair service received the following number of calls per day over a period of 30 days.

     6     5     6     9     7     4     2     4     7     8
     3     4     9     8     2     3     5     9     7     8
     9     7     5     6     7     7     4     6     2     4

Using a frequency table, find the average number of calls per day. Round your answer to 1 decimal place.

Solution:

So, the average number of calls per day is 5.8.

Displaying Data

Statistical graphs are often used to display the data values of a random sample. In the following sections, we will consider bar charts, pie charts and line graphs.

Bar Charts

Bar charts are often used to present data in a pictorial form to illustrate the information collected and highlight important points. They are especially useful to depict monthly car production, monthly sales, quarterly profit, average annual rainfall etc. A bar chart provides a useful comparison of data over time. The height of each bar shows the total amount of the item of interest for each month (or year).
Bar charts are drawn with parallel bars placed vertically (or horizontally). The width of each bar and the spacing between the bars are kept the same to avoid giving a misleading representation. The height of the bar is drawn to scale to represent the amount of the item.

Example 8

The yearly production of cars by a particular company is recorded as follows:

Draw a bar chart to display this information.

Solution:

Thus, bars of equal width whose heights are in the ratio of 4 : 5 : 8 will represent the company's yearly production.

Pie Charts

Pie charts are useful to compare different parts of a whole amount. They are often used to present financial information. E.g. A company's expenditure can be shown to be the sum of its parts including different expense categories such as salaries, borrowing interest, taxation and general running costs (i.e. rent, electricity, heating etc).A pie chart is a circular chart in which the circle is divided into sectors. Each sector visually represents an item in a data set to match the amount of the item as a percentage or fraction of the total data set.

Example 9

A family's weekly expenditure on its house mortgage, food and fuel is as follows:

Draw a pie chart to display the information.

Solution:

We can find what percentage of the total expenditure each item equals.
Percentage of weekly expenditure on:

To draw a pie chart, divide the circle into 100 percentage parts. Then allocate the number of percentage parts required for each item.

Line Graphs

A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time.

Example 10

A cylinder of liquid was heated. Its temperature was recorded at ten-minute intervals as shown in the following table.

a. Draw a line graph to represent this information.

b. Estimate the temperature of the cylinder after 25 minutes of heating.

Solution:

b. The estimated temperature after 25 minutes of heating is 52°C.

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Probability

Probability is an interesting branch of mathematics that is widely used in genetics, insurance, finance, medicine, sociological surveys, marketing and science. We use probability to measure the chance or likelihood of an event or events occurring in the future.

Events

An event is something that may or may not occur at some time or during some period in the future.
When we talk about events in terms of chances, we could make statements such as:
"I will probably play tennis this summer"
"It isn't likely that I will be invited to play in the Australian Open Tennis Tournament"
"My chances of winning Tattslotto are not very good"
We could describe the expected occurrence of an event (or events) with the words certain, probable, fifty-fifty, improbable and impossible. These words tell us something about whether or not the event is expected to occur; but the statements are very vague. We study and apply probability to enable us to better quantify or measure the chance of an event (or events) occurring in the future.

The Probability Scale

If an event is impossible, its probability is 0. If an event is certain to occur, its probability is 1. The probability of any other event is between these two values. That is:

Probability Experiment

A probability experiment involves performing a number of trials to enable us to measure the chance of an event occurring in the future. A trial is a process by which an outcome is noted.
For example, consider the following three experiments:
Experiment: Roll a die two hundred times noting the outcomes.
Event of interest: A six faces upwards.
Trial: Roll the die once.
Number of trials: 200
Outcomes: 1, 2, 3, 4, 5 or 6

Experiment: Toss a coin seventy times noting the outcomes.
Event of interest: A tail faces upwards.
Trial: Toss the coin once.
Number of trials: 70
Outcomes: Head or Tail

Experiment: Spin a spinner one hundred times noting the outcomes.
Event of interest: The spinner stops on the number 3.
Trial: Spin the spinner once.
Number of trials: 100
Outcomes: 1, 2, 3, 4

The Sample Space

The sample space of an experiment is the set of all possible outcomes of any trial of the experiment to be conducted.
For example, if an unbiased coin is tossed, then the two possible outcomes are 'head' and 'tail'. The set of all possible outcomes is therefore {H, T}. This is called the sample space of the experiment and is denoted by S.

An element of the sample space is called a sample point.
For the example under consideration, the sample points are H and T.

Activity 16.1

To investigate the probability of obtaining a head (or tail) when a coin is tossed.
Experiment - Toss the coin 100 times and complete the following table.

Outcome	Tally of outcomes	Total number of trials in which the outcome occurred
Tail Head
Sum

If the coin is unbiased (i.e. well balanced), we say it is just as likely to turn up heads as tails. In the activity under consideration, we would expect about half of the coin tosses to turn up heads. Does this agree with your result?

Long Run Proportion

Long run proportion is defined as the ratio of favourable outcomes to the total number of trials in an experiment after conducting a very large number of trials.

When we express the 50 chances of tossing a head (or tail) to the 100 coin tosses as a ratio, we are describing the chances of tossing a head (or tail) in terms of a probability or long run proportion.

From the preceding discussion, we can conclude that:

The concept of long run proportion enables us to determine the probability (or chance) of an event.

Predictions

We use the data collected in surveys to make predictions.

Example 1

A survey was conducted using 200 randomly selected students from across the state to investigate which of 5 sports is played more often? The question put to each student was which sport out of basketball, cricket, football, netball and tennis do you play more often? The tabulated results appear below.

Sport	Number of students
Basketball Cricket Football Netball Tennis	34 10 44 60 52
Sum	200

If the results give a fair indication of how all students in the state would have answered the survey question, calculate the probability that any randomly selected student from the state plays the following sport the most?

a. Basketball
b. Cricket
c. Football
d. Netball
e. Tennis

Solution:

That is, the sum of the probabilities of all events is equal to 1.

Further Probability

Recall that:

The sample space (S) of an experiment is the set of all possible outcomes of any trial of the experiment to be conducted.

An event (E) is a subset of the sample space. That is, an event is a subset of all possible outcomes. We refer to this subset of outcomes as favourable outcomes.

The probability of event E occurring is given by

This is often written as:

This result holds only if the outcomes of an experiment are equally likely.

Note:

The events are denoted by capital letters A, B, C, D, E, ...

Example 2

A die is rolled. Find:
a. the sample space for this experiment
b. the probability of obtaining a number less than 4
c. the probability of obtaining a square number

Solution:

Example 3

A die is rolled. Find the probability of obtaining:
a. a 7
b. a number less than or equal to 6

Solution:

Note:

It is certain that event A will occur as it contains all 6 possible outcomes.
7 is not an outcome of rolling a die as it is not possible.

Example 4

A pack of 52 playing cards consists of four suits, i.e. clubs, spades, diamonds and hearts. Each suit has 13 cards which are the 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king and the ace card. Clubs and spades are of black colour whereas diamonds and hearts are of red colour. So, there are 26 red cards and 26 black cards.
Find the probability of drawing from a well-shuffled pack of cards:
a. a black card
b. the queen of diamonds
c. a king

Solution:

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Equations

Equations enable us to describe complex problems in simple terms. They are built with numbers, pronumerals and an equal sign.It is clear that the number sentence
5 + 10 = 15
is an equation.
If the value of the pronumeral x is 5, then x can take the place of 5; and we can write this equation as
x + 10 = 15
This is an equation containing the pronumeral x. The value of the pronumeral x is 5.

Example 1

Describe each of the following equations in words:

Solution:

Example 2

Write an equation to represent each of the following statements:
a. When I add 7 to a number, the answer is 16.
b. When I subtract 8 from a number, the answer is 23.
c. When I multiply a number by 9, the answer is 27.
d. When I divide a number by 5, the answer is 29.

Solution:

Equations and a Pair of Scales

Recall that:

An equation is a statement that contains an equal sign.

Consider the simple equation
x = 5
Visualise this equation as a balanced pair of scales with x and 5 measured in kilograms.

If we add 3 kg to the scale on the left-hand side, the scales will balance as long as we add 3 kg to the scale on the right-hand side. That is, x + 3 = 8.

Also, if we subtract the same weight, say 3 kg, from each side of the balance, the scales will remain balanced. That is, x – 3 = 2.

If we double the weight in the scale on the left-hand side, the scales will balance as long as we double the weight in the scale on the right-hand side. That is, 2x = 10.

Also, if we halve the weight in each scale of the balance, the scales will remain balanced.

Solving Equations

Solving an equation means to find the value of a pronumeral that makes a statement true.

In the preceding section, we observed that:

An equation behaves like a pair of balanced scales. The scales remain balanced as long as we do the same thing to both scales.
This suggests that to solve an equation, we can do the same thing to both sides of an equation. That is:

The same number can be subtracted from both sides of an equation.
The same number can be added to both sides of an equation.
Both sides of an equation can be divided by the same number.
Both sides of an equation can be multiplied by the same number.

We will now consider equations involving addition, subtraction, multiplication and division.
Operations such as +, –, × and ÷ are used to build an equation. To solve an equation, we use inverse (i.e. opposite) operations such that the pronumeral is the only term remaining on the left-hand side.

Equations Involving Addition

The inverse operation of + is –. So, to solve an equation involving addition, we undo the addition by subtracting the same number from both sides.

Example 3

Solve the equation x + 6 = 14.

Solution:

Note:

6 is added to x. So, we undo the addition by subtracting 6 from both sides.

Check:

Equations Involving Subtraction

The inverse operation of – is +. So, to solve an equation involving subtraction, we undo the subtraction by adding the same number to both sides.

Example 4

Solve the equation x – 9 = 17.

Solution:

Note:

9 is subtracted from x. So, we undo the subtraction by adding 9 to both sides.

Check:

Equations Involving Multiplication

The inverse operation of × is ÷. So, to solve an equation involving multiplication, we divide both sides of the equation by the same number.

Example 5

Solve the equation 8x = 72.

Solution:

Note:

x is multiplied by 8. So, we undo the multiplication by dividing both sides by 8.

Check:

Equations Involving Division

The inverse operation of ÷ is ×. So, to solve an equation involving division, we multiply both sides of the equation by the same number.

Example 6

Solution:

Note:

x is divided by 6. So, we undo the division by multiplying both sides by 6.

Check:

Problem Solving 1

Linear equations help us to solve word problems. First, we assume that the number we are trying to find is represented by a pronumeral. Then the problem given in words is translated into an equation which is solved using the methods we have learned for solving equations. Finally, we write the answer in words.

Example 7

A number is added to 85 and the result is 172. Find the number.

Solution:

Let the number be x.
So, the number is 87.

Check:

Equations Involving Two or More Operations

To solve an equation involving two or more operations, start by carrying out an inverse operation on the number that is furthest away from x.

Example 8

Solve 7x + 4 = 25.

Solution:

The number 4 is furthest away from x. The inverse of + 4 is – 4. So, subtract 4 from both sides.

Problem Solving 2

Example 9

If three times a number, when diminished by 4, equals 17, what is the number?

Solution:

Let x be the number.
Three times x is 3x, and diminishing this by 4 gives 3x – 4, which we are told equals 17.

So, the number is 7.

Equations with the Pronumeral on Both Sides

When a pronumeral is on both sides of an equation, remove the pronumeral term from the right-hand side of the equation by using inverse operations. Then continue to use inverse operations to solve the equation for the pronumeral.

Example 10

Solution:

Problem Solving 3

Example 11

If 7 less than three times a number is 9 more than the number, what is the number?

Solution:

Let x be the number.

Seven less than three times x is 3x – 7, and 9 more than x is x + 9.

So, the number is 8.

Check:

Equations Containing Brackets

To solve the equation containing brackets, we may proceed as follows:

Remove the brackets by using the Distributive Law.
Collect the pronumeral terms on the left-hand side of the equation and the numerical terms on the right-hand side of the equation by doing the same thing to both sides of the equation.

Example 12

Solution:

Example 13

Solution:

Problem Solving 4

Example 14

Find the width of a rectangular paddock whose length is 60 m and perimeter is 220 m.

Solution:

So, the width of the paddock is 50 m.

Equations Containing Fractions

To solve equations containing fractions:

Find the lowest common multiple of the denominators which is known as the lowest common denominator (LCD).
Remove the fractions by multiplying both sides of the equation by the LCD.
Solve the equation for the unknown pronumeral by performing the same operations to both sides of the equation.

Example 15

Solution:

Lowest common multiple of 3 and 1 is 3. So, we multiply both sides by 3 to obtain:

Example 16

Solution:

Lowest common multiple of 8 and 3 is 24. So, we multiply both sides by 24 to obtain:

Problem Solving 5

Read the problem carefully and describe it by an equation. Then solve the equation and write the answer as a sentence.

Example 17

The sum of three consecutive numbers is 87. What are the numbers?

Solution:

Let the smallest number be n.
The word consecutive means one after the other. So, the next two consecutive numbers after n will be n + 1 and n + 2.

So, the three consecutive numbers are 28, 29 and 30.

Example 18

Use the information given in the following diagram to form an equation and then solve it to find the value of the pronumeral.

Solution:

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Indices

We know that:
5 × 5 = 25The product 5 × 5 can be written as 5².

5 × 5 is known as the expanded form (or factor form) of 25 and 5² is known as the index form of 25.
Generally when a number is multiplied by itself any number of times, the expression is simplified by using the index notation.

Note:

2 is called the base.
3 is called the index or power (or exponent) because it indicates the power to which the base, 2, is raised.
8 is the basic numeral (or number).
2³ is read as '2 to the power 3' or simply '2 cubed'.

That is:

Example 1

Write 4³ as a number.

Solution:

Note the following:

64 = 4³
3 is the power (or index or exponent)
4 is the base number
64 is a basic numeral or number
4³ is the index form (or power form) of 64
4 × 4 × 4 is the expanded form of 64
For 64 = 4 × 4 × 4 = 4³, the base number 4 appears three times as a factor of the basic numeral (or number) 64
4³ is read as '4 to the power 3' or simply '4 cubed'

Example 2

Write each of the following expanded forms in index form:

Solution:

Example 3

Write each of the following in expanded form:

Solution:

Example 4

Find the value of the following:

Solution:

Example 5

Write 16 in index form using base 2.

Solution:

Example 6

Write the following numbers as a product of prime factors:

Solution:

Index Laws

In the following sections, we will consider the six index laws.

Index Law for Multiplication

We know that:

In general:

This formula tells us that when multiplying powers with the same base, add the indices.
This is the first index law and is known as the Index Law for Multiplication.

Example 7

Solution:

Index Law for Division

We know that:

In general:

This formula tells us that when dividing powers with the same base, the index in the denominator is subtracted from the index in the numerator.
This is the second index law and is known as the Index Law for Division.

Example 8

Solution:

Note:

Simplify the numerical coefficients first, and then apply the index law.

Power of Zero

In general:

This formula tells us that any number, except 0, raised to the power zero has a numerical value of 1.
This is the third index law and is known as the Power of Zero.

Example 9

Solution:

Index Law for Powers

We know that:

In general:

This formula tells us that when a power of a number is raised to another power, multiply the indices.
This is the fourth index law and is known as the Index Law for Powers.

Example 10

Solution:

Note:

Always remove brackets first.

Example 11

Simplify each of the following:

Solution:

Index Law for Powers of Products

We know that:

In general:

This formula tells us that when a product is raised to a power, every factor of the product is raised to the power.
This is the fifth index law and is known as the Index Law for Powers of Products.

Example 12

Simplify each of the following:

Solution:

Example 13

Solution:

Index Law for Powers of Quotients

We know that:

In general:

This formula tells us that when a quotient is raised to a power, both the numerator and denominator are raised to the power.
This is the sixth index law and is known as the Index Law for Powers of Quotients.

Example 14

Simplify each of the following:

Solution:

Example 15

Solution:

Physics

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INTRODUCTION OF PHYSICS

Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.

Physics is one of the oldest academic disciplines, perhaps the oldest through its inclusion of astronomy. Over the last two millennia, physics was a part ofnatural philosophy along with chemistry, certain branches of mathematics, andbiology, but during the Scientific Revolution in the 16th century, the natural sciences emerged as unique research programs in their own right. Physics intersects with many interdisciplinary areas of research, such as biophysics and quantum chemistry, and the boundaries of physics are not rigidly defined. New ideas in physics often explain the fundamental mechanisms of other sciences, while opening new avenues of research in areas such as mathematics and philosophy.

Physics also makes significant contributions through advances in new technologies that arise from theoretical breakthroughs. For example, advances in the understanding ofelectromagnetism or nuclear physics led directly to the development of new products which have dramatically transformed modern-day society, such as television, computers, domestic appliances, and nuclear weapons; advances in thermodynamics led to the development of industrialization; and advances in mechanics inspired the development of calculus.

Though physics deals with a wide variety of systems, certain theories are used by all physicists. Each of these theories were experimentally tested numerous times and found correct as an approximation of nature (within a certain domain of validity). For instance, the theory of classical mechanics accurately describes the motion of objects, provided they are much larger than atoms and moving at much less than the speed of light. These theories continue to be areas of active research, and a remarkable aspect of classical mechanics known as chaos was discovered in the 20th century, three centuries after the original formulation of classical mechanics by Isaac Newton (1642–1727).

These central theories are important tools for research into more specialized topics, and any physicist, regardless of his or her specialization, is expected to be literate in them. These include classical mechanics, quantum mechanics, thermodynamics and statistical mechanics, electromagnetism, and special relativity.

Mathematics is the language used for compact description of the order in nature, especially the laws of Physics. This was noted and advocated by Pythagoras, Plato, Galileo,and Newton.

Physics theories use Mathematics to obtain order and provide precise formulas, precise or estimated solutions, quantitative results and predictions. Experiment results in physics are numerical measurements. Technologies based on Mathematics, likecomputation have made computational physics an active area of research.

The distinction between Mathematics and Physics is clear-cut, but not always obvious, especially in Mathematical Physics.

Ontology is a prerequisite for Physics, but not for Mathematics. It means Physics is ultimately concerned with descriptions of the real world, while Mathematics is concerned with abstract patterns, even beyond the real world. Thus Physics statements are synthetic, while Math statements are analytic. Mathematics contains hypothesis, while Physics contains theories. Mathematics statements have to be only logically true, while predictions of Physics statements must match observed and experimental data.

The distinction is clear-cut, but not always obvious. For example, Mathematical Physics is the application of Mathematics in Physics. Its methods are Mathematical, but its subject is Physical. The problems in this field start with a "Math model of a Physical situation" and a "Math description of a Physical law". Every math statement used for solution has a hard-to-find Physical meaning. The final Mathematical solution has an easier-to-find meaning, because it is what the solver is looking for.

Physics is a branch of fundamental science, not practical science. Physics is also called "the fundamental science" because the subject of study of all branches of natural science like Chemistry, Astronomy, Geology and Biology are constrained by laws of physics. For example, Chemistry studies properties, structures, and reactions of matter (chemistry's focus on the atomic scaledistinguishes it from physics). Structures are formed because particles exert electrical forces on each other, properties include physical characteristics of given substances, and reactions are bound by laws of physics, like conservation of energy, mass and charge.

Physics is applied in industries like engineering and medicine.

Applied physics is a general term for physics research which is intended for a particular use. An applied physics curriculum usually contains a few classes in an applied discipline, like geology or electrical engineering. It usually differs from engineering in that an applied physicist may not be designing something in particular, but rather is using physics or conducting physics research with the aim of developing new technologies or solving a problem.

The approach is similar to that of applied mathematics. Applied physicists can also be interested in the use of physics for scientific research. For instance, people working on accelerator physics might seek to build better particle detectors for research in theoretical physics.

Physics is used heavily in engineering. For example, statics, a subfield of mechanics, is used in the building of bridges and other structures. The understanding and use of acoustics results in better concert halls; similarly, the use of optics creates better optical devices. An understanding of physics makes for more realistic flight simulators, video games, and movies, and is often critical in forensic investigations.

With the standard consensus that the laws of physics are universal and do not change with time, physics can be used to study things that would ordinarily be mired in uncertainty. For example, in thestudy of the origin of the Earth, one can reasonably model Earth's mass, temperature, and rate ofrotation, over time. It also allows for simulations in engineering which drastically speed up the development of a new technology.

But there is also considerable interdisciplinarity in the physicist's methods and so many other important fields are influenced by physics, e.g. the fields of econophysics and sociophysics.

Theorists seek to develop mathematical models that both agree with existing experiments and successfully predict future results, while experimentalists devise and perform experiments to test theoretical predictions and explore new phenomena. Although theory and experiment are developed separately, they are strongly dependent upon each other. Progress in physics frequently comes about when experimentalists make a discovery that existing theories cannot explain, or when new theories generate experimentally testable predictions, which inspire new experiments.

Physicists who work at the interplay of theory and experiment are called phenomenologists. Phenomenologists look at the complex phenomena observed in experiment and work to relate them to fundamental theory.

Theoretical physics has historically taken inspiration from philosophy; electromagnetism was unified this way. Beyond the known universe, the field of theoretical physics also deals with hypothetical issues, such as parallel universes, a multiverse, and higher dimensions. Theorists invoke these ideas in hopes of solving particular problems with existing theories. They then explore the consequences of these ideas and work toward making testable predictions.

Experimental physics informs, and is informed by, engineering and technology. Experimental physicists involved in basic research design and perform experiments with equipment such as particle accelerators and lasers, whereas those involved in applied research often work in industry, developing technologies such as magnetic resonance imaging (MRI) andtransistors. Feynman has noted that experimentalists may seek areas which are not well explored by theorists.

Physics involves modeling the natural world with theory, usually quantitative. Here, the path of a particle is modeled with the mathematics of calculus to explain its behavior: the purview of the branch of physics known as mechanics.

Physics covers a wide range of phenomena, from elementary particles (such as quarks, neutrinos and electrons) to the largest superclusters of galaxies. Included in these phenomena are the most basic objects composing all other things. Therefore physics is sometimes called the "fundamental science". Physics aims to describe the various phenomena that occur in nature in terms of simpler phenomena. Thus, physics aims to both connect the things observable to humans to root causes, and then connect these causes together.

For example, the ancient Chinese observed that certain rocks (lodestone) were attracted to one another by some invisible force. This effect was later called magnetism, and was first rigorously studied in the 17th century. A little earlier than the Chinese, the ancient Greeksknew of other objects such as amber, that when rubbed with fur would cause a similar invisible attraction between the two. This was also first studied rigorously in the 17th century, and came to be called electricity. Thus, physics had come to understand two observations of nature in terms of some root cause (electricity and magnetism). However, further work in the 19th century revealed that these two forces were just two different aspects of one force –electromagnetism. This process of "unifying" forces continues today, and electromagnetism and the weak nuclear force are now considered to be two aspects of the electroweak interaction. Physics hopes to find an ultimate reason (Theory of Everything) for why nature is as it is (see section Current researchbelow for more information)

Contemporary research in physics can be broadly divided into condensed matter physics; atomic, molecular, and optical physics;particle physics; astrophysics; geophysics and biophysics. Some physics departments also support research in Physics education.

Since the twentieth century, the individual fields of physics have become increasingly specialized, and today most physicists work in a single field for their entire careers. "Universalists" such as Albert Einstein (1879–1955) and Lev Landau (1908–1968), who worked in multiple fields of physics, are now very rare.

Condensed matter physics is the field of physics that deals with the macroscopic physical properties of matter. In particular, it is concerned with the "condensed" phases that appear whenever the number of constituents in a system is extremely large and the interactions between the constituents are strong.

The most familiar examples of condensed phases are solids and liquids, which arise from the bonding and electromagnetic force between atoms. More exotic condensed phases include the superfluidand the Bose–Einstein condensate found in certain atomic systems at very low temperature, the superconducting phase exhibited by conduction electrons in certain materials, and the ferromagnetic and anti ferromagnetic phases of spins on atomic lattices.

Condensed matter physics is by far the largest field of contemporary physics. Historically, condensed matter physics grew out of solid-state physics, which is now considered one of its main subfields. The term condensed matter physics was apparently coined by Philip Anderson when he renamed his research group — previously solid-state theory — in 1967.

In 1978, the Division of Solid State Physics at the American Physical Society was renamed as the Division of Condensed Matter Physics. Condensed matter physics has a large overlap with chemistry, materials science, nanotechnology and engineering.

Atomic, molecular, and optical physics (AMO) is the study of matter-matter and light-matter interactions on the scale of single atomsand molecules. The three areas are grouped together because of their interrelationships, the similarity of methods used, and the commonality of the energy scales that are relevant. All three areas include both classical, semi-classical and quantum treatments; they can treat their subject from a microscopic view (in contrast to a macroscopic view).

Atomic physics studies the electron shells of atoms. Current research focuses on activities in quantum control, cooling and trapping of atoms and ions, low-temperature collision dynamics and the effects of electron correlation on structure and dynamics. Atomic physics is influenced by the nucleus (see, e.g., hyperfine splitting), but intra-nuclear phenomenon such as fission and fusion are considered part of high energy physics.

Molecular physics focuses on multi-atomic structures and their internal and external interactions with matter and light. Optical physicsis distinct from optics in that it tends to focus not on the control of classical light fields by macroscopic objects, but on the fundamental properties of optical fields and their interactions with matter in the microscopic realm.

Particle physics is the study of the elementary constituents of matter and energy, and theinteractions between them. It may also be called "high energy physics", because many elementary particles do not occur naturally, but are created only during high energycollisions of other particles, as can be detected in particle accelerators.

Currently, the interactions of elementary particles are described by the Standard Model. The model accounts for the 12 known particles of matter (quarks and leptons) that interact via thestrong, weak, and electromagnetic fundamental forces. Dynamics are described in terms of matter particles exchanging gauge bosons (gluons, W and Z bosons, and photons, respectively). The Standard Model also predicts a particle known as the Higgs boson, the existence of which has not yet been verified; as of 2010, searches for it are underway in theTevatron at Fermilab and in the Large Hadron Collider at CERN.

Astrophysics and astronomy are the application of the theories and methods of physics to the study of stellar structure, stellar evolution, the origin of the solar system, and related problems of cosmology. Because astrophysics is a broad subject, astrophysicists typically apply many disciplines of physics, including mechanics, electromagnetism, statistical mechanics, thermodynamics, quantum mechanics, relativity, nuclear and particle physics, and atomic and molecular physics.

The discovery by Karl Jansky in 1931 that radio signals were emitted by celestial bodies initiated the science of radio astronomy. Most recently, the frontiers of astronomy have been expanded by space exploration. Perturbations and interference from the earth’s atmosphere make space-based observations necessary for infrared, ultraviolet, gamma-ray, and X-ray astronomy.

Physical cosmology is the study of the formation and evolution of the universe on its largest scales. Albert Einstein’s theory of relativity plays a central role in all modern cosmological theories. In the early 20th century, Hubble's discovery that the universe was expanding, as shown by the Hubble diagram, prompted rival explanations known as the steady state universe and the Big Bang.

The Big Bang was confirmed by the success of Big Bang nucleosynthesis and the discovery of the cosmic microwave background in 1964. The Big Bang model rests on two theoretical pillars: Albert Einstein's general relativity and the cosmological principle. Cosmologists have recently established the ΛCDM model of the evolution of the universe, which includes cosmic inflation, dark energyand dark matter.

Numerous possibilities and discoveries are anticipated to emerge from new data from the Fermi Gamma-ray Space Telescope over the upcoming decade and vastly revise or clarify existing models of the Universe. In particular, the potential for a tremendous discovery surrounding dark matter is possible over the next several years.Fermi will search for evidence that dark matter is composed of weakly interacting massive particles, complementing similar experiments with the Large Hadron Collider and other underground detectors.

IBEX is already yielding new astrophysical discoveries: "No one knows what is creating the ENA (energetic neutral atoms) ribbon" along the termination shock of the solar wind, "but everyone agrees that it means the textbook picture of the heliosphere — in which the solar system's enveloping pocket filled with the solar wind's charged particles is plowing through the onrushing 'galactic wind' of the interstellar medium in the shape of a comet — is wrong."

Research in physics is continually progressing on a large number of fronts.

In condensed matter physics, an important unsolved theoretical problem is that of high-temperature superconductivity. Many condensed matter experiments are aiming to fabricate workable spintronics and quantum computers.

In particle physics, the first pieces of experimental evidence for physics beyond the Standard Model have begun to appear. Foremost among these are indications that neutrinos have non-zero mass. These experimental results appear to have solved the long-standing solar neutrino problem, and the physics of massive neutrinos remains an area of active theoretical and experimental research. Particle accelerators have begun probing energy scales in theTeV range, in which experimentalists are hoping to find evidence for the Higgs boson andsupersymmetric particles.

Theoretical attempts to unify quantum mechanics and general relativity into a single theory ofquantum gravity, a program ongoing for over half a century, have not yet been decisively resolved. The current leading candidates are M-theory, superstring theory and loop quantum gravity.

Many astronomical and cosmological phenomena have yet to be satisfactorily explained, including the existence of ultra-high energy cosmic rays, the baryon asymmetry, theacceleration of the universe and the anomalous rotation rates of galaxies.

Although much progress has been made in high-energy, quantum, and astronomical physics, many everyday phenomena involving complexity, chaos, or turbulence are still poorly understood. Complex problems that seem like they could be solved by a clever application of dynamics and mechanics remain unsolved; examples include the formation of sandpiles, nodes in trickling water, the shape of water droplets, mechanisms ofsurface tension catastrophes, and self-sorting in shaken heterogeneous collections.

These complex phenomena have received growing attention since the 1970s for several reasons, including the availability of modern mathematical methods and computers, which enabled complex systems to be modeled in new ways. Complex physics has become part of increasingly interdisciplinary research, as exemplified by the study of turbulence in aerodynamics and the observation of pattern formation in biological systems

Algebra

2012-07-10T02:57:00.001+08:00

Algebra Questions With Answers

How many x intercepts does the graph of the function f(x) = x³ + 4 have?

A. 0 B. 1 C. 2 D. 3
If Log_b(5) = c, then b^4c =

A. 4 B. 5 C. 25 D. 625
How many points of intersection do the graphs of the functions f(x) = x² and g(x) = 2^xhave?

A. 1 B. 2 C. 3 D. 4
The range of the function f(x) = -x² + 2x - 5 is given by

A. (0 , infity) B. (-infinity , 4] C. (-infinity , -4] D. (-infinity , -5]
Which of these functions does not have an inverse?

A. f(x) = |x - 5| B. h(x) = 2x - 9

C. k(x) = ln(- x + 4) D. j(x) = x^1/3
The solution set of the inequality (x + 5)⁴(x + 3) ≥ 0 is given by interval

A. [-3 , infinity) B. {-5} U [-3 , infinity) C. (-infinity , infinity) D. (-3 , infinity)
The range of the function h(x) = -|x - 5| - 2 is give by

A. (-infinity , 0) B. (-infinity , -2) C. (-infinity , -2] D. (-infinity , 2]
5 ln e² - 2 ln (7e) + ln 49 =

A. 8 B. 3 C. 4 ln 7 D. 8e - 4 ln 7
If f(x) = x / (- x + 4) and g(x) = 2x + 2, then (f ₀ g)(1) =

A. 4/3 B. 4 1/3 C. 4 D. undefined
√(x² + 6x + 9) =

A. x + √(6x) + 3 B. x + 3 C. |x + 3| D. (x + 3)²
If f(x) = x² + 1 for x ≤ 0, then f^-1(x) =

A. x -1 B. √(x - 1) C. √(x + 1) D. - √(x - 1)
The domain of f(x) = ln(2 - x) + ln(3 - x) is given by

A. (-infinity , 2) B. (2 , 3) C. (-infinity , 3) D. (-infinity , 2]
The solution set of the inequality (x - 1)² ≤ 0 is given by

A. (-infinity , infinity) B. (-infinity , 1] C. {1} D. [1 , infinity)
The solution to the equation 2 (4^x+1) = 4 (8^{x + 2})

A. -5 B. 1 C. 0 D. no solution
The number of solutions of the equation | - x² + 4x + 1 | = 5 is equal to

A. 1 B. 2 C. 3 D. 4
Using the absolute value concept, the statement

The distance from x to -6 is not less than 7

may be written as follows

A. |x + 6| ≥ 7 B. |x + 6| > 7 C. |x + 7| ≥ -6 D. |x + 7| ≥ 6
If f(2x + 3) = 2x - 1, then f(3x + 4) =

A. 6x + 7 B. 3x C. 6x + 11 D. 9x + 15
-x² + x - 3 =

A. -(x + 1/2)²-11/4 B. (x - 1/2)²-11/4 C. -(x - 1/2)²-11/4 D. -(x - 1/2)²+11/4
How many solutions do the equation 1/x + 1/(x + 1) = 1/3 have?

A. 0 B. 1 C. 2 D. 3
If the whole graph of function f lies in quadrant IV, then the graph of the inverse of lies in quadrant

A. I B. II C. III D. IV

Answers to the Above Questions
1. B
2. D
3. B
4. C
5. A
6. B
7. C
8. A
9. D
10. C
11. D
12. A
13. C
14. A
15. C
16. A
17. B
18. C
19. C
20. D

Algebra

2012-07-10T02:46:00.001+08:00

Algebra Questions With Answers

Write 230,000,000,000 in scientific notation.
Evaluate: 30 - 12÷3×2 =
Evaluate: |4 - 8(3 - 12)| - |5 - 11| =
Evaluate: -18 + 4(6 ÷ 2)²
Evaluate: 11 + sqrt(-4 + 6×4÷3)
Simplify: 12x³ - 3(2x³ + 4x -1) - 5x + 7
Simplify: (x⁴ / x³)³
Simplify: (3x²y^-2)³ / (9xy³)³
Simplify: (2x^-3y⁴)³(x³ + y)⁰ / (4xy^-2)³
Write as a mathematical inequality:"9 is less than the product of M and N".
Find the slope of the line perpendicular to the line y = (1/3)x - 7
Write an equation of the line with slope -3 and y-intercept (0 , -5).
Solve the equation: -5x + 20 = 25
Solve the inequality: -3x + 4 < -8
Solve the equation: 2x² - 32 = 0
Solve the equation: -0.25x + 1.3 = -0.55x - 0.2
Solve the equation: -0.25x² + 1.5 = -10.75
What is the slope of a line perpendicular to the line x = -3?
What is the slope of a line parallel to the line x = 5?
What is the slope of a line perpendicular to the line y = 6?

----------------------------------------------------------------------------------------------------------------------------------

Answers to the Above Questions

2.3 × 10¹¹
22
70
18
13
6x³ -17x + 10
x³
(1/27)(x³ / y¹⁵)
(1/8)(y¹⁸ / x¹²)
9 < M × N
-3
y = -3x - 5
-1
x > 4
-4 , 4
-5
-7 , 7
0
undefined
undefined

Algebraic Formulas

2012-07-09T23:53:00.000+08:00

Indices Formulas:

The formulas involving relations between variables and their powers or powers and indices are:

If , then , x=y. ( Provided That : )

If, then , a=b. ( Provided That : )

LAW AND PRINCIPLE IN PHYSICS

2012-07-09T00:08:00.004+08:00

Charles's law

Charles's law (also known as the law of volumes) is an experimental gas law which describes how gases tend to expand when heated. It was first published by French natural philosopher Joseph Louis Gay-Lussac in 1802, although he credited the discovery to unpublished work from the 1780s by Jacques Charles. The law was independently discovered by British natural philosopher John Dalton by 1801, although Dalton's description was less thorough than Gay-Lussac's. The basic principles had already been described a century earlier by Guillaume Amontons.

Taylor Buchanan was the first to demonstrate that the law applied generally to all gases, and also to the vapours of volatile liquids if the temperature was more than a few degrees above the boiling point. His statement of the law can be expressed mathematically as:

where V₁₀₀ is the volume occupied by a given sample of gas at 100 °C; V₀ is the volume occupied by the same sample of gas at 0 °C; and k is a constant which is the same for all gases at constant pressure. Gay-Lussac's value for k was ¹⁄_2.6666, remarkably close to the present-day value of ¹⁄_2.7315.

Volume and temperature are directly proportional, as long as the pressure stays at a constant.

A modern statement of Charles' law is:

At constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature on the absolute temperature scale(i.e. the gas expands as the temperature increases).

which can be written as:

where V is the volume of the gas; and T is the absolute temperature. The law can also be usefully expressed as follows:

The equation shows that, as absolute temperature increases, the volume of the gas also increases in proportion

Bernoulli's principle

In fluid dynamics, Bernoulli's principle states that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.Bernoulli's principle is named after the Dutch-Swissmathematician Daniel Bernoulli who published his principle in his book Hydrodynamica in 1738.

Bernoulli's principle can be applied to various types of fluid flow, resulting in what is loosely denoted as Bernoulli's equation. In fact, there are different forms of the Bernoulli equation for different types of flow. The simple form of Bernoulli's principle is valid for incompressible flows (e.g. most liquid flows) and also for compressible flows (e.g. gases) moving at low Mach numbers. More advanced forms may in some cases be applied to compressible flows at higher Mach numbers (see the derivations of the Bernoulli equation).

Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamlineis the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. Thus an increase in the speed of the fluid occurs proportionately with an increase in both its dynamic pressure and kinetic energy, and a decrease in its static pressure and potential energy. If the fluid is flowing out of a reservoir, the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit volume (the sum of pressure and gravitational potential ρ g h) is the same everywhere.

Bernoulli's principle can also be derived directly from Newton's 2nd law. If a small volume of fluid is flowing horizontally from a region of high pressure to a region of low pressure, then there is more pressure behind than in front. This gives a net force on the volume, accelerating it along the streamline.

Fluid particles are subject only to pressure and their own weight. If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure; and if its speed decreases, it can only be because it has moved from a region of lower pressure to a region of higher pressure. Consequently, within a fluid flowing horizontally, the highest speed occurs where the pressure is lowest, and the lowest speed occurs where the pressure is highest.

Boyle's law

Boyle's law (sometimes referred to as the Boyle–Mariotte law) is one of many gas lawsand a special case of the ideal gas law. Boyle's law describes the inversely proportional relationship between the absolute pressure and volume of a gas, if the temperature is kept constant within a closed system. The law was named after chemist and physicist Robert Boyle, who published the original law in 1662.The law itself can be stated as follows:

For a fixed amount of an ideal gas kept at a fixed temperature, P [pressure] and V [volume] are inversely proportional (while one doubles, the other halves).

Equation

The mathematical equation for Boyle's law is:

where:

p denotes the pressure of the system.

V denotes the volume of the gas.

k is a constant value representative of the pressure and volume of the system.

So long as temperature remains constant the same amount of energy given to the system persists throughout its operation and therefore, theoretically, the value of k will remain constant. However, due to the derivation of pressure as perpendicular applied force and the probabilistic likelihood of collisions with other particles through collision theory, the application of force to a surface may not be infinitely constant for such values of k, but will have a limit when differentiating such values over a given time.

Forcing the volume V of the fixed quantity of gas to increase, keeping the gas at the initially measured temperature, the pressure pmust decrease proportionally. Conversely, reducing the volume of the gas increases the pressure.

Boyle's law is used to predict the result of introducing a change, in volume and pressure only, to the initial state of a fixed quantity of gas. The before and after volumes and pressures of the fixed amount of gas, where the before and after temperatures are the same (heating or cooling will be required to meet this condition), are related by the equation:

Here P₁ and V₁ represent the original pressure and volume, respectively, and P₂ and V₂ represent the second pressure and volume.

Boyle's law, Charles's law, and Gay-Lussac's law form the combined gas law. The three gas laws in combination with Avogadro's lawcan be generalized by the ideal gas law.

Archimedes' principle

Archimedes' principle is a law of physics stating that the upward force ( buoyancy ) exerted on a body immersed in a fluid is equal to the weight of the amount of fluid the body displaces. In other words, an immersed object is buoyed up by a force equal to the weight of the fluid it actually displaces. Archimedes' principle is an important and underlying concept in the field of fluid mechanics. This principle is named after its discoverer, Archimedes of Syracuse.

Assuming Archimedes' principle to be reformulated as follows,

then inserted into the quotient of weights, which has been expanded by the mutual volume

yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes:

(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)

Example: If you drop wood into water, buoyancy will keep it afloat.

Example: A helium balloon in a moving car. In increasing speed or driving a curve, the air moves in the opposite direction of the car's acceleration. The balloon however, is pushed due to buoyancy "out of the way" by the air, and will actually drift in the same direction as the car's acceleration. When an object is immersed in a liquid the liquid exerts an upward force which is known as buoyant force and it is proportional to the weight of displaced liquid. The sum force acting on the object, then, is proportional to the difference between the weight of the object ('down' force) and the weight of displaced liquid ('up' force), hence equilibrium buoyancy is achieved when these two weights (and thus forces) are equal.

Ohm's law

Ohm's law states that the current through a conductor between two points is directlyproportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance one arrives at the usual mathematical equation that describes this relationship:

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current

Pascal's law

Pascal's law or the principle of transmission of fluid-pressure is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure ratio (initial difference) remains the same. The law was established by French mathematician Blaise Pascal

Pascal's principle is defined

A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid

This principle is stated mathematically as:

where

is the hydrostatic pressure (given in pascals in the SI system), or the difference in pressure at two points within a fluid column, due to the weight of the fluid;

ρ is the fluid density (in kilograms per cubic meter in the SI system);

g is acceleration due to gravity (normally using the sea level acceleration due to Earth's gravity inmetres per second squared);

is the height of fluid above the point of measurement, or the difference in elevation between the two points within the fluid column (in metres in SI).

The intuitive explanation of this formula is that the change in pressure between two elevations is due to the weight of the fluid between the elevations. Note that the variation with height does not depend on any additional pressures. Therefore Pascal's law can be interpreted as saying that any change in pressure applied at any given point of the fluid is transmitted undiminished throughout the fluid.

assignment for additional mathematics

2012-07-08T01:38:00.000+08:00

CONTENT

Acknowledgement

Objectives

Introduction

Part 1

Part 2

Part 3

Further Exploration

Conclusion

Reflection

ACKNOWLEDGEMENT

First of all, I would like to say thank you, for giving me the strength to do this project work.

Not forgotten my parents for providing everything, such as money, to buy anything that are related to this project work and their advise, support which are the most needed for this project.Internet, books, computers and all that. They also supported meand encouraged me to complete this task so that I will not procrastinate in doing it.

Then I would like to thank my teacher forguiding me and my friends throughout this project. We had some difficulties in doing this task, but she taught us patiently until weknew what to do. She tried and tried to teach us until weunderstand what we supposed to do with the project work. Last but not least, my friends who were doing this project with me and sharing our ideas. They were helpful that when we combined and discussed together, we had this task done.

OBJECTIVES

- The aims of carrying out this project work are:

- to apply and adapt a variety of problem-solving strategies tosolve problems.

- to improve thinking skills.

- to promote effective mathematical communication.

- to develop mathematical knowledge through problem solving in a way that increases students‟ interest and confidence.

- to use the language of mathematics to express mathematicalideas precisely.

- to provide learning environment that stimulates and enhanceseffective learning.

- to develop positive attitude towards mathematics

INTRODUCTION

We as additional mathematic learner has been asked to do project about solving problem using additional mathematic.This year we have been given a choice of choosing between two topic.The topics are sketching track field and assorted muffin.We must choose one of the topic and finish the assignment according to our priority in answering the question.This project can be done individual or group,and with pleasant i choose to do individualy.When this project is done

- Experienced a challenging classroom environment, interesting and meaningful, then the improve thinking skills

- Experience a classroom environment where knowledge and skills used in a meaningful way in solving real-life problems

- Experience completing projects to stimulate and enhance effective learning

- Acquire mathematical skills effectively through oral and written, and using the language of mathematics to express mathematical ideas and accurate

- Mathematical knowledge and skills through problem solving in ways that increase their interest and confidence

- Prepare ourselves to meet our future in the field of endeavor

- Realize that mathematics is an important and powerful tool in solving problems in life and to develop positive attitudes towards mathematics

- Train ourselves to appreciate the intrinsic value of mathematics and be more creative and innovative

- Realize the importance and beauty of mathematics

Part 1

i)Show that an athlete running in the first lane will cover 400 metres in one lap of the track.

r = 38.2

= 180° X π/180°

= π

s=r θ

s = 38.2 X π

s = 120 M

s=r θ

s = 38.2 X π

s = 120 M

Total Perimeter = 120 M + 120 M + 80 M + 80 M = 400 M

Part 2

(i) Calculate the distance of running in the second lane until the eight lane.

Complete the table 1.

Lane	Diameter (m)	Perimeter / Running Distance (m)
1	76.4	400
2	78.4	406.3
3	80.4	412.58
4	82.4	418.87
5	84.4	425.15
6	86.4	431.43
7	88.4	437.72
8	90.4	444

Table 1

(i)Used another method to find the running distance / perimeter.

a=400

d=6.3

Tn=a+(n-1)d

Lane	Diameter (m)	Calculation	Perimeter (m)
1	76.4	400+(1-1)6.3	400
2	78.4	400+(2-1)6.3	406.3
3	80.4	400+(3-1)6.3	421.6
4	82.4	400+(4-1)6.3	418.9
5	84.4	400+(5-1)6.3	425.2
6	86.4	400+(6-1)6.3	431.5
7	88.4	400+(7-1)6.3	437.8
8	90.4	400+(8-1)6.3	444.1

Table 2

Part 3

(a) (i)Calculate the difference distance from the starting position for each lane . The running distance of each athlete should 400 metre for a lap.

Complete Table 2.

Comparision between lanes	Calculation	Difference distance (m)
Lane 1 & Lane 2	406.3 – 400	6.3
Lane 2 & Lane 3	412.58 – 406.30	6.3
Lane 3 & Lane 4	418.86 – 412.58	6.3
Lana 4 & Lane 5	425.15 – 418.86	6.3
Lane 5 & Lane 6	431.43 – 425.15	6.3
Lane 6 & Lane 7	437.71 – 431.43	6.3
Lane 7 & Lane 8	443 – 437.71	6.3

Table 3

(ii)Used another method to find the difference distance between the lane

a=406.3-400

a=6.3

Lane	Diameter Difference (m)	Calculation	Arc Difference (m)
Lane 1 & Lane 2	2	2 X 3.141	6.282
Lane 2 & Lane 3	2	2 X 3.141	6.282
Lane 3 & Lane 4	2	2 X 3.141	6.282
Lana 4 & Lane 5	2	2 X 3.141	6.282
Lane 5 & Lane 6	2	2 X 3.141	6.282
Lane 6 & Lane 7	2	2 X 3.141	6.282
Lane 7 & Lane 8	2	2 X 3.141	6.282

Table 4

(b)Sketch the athletic track plan with eight lanes, showing the starting position each lane.

Further Question

1) To construct a 400 (m) track in your school field,it is suggested to have at least four lanes. Suggest the minimum area for track. Give your reasons.

diameter 4 lanes-83.4 M

Area = Length x Witdh

Length=75.4+4+4

=83.4

Witdh= 80+4+4+ (75.4/2)+(75.4/2)

=163.4

83.4 x 163.4 =13627.56m²

2) Calculate the area of the school field.How many maximum lanes can be constructed based on your findings ?

Hence ,if your school wish to construct the eight lanes track,find the approximate minimum area needed. Give your reasons.

Area field

Length x Witdh

=200m x 500m

=50000m²

Minimum area for 8 lanes= length x Witdh

Length= 75.4 + 8 + 8 =91.4

Witdh=80 + (75.4/2) + (75.4/2) + 8 +8 =171.4

=171.4 X 91.4= 15665.96 m²

Field for can build 3 track 8 lanes because 15665.96 m² x 3 =46997.88 m²

3) For the standard sport even, a site for discus throw with a sector shape will be built in the field of the athletic track. You can find the information from internet , sport magazines and sport teachers.

a) Find the approximate area for the site.

Area = Length x Witdh

Length= 50 + 2.5/2 = 51.25 m

Witdh= 2.5+ 0.75 + 0.75= 4

Area= 51.25 x 4 = 205

b) Using the suitable scale, draw and label the diagram of the site.Determine the standard angle of the center of the sector.

c) Suggest the range of the distance for the score obtained ( score1, score2, score3 )

50 / 3 = 16.66666667

d) Calculate the area covered representing each point / score.

Area sector 1= 1/2 (16.667)² x ( 34.92 x (22/7 /180)

=84.65185469

Area sector 2 = 1/2 (16.667 + 16.667)² x (34.92 x (22/7 /180)

= 338.7437081 – 84.65185649

= 254.0918534

Area sector 3 = 1/2 (16.667 + 16.667 +16.667)² x (34.92 x (22/7 /180)

=762.1733432 – 254.0918534 – 84.65185469

= 423.4296351

CONCLUSION

After doing research, answering the questions, plan a table and some problem solving,we saw that usage of circular measure is important in our daily activity. It is not justwidely use in the sport but also in certain job like the engineer used this method to solve the problem. We learnt a lot of lesson from this Additional Mathematics Project Work such as measure skill using different method. Without this,the athletes cannot used the track effectively because of the measure is not accurate.We would like to thanks the one who contribute the idea of circular measure and progression to help us a lot in our activity everyday.

A. f(x) = \|x - 5\|	B. h(x) = 2x - 9


C. k(x) = ln(- x + 4)	D. j(x) = x^1/3

physics and mathematics

The Straight Line

Question 1

Answer: 3

Question 2

Answer: 4

AC

=

AB

EC

ED

AC

=

AB

becomes

40

x

EC

ED

10

25

Question 3

Answer: 2

Question 4

Answer: 700

Question 5

Answer: 1/4, 0.25, or .25

Question 6

Answer: See coordinate plane grid

EXAMPLE OF QUESTIONS

Ratios

Note:

Simplification of Ratios

Example 1

Solution:

Note:

Scale Factor

Equivalent Ratios

In general:

Finding the Ratio of Two Quantities

Example 2

Solution:

Note:

Dividing a Quantity in a Given Ratio

Check:

Example 3

Solution:

Example 4

Solution:

Proportion

Example 5

Solution:

Example 6

Solution:

LOGARITHMS

LINEAR GRAPHS

Relations

Consider the following sentence:

This suggests the following definition:

Domain

Range

Note:

Example 1

Solution:

Functions

Domain of a Function

Range of a Function

Example 2

Solution:

Gradient of a Straight Line

Note:

Example 3

Solution:

Note:

Example 4

Solution:

Note:

Applications of Gradients

Example 5

Solution: