Significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact.

The significant figures are 3597 and the third significant figure is 9. The number next to the third significant is more than 4, hence 1 is added to the third significant figure resulting in 0.00360.

Express the decimal in the form a x 10^b where a is a whole number

0.064 = 0.⌒0⌒6⌒4 = 64 x 10^-3

Note: the power is negative since the decimal is moved to the right.

${\left(0.064\right)}^{-\frac{1}{3}}$ = ${\left(\mathrm{64\; x}{10}^{-3}\right)}^{-\frac{1}{3}}$

64 = 4³

${\left(0.064\right)}^{-\frac{1}{3}}$ = ${(4^{3}x{10}^{-3})}^{-\frac{1}{3}}$

The power outside affects each of the powers in the bracket.

${\left(0.064\right)}^{-\frac{1}{3}}$ = $4^{\mathrm{3\; x\; -}\frac{1}{3}}x{10}^{\mathrm{-3\; x\; -}\frac{1}{3}}$

Notes:

1. The 3 cancel each other.

2. -1 x -1 = 1

${\left(0.064\right)}^{-\frac{1}{3}}$ = 4^-1 x 10¹

Note:

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

4^-1 = $\frac{1}{4^1}$ = $\frac{1}{4}$

${\left(0.064\right)}^{-\frac{1}{3}}$ = $\frac{1}{4}$ x 10 = $\frac{5}{2}$

$\frac{\mathrm{y\; +\; 1}}{2}$ - $\frac{\mathrm{2y\; -\; 1}}{3}$ = 4

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 2 and 3. Th L.C.M is 6

6 x $\frac{\mathrm{y\; +\; 1}}{2}$ - $\frac{\mathrm{2y\; -\; 1}}{3}$ x 6 = 4 x 6

3(y + 1) - 2(2y - 1) = 24

3y + 3 - 4y + 2 = 24

3y - 4y + 5 = 24

-y = 24 - 5

-y = 19

Divide both sids by -1

y = $\frac{19}{-1}$ = -19

a² - b² = (a + b)(a - b) known as difference of two squares.

(27.63)² - (12.37)² = (27.63 + 12.37)(27.63 - 12.37)

(27.63)² - (12.37)² = (40)(15.26)

(27.63)² - (12.37)² = 610.4 ≈ 610 (three significant figures)

Note: 0 in the middle or end of a number is significant.

4 (mod 8) = 8 x 1 + 4 = 12

Since y is greater than or equal to 10 (10 ≤ y, thus 10 is less than or equal to y), the minimum value should be 17 (7 + 10)

4 (mod 8) = 8 x 2 + 4 = 20

20 is more than the minimum value (17)

7 + y = 20

y = 20 - 7

y = 13

µ = {x : 0 < x ≤ 10}

x : 0 < x ≤ 10 read as x is such that 0 is less than x and x is also less than or equal to 10

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

T = {prime numbers}

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Note: 1 is not a prime number.

T = {1, 2, 3, 5, 7}

Note: the elements of T must be found in the universal set µ since it is a subset.

T^' (T compliment) are list of elements found in the universal set but not in the set T

T^' = {1, 4, 6, 8, 9, 10}

M = {odd numbers}

Odd numbers are numbers not divisible by 2. Thus there is a remainder when divided by 2.

M ={1, 3, 5, 7, 9}

Note: the elements of M must be found in the universal set µ since it is a subset.

M^' (M compliment) are list of elements found in the universal set but not in the set M

M^' = {2, 4, 6, 8, 10}

Intersection (∩) are elements that can be found in both sets

T^' = {1, 4, 6, 8, 9, 10}

M^' = {2, 4, 6, 8, 10}

T^' ∩ M^' are the elements that can be found in both T^' and M^'

T^' ∩ M^' = {4, 6, 8, 10}

9 = 3²

log₃9 = log₃3²

log^ba = bloga

log₃3² = 2log₃3

log_aa = 1

log₃3 = 1

2log₃3 = 2 x 1 = 2

∴ log₃9 = 2

8 = 2³

log₂8 = log₂2³

log₂2³ = 3log₂2

log₂2 = 1

log₂2³ = 3 x 1 = 3

∴ log₂8 = 3

Change both bases to base 10 to determine the value of y.

23_{y = 2 x y¹ + 3 x y⁰}

Any number raised to the power 0 is 1.

23_{y = 2 x y + 3 x 1}

23_{y = 2y + 3}

1111_two = 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2⁰

1111_two = 8 + 4 + 2 + 1

1111_two = 15

2y + 3 = 15

2y = 15 - 3

2y = 12

Divide both sides by 2

y = $\frac{12}{2}$ = 6

For Arithmetic Progression (A.P), the common difference is the same.

Common difference = Succeeding - Preceding

6, p, 14, ...

14 - p = p - 6

Solve for p

14 + 6 = p + p

20 = 2p

Divide both sides by 2.

p = $\frac{20}{2}$ = 10

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$

Express each surds to be in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{28}}$ = $\sqrt{\mathrm{4\; x\; 7}}$

$\sqrt{\mathrm{50}}$ = $\sqrt{\mathrm{25\; x\; 2}}$

$\sqrt{\mathrm{72}}$ = $\sqrt{\mathrm{36\; x\; 2}}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{4\; x\; 7}}$ - 3$\sqrt{\mathrm{25\; x\; 2}}$ + $\sqrt{\mathrm{36\; x\; 2}}$

From $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x $\sqrt{4}$ x $\sqrt{7}$ - 3 x $\sqrt{\mathrm{25}}$ x $\sqrt{2}$ + $\sqrt{\mathrm{36}}$ x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x 2 x $\sqrt{7}$ - 3 x 5 x $\sqrt{2}$ + 6 x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 15$\sqrt{2}$ + 6$\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 9$\sqrt{2}$

m : n = 2 : 1

Comparing the left hand and right hand sides, m = 2 and n = 1

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$

Substitute the value of m and n into the given relation.

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{\mathrm{3\; x}{2}^2-\mathrm{2\; x}{1}^2}{2^2+\mathrm{2\; x\; 1}}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{\mathrm{3\; x\; 4}-\mathrm{2\; x\; 1}}{4+2}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{12-2}{4+2}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{10}{6}$

$\frac{3{\mathrm{m}}^2-2{\mathrm{n}}^2}{{\mathrm{m}}^2+\mathrm{mn}}$ = $\frac{5}{3}$

H varies directly as p

H ∝ p

H varies inversely as the square of y

H ∝ $\frac{1}{{\mathrm{y}}^2}$

Combining the two

H ∝ $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

Introduce a constant k to remove the proportionality sign.

H = k x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

Use the given values to find k.

H = 1

p = 8

y = 2

1 = k x $\frac{8}{2^2}$

1 = k x $\frac{8}{4}$

1 = k x 2

Divide both sides by 2

k = $\frac{1}{2}$

Substitute the value of k into the general relation.

H = k x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

H = $\frac{1}{2}$ x $\frac{\mathrm{p}}{{\mathrm{y}}^2}$

H = $\frac{\mathrm{1\; x\; p}}{\mathrm{2\; x}{\mathrm{y}}^2}$

H = $\frac{\mathrm{p}}{2{\mathrm{y}}^2}$

4x² - 16x + 15 = 0

Notes:

1. a x c = 4 x 15 = 60

2. Split b (-16) such that two numbers when you sum you will get b (-16) and when you multiply you will get ac (60). The two numbers are -10 and -6

3. -16x = -10x - 6x

4x² -10x - 6x + 15 = 0

2x(2x - 5) - 3(2x - 5) = 0

(2x - 5)(2x - 3) = 0

When 2x - 5 = 0, x = $\frac{5}{2}$ = 2$\frac{1}{2}$

When 2x - 3 = 0, x = $\frac{3}{2}$ = 1$\frac{1}{2}$

Change the decimal numbers to the form a x 10ⁿ, where a is a whole number.

0.42 = 0.⌒4⌒2 = 42 x 10^-2

2.5 = 2.⌒5 = 25 x 10^-1

0.5 = 0.⌒5 = 5 x 10^-1

2.05 = 2.⌒0⌒5 = 205 x 10^-2

Note: the power is negative since the decimal is moved to the right.

0.42 ÷ 2.5 = $\frac{\mathrm{42\; x}{10}^{-2}}{\mathrm{25\; x}{10}^{-1}}$

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

0.42 ÷ 2.5 = $\frac{42}{25}$ x 10^-2-(-1)

0.42 ÷ 2.5 = $\frac{42}{25}$ x 10^{-2 + 1}

0.42 ÷ 2.5 = $\frac{42}{25}$ x 10^-1

0.5 x 2.05 = 5 x 10^-1 x 205 x 10^-2

a^m x aⁿ = a^{m + m}

0.5 x 2.05 = 5 x 205 x 10^{-1 + -2}

0.5 x 2.05 = 5 x 205 x 10^-3

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = $\frac{42}{25}$ x 10^-1 ÷ (5 x 205 x 10^-3)

Every number is being divided by 1

5 x 205 x 10^-3 = $\frac{\mathrm{5\; x\; 205\; x}{10}^{-3}}{1}$

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = $\frac{42}{25}$ x 10^-1 ÷ $\frac{\mathrm{5\; x\; 205\; x}{10}^{-3}}{1}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = $\frac{42}{25}$ x 10^-1 x $\frac{1}{\mathrm{5\; x\; 205\; x}{10}^{-3}}$

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = $\frac{\mathrm{42\; x\; 1}}{\mathrm{25\; x\; 5\; x\; 205}}$ x 10^-1-(-3)

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = $\frac{42}{25625}$ x 10^{-1 + 3}

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = 0.0016390 x 10²

0.0016390 = 1.639 x 10^-3 (Standard form)

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = 1.639 x 10^-3 x 10²

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = 1.639 x 10^{-3 + 2}

$\frac{\mathrm{0.42\; ÷\; 2.5}}{\mathrm{0.5\; x\; 2.05}}$ = 1.639 x 10^-1

Notes:

A standard form should be in the form a.bcde x 10ⁿ

The decimal point must be after the first non-zero digit.

The n is the number of times the decimal point has to be moved to be right after the first non-zero digit.

If it is moved to the left, the n is positive and if to the right, n is negative.

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27

27 = 3³

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀3³

loga^b = bloga

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+3 x $\frac{2}{3}$log₁₀3

Note:

3 x $\frac{2}{3}$ = 2

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+2log₁₀3

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-log₁₀3

loga - logb = log$\frac{\mathrm{a}}{\mathrm{b}}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀$\frac{6}{3}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀2

100% represents the cost price.

Selling Price % = Cost Price % + Profit %

Profit % = 15%

Selling Price % = 100% + 15%

Selling Price % = 115%

Selling Price = ₦6,900

If 115% = ₦6,900

100% = $\frac{\mathrm{₦6900\; x\; 100\%}}{\mathrm{115\%}}$ = ₦6,000

Cost price = ₦6,000

Profit = Selling Price - Cost Price

Profit when sold at ₦6600 = ₦6600 - ₦6,000

Profit when sold at ₦6600 = ₦600

If ₦6,000 = 100%

₦600 = $\frac{\mathrm{₦600\; x\; 100\%}}{\mathrm{₦6000}}$ = 10%

3p = 4q ----- eqn 1

9p = 8q - 12 ---- eqn 2

From ---- eqn 1

Make p the subject by dividing both sides by 3.

p = $\frac{\mathrm{4q}}{3}$ ------- eqn 3

Substitute p into ---- eqn 2

9 x $\frac{\mathrm{4q}}{3}$ = 8q - 12

3 x 4q = 8q - 12

12q = 8q - 12

12q - 8q = - 12

4q = -12

Divide both sides by 4

q = $\frac{-12}{4}$ = -3

Substitute q = -3 into ---- eqn 3.

p = $\frac{\mathrm{4\; x\; -3}}{3}$

p = -4

pq = p x q

pq = -4 x -3

pq = 12

0.25 = $\frac{25}{100}$ = $\frac{1}{4}$ = $\frac{1}{2^2}$

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

$\frac{1}{2^2}$ = 2^-2

(0.25)^y = (2^-2)^y

(0.25)^y = 2^{-2 x y}

(0.25)^y = 2^-2y

2^-2y = 32

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

2^-2y = 2⁵

Since the bases are the same, the powers are also the same.

-2y = 5

Divide both sides by -2

y = $\frac{5}{-2}$ = -$\frac{5}{2}$

Number of samples = 8 + 4 = 12

Number of possible selection = 8

P(Boy) = $\frac{8}{12}$ = $\frac{2}{3}$

x² - 5x - 14

Notes:

Split -5 in such a way that two numbers when you multiply you will get -14 and when you add you will get -5.

The numbers are -7 and 2

-5x = -7x + 2x

x² -7x + 2x - 14

x(x - 7) + 2(x - 7)

(x - 7)(x + 2)

x² - 9x + 14

Notes:

Split -9 in such a way that two numbers when you multiply you will get 14 and when you add you will get -9.

The numbers are -7 and -2

-9x = -7x - 2x

x² -7x - 2x + 14

x(x - 7) - 2(x - 7)

(x - 7)(x - 2)

$\frac{{\mathrm{x}}^2-\mathrm{5x}-14}{{\mathrm{x}}^2-\mathrm{9x}+14}$ = $\frac{\mathrm{(x\; -\; 7)(x\; +\; 2)}}{\mathrm{(x\; -\; 7(x\; -\; 2)}}$

(x - 7) cancels out.

$\frac{{\mathrm{x}}^2-\mathrm{5x}-14}{{\mathrm{x}}^2-\mathrm{9x}+14}$ = $\frac{\mathrm{x\; +\; 2}}{\mathrm{x\; -\; 2}}$

A function is undefined when the denominator is 0

p² - p = 0

p(p - 1) = 0

p = 0 or p - 1 = 0, p = 1

π = $\frac{22}{7}$

Total surface area cylinder = 165 cm²

2 x πr² + 2πrh = 165

Radius (r) = $\frac{\mathrm{Diameter}}{2}$ = $\frac{7}{2}$ = 3.5 cm

2 x $\frac{22}{7}$ x 3.5² + 2 x $\frac{22}{7}$ x 3.5 x h = 165

Get rid of the fractions by multiplying both sides by the denominator 7.

2 x 22 x 3.5² + 2 x 22 x 3.5 x h = 165 x 7

539 + 154h = 1155

154h = 1155 - 539

154h = 616

Divide both sides by 154

h = $\frac{616}{154}$ = 4 cm

$\sqrt{a}$ = $a^{\frac{1}{2}}$

$\sqrt{\mathrm{64}}$ = ${\mathrm{64}}^{\frac{1}{2}}$

64 = 2⁶

$\sqrt{\mathrm{64}}$ = $2^{\mathrm{6\; x}\frac{1}{2}}$ = 2³

2a = 2³

Since the bases are the same, the powers are also the same.

a = 3

$\frac{\mathrm{b}}{\mathrm{a}}$ = 3

b = 3 x a

b = 3 x 3

b = 9

a² + b² = 3² + 9²

a² + b² = 9 + 81

a² + b² = 90

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

Opposite side of 52° is 32 cm

Adjacent side of 52° is |XZ|

tan 52° = $\frac{32}{\mathrm{|XZ|}}$

|XZ| X tan 52° = 32

Divide both sides by tan 52°

|XZ| = $\frac{32}{\mathrm{tan\; 52°}}$ = 25 cm

log_x8 = log_x8

8 = 2³

log_x8 = log_x2³

loga^b = bloga

log_x8 = 3log_x2

But log_x2 = 0.3

log_x8 = 3 x 0.3

log_x8 = 0.9

Length of the arc = $\frac{\mathrm{72°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 3.5 cm

Length of the arc = 4.4 cm

lb = $\frac{1}{2}$(a + b)h

Get rid of the fraction by multiplying both sides by the denominator 2

lb x 2 = h(a + b)

2lb = ah + bh

2lb - bh = ah

Factorize b out.

b(2l - h) = ah

Divide both sides by (2l - h)

b = $\frac{\mathrm{ah}}{\mathrm{2l\; -\; h}}$

100% represents the selling price of the house.

Percentage received by Eric = 100% - 8% = 92%

Amount received by Eric = $117760

If 92% = $117760

100% = $\frac{\mathrm{\$117760\; x\; 100\%}}{\mathrm{92\%}}$ = $128,000

Length of the arc = 22 cm

22 = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 15

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 360 and 7. The L.C.M is 360 x 7

22 x 360 x 7 = θ x 2 x 22 x 15

55440 = 660θ

Divide both sides by 660

θ = $\frac{55440}{660}$ = 84°

Length = 15 cm

Width = x cm

If the sides are double

New length = 15 cm X 2 = 30 cm

New width = 2 x x = 2x

Area of a rectangle = Length x Width

Area of the new rectangular board = 30 cm x 2xcm = 60x cm²

∆ SOT is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ OST = ∠ OTS = 50°

∠ OST + ∠ OTS + ∠ SOT = 180°(Sum of angles in a triangle)

50° + 50° + ∠ SOT = 180°

100° + ∠ SOT = 180°

∠ SOT = 180° - 100°

∠ SOT = 80°

∠ SOT = ∠ ROP = 80° (Vertically opposite angles)

∠ ROP + ∠ OPQ = 180° (Co-interior angles)

80° + ∠ OPQ = 180°

∠ OPQ = 180° - 80°

∠ OPQ = 100°

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

2x + 2y = 2(x + y)

2x - 2y = 2(x - y)

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

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

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

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

Number of sides of the polygon = 5 (since 5 angles are given)

Sum of angles = (5 - 2) x 180°

Sum of angles = 3 x 180°

Sum of angles = 540°

3x° + 2x° + 4x° + 3x° + 6x° = 540°

18x° = 540°

Divide both sides by 18

x = $\frac{540}{18}$ = 30

The smallest angle = 2x°

The smallest angle = 2 x 30°

The smallest angle = 60°

Difference colours = First ball is white and the second ball is blue or the first ball is blue and the second ball is white.

P(Different colours) = P(White first) and P(Blue second) or P(Blue first) and P(White second)

And means multiplication (x) and or means addition (+)

Number of samples = 2 + 3 = 5

Number of white balls = 2

P(White) = $\frac{2}{5}$

Number of blue balls = 3

P(Blue) = $\frac{3}{5}$

P(Different colours) = $\frac{2}{5}$ x $\frac{3}{5}$ + $\frac{3}{5}$ x $\frac{2}{5}$

P(Different colours) = $\frac{\mathrm{2\; x\; 3}}{\mathrm{5\; x\; 5}}$ + $\frac{\mathrm{3\; x\; 2}}{\mathrm{5\; x\; 5}}$

P(Different colours) = $\frac{6}{25}$ + $\frac{6}{25}$ = $\frac{\mathrm{6\; +\; 6}}{25}$ = $\frac{12}{25}$

Gradient (m) = $\frac{3}{4}$

Point (1, -5) → x₁ = 1 and y₁ = -5

y -(-5) = $\frac{3}{4}$(x - 1)

Note: -- = + or -(-) = +

-(-5) = + 5

y + 5 = $\frac{3}{4}$(x - 1)

Get rid of the fraction by multiplying both sides by the denominator 4

4 x y + 5 x 4 = 3(x - 1)

4y + 20 = 3x - 3

0 = 3x - 3 - 20 - 4y

0 = 3x - 4y - 23

3x - 4y - 23 = 0

Let x = the length of the ladder.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

x² = 6² + 8²

x² = 36 + 64

x² = 100

Take square root of both sides.

x = $\sqrt{\mathrm{100}}$ = 10

The ladder is 10 m long.

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan x = $\frac{3}{4}$ means the opposite side of x is 3 and the adjacent side is 4

Use this information to draw a right-angle triangle and calculate th hypotenuse.

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypotenuse (c)

c² = 3² + 4²

c² = 9 + 16

c² = 25

Take square root of both sides.

c = $\sqrt{\mathrm{25}}$ = 5

SOH

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

Opposite side of x = 3

Hypothenuse = 5

sin x = $\frac{3}{5}$

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

Adjacent side of x = 4

cos x = $\frac{4}{5}$

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ (2 x $\frac{3}{5}$)

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ $\frac{\mathrm{2\; x\; 3}}{5}$

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ $\frac{6}{5}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ x $\frac{5}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$

Let θ = the angle of depression.

tan θ = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

The opposite side of θ = 20 m

The adjacent side of θ = 75 m

tan θ = $\frac{\mathrm{20\; m}}{\mathrm{75\; m}}$

tan θ = $\frac{20}{75}$

Take tan^-1 of both sides.

θ = tan^-1($\frac{20}{75}$) = 14.93° ≈ 15° (nearest degree)

The angle subtended by the arc is ∠ ZOY

∠ ZOY = 2 x ∠ ZXY

∠ ZOY = 2 x 70° = 140°

Radius (r) = 18 cm

π = $\frac{22}{7}$

Length of arc ZY = $\frac{140}{360}$ x 2 x $\frac{22}{7}$ x 18 cm = 44 cm

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

Angles subtended by the same arc are the same.

∠ QSR = ∠ QPR = 52°

y = 52°

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

∠ QPR + ∠ PRQ + ∠ PQR = 180°

52° + x + 70° = 180°

x + 122° = 180°

x = 180° - 122°

x = 58°

Mean (x) = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean (x) = $\frac{\mathrm{2\; +\; 4\; +\; 7\; +\; 8\; +\; 9}}{5}$ = $\frac{30}{5}$ = 6

Σfx² = 1 x 2² + 1 x 4² + 1 x 7² + 1 x 8² + 1 x 9²

Σfx² = 4 + 16 + 49 + 64 + 81 = 214

Σf = 1 + 1 + 1 + 1 + 1 = 5

Variance = $\frac{214}{5}$ -6² = 6.8

a₁ = -20

n = 4

a₄ = 37

-20 + (4 - 1) x d = 37

-20 + 3d = 37

3d = 37 + 20

3d = 57

Divide both sides by 3

d = $\frac{57}{3}$ = 19

∠ QFG + ∠ EFG = 180°

105° + ∠ EFG = 180°

∠ EFG = 180° - 105°

∠ EFG = 75°

∠ FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)

50° + m + 75° = 180°

m + 125° = 180°

m = 180° - 125°

m = 55°

n = ∠ FGS

∠ FGS = 180° - 105°(Corresponding angle to line RS)

∠ FGS = 75°

n = ∠ FGS = 75°

Number of samples = 5 + 6 + 7 = 18

Number of possible selection for green pencil = 6

P(Green) = $\frac{6}{18}$ = $\frac{1}{3}$

Angle of sector of orange = 360° - (100° + 60° + 120°)

Angle of sector of orange = 360° - 280°

Angle of sector of orange = 80°

Angle of sector of apple = 120°

Number of oranges = 60

If 80° = 60

120° = $\frac{\mathrm{60\; x\; 120°}}{\mathrm{80°}}$ = 90

There are 90 apples

The mode is the one which occurs the most/ the one with the highest frequency/number of students

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

8, 8, 10, 11, 13, 13, 13, 14, 14, 15, 16, 17, 18, 18

There are two numbers 13 and 14 at the middle.

Median = $\frac{\mathrm{13\; +\; 14}}{2}$ = 13.5

Mean = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{8\; +\; 8\; +\; 10\; +\; 11\; +\; 13\; +\; 13\; +\; 13\; +\; 14\; +\; 14\; +\; 15\; +\; 16\; +\; 17\; +\; 18\; +\; 18}}{14}$

Mean = $\frac{188}{14}$ = 13.4286 ≈ 13.4

7 students scored more than 13.4 (14, 14, 15, 16, 17, 18 and 18)

100% represents the cost price.

Selling Price% = Cost Price% + Profit%

Profit% = 22.5%

Selling Price% = 100% + 22.5% = 122.5%

Selling Price = GH₵ 58,000.00

If 122.5% = GH₵ 58,000.00

100% = $\frac{\mathrm{GH₵\; 58000\; x\; 100\%}}{\mathrm{122.5\%}}$ = GH₵ 47346.94

Cost Price = GH₵ 47,346.94

Profit = Selling Price - Cost Price

Profit if the car had been sold at GH₵ 61,200.00 = GH₵ 61,200.00 - GH₵ 47,346.94 = GH₵ 13853.06

If GH₵ 47346.94 = 100%

GH₵ 13853.06 = $\frac{\mathrm{100\%\; x\; GH₵\; 13853.06}}{\mathrm{GH₵\; 47346.94}}$ = 29.2586%

Percentage profit ≈ 29.26% (two decimal places).