For a standard form, the decimal point must be after the first non-zero digit.

It is expressed in the form a.bcd x 10ⁿ.

Where n is the number of moves of the decimal point to be after the first non-zero digit.

In order to know the value of n, you have to count how many digits you have to move the decimal point over.

When you move the decimal point to the right, the n is negated but if you have to move the decimal point to the left, the n is positive.

0.01723 = 0.⌒0⌒1.723 = 1.723 x 10^-2

3x² + 2x - 7

Substitute x = -3 into the expression

3(-3)² + 2(-3) - 7

- x - = +

a² = a x a

(-3)² = - 3 x -3 = 9

3(-3)² + 2(-3) - 7 = 3 x -3 x -3 + 2 x -3 - 7

3(-3)² + 2(-3) - 7 = 3 x 9 - 6 - 7

3(-3)² + 2(-3) - 7 = 27 - 13

3(-3)² + 2(-3) - 7 = 14

y-axis as mirror/reflection in y-asis

$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ → $\left(\begin{array}{c}\mathrm{-x}\\ \mathrm{y}\end{array}\right)$

Thus the x is negated.

$\left(\begin{array}{c}-2\\ 2\end{array}\right)$ → $\left(\begin{array}{c}\mathrm{-(-2)}\\ 2\end{array}\right)$ → $\left(\begin{array}{c}2\\ 2\end{array}\right)$ → (2,2)

The 25^o is at the south between the south and the east.

Between each cardinal point is 90^o.

The angle from the east to the S25^oE = 90 - 25 = 65

Bearing starts from the north. Hence the bearing of S25^oE is from the north to the S25^oE line.

Bearing = 90 + 65 = 155^o

m = $\frac{\mathrm{b}}{2}$ - t

Get rid of the fraction by multiplying both sides of the equation by the denominator (2)

2 x m = b - 2 x t
2m = b - 2t
2m + 2t = b
2t = b - 2m

Divide both sides by 2 so that t can stand alone.

t = $\frac{\mathrm{b\; -\; 2m}}{2}$

(x - 2)(x + 4)

Take each express in the first bracket to multiply each expression in the second bracket and simplify the expressions.

(x - 2)(x + 4) = x(x + 4) - 2(x + 4)
(x - 2)(x + 4) = x x x + x x 4 - 2 x x - 2 x 4
(x - 2)(x + 4) = x² + 4x - 2x - 8
(x - 2)(x + 4) = x² + 2x - 8

E only = 15 - 6 = 9
F only = 20 - 6 = 14

n(E∪F) is the addition of all the numbers in the rectangle.

n(E∪F) = 9 + 6 + 14 = 29

Method I

Let x = the total land

Of means multiplication

Tony shared $\frac{2}{3}$ of his plot of land

Land shared = $\frac{2}{3}$ x x = $\frac{\mathrm{2x}}{3}$

Plot of land received by each = $\frac{\mathrm{2x}}{3}$ ÷ 3

Every number is divisible by 1.

3 = $\frac{3}{1}$

Plot of land received by each = $\frac{\mathrm{2x}}{3}$ ÷ $\frac{3}{1}$

Reciprocate (the numerator becomes denominator and the donominator becomes the numerator) the fraction after the division sign and change the division sign to multiplication.

Plot of land received by each = $\frac{\mathrm{2x}}{3}$ x $\frac{1}{3}$

Plot of land received by each = $\frac{\mathrm{2x\; x\; 1}}{\mathrm{3\; x\; 3}}$

Plot of land received by each = $\frac{\mathrm{2x}}{9}$

Fraction = $\frac{\mathrm{Number}}{\mathrm{Total}}$

Fraction of plot of land received by each = $\frac{\mathrm{2x}}{9}$ ÷ x = $\frac{2}{9}$

Method II

Fraction for the entire plot of land = 1

Of means multiplication

Tony shared $\frac{2}{3}$ of his plot of land

Fraction of land shared = $\frac{2}{3}$ x 1 = $\frac{2}{3}$

Fraction received by each = $\frac{2}{3}$ ÷ 3

Every number is divisible by 1.

3 = $\frac{3}{1}$

Fraction received by each = $\frac{2}{3}$ ÷ $\frac{3}{1}$

Reciprocate (the numerator becomes denominator and the donominator becomes the numerator) the fraction after the division sign and change the division sign to multiplication.

Fraction received by each = $\frac{2}{3}$ x $\frac{1}{3}$

Fraction received by each = $\frac{\mathrm{2\; x\; 1}}{\mathrm{3\; x\; 3}}$

Fraction received by each = $\frac{2}{9}$

Total oranges shared = 20
Ama's share = 12
Araba's share = 20 - 12
Araba's share = 8

Araba:Ama = 8:12
Araba:Ama = 2:3

Note: 4 divides 8, 2 times and 12, 3 times.

3(x - 2) - $\frac{\mathrm{x}}{3}$ + 6 > 0

Get rid of the fraction by multiplying both sides of the inequality by the denominator (3).

3 x 3(x - 2) - 3 x $\frac{\mathrm{x}}{3}$ + 3 x 6 > 3 x 0
9(x - 2) -x + 18 > 3 x 0
9 x x -2 x 9 -x > 0 - 18
9x - 18 -x >-18
9x -x > -18 + 18
8x > 0

Divide both sides by 8 to make x stand alone.

x > $\frac{0}{8}$
x > 0

Note: 0 divided by any number is 0.

You can substite x = 0 into the equation to confirm you get 0.

3(0 - 2) - $\frac{0}{3}$ + 6 > 0

3(- 2) - 0 + 6 > 0
3 x -2 + 6 > 0
-6 + 6 > 0
0 > 0

The modal mark is the mark which appeared the most. Observe the leaf in each of the stems to determine which number appeared the most.

5 appeared 3 times in the stem 4 more than the rest. The mark is obtained by joining the leaf to the stem.

The modal mark = Stem 4 and leaf 5 hence 45

The leaf represents each mark. Hence count how many leaves to determine the number of students who took the test.

Total number of students who took the test = 5 + 3 + 4 + 4 + 3 + 2
Total number of students who took the test = 21

The median is the middle number. The median mark is the middle mark. Locate the middle leaf and join its stem with the leaf to obtain the median mark.

The middle leaf is 5 (the 11th position) at stem 4, hence the median mark is 45

Even number: number divisible by 2

Prime number: is a natural number greater than 1 that has only two factors (1 and the number itself).

2 is an even number because it can divide itself to obtain 1 and it is also a prime number because no other number can divide it except 1 and itself.

2n - 5 = $\frac{1}{2}$n

Get rid of the fraction by multiplying both sides of the equation by the denominator (2).

2 x 2n - 2 x 5 = 2 x $\frac{1}{2}$n

4n - 10 = 1 x n
4n - 10 = n
4n - n = 10
3n = 10

Divide both sides by 3 so that n can stand alone

n = $\frac{10}{3}$ = 3$\frac{1}{3}$

Simple Interest = $\frac{\mathrm{Principal\; x\; Time\; x\; Rate}}{100}$

Principal = GH₵ 350.00
Time = 4 years
Rate = 5

Simple Interest = $\frac{\mathrm{GH₵\; 350\; x\; 4\; x\; 5}}{100}$

Simple Interest = $\frac{\mathrm{GH₵\; 350\; x\; 20}}{100}$

Simple Interest = $\frac{\mathrm{GH₵\; 350}}{5}$

Note: 20 divides itself once and 100, 5 times.

Simple Interest = GH₵ 70.00

Note: 5 divides itself once and 350, 70 times.

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = -7
Y₁ = -3
X₂ = -4
X₁ = -5

Gradient = $\frac{\mathrm{-7-(-3)}}{\mathrm{-4-(-5)}}$ = $\frac{\mathrm{-7+3}}{\mathrm{-4\; +\; 5}}$ = $\frac{-4}{1}$ = -4

If $\frac{5}{8}$ = 1 kilometre
240 mile = 1 kilometre x 240 mile ÷ $\frac{5}{8}$ mile

Change the division sign to multiplication and reciprocate the fraction.

240 mile = 1 kilometre x 240 x $\frac{8}{5}$

240 mile = 1 kilometer x 48 x 8

5 divides itself once and 240, 48 times.

240 mile = 384 kilometres

Dividing 240 by 5

Change the mixed fractions to improper fractions.

22$\frac{1}{2}$ = $\frac{\mathrm{22\; x\; 2\; +\; 1}}{2}$

22$\frac{1}{2}$ = $\frac{\mathrm{44\; +\; 1}}{2}$

22$\frac{1}{2}$ = $\frac{45}{2}$

7$\frac{1}{2}$ = $\frac{\mathrm{7\; x\; 2\; +\; 1}}{2}$

7$\frac{1}{2}$ = $\frac{\mathrm{14\; +\; 1}}{2}$

7$\frac{1}{2}$ = $\frac{15}{2}$

Total time spent to school = $\frac{45}{2}$ minutes + $\frac{15}{2}$ minutes

Total time spent to school = $\frac{\mathrm{45\; +\; 15}}{2}$ minutes

Total time spent to school = $\frac{60}{2}$ minutes

Total time spent to school = 30 minutes

Method I

The entire minutes spent to school is 100%.

If 30 minutes = 100%
$\frac{45}{2}$ minutes = $\frac{45}{2}$ x 100% ÷ 30

2 divides itself once and 100, 50 times.

$\frac{45}{2}$ minutes = $\frac{\mathrm{45\; x\; 50\%}}{30}$

$\frac{45}{2}$ minutes = 15 x 5%

Note: The zero in 50 and 30 cancel each other. 3 divides itself once and 45, 15 times.

$\frac{45}{2}$ minutes = 75%

Method II

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Percentage (walking) = $\frac{45}{2}$ ÷ 30 x 100%

Every number is being divided by 1

30 = $\frac{30}{1}$

Percentage (walking) = $\frac{45}{2}$ ÷ $\frac{30}{1}$ x 100%

Change the division sign to multiplication and reciprocate the fraction.

Percentage (walking) = $\frac{45}{2}$ x $\frac{1}{30}$ x 100%

Note: The zero in 30 and 100 cancels each other. 3 divides itself once and 45, 15 times. 2 divides itself once and 10, 5 times.

Percentage (walking) = 15 x 1 x 5%
Percentage (walking) = 75%

Since Kobby is 5 years older than his brother, his brother's age is 5 years less than that of Kobby's age.

Kobby is 13 years now.

Brother's age = 13 - 5
Brother's age = 8

Note: If the mapping has equal interval (difference) for the y and x interval of 1, you can use the general linear sequence formula to find the rule of the mapping.

U_n = U₁ + (n-1)d

In the mapping, U_n = y and n = x and U₁, the first value of y

U₁ = 4

y = 4 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 2-4 = 0-2 = -2 - 0 = -4 -(-2) = -2

We can substitute our d and simplify the expression for y

y = 4 + (x-1)d

y = 4 + (x-1)(-2)

y = 4 + -2 x x -1 x -2

y = 4 + -2x + 2

y = 4 + 2 -2x

y = 6 + -2x

y = -2x + 6

$\frac{9}{5}$,1.88,$\frac{15}{8}$

Method I

Change the fractions to decimal and compare them.

9 divided by 5

15 divided by 8

$\frac{9}{5}$ = 1.8

$\frac{15}{8}$ = 1.875

Ascending means from the smallest to the largest.

Arranging 1.8,1.88,1.875 ascending order = 1.8,1.875,1.88

Replace the decimals by their corresponding fractions in 1.8,1.875,1.88.

Ascending order = $\frac{9}{5}$,$\frac{15}{8}$,1.88

Method II

Change the decimal to fraction and compare the fractions

Every number is being divided by 1.

1.88 = $\frac{1.88}{1}$

Count how many digits (n) are after the decimal point and multiply both the numerator and the denominator by 1 joined by the n number of zeros.

In 1.88, there are two digits (8 and 8) after the decimal point and so we will multiply both the numerator and the denominator by 100. Multiplying the numerator by 100 changes the decimal to a whole number.

$\frac{1.88}{1}$ = $\frac{\mathrm{1.88\; x\; 100}}{\mathrm{1\; x\; 100}}$ = $\frac{188}{100}$ = $\frac{47}{25}$

4 divides 188, 47 times and 100, 25 times.

Now compare the fractions.

$\frac{9}{5}$,1.88,$\frac{15}{8}$

1.88 = $\frac{47}{25}$

$\frac{9}{5}$,$\frac{47}{25}$,$\frac{15}{8}$

To compare the fractions, the denominators must be the same. Hence find the equivalent fractions for each of the fractions having the same denominator. To do so follow the steps below.

1. Find the Lowest Common Factor (L.C.M) of the denominators. The L.C.M of 5,25 and 8 is 200.

2. Multiply the numerator and denominator of each fractions by how many times the denominator will divide the L.C.M

For the fraction with denominator 5, we have to multiply both the numerator and the denominator by 40 since 5 x 40 = 200
For the fraction with denominator 25, we have to multiply both the numerator and the denominator by 8 since 25 x 8 = 200
For the fraction with denominator 8, we have to multiply both the numerator and the denominator by 25 since 8 x 25 = 200

Equivalent Fractions

$\frac{9}{5}$ = $\frac{\mathrm{9\; x\; 40}}{\mathrm{5\; x\; 40}}$ = $\frac{360}{200}$

$\frac{47}{25}$ = $\frac{\mathrm{47\; x\; 8}}{\mathrm{25\; x\; 8}}$ = $\frac{376}{200}$

$\frac{15}{8}$ = $\frac{\mathrm{15\; x\; 25}}{\mathrm{8\; x\; 25}}$ = $\frac{375}{8}$

Now since the equivalent fractions have the same denominator, you can compare their numerators.

Ascending order of the numerators = 360, 375, 376.

Now replace the order of the numerators by the fractions or decimal that those numerators were obtained from.

360 was obtained from $\frac{9}{5}$

375 was obtained from $\frac{15}{8}$

376 was obtained from 1.88

Ascending order = $\frac{9}{5}$,$\frac{15}{8}$,1.88

a² = a x a

(x - 1)² = (x - 1) x (x - 1)

Expand the bracket by taking each expressions in the first bracket to multiply the second bracket and simplify.

(x - 1)² = x x (x - 1) - 1 x (x - 1)
(x - 1)² = x x x - 1 x x - 1 x x - 1 x -1
(x - 1)² = x² - x - x + 1

Note: - x - = +

(x - 1)² = x² -2x + 1

(x - 1)² - 1 = x² -2x + 1 - 1
(x - 1)² - 1 = x² -2x + 0
(x - 1)² - 1 = x² -2x

Note: (expression)² = The first number squared + 2 x the first number and the second number + second number squared.

(x - 1)² = x² + 2 x x x (-1) + (-1)²
(x - 1)² = x² - 2x + 1

Note: (-1)² = -1 x -1 = 1 (- x - = +)

Probability = $\frac{\mathrm{Number\; of\; possible\; selection}}{\mathrm{Number\; of\; samples}}$

Number of good oranges = 20 - 6 = 14

Number of samples = 20

Probability (Good oranges) = $\frac{14}{20}$ = $\frac{7}{10}$

Note: 2 divides 14, 7 times and 20, 10 times.

r = 80^o (Corresponding angles)

CORRESPONDING ANGLES

r = 80^o (Corresponding angles)

CORRESPONDING ANGLES

r + 5k = 180 (Sum of angles on a straight line is 180)
But r = 80
80 + 5k = 180
5k = 180 - 80
5k = 100
Divide both sides by 5 so that k stands alone.

k = $\frac{100}{5}$ = 20

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$

Change the mixed fraction to improper fraction.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $($ $\frac{3}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $($ $\frac{\mathrm{3\; x\; 3\; -\; 1\; x\; 5}}{6}$ $)$ x $\frac{9}{10}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $($ $\frac{\mathrm{9\; -\; 5}}{6}$ $)$ x $\frac{9}{10}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $($ $\frac{4}{6}$ $)$ x $\frac{9}{10}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $\frac{4}{6}$ x $\frac{9}{10}$

Note: 2 divides 4, 2 times and 6, 3 times

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $\frac{2}{3}$ x $\frac{9}{10}$

Note: 2 divides itself once and 10, 5 times. 3 divides itself once and 9, 3 times

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $\frac{1}{1}$ x $\frac{3}{5}$

$($ 1$\frac{1}{2}$ - $\frac{5}{6}$ $)$ x $\frac{9}{10}$ = $\frac{3}{5}$

2pq + 6p -6q - 18

Bring out the common terms.

6p = 2p x 3
18 = 3 x 6

2pq + 2p x 3 -6q - 3 x 6

2p(q + 3) - 6(q + 3)
(2p - 6)(q + 3)

Factorize (2p - 6) further

2p - 6
6 = 2 x 3
2p -3 x 2
2(p - 3)

(2p - 6)(q + 3) = 2(p - 3)(q + 3)

The area of the shaded portion = Area of the square - Area of the circle

Area of square = Length x Length
Area of square = 2 cm x 2 cm
Area of square = 4 cm²

Area of a circle = πr²
Where r = radius
Diameter of the circle = 2 cm
Radius is half the diameter
Radius = $\frac{\mathrm{2\; cm}}{2}$ = 1 cm

π = $\frac{22}{7}$

r = 1 cm
r² = r x r

Area of the circle = $\frac{22}{7}$ x 1 cm x 1 cm = 3.14 cm²

Area of the shaded portion = 4 cm² - 3.14 cm²
Area of the shaded portion = 0.86 cm²

22 divided by 7

3.1428 ≈ 3.14 (2 decimal places)

Area of a circle = πr²
Where r = radius
Diameter of the circle = 2 cm
Radius is half the diameter
Radius = $\frac{\mathrm{2\; cm}}{2}$ = 1 cm

π = $\frac{22}{7}$

r = 1 cm
r² = r x r

Area of the circle = $\frac{22}{7}$ x 1 cm x 1 cm = 3.14 cm²

22 divided by 7

3.1428 ≈ 3.14 (2 decimal places)

The more the workers, the lesser the time to complete the task. Hence number of workers is inversely proportional to time.

Let w = workers, t = time to complete task

w ∝ $\frac{1}{\mathrm{t}}$

We can introduce a constant to remove the proportionality.

Let k = proportionality constant.

w = k x $\frac{1}{\mathrm{t}}$

w = $\frac{\mathrm{k}}{\mathrm{t}}$

Make k the subject.

Multiply both sides by t

k = w x t

The value of k is the same for all scenarios since they work at the same rate.

k value for 1 man is the same for the 6 men.

Let t = time taken for the 6 men to complete the task.

1 x 12 = 6 x t
Divide both sides by 6 so that t stands alone

t = $\frac{\mathrm{1\; x\; 12}}{6}$ = $\frac{12}{6}$ = 2 hours

It will take 2 hours for the 6 men to complete the work at the same rate.

If 1 m = 100 cm
12.03 m = $\frac{\mathrm{12.03\; m\; x\; 100\; cm}}{\mathrm{1\; m}}$ = 1203 cm

Note: When multiplying a decimal number by 1n number of zeros, you move the decimal point to the right n times.

12.03 x 100 = 12.⌒0⌒3. = 1203

Trapezium

A trapezium is a quadrilateral in which one pair of opposite sides is parallel.

Method I

Profit % = $\frac{\mathrm{Profit}}{\mathrm{Cost\; Price}}$ x 100%

Selling Price = Cost Price + Profit
Profit = Selling Price - Cost Price

Profit % = $\frac{\mathrm{Selling\; Price\; -\; Cost\; Price}}{\mathrm{Cost\; Price}}$ x 100%

Let c = Cost Price

Profit % = $\frac{\mathrm{Selling\; Price\; -\; c}}{\mathrm{c}}$ x 100%

Profit % = 20
Selling Price = GH₵ 48.00

20 = $\frac{\mathrm{48\; -\; c}}{\mathrm{c}}$ x 100

20 = $\frac{\mathrm{48\; x\; 100\; -\; c\; x\; 100}}{\mathrm{c}}$

20 = $\frac{\mathrm{4800\; -\; 100c}}{\mathrm{c}}$

Multiply both sides by c

20 x c = 4800 - 100c
20c + 100c = 4800
120c = 4800

Divide both sides by 120 for c to stand alone.

c = $\frac{4800}{120}$ = 40

Note: The zeros in 4800 and 120 cancel each other. 12 divides itself once and 480, 40 times.

Cost Price = GH₵ 40

Method II

Selling Price = Cost Price + Profit
Selling Price % = Cost Price % + Profit %
Cost Price % = 100%
Selling Price % = 100% + Profit %

Profit % = 20%

Selling Price % = 100% + 20%
Selling Price % = 120%

Selling Price = GH₵ 48

If 120% = GH₵ 48
100% = $\frac{\mathrm{GH₵\; 48\; x\; 100\%}}{\mathrm{120\%}}$ = GH₵ 40

Let $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ = the translation vector.

$\left(\begin{array}{c}2\\ 6\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ = $\left(\begin{array}{c}5\\ 4\end{array}\right)$

$\left(\begin{array}{c}\mathrm{2\; +\; x}\\ \mathrm{6\; +\; y}\end{array}\right)$ = $\left(\begin{array}{c}5\\ 4\end{array}\right)$

The top at the left equals the top at the right and the bottom at the left equals the bottom at the right.

2 + x = 5
x = 5 - 2
x = 3

6 + y = 4
y = 4 - 6
y = -2

$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

Cement:Sand = 2:5

Total ratios = 2 + 5 = 7

Ratio for cement = 2

If 7 = 35 kg
2 = $\frac{\mathrm{2\; x\; 35\; kg}}{7}$ = 2 x 5 kg = 10 kg

Integers

These are the set of all whole numbers but it includes a negative set of natural numbers also. Hence whole numbers and natural numbers are subsets of set of integers. "Z" represents integers and the set of integers include, Z = {...,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,...}

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

Law of division of indices

$\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

a⁰ = 1

$\frac{5^{7}x{5}^{-4}}{5^3}$ = $\frac{5^{\mathrm{7+-4}}}{5^3}$

$\frac{5^{7}x{5}^{-4}}{5^3}$ = $\frac{5^{\mathrm{7-4}}}{5^3}$

$\frac{5^{7}x{5}^{-4}}{5^3}$ = $\frac{5^3}{5^3}$ = 1

Note: 5³ divides 5³ once.

Alternatively, the law of division could be applied.

$\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

$\frac{5^3}{5^3}$ = 5^3-3

$\frac{5^3}{5^3}$ = 5⁰ = 1

5⁰ = 1 (a⁰ = 1)

Volume of a cylinder = πr²h
Where r = radius, h = height

π = $\frac{22}{7}$

r² = r x r
r = 7 cm
h = 5 cm

Volume of a cylinder = $\frac{22}{7}$ x 7 cm x 7 cm x 5 cm

Volume of a cylinder = 22 x 1 cm x 7 cm x 5 cm

Volume of a cylinder = 770 cm³

q = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$

r = $\left(\begin{array}{c}3\\ -2\end{array}\right)$

q - r = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ - $\left(\begin{array}{c}3\\ -2\end{array}\right)$

q - r = $\left(\begin{array}{c}\mathrm{-2\; -\; 3}\\ \mathrm{3-(-2)}\end{array}\right)$

q - r = $\left(\begin{array}{c}-5\\ \mathrm{3+2}\end{array}\right)$

q - r = $\left(\begin{array}{c}-5\\ 5\end{array}\right)$