∆ MNT is an isosceles triangle (Two sides equal).

Isosceles triangle has the base angles the equal.

∠ MNT = ∠ MTN = 32°

∠ LMN = ∠ MNT + ∠ MTN (Sum of two opposite interior angles equal to exterior angle)

∠ LMN = 32° + 32° = 64°

∠ MLN = $\frac{1}{2}$ x ∠MON

|OM| = |ON| (Radii)

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

∠ OMN = ∠ ONM = 20°

∠MON = 180° - (20° + 20°) (Sum of interior angles of a triangle is 180°)

∠MON = 180° - 40°

∠MON = 140°

∠ MLN = $\frac{1}{2}$ x ∠MON = $\frac{1}{2}$ x 140° = 70°

∠ MLN + ∠ LMN + ∠ LNM = 180°(Sum of interior angles of a triangle is 180°)

64° + 70° + ∠ LNM = 180°

134° + ∠ LNM = 180°

∠ LNM = 180° - 134°

∠ LNM = 46°

Pentagon has five sides hence n = 5.

Sum of the interior angles = (5 - 2) x 180°

Sum of the interior angles = 3 x 180°

Sum of the interior angles = 540°

y° + 2x° + 3x° + 2x° + y° = 540°

2y + 7x = 540 ---- eqn 1

But y = $\frac{\mathrm{3x}}{2}$

Make x the subject and substitute it into --- eqn 1

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

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

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

2y = 3x

Divide both sides by 3.

x = $\frac{\mathrm{2y}}{3}$ --- eqn 2

Substitute --- eqn 2 into --- eqn 1

2y + 7 x $\frac{\mathrm{2y}}{3}$ = 540

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

3 x 2y + 3 x 7 x $\frac{\mathrm{2y}}{3}$ = 540 x 3

6y + 14y = 1,620

20y = 1,620

Divide both sides by 20

y = $\frac{1620}{20}$ = 81

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

Sum of the fractions = (1 + 2 + 3 + 4 + 5)[$\frac{1}{2}$ + $\frac{2}{3}$ + $\frac{3}{4}$ + $\frac{4}{5}$ + $\frac{5}{6}$ ]

The L.C.M of 2, 3, 4, 5 and 6 is 60.

Sum of the fractions = 15[$\frac{\mathrm{30x\; 1\; +\; 20\; x\; 2\; +\; 15\; x\; 3\; +\; 12\; x\; 4\; +\; 10\; x\; 5}}{60}$]

Sum of the fractions = 15[$\frac{\mathrm{30\; +\; 40\; +\; 45\; +\; 48\; +\; 50}}{60}$]

Sum of the fractions = 15[$\frac{213}{60}$]

Sum of the fractions = 15$\frac{213}{60}$ = $\frac{\mathrm{15\; x\; 60\; +\; 213}}{60}$ = $\frac{1113}{60}$

Number of numbers = 5

Mean = $\frac{1113}{60}$ ÷ 5

Every number is being divided by 1 but not written.

5 = $\frac{5}{1}$

Mean = $\frac{1113}{60}$ ÷ $\frac{5}{1}$

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.

Mean = $\frac{1113}{60}$ x $\frac{1}{5}$

Mean = $\frac{\mathrm{1113\; x\; 1}}{\mathrm{60\; x\; 5}}$

Mean = $\frac{1113}{300}$ = 3.71

Mean (x̄) = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{38}{16}$ = 2.375 ≈ 2.4 (one decimal place)

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°

The regular time to reach destination = $\frac{\mathrm{40\; km}}{\mathrm{50\; km/h}}$ = 0.8 hours

18 minutes = $\frac{18}{60}$ hours = 0.3 hours

Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.

The ferry must cover the 40 km in 0.5 hours.

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$ = $\frac{\mathrm{40\; km}}{\mathrm{0.5\; h}}$ = 80 km/h

100% represents the original price

% Paid = 100% - %Discount

% Paid = 100% - %20

% Paid = 80%

Amount paid = $ 525.00

If 80% = $ 525

100% = $\frac{\mathrm{100\%\; x\; \$\; 525}}{\mathrm{80\%}}$ = $ 656.25

A function is undefined when the denominator is 0.

x + 1 = 0

x = 0 - 1

x = -1

⊗ 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

x : y : z = 2 : 3 : 4

x = 2
y = 3
z = 4

Substitute the values into the equation $\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{\mathrm{9\; x\; 2\; +\; 3\; x\; 3}}{\mathrm{6\; x\; 4\; -\; 2\; x\; 3}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{\mathrm{18\; +\; 9}}{\mathrm{24\; -\; 6}}$

$\frac{\mathrm{9x\; +\; 3y}}{\mathrm{6z\; -\; 2y}}$ = $\frac{27}{18}$ = 1.5

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°

tan θ = $\frac{3}{4}$

The opposite side to θ is 3.

The adjacent side to θ is 4.

Draw the right angle and find the other side.

From pythagoras theorem

c² = 3² + 4²

c² = 9 + 16

c² = 25

Take the square root of both sides.

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

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

The adjacent side to θ is 4.

The hypotenuse side to θ is 5.

cos θ = $\frac{4}{5}$

Use the sign of θ in the various quadrants.

cos θ in the range 180° < θ < 270° is negative.

cos θ = - $\frac{4}{5}$

If a and b are the roots, (x - a)(x - b) = 0

Change the mixed fraction to improper fraction.

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

For the roots -$\frac{1}{2}$ and $\frac{3}{2}$,

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

-- = +

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

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

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 2 and 4.

The L.C.M is 4.

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

4x² - 4x - 3 = 0

Draw an imaginary third parallel line.

∠ QPO = ∠ TOP = x° (Alternate angles)

∠ RSO = ∠ TOS = 68° (Alternate angles)

x° + 68° + 246° = 360° (Sum of angles at centre)

x° + 314° = 360°

x° = 360° - 314°

x° = 46°

x = 46

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 = $\frac{11}{15}$

Equilateral triangle has all three sides equal.

When you divide it, you will get two equal right angle triangles with base length being half of the side of the equilateral triangle.

Let x = the side of the equal lateral triangle.

From Pythagorean theorem,

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

(10$\sqrt{3}$)² + ${\left(\frac{\mathrm{x}}{2}\right)}^2$ = x²

Note: The square cancels the square root.

10² x 3 + $\frac{{\mathrm{x}}^2}{2^2}$ = x²

100 x 3 + $\frac{{\mathrm{x}}^2}{4}$ = x²

Get rid of the fraction by multiplying each sides by the denominator (4).

4 x 300 + x² = 4 x x²

1200 + x² = 4x²

1200 = 4x² - x²

1200 = 3x²

Divide both sides by 3

x² = $\frac{1200}{3}$

x² = 400

Take square root of both sides.

x = $\sqrt{\mathrm{400}}$ = 20

∴ each side of the equilateral triangle is 20 cm.

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

Perimeter of the equilateral = 20 cm + 20 cm + 20 cm

Perimeter of the equilateral = 60 cm

If α and β are the roots, then

(x - α)(x - β) = 0

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

Note: -- = +

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

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

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

x² + $\frac{\mathrm{x}}{3}$ - $\frac{\mathrm{x}}{2}$ - $\frac{1}{6}$ = 0

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

The L.C.M of 3, 2 and 6 is 6.

6 x x² + 6 x $\frac{\mathrm{x}}{3}$ - 6 x $\frac{\mathrm{x}}{2}$ - 6 x $\frac{1}{6}$ = 0 x 6

6x² + 2x - 3x - 1 = 0

6x² - x - 1 = 0

75.0785 - 34.624 + 9.83 = 50.2845

75.0785 - 34.624 + 9.83 ≈ 50.28 (two decimal places)

To be parallel, two lines must have the same gradient.

For a general equation of a line, y = mx + c

m = gradient and c = y intercept.

Express the given equation to be in the form y = mx + c and compare to find the value of m, the gradient.

2y = 3(x - 2)

2y = 3x - 6

Divide both sides by 2

y = $\frac{\mathrm{3x}}{2}$ - $\frac{6}{2}$

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

Comparing the equation with the general equation y = mx + c, the gradient is $\frac{3}{2}$ and the y intercept is -3

Use the given point to calculate the gradient with the point (x, y) and make y the subject.

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

Gradient using (x, y) and (2, 3)

Gradient = $\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$

But the gradient = $\frac{3}{2}$

$\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$ = $\frac{3}{2}$

Cross multiply.

2 x (y - 3) = 3 x (x - 2)

2y - 6 = 3x - 6

2y = 3x - 6 + 6

2y = 3x

Divide both sides by 2.

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

Method I

The angle for the entire circle is 360°

If 140° = 40 cm²
360° = $\frac{\mathrm{40\; x\; 360}}{140}$ = 102.86 cm²

Method II

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x πr²

But πr² = area of the circle

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x area of circle

Area of a sector = 40 cm²
θ = 140

40 = $\frac{140}{360}$ x area of circle

Make area of circle the subject.

Multiply both sides by 360

40 x 360 = 140 x area of circle

Divide both sides by 140

$\frac{\mathrm{40\; x\; 360}}{140}$ = area of circle

Area of circle = 102.86 cm²

∧ is logical conjunction (and)

→ is then or implies

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

Let x = number of students who liked both Music and Movies.

Coverig Music,

12 + 15 - x = 20

27 - 20 = x

7 = x

x = 7

Number of students who liked both Music and Movies = 7

Number of students = 20

P(Both Music and Movies) = $\frac{7}{20}$

|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°

1. Find the number of times the cyclist complete one cycle by dividing the distance overed by the circumference.

Number of completed cycles = $\frac{\mathrm{302\; km}}{\mathrm{9\; km}}$ = 33.555

Number of completed cycles = 33

Note: completed cycle is when the cyclist reaches the starting point.

2. Distance from starting point = Distance covered - Number of completed cycles x the circumference

Distance from starting point = 302 km - 33 x 9 km

Distance from starting point = 302 km - 297 km

Distance from starting point = 5 km

168 = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

168 x n = (n - 2) x 180

168n = 180n - 360

360 = 180n - 168n

360 = 12n

Divide both sides by 12.

n = $\frac{360}{12}$

n = 30

Let y = Juliet's age.

Badu is four times as old as Juliet

Badu's age = 4 x Juliet's age

Badu's age = 4 x y

Badu's age = 4y

Badu's age 10 years time = Badu's age now + 10 years.

Badu's age 10 years time = 4y + 10

Juliet's age in 10 years time = Juliet's age now + 10

Juliet's age in 10 years time = y + 10

In 10 years Badu will be twice as old as Juliet

Badu's age in 10 years will be 2 x Juliet's age

4y + 10 = 2 x (y + 10)

4y + 10 = 2 x y + 2 x 10

4y + 10 = 2y + 20

4y - 2y = 20 - 10

2y = 10

Divide both sides by 2

y = $\frac{10}{2}$ = 5

∴ Juliet is 5 years old now

$\frac{2}{3}$(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

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{9\; x\; (2x\; +\; 1)\; -\; 2\; x\; (3x\; -\; 7)\; -\; 1\; x\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; +\; 9\; -\; 6x\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; -\; 6x\; +\; 9\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{12x\; +\; 18}}{18}$

Factorize 6 out at numerator.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{6(2x\; +\; 3)}}{18}$

Note: 6 divides itself 1 time and 18, 3 times.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{1\; x\; (2x\; +\; 3)}}{3}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 3}}{3}$

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 = $\frac{2}{3}$

$\frac{9}{28}$ - $\frac{2}{7}$ = $\frac{\mathrm{1\; x\; 9\; -\; 4\; x\; 2}}{28}$

$\frac{9}{28}$ - $\frac{2}{7}$ = $\frac{\mathrm{9\; -\; 8}}{28}$

$\frac{9}{28}$ - $\frac{2}{7}$ = $\frac{1}{28}$

$\frac{9}{28}$ - $\frac{2}{7}$ = 0.0357142857142857 ≈ 0.036 (Two significant figures)

Significant figures

Significant figures (or significant digits) are the number of digits important to determine the accuracy and precision of measurement, such as length, mass, or volume.

Significant digits in math convey the value of a number with accuracy. They are considered substantial figures that contribute to the precision of a number.

Rules to determine the number of significant figures in a number

1. Leading zeros are not significant
2. Non-zeros are significant
3. Zeros in between non-zero digits are significant
4. Trailing zeros to the right of the decimal point are significant

Examples

How to Round Off Significant Figures

To round off significant figures, we have to omit one or more digits from the right side of the number until we reach the number of significant digits that we want to round it off to.

First, we have to look at the digit on the right end of the number (to the right of the digit we want to round it off to).

If the digit is lower than 5, the number is rounded off to the lower number.

If the digit is greater than or equal to 5, the number is rounded up to the higher number.

If the number to be rounded off is a whole number, the remaining significant figures are replaced by 0.

Rounding off 0.0357142857142857

The significant figures are 3,5,7,1,4,2,8,5,7,1,4,2,8,5,7.

The second significant digit is 5 and the digit next to it (7) is greater than 5 hence we have to round up the 5 to 6 by simply adding 1 to the 5.

0.0357142857142857 = 0.036 (Two significant figures).

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}

-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.

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

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

Choose any two points from the table.

P₁(6.20, 3.90) and P₂(6.85, 5.20)

Gradient = $\frac{\mathrm{5.2-3.9}}{\mathrm{6.85-6.20}}$ = $\frac{1.3}{0.65}$ = 2

Angles subtended by the diameter (or semi-circle) is 90°.

∠YXW = 90°

26° + ∠ZXW = ∠YXW = 90°

26° + ∠ZXW = 90°

∠ZXW = 90° - 26°

∠ZXW = 64°

Base angles of isosceles triangle are the same.

∠WZX = ∠XWZ

∠WZX + ∠XWZ + ∠ZXW = 180°(Sum of interior angles of a triangle)

∠WZX + ∠XWZ + 64° = 180°

∠WZX + ∠XWZ = 180° - 64°

∠WZX + ∠XWZ = 116°

∠WZX = ∠XWZ = $\frac{\mathrm{116°}}{2}$ = 58°

Sum of two opposite interior angles of a triangle is equal to it's angle exterior.

∠XYZ + ∠YXZ = ∠WZX

∠XYZ + 26° = 58°

∠XYZ = 58° - 26°

∠XYZ = 32°

If a and b are the roots, (x - a)(x - b) = 0

For the roots $\frac{1}{2}$ and -3,

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

-- = -(-) = +

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

Expand the brackets

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

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

Get rid of the fractions by multiplying both sides by 2

2x² + 6x - x - 3 = 0

2x² + 5x - 3 = 0

Now compare the equation to the general equation to find each of the variables.

px² + qx + r = 0

p = 2
q = 5
r = -3

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)

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 = 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

For the equation to be undefined, the denominator must be 0.

2x² - 13x + 15 = 0

Solve for the value of x

2 x 15 = 30

Two numbers when multiplied the result will be 30 and when added the result will be - 13.

The numbers are -10 and - 3.

Express - 13x as -10x - 3x

2x² - 10x - 3x + 15 = 0

Factorize out

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

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

x - 5 = 0
x = 0 + 5
x = 5

2x-3 = 0
2x = 3

Divide both sides of the equation by 2

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

48 = 3 x 16
6 = 3 x 2

$\sqrt{\mathrm{ab}}$ = $\sqrt{a}$ x $\sqrt{b}$

$\sqrt{\mathrm{48}}$ = $\sqrt{\mathrm{16\; x\; 3}}$ = $\sqrt{\mathrm{16}}$ x $\sqrt{3}$

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

$\sqrt{6}$ = $\sqrt{\mathrm{3\; x\; 2}}$ = $\sqrt{3}$ x $\sqrt{2}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{16}}\mathrm{x}\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{\sqrt{3}\mathrm{+\; 4}\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{5\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\sqrt{3}$ cancels each other.

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{5}{\sqrt{2}}$

$\sqrt{2}$ = 2^½

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

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = 5 x 2^-½

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

5 x 2^-½ = 5 x 2^-1 x 2^½

5 x 2^-½ = $\frac{5}{2}$ x 2^½

2^½ = $\sqrt{2}$

5 x 2^-½ = $\frac{5}{2}$$\sqrt{2}$

An expression is undefined when the denominator equals 0.

6x(x + 1) = 0

x + 1 = 0

x = -1

6x = 0

x = $\frac{0}{6}$ = 0

{0, -1}

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

$\frac{\mathrm{4m\; +\; 3n}}{\mathrm{4m\; -\; 3n}}$ = $\frac{5}{2}$

Method I

The numerator at the left equals the numerator at the right and the denominator at the left equals the denominator at the right.

4m + 3n = 5 ---- eq. 1

4m - 3n = 2 ---- eq. 2

Let's get rid of n by adding --- eq. 1 and 2

8m = 7

Divide both sides by 8

m =$\frac{7}{8}$

Get rid of m by subtracting -- eq. 2 from eq. 1

3n--3n = 3

-- = +

3n + 3n = 3

6n = 3

Divide both sides by 6

n = $\frac{3}{6}$ = $\frac{1}{2}$

Note: alternatively, you can substitute the value of m into any of the equations and solve for n.

m : n = $\frac{7}{8}$ : $\frac{1}{2}$

m : n = $\frac{7}{8}$ ÷ $\frac{1}{2}$

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

m : n = $\frac{7}{8}$ x $\frac{2}{1}$

Note: 2 divides itself 1 time and 8, 4 times.

m : n = $\frac{7}{4}$

⇒ m : n = 7 : 4

Method II

Cross multiply.

$\frac{\mathrm{4m\; +\; 3n}}{\mathrm{4m\; -\; 3n}}$ = $\frac{5}{2}$

2 x (4m + 3n) = 5 x (4m - 3n)

2 x 4m + 2 x 3n = 5 x 4m - 5 x 3n

8m + 6n = 20m - 15n

6n + 15n = 20m - 8m

21n = 12m

Divide both sides by n in order to get m : n

21 = $\frac{\mathrm{12m}}{\mathrm{n}}$

Divide both sides by 12

$\frac{21}{12}$ = $\frac{\mathrm{m}}{\mathrm{n}}$

Note: 3 divides 21, 7 times and 12, 4 times.

$\frac{7}{4}$ = $\frac{\mathrm{m}}{\mathrm{n}}$

∴ m : n = 7 : 4

Inscribed angles subtended by the same arc are equal.

∠ PMQ = ∠ QNP = 57^o

∠ MPN = ∠ MQN = 43^o

Sum of interior angles of a triangle is 180^o

57^o + ∠ MPN + ∠ PON = 180^o

∠ MPN = 43^o

57^o + 43^o + ∠ PON = 180^o

100^o + ∠ PON = 180^o

∠ PON = 180^o - 100^o

∠ PON = 80^o

Sum of angles on a straight line is 180^o

y + ∠ PON = 180^o

But ∠ PON = 80^o

y + 80^o = 180^o

y = 180^o - 80^o

y = 100^o

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

Get rid of the fraction by multiplying each sides by the denominator (3U + 2).

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

3xU + 2x = 2U - 3

3xU - 2U = -2x - 3

Factorize U out.

U(3x - 2) = -2x - 3

Divide both sides by (3x - 2).

U = $\frac{\mathrm{-2x\; -\; 3}}{\mathrm{3x\; -\; 2}}$

Factorize (-1) out.

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

-1 cancels -1 and you can rearrange -3x + 2 as 2 - 3x

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

Alternatively,

3xU + 2x = 2U - 3

2x + 3 = 2U - 3xU

Factorize U out.

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

Divide both sides by (2 - 3x)

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

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

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%