$\frac{\frac{1}{4}x2\frac{1}{2}}{12 ÷ 1\frac{1}{2}}$ = $\frac{\frac{1}{4}x2\frac{1}{2}}{12 ÷ 1\frac{1}{2}}$

$\frac{1}{4}x2\frac{1}{2}$ = $\frac{1}{4}x2\frac{1}{2}$

Change the mixed fraction to improper fraction.

2$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 2\; +\; 1}}{2}$ = $\frac{\mathrm{4\; +\; 1}}{2}$ = $\frac{5}{2}$

$\frac{1}{4}x2\frac{1}{2}$ = $\frac{1}{4}x\frac{5}{2}$

$\frac{1}{4}x2\frac{1}{2}$ = $\frac{\mathrm{1\; x\; 5}}{\mathrm{4\; x\; 2}}$ = $\frac{5}{8}$

$12 ÷ 1\frac{1}{2}$ = $12 ÷ 1\frac{1}{2}$

Change the mixed fraction to improper fraction.

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

$12 ÷ 1\frac{1}{2}$ = 12 ÷ $\frac{3}{2}$

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.

$12 ÷ 1\frac{1}{2}$ = 12 x $\frac{2}{3}$ = 4 x 2 = 8

$\frac{\frac{1}{4}x2\frac{1}{2}}{12 ÷ 1\frac{1}{2}}$ = $\frac{5}{8}$ ÷ 8

Every number is being divided by 1

8 = $\frac{8}{1}$

$\frac{\frac{1}{4}x2\frac{1}{2}}{12 ÷ 1\frac{1}{2}}$ = $\frac{5}{8}$ ÷ $\frac{8}{1}$

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.

$\frac{\frac{1}{4}x2\frac{1}{2}}{12 ÷ 1\frac{1}{2}}$ = $\frac{5}{8}$ x $\frac{1}{8}$ = $\frac{\mathrm{5\; x\; 1}}{\mathrm{8\; x\; 8}}$ = $\frac{5}{64}$

3 x 9 ^{1 + x} = 27^-x

Concept

A^b x A^c = A^b+c

Thus if the bases are the same, we can add the indexes (superscripts)

3 can go into the other numbers(9,27) hence we change then to have base 3

9 = 3¹x 3¹ = 3^{1 + 1} = 3²

27 = 3¹x 3¹ x 3¹ =3¹⁺¹⁺¹ = 3³

⇒ 3 x 3^{2(1 + x)} = 3^3(-x)

But 3 = 3¹, thus every number is raised to the power 1
⇒ 3¹ x 3^{2(1 + x)} = 3^3(-x)

Applying A^b x A^c = A^{b + c}

3 ¹ x 3^{2(1+ x)} = 3^{1 + 2(1 + x )}

3^{1+2(1+ x)} = 3^3(-x)

Concept
If A^b = A^c ⇒ b = c

Thus if the base at the left side of the equation, is the same as the right side of the equation, then the powers are the same

∴ 1 + 2(1 + x ) = 3(-x)

Concept

a(b + c ) = a x b + a x c

a(b - c ) = a x b - a x c

Thus the number outside the bracket multiplies each of the number in the bracket and the sign ( + or - ) is(are) maintained

-x = -1 x x

3(-x) = 3 x (-1) x x = -3x

1 + 2(1 + x ) = 1 + 2 x 1 + 2 x x

1 + 2(1 + x ) = 1 + 2 + 2x

1 + 2(1 + x ) = 3 + 2x

1 + 2(1+x) = 3(-x)

⇒ 3 + 2x = -3x

Grouping the x at the left and the rest at the right

Concept
If the sign is positive(+) and crosses to the other side of the equation the sign changes to negative ( - ) and the vice versa

2x + 3x = -3

5x = -3

Concept
Divide both sides by 5

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

Use a venn diagram to illustrate

In a class of 39 student ⇒ n(U)=39
let F represent students who offer Fante ⇒ n(F)= 25
Let T represent students who offer Twi ⇒ n(T)= 19
Five student do not offer any of the two languages ⇒ 5 are outside the sets F and T
Let n represent number of student who offer both languages

NB: You need to substract the number of the students who offer both languaes from each of the languages to get the students who offer only that languages.

When we add all the parts in the venn diagram we are to get the total number of students n(u) = 39

(25 - n)+ n + (19 - n) + 5 =39
25 - n + n + 19 -n + 5 = 39
-n + 49 = 39
-n = 39 - 49
-n = -10

Divide both sides by -1

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

n = 10
Only Twi = 19 - n
⇒ Only Twi = 19 - 10 = 9

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4-\sqrt{2}}{\sqrt{2}}$

Get rid of the denominator surd $\sqrt{2}$ by multiplying both the numerator and denominator by the denominator surd.

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{(4-\sqrt{2})x\sqrt{2}}{\sqrt{2}x\sqrt{2}}$

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4x\sqrt{2}-\sqrt{2}x\sqrt{2}}{\sqrt{2}x\sqrt{2}}$

Notes:

1. $\sqrt{a}$ x $\sqrt{a}$ = a

2. $\sqrt{2}$ x $\sqrt{2}$ = 2

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4\sqrt{2}-2}{2}$

The denominator affects each of the terms in the numerator.

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4\sqrt{2}}{2}$ - $\frac{2}{2}$

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = 2$\sqrt{2}$ - 1

Mary : Charity = 4 : 5

The difference in their ratios represent how much more or less one gets.

The sum of their ratios represents the total profit made.

Note: difference means subtraction.

Difference in ratios = 5 - 4 = 1

Sum of ratios = 4 + 5 = 9

Mary received GH₵ 5,000.00 less than Charity

The difference in ratios (1) represents how much less Mary received.

If 1 = GH₵ 5,000

9 = $\frac{\mathrm{9\; x\; GH₵\; 5,000}}{1}$ = GH₵ 45,000

∴ The profit made is GH₵ 45,000

Cost of car = ₦5,000,000.00

Deposit = ₦3,000.000.00

Rate = 8% compound intesrest per annum = $\frac{8}{100}$ = 0.08

Balance = ₦5,000,000.00 - ₦3,000,000.00 = ₦2,000,000.00

Hence the interest will only be charged on the ₦2,000,000.00 balance after the first year

Fisrt year

Interest = Principal x Rate x Time

Time = 1 (Compound interest is calculated year by year)

Interest = 0.08 x ₦2,000,000 = ₦160,000

Total amount to be paid by the end of the first year
⇒ ₦2,000,0000 + ₦160,000 = ₦2,160,000

Since he pays ₦1000,000 at each year

Amount left to pay the subsequent year
⇒ ₦2,160,000 - ₦1,000,000 = ₦1,160,000

Second Year

Interest = 0.08 x ₦1,160,000 = ₦92,800

Amount to pay
⇒ ₦92,800 + ₦1,160,000 = ₦1,250,8000

He pays another ₦1,000,000 by the end of the second year.

Hence amount to pay after the second year
⇒ ₦1,252,800 - ₦1,000,000.00 = ₦252,800

NB: For compound interest, the new principal amount is the interest + the previous amount to be paid

y ∝ $\frac{1}{{\mathrm{x}}^2}$

NB: ∝ is the symbol for proportional
inversely proportional is 1 over the expression

Introduce a constant to remove the proportionality sign.

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

Use the known values to calculate the constant(k) to get the general expression

y = 100 and x = 3

100 = k x $\frac{1}{3^2}$

100 = k x $\frac{1}{9}$

Get rid of the fraction by multiplying both sides of the equation by the denominator 9

100 x 9 = k x $\frac{1}{9}$ x 9

100 x 9 = k x 1

k = 900

Substitute the value of k to get the general equation.

y = 900 x $\frac{1}{{\mathrm{x}}^2}$ = $\frac{900}{{\mathrm{x}}^2}$

When y = 25

25 = $\frac{900}{{\mathrm{x}}^2}$

Get rid of the fraction by multiplying both sides of the equation by the denominator x²

25 x x² = $\frac{900}{{\mathrm{x}}^2}$ x x²

25 x x² = 900

Divide both sides by 25

x² = $\frac{900}{25}$

x² = 36

Take square root of both sides to get rid of the square.

x = $\sqrt{\mathrm{36}}$ = 6

t - $\frac{9}{5}$ = -1$\frac{1}{15}$

Change the mixed fraction to improper fraction.

1$\frac{1}{15}$ = $\frac{\mathrm{1\; x\; 15\; +\; 1}}{15}$

1$\frac{1}{15}$ = $\frac{\mathrm{15\; +\; 1}}{15}$

1$\frac{1}{15}$ = $\frac{16}{15}$

t - $\frac{9}{5}$ = -$\frac{16}{15}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 5 and 15. The L.C.M is 15

15 x t - 15 x $\frac{9}{5}$ = -$\frac{16}{15}$ x 15

15t - 3 x 9 = -16

15t - 27 = -16

15t = 27 - 16

15t = 11

Divide both sides by 15

t = $\frac{11}{15}$

Concept

(a + b)² = a² + 2 x a x b + b² = a² + 2ab + b²
(a - b)² = a²-2 x a x b + b² = a² -2ab + b²

Alternatively
(a + b)² = (a + b)x(a + b) [Expand and simplify]
(a -b)² = (a - b)x(a - b) [Expand and simplify]

(x - 2)² = x² - 2 x x x 2 + 2² = x² -4x + 4
(x + 1)² = x² + 2 x x x 1 + 1² = x² + 2x + 1

(x - 2)² + 3 = (x + 1)² - 6
x² - 4x + 4 + 3 = x² + 2x + 1 - 6
x² - 4x + 7 = x² + 2x - 5
x² - x² - 4x - 2x = -5 - 7
-6x = -12

Divide both sides by -6

x = $\frac{-12}{-6}$ = 2

The amount of money(f) which Dede has was not enough to buy the car
Which implies that Dede's amount is less than the cost of the car.

f(Dede's amount) < (less than) the cost of car (₦3,000,000.00)
but Kofi had enough money(k) to buy the car

This implies Kofi has either the exact amount or more than the cost of the car

⇒k ≥ ₦3,000,000.00 and the cost of the car is less than or equal (≤) to Kofi's amount
f < ₦3,000,000.00 ≤ k
Read as Dede's amount(f) is less than ₦3,000,000.00 and ₦3,000,000.00 is less than or equal to Kofi's amount(k)

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z(z\; -\; 3)\; -\; (z\; -\; 2)(z\; +\; 2)}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z\; x\; z\; -\; 3\; x\; z\; -\; [z(z\; +\; 2)\; -\; 2(z\; +\; 2)]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [z\; x\; z\; +\; z\; x\; 2\; -\; 2\; x\; z\; -2\; x\; 2]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [}{\mathrm{z}}^2\mathrm{+\; 2z\; -\; 2z\; -\; 4]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [}{\mathrm{z}}^2\mathrm{-\; 4]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

Notes:
1. -[+] = - x + = -
2. -[z²] = - z²
3. -[-] = - x - = +
4. -[-4] = + 4

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -}{\mathrm{z}}^2\mathrm{+\; 4}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-}{\mathrm{z}}^2\mathrm{4\; -\; 3z}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{4\; -\; 3z}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$

Notes:

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

2. a^-1 = $\frac{1}{{\mathrm{a}}^1}$ = $\frac{1}{\mathrm{a}}$

3. x^-1 = $\frac{1}{{\mathrm{x}}^1}$ = $\frac{1}{\mathrm{x}}$

4. y^-1 = $\frac{1}{{\mathrm{y}}^1}$ = $\frac{1}{\mathrm{y}}$

x^-1 + y^-1 = $\frac{1}{\mathrm{x}}$ + $\frac{1}{\mathrm{y}}$

x^-1 + y^-1 = $\frac{\mathrm{1\; x\; y\; +\; 1\; x\; x}}{\mathrm{xy}}$

x^-1 + y^-1 = $\frac{\mathrm{y\; +\; x}}{\mathrm{xy}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ ÷ (x + y)

Every number or expression is being divided by 1 but not written

x + y = $\frac{\mathrm{x\; +\; y}}{1}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ ÷ $\frac{\mathrm{x\; +\; y}}{1}$

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.

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{\mathrm{x\; +\; y}}{\mathrm{xy}}$ x $\frac{1}{\mathrm{x\; +\; y}}$

x + y cancels each other.

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{1}{\mathrm{xy}}$ x $\frac{1}{1}$

$\frac{{\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}}{\mathrm{x\; +\; y}}$ = $\frac{1}{\mathrm{xy}}$

Let c = length of the chord.

Method I

Using the cosine rule

c² = 5² + 5² - 2 x 5 x 5 cos70°

c² = 25 + 25 - 50cos70°

c² = 50 - 50cos70°

c² = 32.9

Take the square root of both sides to get rid of the square.

c = $\sqrt{\mathrm{32.9}}$ = 5.7359 cm ≈ 5.7 cm (one decimal place)

Method II

Using the sine rule

Since the radii are the same, the triangle the chord makes with the circle is an isosceles triangle (has two sides equal).

The base angles of an isosceles triangle are the same.

Let y = base angle of the isosceles triangle.

y + y + 70° = 180° (Sum of angles in a triangle).

2y = 180° - 70°

2y = 110°

Divide both sides by 2.

y = $\frac{\mathrm{110°}}{2}$ = 55°

Each of the base angles is 55°

Applying the sine rule,

$\frac{\mathrm{sin\; 70°}}{\mathrm{c}}$ = $\frac{\mathrm{sin\; 55°}}{5}$

Cross multiply.

c x sin 55° = 5 x sin 70°

Divide both sides by sin 55°

c = $\frac{\mathrm{5\; x\; sin\; 70°}}{\mathrm{sin\; 55°}}$ = 5.7358 cm ≈ 5.7 cm (one decimal place)

Base area is a square.

Area of a square = Length x Length

Base area of the pyramid = 10 cm x 10 cm = 100 cm²

Volume = 1400 cm³

Let h = the height of the pyramid.

1400 = $\frac{1}{3}$ x 100 x h

Get rid of the fraction by multiplying both sides of the equation by the denominator 3

1400 x 3 = $\frac{1}{3}$ x 100 x h x 3

1400 x 3 = 100 x h

Divide both sides by 100

h = $\frac{\mathrm{1400\; x\; 3}}{100}$ = 42 cm

∠ POR = 2 x ∠ PQR = 2 x 45° = 90°

Radius = 12 cm

π = $\frac{22}{7}$

Length of arc PR = $\frac{\mathrm{90°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 12 cm

Length of arc PR = 18.857 cm ≈ 19 cm (nearest cm)

w + x + y + z = 180° (Sum of angles on a straight line)

But (w + x + y) = 140°

140° + z = 180°

z = 180° - 140°

z = 40°

(x + y + z) = 130°

x + y + z = 130°

x + y = 130° - z

But z = 40°

x + y = 130° - 40°

x + y = 90°

Pentagon has five sides.

n = 5

Sum of angles = (5 - 2) x 180°

Sum of angles = 3 x 180° = 540°

x° + (x + 5)° + (x + 10)° + (x + 15)° + (x + 20)° = 540°

x + x + 5 + x + 10 + x + 15 + x + 20 = 540

5x + 50 = 540

5x = 540 - 50

5x = 490

Divide both sides by 5

x = $\frac{490}{5}$ = 98

∠ PQR : ∠ QPS = 3 : 7

∠ PQR + ∠ QPS = 180° (Respective consecutive angles are supplementary)

Sum of ratios = 3 + 7 = 10

The sum of ratios represents the sum of the angles.

If 10 = 180°

3 = $\frac{\mathrm{3\; x\; 180°}}{10}$ = 54°

∴ ∠ PQR = 54°

If 10 = 180°

7 = $\frac{\mathrm{7\; x\; 180°}}{10}$ = 126°

∴ ∠ QPS = 126°

Alternatively,

∠ QPS = 180° - ∠ PQR

∠ QPS = 180° - 54° = 126°

∠ PSR = ∠ PQR = 54° (Opposite angles are equal)

∠ QSR = $\frac{\mathrm{\angle \; PSR}}{2}$ = $\frac{\mathrm{54°}}{2}$ = 27°

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

Y₂ = 6
Y₁ = 4
X₂ = -2
X₁ = 1

Gradient = $\frac{\mathrm{6\; -\; 4}}{\mathrm{-2\; -\; 1}}$ = $\frac{2}{-3}$ = -$\frac{2}{3}$

Distance = $\sqrt{{\left(\mathrm{8\; -\; 3}\right)}^2+{\left(\mathrm{10\; --\; 2}\right)}^2}$

Note: - - = +

Distance = $\sqrt{{\left(5\right)}^2+{\left(\mathrm{10\; +\; 2}\right)}^2}$

Distance = $\sqrt{5^2+{12}^2}$

Distance = $\sqrt{\mathrm{25\; +\; 144}}$

Distance = $\sqrt{169}$ = 13 units

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = $\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$

cos 65° = cos (90° - 25°) = sin 25°

sin 35° = sin (90° - 55°) = cos 55°

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = $\frac{\mathrm{sin\; 25°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{cos\; 55°}}{\mathrm{cos\; 55°}}$

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = 1 + 1 = 2

Let h = height of the tree

h = y + 2 m

tan 30° = $\frac{\mathrm{y}}{12\sqrt{3}}$

Get rid of the fraction by multiplying both sides by the denominator 12$\sqrt{3}$

y = 12$\sqrt{3}$ x tan 30° = 12 m

h = 12 m + 2 m = 14 m

Bearing starts from the North clockwise.

312° = 270° + 42°

Samples = {1, 2, 3, 4, 5, 6}

Number of samples = 6

Number of possible selection/Number of 3 = 1

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

Mean (x̄) = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{240}{15}$ = 16 years

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.

Method I

Σf = 5 + 6 + 3 + 1 = 15

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{15\; +\; 1}}{2}$ = $\frac{16}{2}$ = 8 position

5 position for 15 years

5 + 6 = 11 position for 16 years

∴ The 8^th position falls under 16 years

Method II

List the ages and pick the middle age.

Note: you list each age n times. Where n is the No. of students.

15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 18

x = $\frac{\mathrm{mn}}{3}$ ------ eqn 1

Substitute m = $\frac{\mathrm{v}}{\mathrm{y}}$ into ------ eqn 1

x = $\frac{\mathrm{v}}{\mathrm{y}}$ x n ÷ 3

x = $\frac{\mathrm{v\; x\; n}}{\mathrm{y}}$ ÷ 3

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ ÷ 3

Every number is being divided by 1 but it is not written.

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

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ ÷ $\frac{3}{1}$

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.

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ x $\frac{1}{3}$

x = $\frac{\mathrm{vn\; x\; 1}}{\mathrm{y\; x\; 3}}$

x = $\frac{\mathrm{vn}}{\mathrm{3y}}$

Calculate the value of 123_five in base 10 and change the base 10 to base six to get it's value in base six (M).

123_five = 1 x 5² + 2 x 5¹ + 3 x 5⁰

123_five = 25 + 10 + 3 = 38

38 base 10 to base 6

$$\begin{array}{cc}\mathrm{}& 38\\ 6& \mathrm{6\; r\; 2}\\ 6& \mathrm{1\; r\; 0}\\ 6& \mathrm{0\; r\; 1}\end{array}$$

You list the remainders from the bottom to the top.

38_ten = 102_six

M = 102

Let h = height of triangle XYZ

h² + 4² = 5²

h² + 16 = 25

h² = 25 - 16

h² = 9

Take square root of both sides.

h = $\sqrt{9}$ = 3 cm

Base = 8 cm

Height = 3 cm

Area of triangle XYZ = $\frac{1}{2}$ x 8 cm x 3 cm = 12 cm²

Total surface ara = 2(12 x 8 + 12 x 3 + 3 x 8) cm²

Total surface ara = 2 x 156 cm² = 312 cm²

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{6\; cm}}{2}$ = 3 cm

π = $\frac{22}{7}$

Volume = $\frac{4}{3}$ x $\frac{22}{7}$ x 3³ = 113.143 cm³ ≈ 113 cm³

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43° (Vertically opposite angles)

∠ OBC = ∠ OCB = y = 43°

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43°

∠ OBC = ∠ OCB = y = 43°

∠ OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)

∠ OBC = ∠ OCB = 43°

43° + 43° + ∠ BOC = 180°

86° + ∠ BOC = 180°

∠ BOC = 180° - 86°

∠ BOC = 94°

x = $\frac{\mathrm{BOC}}{2}$ = $\frac{\mathrm{94°}}{2}$ = 47°

nth term = ar^n-1

First term (a) = $\frac{1}{2}$

Common ratio (r) = $\frac{{\mathrm{a}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n\; -\; 1}}}$

Common ratio (r) = $\frac{1}{4}$ ÷ $\frac{1}{2}$

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.

Common ratio (r) = $\frac{1}{4}$ x $\frac{2}{1}$ = $\frac{1}{2}$

Note: 2 divides itself one time and 4, 2 times.

nth term = ar^n-1

nth term = $\frac{1}{2}$ x ${\left(\frac{1}{2}\right)}^{\mathrm{n\; -\; 1}}$

$\frac{1}{2}$ = ${\left(\frac{1}{2}\right)}^1$ (Every number is raised to the power 1 but not written)

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

nth term = ${\left(\frac{1}{2}\right)}^{\mathrm{1\; +\; n\; -\; 1}}$

nth term = ${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{c}}$ = $\frac{{\mathrm{a}}^{\mathrm{c}}}{{\mathrm{b}}^{\mathrm{c}}}$

nth term = $\frac{1^{\mathrm{n}}}{2^{\mathrm{n}}}$

Note:
1. 1 raised to the power of any number is 1
2. 1ⁿ = 1

nth term = $\frac{1}{2^{\mathrm{n}}}$

Wednesday → Went, day 1

Thursday → No walk, day 2

Friday → No walk, day 3

Saturday → No walk, day 4

Sunday → No walk, day 5

Monday → Will go for a walk, day 6

Notes:

1. Interval for the minutes axis = 2.45 - 1.95 = 0.5

2. Each small grid = $\frac{0.5}{10}$ = 0.05

3. Interquartile range of the distribution = Q₃ − Q₁

Q₃ position = $\frac{\mathrm{3\; x\; \Sigma f}}{4}$

Q₃ position = $\frac{\mathrm{3\; x\; 50}}{4}$ = 37.5 position

Corresponding minutes for 37.5 is 2.95 + 5 x 0.05 = 3.2 minutes

Q₃ = 3.2 minutes

Q₁ position = $\frac{\mathrm{1\; x\; \Sigma f}}{4}$

Q₁ position = $\frac{\mathrm{1\; x\; 50}}{4}$ = 12.5 position

Corresponding minutes for 12.5 is 2.45 minutes

Q₃ = 3.2 minutes

Q₁ = 2.45

Interquartile range of the distribution = 3.2 - 2.45 = 0.75

At least 3 minutes means 3 minutes or more.

Number of customers who waited for at least 3 minutes = 50 - 32 = 18

Notes:

1. Interval for the minutes axis = 2.45 - 1.95 = 0.5

2. Each small grid = $\frac{0.5}{10}$ = 0.05

3. 3 minutes is 1 grid (0.05) after 2.95. 3 = 2.95 + 0.05

4. The corresponding number of customers for the 3 minutes is 32

(4x² + 3x - 1)(3x + 1) = (4x² + 3x - 1)(3x + 1)

(4x² + 3x - 1)(3x + 1) = 3x(4x² + 3x - 1) + 1(4x² + 3x - 1)

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² - 3x + 4x² + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² + 4x² - 3x + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 13x² + 0x - 1

Method I

If α and β are the roots, then

(x - α)(x - β) = 0

(x - $\frac{1}{2}$)(x - -$\frac{3}{2}$) = 0

Note: -- = +

(x - $\frac{1}{2}$)(x + $\frac{3}{2}$) = 0

x(x + $\frac{3}{2}$) -$\frac{1}{2}$(x + $\frac{3}{2}$) = 0

x² + $\frac{\mathrm{3x}}{2}$ - $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = 0

x² + $\frac{\mathrm{3x\; -\; x}}{2}$ - $\frac{3}{4}$ = 0

x² + $\frac{\mathrm{2x}}{2}$ - $\frac{3}{4}$ = 0

x² + x - $\frac{3}{4}$ = 0

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

4 x x² + 4 x x - $\frac{3}{4}$ x 4 = 0 x 4

4x² + 4x - 3 = 0

Method II

x² - (sum of roots)x + product of roots = 0

Sum of roots = $\frac{1}{2}$ + -$\frac{3}{2}$

Sum of roots = $\frac{1}{2}$ - $\frac{3}{2}$

Sum of roots = $\frac{\mathrm{1\; -\; 3}}{2}$ = $\frac{-2}{2}$ = -1

Product of roots = $\frac{1}{2}$ x - $\frac{3}{2}$ = - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = - $\frac{3}{4}$

x² - (-1)x + - $\frac{3}{4}$ = 0

-(-) or -- = +

x² + x - $\frac{3}{4}$ = 0

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

4 x x² + 4 x x - $\frac{3}{4}$ x 4 = 0 x 4

4x² + 4x - 3 = 0

Let the number = n

Two times = 2 x n = 2n

one-third = $\frac{1}{3}$

one-third of the number = $\frac{1}{3}$ x n = $\frac{\mathrm{n}}{3}$

Two times a number added to one-third of the number = 2n + $\frac{\mathrm{n}}{3}$

Gives means =

2n + $\frac{\mathrm{n}}{3}$ = 5$\frac{1}{6}$

Change the mixed fraction to improper fraction.

5$\frac{1}{6}$ = $\frac{\mathrm{6\; x\; 5\; +\; 1}}{6}$ = $\frac{\mathrm{30\; +\; 1}}{6}$ = $\frac{31}{6}$

2n + $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$

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

6 x 2n + 6 x $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$ x 6

12n + 2 x n = 31

12n + 2n = 31

14n = 31

Divide both sides by 14

n = $\frac{31}{14}$ = 2$\frac{3}{14}$

∴ the number is 2$\frac{3}{14}$

Mean (x̄) = $\frac{\mathrm{5\; +\; 8\; +\; 3\; +\; 7\; +\; 2}}{5}$ = $\frac{25}{5}$ = 5

|(x - x̄)| for 5 = 5 - 5 = 0

|(x - x̄)| for 8 = 8 - 5 = 3

|(x - x̄)| for 3 = 3 - 5 = |-2| = 2

|(x - x̄)| for 7 = 7 - 5 = 2

|(x - x̄)| for 2 = 2 - 5 = |-3| = 3

f = 1 for each

Σf|(x - x̄)| = 1 x 0 + 1 x 3 + 1 x 2 + 1 x 2 + 1 x 3 = 10

Σf = 1 + 1 + 1 + 1 + 1 = 5

Mean Deviation = $\frac{10}{5}$ = 2

1 out of every 8 persons is a graduate

P(Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) and P(Etteh being Graduate)

And means multiplication (x)

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) x P(Etteh being Graduate)

P(Yonni being Graduate) = $\frac{1}{8}$

P(Etteh being Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{1}{8}$ x $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{\mathrm{1\; x\; 1}}{\mathrm{8\; x\; 8}}$ = $\frac{1}{64}$

The angle between a tangent and the radius is 90° and not complementary.

Two angles are complementary if the sum of the angles is equal to 90º (ninety degrees). For example, an angle that is 30º and an angle that is 60º are two complementary angles.

Use points on the graph to substitute the general equation y = ax² + bx + c.

Note: it is easier to use points that either x = 0 or y = 0

When x = -1, y = 0

0 = a(-1)² + b(-1) + c

a - b + c = 0 ------- eqn 1

When x = 2, y = 0

0 = a(2)² + b(2) + c

4a + 2b + c = 0 ------- eqn 2

When y = 2, x = 0

2 = a(0)² + b(0) + c

c = 2

Substitute c = 2 into ----- eqn 1

a - b + 2 = 0

a - b = -2 -------- eqn 3

Substitute c = 2 into ----- eqn 2

4a + 2b + 2 = 0

4a + 2b = - 2

Divide both sides by 2

2a + b = - 1 ------ eqn 4

Get rid of b. ----- eqn 3 + ----- eqn 4

3a = -3

Divide both sides by 3.

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

Substitute a = -1 into ---- eqn 4

2(-1) + b = -1

-2 + b = -1

b = -1 + 2

b = 1

Substitute a = -1, b = 1 and c = 2 into the general equation y = ax² + bx + c

y = ax² + bx + c

y = (-1)x² + 1(x) + 2

y = -x² + x + 2

y = 2 + x - x²

Notes:

1. log a^b = b log a

2. ${\mathrm{log\; Z}}^{\frac{1}{3}}$ = $\frac{1}{3}$ x log Z

3. log y³ = 3 log y

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{log\; x\; -\; }\frac{1}{3}\mathrm{log\; Z}}{\mathrm{3log\; y}}$

log x = 0.3030, log y = 0.4777 and log Z = 0.8451

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{0.3030\; -\; }\frac{1}{3}\mathrm{x\; 0.8451}}{\mathrm{3\; x\; 0.4777}}$

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = 0.01486 ≈ 0.0149

~ means not.

⇒ means implies.

-5 = $\frac{-5}{1}$ hence is a rational number

$\sqrt{4}$ = 2 = $\frac{2}{1}$ hence is a rational number

3$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 4\; +\; 3}}{4}$ = $\frac{\mathrm{12\; +\; 3}}{4}$ = $\frac{15}{4}$ hence is a rational number

$\sqrt{\mathrm{90}}$ = 9.4868 hence not a rational number since can't be expressed as a fraction of two integers.

∆ BCD is isosceles since two sides are equal.

The base angles of isosceles triangles are the same.

∠ ECD = ∠ DBE = 35° (Base angles are equal)

∠ EDB = $\frac{1}{2}$ x ∠ CDB

∠ ECD + ∠ DBE + ∠ CDB = 180°(Sum of angles in a triangle)

35° + 35° + ∠ CDB = 180°

70° + ∠ CDB = 180°

∠ CDB = 180° - 70°

∠ CDB = 110°

∠ EDB = $\frac{1}{2}$ x 110° = 55°

∠ ABD = 90°(Angle subtended by the diameter)

∠ BAD + ∠ EDB + ∠ ABD = 180°(Sum of angles in a triangle)

m + 55° + 90° = 180°

m + 145° = 180°

m = 180° - 145°

m = 35°

∆ BCD is isosceles since two sides are equal.

The base angles of isosceles triangles are the same.

∠ ECD = ∠ DBE = 35° (Base angles are equal)

∠ ACD = 90°(Angle subtended by a diameter)

∠ ACE + ∠ ECD = 90°

n + 35° = 90°

n = 90° - 35°

n = 55°