MathsAnimation

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noreply@blogger.com (fatah rohmani) — Sun, 20 Sep 2026 10:49:03 +0000

Maths Interactive Animations and Games

noreply@blogger.com (fatah rohmani) — Tue, 12 Aug 2025 22:23:00 +0000

1) How to Find Gradient and Y intercept Based on Two Points. Click:

👉🏻👉🏻👉🏻 DEMO LINK 👉🏻👉🏻👉🏻 REAL LINK

2) Opposite, Adjacent and Hypotenuse Position of Trigonometry. Click:

👉🏻👉🏻👉🏻 REAL LINK

Matching Object Interactive Worksheet

noreply@blogger.com (fatah rohmani) — Mon, 12 May 2025 13:19:00 +0000

Open this worksheet,click HERE

Trigonometric Ratio, Quadrants Worksheet and GAME

noreply@blogger.com (fatah rohmani) — Sun, 04 May 2025 06:40:00 +0000

Trigonometric Ratio, Quadrants Worksheet and GAME

Download the worksheet 👉👉👉👉👉 HERE

Basic Algebra Equations and Basic Algebra Word Problems, For Beginners

noreply@blogger.com (fatah rohmani) — Wed, 05 May 2021 23:02:00 +0000

Basic Algebra Equations and Basic Algebra Word Problems, For Beginners

Basic Algebra Equations is like 2 apples plus 3 oranges plus 4 apples equal to 6 apples plus 3 oranges. Apples and oranges are variables for basic algebra equation in here. Variables is stands for vary and ables, it's means able to become vary, could be x , y, z, apples, oranges etc. Basic Algebra Equation may contains expand equation from it's original equation, example: 2 baskets of 3 apples and 4 oranges for each basket. So become 6 apples and 8 oranges in total. Same equation with 2 ( 3x + 4y) = 6 x + 8y.

Basic Algebra is just simple operations are like addition, substraction, multiplication or division with variables and constans involved in mathematics or in daily life.

Basic Algebra word problem is like "John has 3 baskets of 2 candies and 4 chocolates, how many candies and chocolates which John has?" Basic Algebra word problem here could be solve with expansion 3(2 candies + 4 chocolates) = 6 candies + 12 chocolates.

Another Basic Algebra word problem is like "Maryam has 7 bags of 3 iphones and 2 laptops, how many iphones and laptops which Maryam has?" Basic Algebra word problem here could be solve with expansion 7(3 iphones + 2 laptops) = 21 iphones + 14 laptops.

Another Basic Algebra equation is like 6x + 9y = 3(2x+3y) because 6x and 9y have common factor, e.g 3. So 3 could be put out of bracket here.

Another Basic Algebra equation is like 10m+ 15n = 5(2m+3n) because 10m and 15n have common factor, e.g 5. So 5 could be put out of bracket here.

Another Basic Algebra word problem is like "A house has 20 rooms, for each room has 1 table, 2 chairs, 2 beds. How many tables, chairs and beds does this house has in total?

Answer of this Basic Algebra equations is 20 ( table + 2 chairs + 2 beds) = 20 tables + 40 chairs + 40 beds in total.

Another Basic Algebra equation is like 9 ( p + 2q - r ) - 3 ( p - q + 3r ) = 9p + 18q - 9r - 3p + 3q - 9r, Simplifying Basic Algebra expression to become 6p + 21q - 18r .

Another Basic Algebra equation is like 7m ( x + 2p - r ) - 2m ( x - p + 3r ) = 7mx + 14mp - 7mr - 2mx + 2mp - 6mr, Simplifying Basic Algebra expression to become 5mx + 16mp - 13mr .

Other Basic Algebra equations is like 3 US dollars + 5 Rupees +10 US dollar -1 Rupee = 13 US dollar + 4 Rupees

Basic Algebra could be found in our daily life, with Basic Algebra examples above. I hope this Basic Algebra article will help you to understand about Basic Algebra in daily life.

Another Basic Algebra example is "Miranda just bought 3 apples with prices 6 US dollars in total. How much price 1 apple is?"

Basic Algebra equations in animation video below:

TRIANGULAR PRISM VOLUME FORMULA | TRIANGULAR PRISM BASE AREA AND ITS CROSS SECTION AREA

noreply@blogger.com (fatah rohmani) — Sat, 30 Jun 2018 22:29:00 +0000

TRIANGULAR PRISM VOLUME FORMULA | TRIANGULAR PRISM BASE AREA AND ITS CROSS SECTION AREA

The diagram shows a solid in the shape of a triangular prism

AC = 5 cm, BC = 4 cm and CD = 7 cm.

Angle ABC = 90°.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

(a) What does the word prism tell you about the solid in the diagram?

Answer:

Constant or uniform cross-sectional area of triangular prism. It means if you cut parallel to the base side, it will be same area with area of the base. The diagram shown below, red area will same with blue area of triangular prism.

(b) Show that AB = 3 cm.

Answer:

By using Phytagoras Formula, 5² = 4² + 3²

It shown that AB = 3 cm.

Loot at triangular prism picture above.

Give the units of your answer.

Answer:

Triangular Prism volume formula is Base Area x Height

= Area of Triangle ABC x Height of Triangular Prism

= ½ x 4 x 3 x 7

= 42 cm squares.

ELECTRICITY AND GAS PRICES FOR FAMILY USE, CHARGED BY ENERGY COMPANY

noreply@blogger.com (fatah rohmani) — Fri, 29 Jun 2018 22:18:00 +0000

An energy company charged these prices in 2013.

Electricity price

23.15 cents per day

plus

13.5 cents for each unit used

Gas price

24.5 cents per day

plus

5.5 cents for each unit used

(Mathematics Cambridge IGCSE Test 0580 s17 qp 41)

(a) i) In 90 days, the Siddique family used 1885 units of electricity.

Calculate the total cost, in dollars, of the electricity they used.

Answer:

90 days x 23.15 cents + 1885 units used x 13.5 cents

= 27531 cents

= $275.31

So, $275.31 is the total cost of Siddique family used the electricity, in dollar, in 90 days.

ii) In 90 days, the gas used by the Khan family cost $198.16 .

Calculate the number of units of gas used.

Answer:

Where m is the total number of units used by Khan family.

$198.16 = 90 days x 24.5 cents + m units used x 5.5 cents

$198.16 = 2205 + 5.5 m

19816 cents - 2205 = 5.5 m

17611 cents = 5.5 m

3202 units = m

So, Khan family used 3202 units in total in 90 days

(b) In 2013, the price for each unit of electricity was 13.5 cents.

Over the next 3 years, this price increased exponentially at a rate of 8% per year. Calculate the price for each unit of electricity after 3 years.

Answer:

This question is similar with compound interest question, because there is a word ‘increased exponentially’. So we have to use compound interest formula to calculate the answer.

The informations given are:

Principal (initial price/capital/investment) = 13.5 cents

Rate = 8% per year.

Time = 3 years.

Amount of price after 3 years is:

A = P (1+ R)^T

A = 13.5 (1+ 8/100)^3

A = 17 cents

(c) Over these 3 years, the price for each unit of gas increased from 5.5 cents to 7.7 cents.

(i) Calculate the percentage increase from 5.5 cents to 7.7 cents.

Answer:

Let say 5.5 cents as 100% or initial value.

5.5 cents → 100%

1 cents → 100% divided by 5.5 = 18.18%

7.7 cents → 7.7 x 18.18% = 140%

So, the percentage increase over 3 years is 140% - 100% = 40%

(ii) Over the 3 years, the 5.5 cents increased exponentially by the same percentage each year to 7.7 cents.

Calculate the percentage increase each year.

Answer:

Because it increased exponentially, we must use compound interest formula to calculate percentage increase from 5.5 cents to 7.7 cents for over 3 years.

Amount after 3 years = P(1+R/100)^3 years

7.7 cents = 5.5 cents(1+R/100)^3 years

7.7/5.5 = (1+R/100)^3

1.4 = (1+R/100)^3

Cuberoot(1.4) = 1+R/100

1.1187 = 1+R/100

0.1187 = R/100

11.87 = R

So, each year, the percentage increase is 11.87 percents.

(d) In 2015, the energy company divided its profits in the ratio

shareholders : bonuses : development = 5 : 2 : 6.

In 2015, its profits were $390 million.

Calculate the amount the company gave to shareholders.

Answer:

The ratio in the simplest form is 5 : 2 : 6.

The total parts of ratio are 13 parts, with 5 parts are given to shareholders.

Let say $390 million as 13 parts or total value/parts.

13 parts → $390 million

1 parts → $390 million divided by 13 parts = $30 million

5 parts → $30 million x 5 parts = $150

So, shareholders will get $150 from the company

(e) The share price of the company in June 2015 was $258.25 .

This was an increase of 3.3% on the share price in May 2015.

Calculate the share price in May 2015.

Answer:

Let say $258.25 as 103.3%, because this price is already increase 3.3% compared to share price of previous month, May 2015.

And the share price in May 2015 as 100%.

103.3% → $258.25

1% → $258.25 divided by 103.3 = $2.5

100% → $2.5 x 100 = $250

So, in May 2015 the price of share was $250

FORMULA OF TRIANGLE AREA WITH A KNOWN ANGLE

noreply@blogger.com (fatah rohmani) — Thu, 28 Jun 2018 22:33:00 +0000

Calculate the area of this triangle.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

Answer:

Based on the formula area of triangle

Where a and b are adjacent sides of angle c. This formula could be applied for any triangle (scalene, right or isosceles triangle).

Area of Triangle

= 1/2 a b sin ⁡c
= 1/2 × 27.6 × 22.3 sin ⁡25°
= 130.1

So the area of triangle is 130.1 cm square.

==================================================

SIMILAR QUESTION (EXAMPLE ONLY/ FICTIF QUESTION)

A Mesothelioma attorney has a triangular land as shown below,

This location will be used for car donation program where people would give some used car for charity. Find the area of this triangular land.

Answer:

Based on the formula area of triangle with a known angle.

Where a and b are adjacent sides of angle c. This formula could be applied for any triangle (scalene, right or isosceles triangle).

Area of Triangle

= 1/2 a b sin ⁡c

= 1/2 × 290 ×340 sin 30°

= 24650 metre squares

So, this Mesothelioma attorney’s triangular land area is metre square.

DAILY AND WEEKLY HORSES EXERCISE SCHEDULE IN MATHEMATICAL FRACTION

noreply@blogger.com (fatah rohmani) — Wed, 27 Jun 2018 22:26:00 +0000

Whisper is a horse.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

(a) Whisper eats three apples every day.

Work out how many apples she eats in 52 weeks.

Answer:

52 weeks are same with 52 x 7 days = 364 days

She eats 3 apples x 364 days = 1092 apples.

(b) Whisper is exercised twice a day.

The first time is for 30 minutes.

The second time is for 1.5 hours.

(i) Write down the fraction of a day that Whisper is exercised.

Write your answer in its simplest form.

Answer:

Total time Whisper do the exercise per day is 30 minutes + 1.5 hours, will be equal to 2

hours.
The fraction of a day that Whisper is exercised is 2 / 24 hours per day, will be equal to
1/12 day in the simplest form.

(ii) Write down the fraction of a day that she is not being exercised.

Answer:

The fraction of a day that she is not being exercised is 1 day is substracted by 1/12 day, will be equal 11/12 day.

Answer:

Because of 429 kg is already rounded to the correct nearest kilogram,
the lower bound before 429 is rounded off is equal to 428.5 or above.
And the upper bound before 429 is rounded off is below than 429.5, not equal to 429.5.
Because, if the value of w was equal to 429.5, it would be rounded off to 430 kg.
So the range value before 429 is rounded off are 428.5 ≤ w < 429.5

======================================================

SIMILAR QUESTION (EXAMPLE QUESTION ONLY)

Bello is a male horse.

(a) Bello eats 0.7 pounds of pasture grass per hour.
Calculate how many pounds of pasture grass he eats in 56 hours.

Answer:

He eats 0.7 pounds of pasture grass x 56 hours = 39.2 pounds of pasture grass.

(b) Bello is exercised three times per week.
In one day, his first time exercise is for 45 minutes and his second time is for 1.5 hours.

(i) Write down the fraction of a week that Bello is exercised.
Give your answer in the simplest form.

Answer:

Total time Bello do the exercise per day is 45 minutes + 1.5 hours, will be equal to 2 hours
and 15 minutes, or 2.25 hours.
The fraction of a day that Bello is exercised is 2.25 / 24 hours per day, will be equal to
9375/100000 day .
The fraction of a week that Bello is exercised is 7 days x 9375/100000, will be equal to

65625/100000 week or 21/32 week

(ii) Write down the fraction of a week that Bello is not being exercised.

Answer:

The fraction of a week that Bello is not being exercised is

1 week is substracted by 21/32 week, will be equal 11/32 week.

(c) Bello weighs 507,000 gr, correct to the nearest gram.
Complete the lower bound and upper bound about his weight, w gr.

Answer:

Because of 507,000 gr is already rounded to the correct nearest gram,
the lower bound before 507,000 is rounded off is equal to 506,999.5 or above.

And the upper bound before 507,000 is rounded off is below than 507,000.5, not equal to

507,000.5.
Because, if the value of w was equal to 507,000.5, it would be rounded off to 507,001 gr.
So the range value before 507,000 is rounded off are 506,999.5 ≤ w < 507,000.5

CARROT, SPORT CAR, CLASSIC CAR AND LUXURY HOME SELLING PRICE (MATH QUESTIONS)

noreply@blogger.com (fatah rohmani) — Tue, 26 Jun 2018 23:07:00 +0000

The farmer grows carrots.

In 2014 the selling price for carrots was $96 per tonne.

In 2015 this selling price increased by 18%.

Work out the increase in the selling price from 2014 to 2015.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

Answer:

The increase in the selling price from 2014 to 2015 is

18% of $96 = $17.28

So, this carrot’s selling price per tonne was increase $17.28 from 2014 to 2015

================================================

SIMILAR QUESTION (EXAMPLE QUESTION ONLY)

1) An attorney bought a sport car.

In 2017 the selling price for a sport car was $30,476.

In 2018 this selling price increased by 15% from the previous year.

Calculate the increase in the selling price of a sport car from 2017 to 2018.

Answer:

The increase in the selling price from 2017 to 2018 is

15% of $30,476 = $4571.4

So, this sport car’s selling price was increase $4571.4 from 2017 to 2018

2) A Criminal Defense attorney bought a classic car.

In 2016 the selling price for a classic car was $43,796.

In 2017 this selling price increased by 13% from the previous year.

Calculate the increase in the selling price of a classic car from 2016 to 2017.

Answer:

The increase in the selling price from 2016 to 2017 is

13% of $43,796 = $5693.48

So, this classic car’s selling price was increase $5693.48 from 2016 to 2017

3) A Criminal Defense attorney bought a luxury home.

In 2015 the selling price for a luxury home was $3,600,000.

In 2016 this selling price increased by 19% from the previous year.

Calculate the increase in the selling price of a luxury home from 2015 to 2016.

Answer:

The increase in the selling price from 2015 to 2016 is

19% of $3,600,000 = $684,000

So, this luxury home’s selling price was increase $684,000 from 2015 to 2016

THE NUMBER OF BUSES NEEDED TO TRAVEL SUPPORTERS OF BURTON CITY FOOTBALL TEAM

noreply@blogger.com (fatah rohmani) — Mon, 25 Jun 2018 22:29:00 +0000

Burton City is a football team.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

Burton City has 1732 supporters who want to travel to the next game.

The football club hires some buses.

Each bus seats 52 supporters.

(i) Work out the number of buses needed to take these supporters to the game.

Answer:

The number of buses needed are 1732/52 = 33.307

So the exact number of buses are at least 34 buses.

(ii) The cost of the buses will be shared equally amongst these supporters.

Each bus costs $198 to hire.

Work out the amount that each supporter must pay.

Give your answer correct to the nearest 10 cents.

Answer:

The amount that each supporter must pay is

(198 x 34)/1732

The answer will be $3.90 or $3 and 90 cents.

======================================================

SIMILAR QUESTION (EXAMPLE QUESTION ONLY)

R United is a football club.

2976 supporters of R United football club want to travel to the next stadion.

The football club rent some caravans.

Each caravan seats 56 supporters.

(i) How many number of caravans needed to travel these supporters to the next stadion.

Answer:

The number of caravans needed are 2976/56 = 53.1428

So the exact number of caravans are at least 53 caravans.

(i) The supporters must pay equally all the cost of caravans

Each caravan costs $189 to rent.

Calculate the nominal amount that each supporter must pay.

Give your answer correct to the nearest 10 cents.

Answer:

The amount that each supporter must pay is

($198 x 53 caravans)/ 2976 supporters

The answer will be $3.53 or $3 and 53 cents for each supporter must pay all caravans.

PEMERATAAN KELAS INKLUSI UNTUK SEKOLAH-SEKOLAH DI INDONESIA, MUNGKINKAH?

noreply@blogger.com (fatah rohmani) — Sun, 24 Jun 2018 22:17:00 +0000

Susah dipercaya, anak-anak dengan label berkebutuhan khusus (special needs) sudah banyak sekarang ini. Yang dulu jarang ditemui. Mungkin karena faktor makanan yang dikonsumsi atau karena hal lain yang sifatnya medis atau psikologis? Saya tidak membahas akan faktor penyebabnya. Tapi saya akan membahas sekolah-sekolah yang menyediakan fasilitas untuk mereka. Salah satunya yaitu kelas inklusi. Di mana anak-anak yang katanya “normal” dan tidak perlu perlakuan khusus disatukan dengan anak-anak yang berkebutuhan khusus (special needs) dalam satu kelas dalam proses belajar mengajar di sekolah.

Saya pernah mengajar di salah satu sekolah yang mempunyai fasilitas kelas seperti itu. Menurut kepala sekolahnya, kelas seperti ini menguntungkan bagi anak-anak yang “normal” dan anak-anak yang berkebutuhan khusus (special needs). Bagi anak-anak yang “normal”, kelas seperti ini membuatnya memiliki pengalaman berempati terhadap sesama manusia lain yang mempunyai keterbatasan. Sedangkan bagi anak-anak yang berkebutuhan khusus (special needs), kelas seperti ini membuat mereka menjadi terbantu dalam hal belajar dan percepatan untuk mengejar ketertinggalannya. Dan di kelas seperti ini, biasanya memiliki minimal 2 guru, ketika satu guru sedang mengajar, maka guru yang lainnya berfungsi sebagai guru bayangan yang mendampingi satu persatu murid yang masih tertinggal proses belajarnya.

Pertanyaannya, seberapa banyakkah kelas-kelas inklusi di Indonesia? Apakah hanya ada di kota-kota besar? Sekolah-sekolah di kota besar pun belum tentu menerapkan kelas-kelas seperti ini.

Apakah kesadaran bahwa “pendidikan untuk semua” itu belum semua sekolah menyadarinya? Juga mengingat bahwa untuk menyediakan kelas-kelas seperti ini membutuhkan biaya yang lebih besar. Dengan data yang menyatakan bahwa anak-anak yang berkebutuhan khusus (special needs) di Indonesia bisa mencapai 4,2 juta jiwa. Angka yang begitu besarnya. Dengan anggaran pendidikan yang disediakan oleh pemerintah, mungkinkah pemerataan kelas-kelas inklusi bisa sampai ke daerah-daerah di Indonesia? Semoga bisa dan mampu.

Mengingat dari sisi kemanusiaan, kita harus prihatin akan kondisi anak-anak seperti itu. Apalagi mendengar bahwa anak dari teman-teman saya juga ada yang berkebutuhan khusus (special needs). Kasihan mendengarnya. Belum lagi mendengar, perlakuan diskriminatif yang dialami mereka di media-media sosial.

CMIIW

Semoga Bermanfaat

THE FARMER INVESTED MONEY WITH COMPOUND INTEREST (CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Sat, 23 Jun 2018 23:14:00 +0000

Three years ago the farmer invested $3750 at a rate of 4% per year compound interest.

(i) Calculate the total value of his investment after the 3 years.
(ii) The farmer wants to spend his investment on buying goats.
Goats cost $126 each.
Work out the maximum number of goats he can buy and how much money is left over.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

Answer:

i) The formula of Compound Interest is

A=P (1+R/100)^T

This formula is applied when for each year we calculate the interest from previous year, until T years. So we calculate the interest compoundly.

The Amount A is the total of money with its interest, after T years.

In this question:

P = $3750
After first year = $3750 + 4% x $3750 = $3900
After second year = $3900 + 4% x $3900 = $4056
After third year = $4056 + 4% x $4056 = $4218.24
Or simply use the formula

A=$3750(1+4/100)^3

A=$4218.24

(ii) Number of goats will be calculated as

=$4218.24/126
=33.49

Exact the maximum number of goats he can buy are 33 goats.

The remaining money after he bought goats is 60.24

THE FARMER HAS FEMALE SHEEP AND LAMBS (CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Sat, 23 Jun 2018 01:11:00 +0000

The farmer has 20 female sheep.

Each sheep has 0, 1, 2 or 3 lambs.

The table shows this information.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

i) Calculate the mean number of lambs per sheep.

Answer:

Mean number of lambs per sheep is

= ( 0x1 + 1x4 + 2x12 + 3x3 ) lambs divided by total sheep

= ( 0x1 + 1x4 + 2x12 + 3x3 ) lambs / 20 sheep

= 1.85

ii) The farmer takes 1 lamb away from each of the sheep with 3 lambs.

These lambs are given to 3 of the 4 sheep that have 1 lamb.

Complete the table after the farmer has done this.

Answer:

1 lamb is taken away from each of the sheep with 3 lambs. So there is no sheep which has 3 sheep.

Then for every 1 lamb which is already taken, will be given to the sheep which only has 1 lamb.

So only one sheep which has 1 lamb.

Finally, the sheep which has 2 lambs are 18 sheeps.

(iii) Explain why the mean number of lambs per sheep does not change.

Answer:

Same total number of sheep and same total number of lambs.

THE RATIO OF HORSES AND COWS AT THE FARM (CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Fri, 22 Jun 2018 03:56:00 +0000

A farmer has 45 horses and 20 cows.
(i) Write this as a ratio horses : cows.
Give your answer in its simplest form.
(ii) The farmer wants the ratio horses : cows to equal 5 : 3.
He keeps his 45 horses but buys some more cows. Work out the number of cows he must buy.

Mathematics Cambridge IGCSE Test (0580 m16 qp 32)

Answer:
(i) The ratio is always expressed in the simplest form
45 : 20
Divided by 5 to become
9:4
(ii) In this case, it must be answered by algebraic expression.
Because we have not know how many cows more to be bought yet, so it is expressed by x.
The algebraic expression is become
45: (20+x)=5:3
With cross multiplication
45x3 = 5(20+x)
135 = 100 + 5x
35 = 5x
7 = x
So the number of cows he must buy are 7.

=====================================================

SIMILAR QUESTION (EXAMPLE QUESTION ONLY)

One of Texas Attorney has 75 horses and 55 buffaloes.

(i) Write this as a ratio horses : buffaloes.

Give your answer in its simplest form.

(ii) The Texas Attorney wants the ratio horses : buffaloes to equal 5 : 4.

He keeps his 75 horses but buys some more buffaloes.

Work out the number of buffaloes Texas Attorney must buy.

Answer:

(i)The ratio is always expressed in the simplest form
75 : 55
Divided by 5 to become
15:11

(ii) In this question, have to processed algebraically.

Because Texas Attorney haven’t know how many buffaloes more to be bought yet, so it is

expressed by x.

The algebraic expression is become

75: (55+x)=5:4

With cross multiplication

75x4 = 5(55+x)

300 = 275 + 5x

25 = 5x
5 = x

So the number of buffaloes TexasAttorney must buy are 5 buffaloes.

Les Privat Cambridge dan IB Hanya Untuk Matematika Saja (Cenderung Berkebutuhan Khusus, Lambat Belajar Dll)

noreply@blogger.com (fatah rohmani) — Thu, 07 Sep 2017 09:55:00 +0000

Les Privat Cambridge dan International Baccalaureate (IB) Hanya Untuk Matematika Saja

Berikut penjelasan tentang IGCSE dari Wikipedia

IGCSE singkatan dari International General Certificate of Secondary Education (Indonesia: Sertifikat Internasional Pendidikan Menengah) adalah ujian internasional untuk siswa sekolah menengah. IGCSE dikembangkan oleh CIE (Cambridge International Examination) pada tahun 1988 dan dijadikan ujian internasional oleh CIE dan London Examinations. IGCSE dikhususkan untuk anak berusia 14-16 tahun, dan untuk menyiapkan siswa ke jenjang berikutnya, seperti AS Level, A Level dan program diploma IB. CIE dan Edexel pun dipercaya oleh sekolah-sekolah dipenjuru dunia sebagai acuan untuk mengadakan ujian internasional. CIE dan Edexel juga membuat penjurusan, seperti penjurusan Ilmu Alam, Ilmu Sosial, Bahasa dan lain-lain. IGCSE membolehkan pembelajaran yang disesuaikan dengan daerahnya. Tergantung pada negara yang bahasa ibunya bukan Bahasa Inggris dan diperbolehkan untuk mengiktui ujian. Pada tahun 2004-2005, paper IGCSE sebanyak 60 mata pelajaran telah diikuti oleh siswa-siswi di 120 negara di seluruh dunia, termasuk Amerika Serikat.

Mathsanimation menyediakan Les Privat Cambridge dan IB, hanya fokus untuk matematika saja.

Mathsanimation berpengalaman menangani siswa-siswa dari sekolah-sekolah:

- BPK Penabur International
- Highfield Secondary School
- Cikal
- Sekolah Pelita Harapan
- Binus International
- ACS Tiara Bangsa
Dll.

Mathsanimation juga menyediakan bimbingan Les Privat Cambridge dan IB untuk siswa-siswa:

- Slow learner (lambat belajar)
- Susah konsentrasi
- Cenderung berkebutuhan khusus

Dengan bantuan animasi, origami dan lain-lain, kita berusaha untuk membantu.

Klik kontak WA 👉👉👉 : https://wa.me/628988616810

Klik my Youtube Channel 👉: mathZanimation

How to Simplify Algebraic Expression With Fruit Picture Worksheet

noreply@blogger.com (fatah rohmani) — Fri, 04 Aug 2017 01:44:00 +0000

How to Simplify Algebraic Expression With Fruit Picture Worksheet

Download this worksheet here

PACMAN GAME TO STUDY CORRESPONDING, ALTERNATE, AND INTERIOR ANGLE

noreply@blogger.com (fatah rohmani) — Fri, 28 Jul 2017 05:46:00 +0000

PACMAN GAME TO STUDY CORRESPONDING, ALTERNATE, AND INTERIOR ANGLE

Download this game here

Trying to answer trigonometry Cambridge A level, with animation

noreply@blogger.com (fatah rohmani) — Wed, 26 Jul 2017 03:06:00 +0000

Trying to answer trigonometry Cambridge A level, with animation
See the video:

To see another blog which answer IGCSE paper is mathematics-exam-paper

Common Tangent Lines to Two Circles

noreply@blogger.com (fatah rohmani) — Wed, 26 Jul 2017 03:04:00 +0000

Common Tangent Lines to Two Circles
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ABSOLUTE VALUE in Pure Mathematics 2 (AS Level CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Tue, 21 Jun 2016 10:12:00 +0000

ABSOLUTE VALUE in Pure Mathematics 2 (AS Level)

Question:

[Taken from Paper 2 Pure Mathematics 2 (P2) Cambridge International Examinations Advanced Subsidiary Level, May/June 2015]

Answer:

DIFFERENTIATON in Pure Mathematics 2 (AS Level CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Wed, 20 Apr 2016 07:40:00 +0000

DIFFERENTIATON in Pure Mathematics 2 (AS Level)
Question:

[Taken from Paper 2 Pure Mathematics 2 (P2) Cambridge International Examinations Advanced Subsidiary Level, May/June 2015]

answer:

INTEGRATION in Pure Mathematics 2 (AS Level CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Wed, 13 Apr 2016 00:58:00 +0000

INTEGRATION in Pure Mathematics 2 (AS Level)
Question:

[Taken from Paper 2 Pure Mathematics 2 (P2) Cambridge International Examinations Advanced Subsidiary Level, May/June 2015]

THE POLYNOMIAL in Pure Mathematics 2 (AS Level CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Tue, 05 Apr 2016 10:24:00 +0000

THE POLYNOMIAL in Pure Mathematics 2 (AS Level)
Question:

[Taken from Paper 2 Pure Mathematics 2 (P2) Cambridge International Examinations Advanced Subsidiary Level, May/June 2015]

TANGENT EQUATION OF CURVE in Pure Mathematics 2 (AS Level CAMBRIDGE PAST PAPER)

noreply@blogger.com (fatah rohmani) — Tue, 29 Mar 2016 09:48:00 +0000

TANGENT EQUATION OF CURVE in Pure Mathematics 2 (AS Level)

Question:

[Taken from Paper 2 Pure Mathematics 2 (P2) Cambridge International Examinations Advanced Subsidiary Level, May/June 2015]