Sum of interior angles = 1260°

(n - 2) x 180° = 1260°

(n - 2) x 180 = 1260

Divide both sides by 180

n - 2 = $\frac{1260}{180}$

n - 2 = 7

n = 7 + 2

n = 9

Using the known sides, tan θ = $\frac{\mathrm{12\; m}}{\mathrm{30\; m}}$ where 12 m is the opposite side of θ and 30 m is the adjacent side of θ

tan θ = $\frac{\mathrm{12\; m}}{\mathrm{30\; m}}$

tan θ = $\frac{12}{30}$

Take tan^-1 of both sides.

θ = tan^-1($\frac{12}{30}$)

θ = 21.80° ≈ 22°(nearest degree)

Set = {13, 14, ...., 30}

Set = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Number of samples = 18

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.

Set of prime numbers from the set = {13, 17, 19, 23, 29}

Number of prime numbers = 5

P(Prime) = $\frac{5}{18}$

Regina's age now = 34.

In n years time, Regina will be (34 + n) years.

Daughter's age now = 5.

In n years time, the daughter will be (5 + n) years.

In n years, Regina will be twice as old as her daughter

Twice means 2 times (2 x)

Regina's age in n years time = 2 x daughter's age in n years time.

34 + n = 2 x (5 + n)

34 + n = 2 x 5 + 2 x n

34 + n = 10 + 2n

34 - 10 = 2n - n

24 = n

∴ n = 24

Reflection in the line x - 2 = 0, x = 2

k = 2

(-7, 8) → (2 x 2 - (-7), 8)

(-7, 8) → (4 + 7, 8)

(-7, 8) → (11, 8)

Note:
1. -- = + or -(-) = +
2. -(-7) = + 7

P is inversely proportional to Q₂

P ∝ $\frac{1}{{\mathrm{Q}}^2}$

Get rid of the proportionality sign (∝) by introducing a constant.

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

P = $\frac{\mathrm{k}}{{\mathrm{Q}}^2}$

When Q = 6, P = 2.25

$\frac{\mathrm{k}}{6^2}$ = 2.25

$\frac{\mathrm{k}}{36}$ = 2.25

Multiply both sides by 36

k = 2.25 x 36 = 81

P = $\frac{\mathrm{k}}{{\mathrm{Q}}^2}$

P = $\frac{81}{{\mathrm{Q}}^2}$

When Q = 3

P = $\frac{81}{3^2}$

P = $\frac{81}{9}$ = 9

y = $\sqrt{\frac{\mathrm{px}}{\mathrm{r}}-{\mathrm{r}}^2\mathrm{x}}$

Get rid of the square root by squaring both sides of the equation.

y² = $\frac{\mathrm{px}}{\mathrm{r}}-{\mathrm{r}}^2\mathrm{x}$

Factorize x out.

y² = x($\frac{\mathrm{p}}{\mathrm{r}}-{\mathrm{r}}^2$)

Every number is being divided by 1

r² = $\frac{{\mathrm{r}}^2}{1}$

y² = x($\frac{\mathrm{p}}{\mathrm{r}}-\frac{{\mathrm{r}}^2}{1}$)

y² = x($\frac{\mathrm{p\; -}\mathrm{r\; x}{\mathrm{r}}^2}{\mathrm{r}}$)

Note: r x r² = r^{1 + 2} = r³

y² = x($\frac{\mathrm{p\; -}{\mathrm{r}}^3}{\mathrm{r}}$)

Multiply both sides by r to get rid of the fraction.

y² x r = x(p - r³)

ry² = x(p - r³)

Divide both sides by (p - r³)

x = $\frac{\mathrm{r}{\mathrm{y}}^2}{\mathrm{p}-{\mathrm{r}}^3}$

The more the workers, the lesser the time.

Hence time (t) is inversely proportional to the number of workers (w).

t ∝ $\frac{1}{\mathrm{w}}$

Remove the proportionality by introducing a constant (k).

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

Use the known information to solve for k

7 workers used 6 days.

6 = k x $\frac{1}{7}$

k = 6 x 7

The general equation is:

t = 6 x 7 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{6\; x\; 7}}{\mathrm{w}}$

Calculate the time (t) for 2 workers.

w = 2

t = $\frac{\mathrm{6\; x\; 7}}{2}$ = 3 x 7 = 21

∴ It will take the two men 21 days to complete the work.

Volume = 264 cm³

Radius (r) = 6 cm

π = $\frac{22}{7}$

264 = $\frac{22}{7}$ x 6² x $\frac{\mathrm{h}}{3}$

264 = $\frac{\mathrm{22\; x\; 36\; x\; h}}{\mathrm{7\; x\; 3}}$

Get rid of the fraction by multiplying both sides by the denominator 7 x 3.

264 x 7 x 3 = 22 x 36 x h

Divide both sides by 22 x 36

h = $\frac{\mathrm{264\; x\; 7\; x\; 3}}{\mathrm{22\; x\; 36}}$ = 7 cm

For a relation to be not defined/undefined, the denominator must be equal to 0

x² + 2x + 1 = 0

Express 2x as two numbers when you multiply you will get c (1) and when you add you will get b (2)

1 x 1 = 1 and 1 + 1 = 2 hence the number is 1

x² + x + x + 1 = 0

x(x + 1) + 1(x + 1) = 0

(x + 1)(x + 1) = 0

When x + 1 = 0, x = -1

Range = 5

Largest value = 50

50 - Least = 5

50 - 5 = Least

45 = Least

Mean = 16

Number of numbers = 10

$\frac{\mathrm{Sum\; of\; numbers}}{10}$ = 16

Get rid of the fraction by multiplying both sides by the denominator 10.

Sum of numbers = 16 x 10 = 160

When k is added, the new sum = 160 + k and the number of numbers become 10 + 1 = 11

Mean for the eleven (11) numbers is 18

$\frac{\mathrm{160\; +\; k}}{11}$ = 18

Get rid of the fraction by multiplying both sides by the denominator 11

160 + k = 18 x 11

160 + k = 198

k = 198 - 160

k = 38

P(Both fail) = P(John fail) and P(James fail)

And means multiplication (x)

P(Both fail) = P(John fail) x P(James fail)

Probability of not = 1 - Probability of is

P(John pass) = $\frac{3}{4}$

P(John fail) = 1 - P(John pass)

P(John fail) = 1 - $\frac{3}{4}$

Every number is being divided by 1.

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

P(John fail) = $\frac{1}{1}$ - $\frac{3}{4}$

P(John fail) = $\frac{\mathrm{4\; x\; 1\; -\; 1\; x\; 3}}{4}$

P(John fail) = $\frac{\mathrm{4\; -\; 3}}{4}$

P(John fail) = $\frac{1}{4}$

P(James pass) = $\frac{3}{5}$

P(James fail) = 1 - P(James pass)

P(James fail) = 1 - $\frac{3}{5}$

P(James fail) = $\frac{1}{1}$ - $\frac{3}{5}$

P(James fail) = $\frac{\mathrm{5\; x\; 1\; -\; 1\; x\; 3}}{5}$

P(James fail) = $\frac{\mathrm{5\; -\; 3}}{5}$

P(James fail) = $\frac{2}{5}$

P(Both fail) = P(John fail) x P(James fail)

P(Both fail) = $\frac{1}{4}$ x $\frac{2}{5}$ = $\frac{1}{10}$

From Pythagorean theorem,

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

c² = 6² + 8²

c² = 100

Take square root of both sides.

c = $\sqrt{\mathrm{100}}$ = 10 cm

Perimeter is the sum of all the sides of a shape.

The perimeter of the rhombus = c + c + c + c

The perimeter of the rhombus = 4c

The perimeter of the rhombus = 4 x 10 cm

The perimeter of the rhombus = 40 cm

OR = OQ = Radius

∆ RQO is an isosceles triangle.

Isosceles triangle has the base angles the same.

∠ ORQ = ∠ OQR = 42°

∠ ROQ + 42° + 42° = 180° (sum of angles of a triangle)

∠ ROQ + 84 = 180

∠ ROQ = 180 - 84

∠ ROQ = 96°

QPR = $\frac{1}{2}$ x ∠ ROQ

QPR = $\frac{1}{2}$ x 96° = 48°

x(3x + 4) = 4

Expand the bracket.

3x x x + 4 x x = 4

3x² + 4x = 4

3x² + 4x - 4 = 0

Notes:

1. a x c = 3 x -4 = -12

2. Split b (4) such that two numbers when you multiply you will get a x c (-12) and when you add you will get b(4)

3. The numbers are 6 and - 2

3x² + 6x - 2x - 4 = 0

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

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

When x + 2 = 0, x = -2

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

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

P = {-2, 0, 2, 4, 6}

Q = {-3, -1, 0, 2, 3, 5}

P ∩ Q = {0, 2}

Method I

P = 1200

r = $\frac{15}{100}$ = 0.15

n = 3

A = 1200(1 - 0.15)³

A = 1200(0.85)³

A = 1200 x 0.85³

A = 736.95

∴ value at the end of three years = GH₵ 736.95

Method II

Depreciation rate = 15%

First Year

Amount = GH₵ 1,200

Depreciated amount = $\frac{15}{100}$ x GH₵ 1200 = GH₵ 180

Value at end of year = GH₵ 1200 - GH₵ 180

Value at end of year = GH₵ 1020

Second Year

Amount = GH₵ 1020

Depreciated amount = $\frac{15}{100}$ x GH₵ 1020 = GH₵ 153

Value at end of year = GH₵ 1020 - GH₵ 153

Value at end of year = GH₵ 867

Third Year

Amount = GH₵ 867

Depreciated amount = $\frac{15}{100}$ x GH₵ 867 = GH₵ 130.05

Value at end of year = GH₵ 867 - GH₵ 130.05

Value at end of year = GH₵ 736.95

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

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (4, 3 and 6). The L.C.M is 12.

12 x $\frac{\mathrm{x\; -\; 2}}{4}$ - 12 x $\frac{\mathrm{2x\; -\; 4}}{3}$ = 12 x $\frac{5}{6}$

3 x (x - 2) - 4(2x - 4) = 2 x 5

3(x - 2) - 4(2x - 4) = 10

3x - 6 - 8x + 16 = 10

Note:
1. - x - = +
2. -4 x -4 = + 16

3x - 8x - 6 + 16 = 10

-5x + 10 = 10

-5x = 10 - 10

-5x = 0

Divide both sides by -5

x = $\frac{0}{-5}$ = 0

Draw an imaginary parallel line to cut t.

t = ∠ CGE + ∠ EGJ

∠ ACD = ∠ BCG = 35° (Vertically opposite angles)

∠ CGE = ∠ BCG = 35° (Alternate angles)

∠ HJG + ∠ IJG = 180° (Angles on a straight line)

∠ HJG = 140°

140° + ∠ IJG = 180°

∠ IJG = 180° - 140°

∠ IJG = 40°

∠ EGJ = IJG = 40° (Alternate angles)

t = ∠ CGE + ∠ EGJ

t = 35° + 40°

t = 75°

⊗ means multiplication

6 x 7 = 42

42 ÷ 8 = 5 remainder 2 (42 = 5 x 8 + 2)

The remainder is the modulo, hence 42 modulo 8 is 2.

∴ y = 2

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.

0.0063075

The leading zeros (0) are not significant.

The significant figures are 6, 3, 0, 7 and 5

The third significant falls on 0 but the next significant to it is more than 4 hence 1 is added to the 0 (0 + 1) making 1.

0.0063075 = 0.00631 (three significant figures)

∧ is used to represent logical AND and ∨ is used to represent logical OR

With ∨, the statement is true if any is true and false if both are false.

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

Pick any two points from the table to calculate the gradient.

Using the first two points, (0, 1) and (2, 2).

Y₂ = 2
Y₁ = 1
X₂ = 2
X₁ = 0

Gradient = $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; -\; 0}}$ = $\frac{1}{2}$

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

Pick any two points from the table to calculate the gradient.

Using the first two points, (0, 1) and (2, 2).

Y₂ = 2
Y₁ = 1
X₂ = 2
X₁ = 0

Gradient = $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; -\; 0}}$ = $\frac{1}{2}$

Using the first point (0, 1)

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

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

2 x y - 1 x 2 = 1(x - 0)

2y - 2 = x

2y = x + 2

a_{n = ar^n-1}

Second term = 9

n = 2

ar^2-1 = 9

ar = 9 ------ eqn 1

Fourth term = 81

n = 4

ar^4-1 = 81

ar³ = 81 ----- eqn 2

Get rid of a by dividing ----- eqn 2 by ----- eqn 1

$\frac{\mathrm{a}{\mathrm{r}}^3}{\mathrm{ar}}$ = $\frac{81}{9}$

Notes:

1. The a cancel each other

2. From law of indices a^m ÷ aⁿ = $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}$ = a^{m - n}

3. r = r¹

r^{3 - 1} = 9

r² = 9

9 = 3 x 3 = 3²

r² = 3²

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

r = 3

∴ common ratio = 3

The ratio of the corresponding sides of the larger and smaller ∆s are the same.

$\frac{\mathrm{|PR|}}{\mathrm{|PY|}}$ = $\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

|PQ| = |PW| + |WQ|

|WQ| = |PQ| - |PW|

Use the known sides to find the |PQ|

$\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

$\frac{\mathrm{15\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{4\; cm}}$

Cross multiply.

|PQ| x 6 cm = 15 cm x 4 cm

Divide both sides by 6 cm

|PQ| = $\frac{\mathrm{15\; cm\; x\; 4\; cm}}{\mathrm{6\; cm}}$ = 10 cm

|WQ| = |PQ| - |PW|

|WQ| = 10 cm - 4 cm

|WQ| = 6 cm

Radius (r) = 3 cm

π = $\frac{22}{7}$

Height (h) = 4 cm

Total surface area = 2 x $\frac{22}{7}$ x 3² + 2 x $\frac{22}{7}$ x 3 x 4 cm²

Total surface area = 132 cm²

Distance covered in one complete revolution = the circumference of the wheel

Distance covered in the 8 complete revolutions = 8 x the circumference of the wheel

Circumference = 2 x π x radius

Distance covered in the 8 complete revolutions = 8 x 2 x π x radius

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{21\; cm}}{2}$ = 10.5 cm

π = $\frac{22}{7}$

Distance covered in the 8 complete revolutions = 8 x 2 x $\frac{22}{7}$ x 10.5 cm

Distance covered in the 8 complete revolutions = 528 cm

Sum of exterior angles of a polygon = 360°

3x + 2x + x + 10 + 4x + 50 + 20 = 360

10x + 80 = 360

10x = 360 - 80

10x = 280

Divide both sides by 10

x = $\frac{280}{10}$ = 28

Find the value of y in order to calculate each angles to determine the type of triangle.

Angles of the triangle are (y + 10)°, (2y - 40)° and (3y - 90)°

Sum of interior angles of a triangle is 180°

y + 10 + 2y - 40 + 3y - 90 = 180

6y - 120 = 180

6y = 180 + 120

6y = 300

Divide both sides by 6

y = $\frac{300}{6}$ = 50

Angle y + 10 = 50 + 10 = 60°

Angle 2y - 40 = 2 x 50 - 40 = 100 - 40 = 60°

3y - 90 = 3 x 50 - 90 = 150 - 90 = 60°

All the angles are the same (60°) hence the triangle is equilateral

Draw the right-angled triangle and determine the third side.

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

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

In sin A = $\frac{3}{5}$, the opposite side of A is 3 and hypothenuse side is 5.

From pythagoras theorem

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

From the triangle, the side facing the right-angle is y, hence y is the hypothenuse (c)

b² + 3² = 5²

b² + 9 = 25

b² = 25 - 9

b² = 16

Take square root of both sides.

b = $\sqrt{\mathrm{16}}$ = 4

TOA

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

Side opposite to A is 3 and adjacent to A is 4

tan A = $\frac{3}{4}$

CAH

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

Hypothenuse side of A is 5

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

tan A - cos A = $\frac{3}{4}$ - $\frac{4}{5}$

tan A - cos A = $\frac{\mathrm{5\; x\; 3\; -\; 4\; x\; 4}}{20}$

tan A - cos A = $\frac{\mathrm{15\; -\; 16}}{20}$

tan A - cos A = -$\frac{1}{20}$

Notes:

1. a log b = log b^a

2. log b + log c = log b x c

3. log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log x³ + log y - log z²

3 log x + log y - 2 log z = log x³ x y - log z²

3 log x + log y - 2 log z = log x³y - log z²

Apply log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log ($\frac{{\mathrm{x}}^3\mathrm{y}}{{\mathrm{z}}^2}$)

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

Law of division of indices

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

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{-\frac{1}{4}+\frac{1}{2}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{\mathrm{-1\; +\; 2}}{4}}$

${\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}$ = ${\mathrm{a}}^{\frac{1}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = $\frac{{\mathrm{a}}^{\frac{1}{4}}}{{\mathrm{a}}^{-\frac{1}{2}}}$

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

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}-\; -\frac{1}{2}}$

Note: -- = +

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{1}{4}+\frac{1}{2}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; x\; 1\; +\; 2\; x\; 1}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{\mathrm{1\; +\; 2}}{4}}$

$\frac{{\mathrm{a}}^{-\frac{1}{4}}x\; {\mathrm{a}}^{\frac{1}{2}}}{{\mathrm{a}}^{-\frac{1}{2}}}$ = ${\mathrm{a}}^{\frac{3}{4}}$

Perimeter is the sum of all the sides of a shape.

A square has the same lengths (L) for all 4 sides.

Perimeter of a square = L + L + L + L = 4L

Area of a square = Length (L) x Length (L) = L²

Area of the square parcel of land = 256 m²

L² = 256

Take square root of both sides.

L = $\sqrt{\mathrm{256}}$ = 16 m

Perimeter of the square parcel of land = 4L

Perimeter of the square parcel of land = 4 x 16 m = 64 m

The rectangular field has the same perimeter.

Perimeter of the rectangular field = 64 m

Rectangle has two of the lengths equal and two of the breadth equal.

Perimeter of a rectangle = Length(L) + Length(L) + Breadth(B) + Breadth (B)

Perimeter of a rectangle = L + L + B + B

Perimeter of a rectangle = 2L + 2B

Perimeter of the rectangular field = 64 m

The length of the rectangular field = 20 m

2 x 20 m + 2B = 64 m

40 m + 2B = 64 m

2B = 64 m - 40 m

2B = 24 m

Divide both sides by 2.

B = $\frac{\mathrm{24\; m}}{2}$ = 12 m

The breadth of the rectangular field = 12 m

Area of a rectangle = Length x Breadth

Area of the rectangular field = 20 m x 12 m = 240 m²

Method I

n = 1 (compounded once every year)

A = P(1 + r)^t

P = $1,200

r = $\frac{5}{100}$ = 0.05

t = 3

A = $1,200(1 + 0.05)³ = $1389.15

Amount earned at the end of the third year = $1389.15

Amount = Principal + Interest

Interest = Amount - Principal

Principal = $1,200

Interest earned at the end of three years = $1389.15 - $1,200

Interest earned at the end of three years = $189.15

Method II

First Year

Principal = $1,200

Time = 1

Rate = 5

Simple Interest = $\frac{\mathrm{\$1200\; x\; 1\; x\; 5}}{100}$ = $60

Amount = Principal + Interest

Amount by the end of the first year = $1200 + $60 = $1260

Second Year

Principal = $1260

Time = 1

Rate = 5

Simple Interest = $\frac{\mathrm{\$1260\; x\; 1\; x\; 5}}{100}$ = $63

Amount by the end of the second year = $1260 + $63 = $1323

Third Year

Principal = $1323

Time = 1

Rate = 5

Simple Interest = $\frac{\mathrm{\$1323\; x\; 1\; x\; 5}}{100}$ = $66.15

Amount by the end of the third year = $1323 + $66.15 = $1389.15

Amount = Principal + Interest

Interest = Amount - Principal

Principal = $1,200

Interest earned at the end of three years = $1389.15 - $1,200

Interest earned at the end of three years = $189.15

|OS| = |OQ| = Radius of the circle

∆ QOS is an isosceles triangle hence the base angles are the same.

∠ OSQ = ∠ OQS

∠ QOS + ∠ OSQ + ∠ OQS = 180°(Sum of angles of a triangle)

108° + ∠ OSQ + ∠ OQS = 180°

∠ OSQ + ∠ OQS = 180° - 108°

∠ OSQ + ∠ OQS = 72°

∠ OSQ = ∠ OQS = $\frac{\mathrm{72°}}{2}$ = 36°

From alternate segment theorem,

Draw an imaginary angle subtended by the arc QS

From the alternate segment theorem, ∠ QTS = ∠ SQR

∠ QTS = $\frac{\mathrm{\angle \; QOS}}{2}$ = $\frac{\mathrm{108°}}{2}$ = 54°

∠ QTS = ∠ SQR = 54°

$\frac{2}{3}$(x + 5) = 1 - $\frac{\mathrm{x\; -\; 7}}{2}$

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

6 x $\frac{2}{3}$(x + 5) = 1 x 6 - $\frac{\mathrm{x\; -\; 7}}{2}$ x 6

2 x 2(x + 5) = 6 - 3(x - 7)

4(x + 5) = 6 - 3x + 21

Notes:
1. - x - = +
2. -3 x -7 = + 21

4x + 20 = 27 - 3x

4x + 3x = 27 - 20

7x = 7

Divide both sides by 7

x = $\frac{7}{7}$ = 1

Total surface area = 2(12 x 7 + 7 x 5 + 12 x 5) cm²

Total surface area = 358 cm²

Percentage for selling price = 100%

Percentage received = 100% - Commission%

Percentage received = 100% - 9%

Percentage received = 91%

Amount received = GH₵ 236,600

If 91% = GH₵ 236,600

100% = $\frac{\mathrm{GH₵\; 236,600\; x\; 100\%}}{\mathrm{91\%}}$ = GH₵ 260,000

∴ Selling price = GH₵ 260,000.00

From alternate segment theorem,

∠ MNP = ∠ QMP = 73°

∠ MNP + ∠ MPN + ∠ PMN = 180°(Sum of angles in a triangle)

73° + ∠ MPN + 58° = 180°

∠ MPN + 131° = 180°

∠ MPN = 180° - 131°

∠ MPN = 49°

U = {Students}

n(U) = 120

M = {Students who passed Mathematics}

n(M) = 90

S = {Students who passed Science}

n(S) = 40

Let x = number of students who passed both subjects.

Covering Mathematics and solving for x.

90 + 40 - x + 5 = 120

135 - x = 120

135 - 120 = x

15 = x

Science only = 40 - x

Science only = 40 - 15

Science only = 25

U = {Students}

n(U) = 120

M = {Students who passed Mathematics}

n(M) = 90

S = {Students who passed Science}

n(S) = 40

Let x = number of students who passed both subjects.

Covering Mathematics and solving for x.

90 + 40 - x + 5 = 120

135 - x = 120

135 - 120 = x

15 = x

Number of samples = 120

Number of students who passed only one subject = 75 + 25 = 100

P(only one subject) = $\frac{100}{120}$ = $\frac{5}{6}$

PS = RS

∆ PSR is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ SPR = ∠ SRP

∠ PSR = 90°

∠ SPR + ∠ SRP + ∠ PSR = 180° (Sum of angles in a triangle)

∠ SPR + ∠ SRP + 90° = 180°

∠ SPR + ∠ SRP = 180° - 90°

∠ SPR + ∠ SRP = 90°

∠ SPR = ∠ SRP = $\frac{\mathrm{90°}}{2}$ = 45°

∠ PQS = ∠ SRP = 45°(Angles subtended by the same arc PS)

Reflection in the line x - 2 = 0, x = 2

k = 2

(-7, 8) → (2 x 2 - (-7), 8)

(-7, 8) → (4 + 7, 8)

(-7, 8) → (11, 8)

Note:
1. -- = + or -(-) = +
2. -(-7) = + 7

Product means multiplication.

Change the numbers to base 10 and multiply and change it back to base seven.

124_seven = 1 x 7² + 2 x 7¹ + 4 x 7⁰

124_seven = 67

23_seven = 2 x 7¹ + 3 x 7⁰

23_seven = 17

124_seven x 23_seven = 67 x 17

124_seven x 23_seven = 1139

1139 base 10 to base 7

$$\begin{array}{cc}\mathrm{}& 1139\\ 7& \mathrm{162\; r\; 5}\\ 7& \mathrm{23\; r\; 1}\\ 7& \mathrm{3\; r\; 2}\\ 7& \mathrm{0\; r\; 3}\end{array}$$

You list the remainders from the bottom to the top.

1139_ten = 3215_seven

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$

When m = 5, n = 3 and r = 2

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{(5^2-3^2)+2^2}{5^2+(3^2-2^2)}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{\mathrm{25\; -\; 9\; +\; 4}}{\mathrm{25\; +\; 9\; -\; 4}}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{20}{30}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{2}{3}$

Ratios = 2 : 4 : 10

2 divides all.

Ratios = 1 : 2 : 5

Sum of ratios represents the total oranges shared.

The smallest ratio represents the least share.

Sum of ratios = 1 + 2 + 5 = 8

The smallest ratio = 1

If 8 = 32

1 = $\frac{\mathrm{1\; x\; 32}}{8}$ = 4

∴ The least share is 4

Sum of items = Mean x Number of items.

Mean = 17.6

Number of girls in the class = t

Sum of the ages of girls = 17.6 x t = 17.6t

When the girls where dismissed, the sum of the ages reduced to 17.6t - (16 + 19 + 20 + 17) and the number of girls reduced to t - 4

Note: 4 girls were dismissed.

New sum of ages = 17.6t - (16 + 19 + 20 + 17)

New sum of ages = 17.6t - 72

New number of girls = t - 4

New mean = $\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$

The new mean age of girls in the class became 0.2 less than the original mean.

Original mean = 17.6

New mean = 17.6 - 0.2 = 17.4

$\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$ = 17.4

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

$\frac{\mathrm{17.6t\; -\; 72}}{\mathrm{t\; -\; 4}}$ x (t - 4) = 17.4 x (t - 4)

17.6t - 72 = 17.4t - 69.6

17.6t - 17.4t = 72 - 69.6

0.2t = 2.4

Divide both sides by 0.2

t = $\frac{2.4}{0.2}$ = 12