The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.

Let θ = angle of depression.

The opposite and the adjacent sides of the angle of depression are known.

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

tan(θ) = $\frac{\mathrm{12\; cm}}{\mathrm{5\; cm}}$

θ = tan^-1$\left(\frac{12}{5}\right)$

θ = 67.380135052^o ≈ 67^o (nearest degree)

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

Y₂ = 4
Y₁ = 6
X₂ = x
X₁ = 3

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

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

But gradient = - $\frac{2}{5}$ = $\frac{-2}{5}$

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

Since the numerator at the left and right are the same, the denominators are also the same.

x - 3 = 5

x = 5 + 3

x = 8

Volume of a cylinder = π x radius x radius x height

1 Litre = 1000 cm³

85 litres = 85 x 1000 cm³

85 litres = 85000 cm³

Volume of kerosene = 85000 cm³

π = $\frac{22}{7}$

Radius = 7 cm

Let d = depth/height of kerosene in the container

85000 cm³ = $\frac{22}{7}$ x 7 cm x 7 cm x d

The denominator 7 cancels out one of the 7s multiplying.

85000 cm³ = 22 x 1 cm x 7 cm x d

85000 cm³ = 154 cm² x d

Divide both sides by 154 cm²

d = $\frac{85000{\mathrm{cm}}^3}{154{\mathrm{cm}}^2}$

d = 551.948 cm ≈ 552 cm

If a polynomial ax² + bx + c is a perfect square if b² = 4ac

For the polynomial x² - 11x + m

a = 1
b = -11
c = m

Since the polynomial x² - 11x + m is a perfect square, b² = 4ac

(-11)² = 4 x 1 x m

121 = 4m

Divide both sides by 4

m = $\frac{121}{4}$

S = $\frac{\mathrm{4\; +\; 9\; +\; 11}}{2}$

S = $\frac{24}{2}$ = 12

Area = $\sqrt{\mathrm{12(12\; -\; 4)(12\; -\; 9)(12\; -\; 11)}}$

Area = $\sqrt{\mathrm{12(8)(3)(1)}}$

Area = $\sqrt{\mathrm{12\; x\; 8\; x\; 3\; x\; 1}}$

Area = $\sqrt{\mathrm{288}}$

Area = 16.970562748 ≈ 17 cm²

Number from a percentage = Percentage x Total number

% means over 100

12% are children

Total number of persons = 175

Number of children = $\frac{12}{100}$ x 175

Number of children = 21

Number of women = Total number of persons - children - men

Number of women = 175 - 21 - 56

Number of women = 98

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

Probability (Woman) = $\frac{\mathrm{Number\; of\; women}}{\mathrm{Number\; of\; persons}}$

Probability (Woman) = $\frac{98}{175}$ = $\frac{14}{25}$

Note: 7 divides 98, 14 times and 175, 25 times.

Let x = actual value

Percentage error = 5%

Multiply both sides by x

(x - 3.99) x 100 = 5 x x

100x - 399 = 5x

100x - 5x = 399

95x = 399

Divide both sides by 95

x = $\frac{399}{95}$

x = 4.2

∴ the actual value is 4.2 m

Distance covered the first time = 80 km

Time covered the first time = 2 hours

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Distance = Speed x Time

Distance covered second time = 2.5 hours x 50 km/h = 125 km

Total distance covered = 80 km + 125 km = 205 km

Total time taken = 2 hours + 2.5 hours = 4.5 hours

Average Speed = $\frac{\mathrm{Total\; Distance}}{\mathrm{Total\; Time}}$

Average Speed = $\frac{\mathrm{205\; km}}{4.5}$

4.5 = $\frac{45}{10}$ = $\frac{9}{2}$

Average Speed = 205 ÷ $\frac{9}{2}$ km/h

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

Average Speed = 205 x $\frac{2}{9}$ km/h

Average Speed = $\frac{\mathrm{205\; x\; 2}}{9}$ km/h

Average Speed = $\frac{410}{9}$ km/h

Average Speed = 45$\frac{5}{9}$ km/h

log x = 2 - 3log 2

Group the logs

log x + 3log 2 = 2

From law of logarithms

nlogm = logmⁿ

3log 2 = log 2³

log x + log 2³ = 2

From law of logarithms

LogA + LogB = Log A x B

log (x x 2³) = 2

2³ = 2 x 2 x 2 = 8

log (x x 8) = 2

log 8x = 2

From law of logarithms

log _a^b = c → a^c = b

Where a is the base and b is the logarithm

The base of the logarithm log 8x is 10

10² = 8x

10² = 10 x 10 = 100

100 = 8x

Divide both sides by 8

x = $\frac{100}{8}$ = $\frac{25}{2}$

Note: 4 divides 100, 25 times and 8, 2 times.

y = x² - 5x + k

From the point (3, 1), x = 3 and y = 1

Substitute the values of x and y into the equation and solve for k

1 = 3² - 5 x 3 + k

1 = 3 x 3 - 15 + k

1 = 9 - 15 + k

1 = -6 + k

1 + 6 = k

7 = k

∴ k = 7

The sum of all the exterior angles of a polygon is always 360^o

When you sum all the exterior angles, you should get 360^o

(2x + 5)⁰ + (2x)⁰ + (x - 20)⁰ + (x)⁰ + (3x + 10)⁰ + (x + 15)⁰ = 360⁰

2x + 5 + 2x + x - 20 + x + 3x + 10 + x + 15 = 360

10x + 10 = 360

10x = 360 - 10

10x = 350

Divide both sides by 10

x = $\frac{350}{10}$ = 35

Sum of angles on a straight line is 180^o

p + ∠ BAC = 180^o

∠ BAC = 180^o - p

q + ∠ BCA = 180^o

∠ BCA = 180^o - q

Sum of interior angles of a triangle is 180^o

s + ∠ BAC + ∠ BCA = 180^o

s + (180^o - p) + (180^o - q) = 180^o

s + 180^o - p + 180^o - q = 180^o

s + 180^o + 180^o - p - q = 180^o

s + 360^o - p - q = 180^o

s = 180^o - 360^o + p + q

s = -180^o + p + q

s = p + q - 180^o

But p + q = 250^o

s = 250^o - 180^o

s = 70^o

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

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

Mean = $\frac{\mathrm{(x\; +\; y)\; +\; (2x\; +\; 3y)\; +\; (2x\; -\; 2y)\; +\; (3x\; -\; 2y)}}{4}$

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

Group like terms

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

Mean = $\frac{\mathrm{8x\; +\; 0}}{4}$

Mean = $\frac{\mathrm{8x}}{4}$ = 2x

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

P(Black) = $\frac{\mathrm{Number\; of\; black\; balls}}{\mathrm{Total\; number\; of\; balls}}$

Number of black balls = 5

Total number of balls = 2 + 6 + 5

Total number of balls = 13

P(Black) = $\frac{5}{13}$

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

In geometry, a locus (plural: loci) is a set of all points whose location satisfies or is determined by one or more specified conditions.

Radius = 8 cm
Slant height = 14 cm

We have to calculate for the vertical height to calculate the volume.

The vertical height, base radius and slant height forms a right-angle triangle and pythagoras theorem could be applied to calculate the height.

From Pythagorean theorem,

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

The slant height is the hypotenuse = 14 cm

Let h = the vertical height.

8² + h² = 14²

h² = 14² - 8²

h² = 196 - 64

h² = 132

Take the square root of both sides.

h = $\sqrt{\mathrm{132}}$ = 11.489

Vertical height = 11.489

π = $\frac{22}{7}$

Volume = $\frac{1}{3}x\frac{22}{7}x{8}^{2}x11.489$

Volume = 770.318 cm³ ≈ 770.32 cm³

Abudu takes 6 days to complete work.

Fraction of work done by Abudu in a day = $\frac{1}{6}$

Efah takes 3 days to complete the same work.

Fraction of work completed by Efah in a day = $\frac{1}{3}$

Fraction of work done by both in a day is the sum of fractions of work done by each.

Fraction of work done by both = $\frac{1}{6}$ + $\frac{1}{3}$

Fraction of work done by both = $\frac{\mathrm{1\; +\; 2}}{6}$

Fraction of work done by both = $\frac{3}{6}$

Fraction of work done by both = $\frac{1}{2}$

Alternatively

Let y the work to be done

Abudu takes 6 days to complete work.

Work completed by Abudu in a day = $\frac{1}{6}$ x y

Work completed by Abudu in a day = $\frac{\mathrm{y}}{6}$

Efah takes 3 days to complete the same work.

Work completed by Efah in a day = $\frac{1}{3}$ x y

Work completed by Efah in a day = $\frac{\mathrm{y}}{3}$

Work completed by both Abudu and Efah in a day is the sum of work done by each in a day.

Work done by both = $\frac{\mathrm{y}}{6}$ + $\frac{\mathrm{y}}{3}$

Work done by both = $\frac{\mathrm{y\; +\; 2y}}{6}$

Work done by both = $\frac{\mathrm{3y}}{6}$

Work done by both = $\frac{\mathrm{y}}{2}$

Fraction of work done in a day = $\frac{\mathrm{Work\; done\; in\; a\; day}}{\mathrm{Total\; work\; to\; be\; done}}$

Total work to be done is y

Work done by both in a day = $\frac{\mathrm{y}}{2}$

Fraction of work done in a day = $\frac{\mathrm{y}}{2}$ ÷ y

Every number is divisible by 1

Fraction of work done in a day = $\frac{\mathrm{y}}{2}$ ÷ $\frac{\mathrm{y}}{1}$

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

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

Fraction of work done in a day = $\frac{\mathrm{y}}{2}$ x $\frac{1}{\mathrm{y}}$

The y cancels each other.

Fraction of work done in a day = $\frac{1}{2}$

P = {x: 1 ≤ x ≤ 6}

Read as x is such that 1 is less than or equal to x and x is less than or equal to 6.

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

Q = {x: 2 < x < 9}

Read as x is such that 2 is less than x and x is less than 9.

Q = {3, 4, 5, 6, 7, 8}

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

(P ∩ Q) = {3, 4, 5, 6}

(P ∩ Q) = {x: 2 < x ≤ 6}

Note: ≤ means less than or equal to, hence the number at the right of it is inclusive.

p² + q² + r² = 50

p = 5

$\sqrt{q}$ = 2

Square both sides to get rid of the square root.

q = 2² = 4

Substitute the known values into the equation.

p² + q² + r² = 50

5² + 4² + r² = 50

25 + 16 + r² = 50

41 + r² = 50

r² = 50 - 41

r² = 9

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

r = $\sqrt{9}$ = 3

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

Radius = 5 cm
Height = 10 cm

π = $\frac{22}{7}$

Total surface area of a closed cylinder = 2 x $\frac{22}{7}$ x 5² + 2 x $\frac{22}{7}$ x 5 x 10

Total surface area of a closed cylinder = 157.143 cm² + 314.286 cm²

Total surface area of a closed cylinder = 471.429 ≈ 471.43 cm²

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

Rotation through 90^o anticlockwise

When rotating a point 90^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-y,x).

In other words, switch x and y and make y negative

Rotation through 270^o anticlockwise

When rotating a point 270^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(y,-x).

This means, we switch x and y and make x negative

Rotation through 180^o anticlockwise

When rotating a point 180^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-x,-y).

So all we do is make both x and y negative

Alternate Angles

Project the parallel lines to see which angles are alternate anges.

x = a + b

a = 35^o (Alternate angles)

b = c (Alternate angles)

130^o + c = 180^o (Sum of angles on a straight line)

c = 180^o - 130^o

c = 50^o

b = 50^o

x = a + b

x = 35^o + 50^o

x = 85^o

If $0.045 = ₦1.00

$18,000.00 = $\frac{\mathrm{₦1.00\; x\; \$18,000.00}}{\mathrm{\$0.045}}$ = ₦ 400,000

Modulo is a math operation that finds the remainder when one integer is divided by another.

The least number which a modulo is obtained is the remainder + the mod.

The least number in which the 3 (remainder) is obtained is 3 + 8 = 11

⇒ 7 + x = 11

x = 11 - 7

x = 4

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

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

The total sales is 100%

Change the mixed fraction to improper fraction

12$\frac{1}{2}$% = $\frac{\mathrm{2\; x\; 12\; +\; 1}}{2}$% = $\frac{\mathrm{24\; +\; 1}}{2}$% = $\frac{25}{2}$%

If $\frac{25}{2}$% = $35,000.00

100% = $35,000.00 x 100% ÷ $\frac{25}{2}$%

% cancels each other.

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.

100% = $35,000.00 x 100 x $\frac{2}{25}$ = $280,000

Let the height of the pole = h

Since we know both the opposite and adjacent to the angle 60°, tan is applied.

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

tan 60° = $\frac{\mathrm{h}}{\mathrm{55\; m}}$

But tan 60° = $\sqrt{3}$

$\sqrt{3}$ = $\frac{\mathrm{h}}{\mathrm{55\; m}}$

Multiply both sides by 55 m

h = $\sqrt{3}$ x 55 m

h = 55$\sqrt{3}$ m

The sum of ratios represents the total members

5 + 2 represents Total members

7 represents the total members of the committee

The ratio for girls is 5 and that of boys is 2.

Total members is 35

If 7 = 35

5 = $\frac{\mathrm{35\; x\; 5}}{7}$ = 25

∴ there are 25 girls

If 7 = 35

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

∴ there are 10 boys

The number of girls remain the same but we now wish to increase the number of boys.

The new ratio for boys is 4

We know the number of girls.

If 5 = 25

4 = $\frac{\mathrm{25\; x\; 4}}{5}$ = 20

The boys must be 20 in order to get 5 : 4 ratio.

∴ 10 more boys must be added to the current 10 boys to get 20 boys in the committee.

For any equation to be not defined (undefined), the denominator must be 0.

$\frac{\mathrm{x\; +\; 1}}{3x^2\mathrm{-\; 12}}$

For the equation to be not defined, 3x² - 12 = 0

3x² - 12 = 0

Divide both sides by 3

x² - 4 = 0

x² = 4

Take square root of both sides.

x = $\sqrt{4}$ = ± 2, thus -2 or 2

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

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

tan x = $\frac{12}{5}$ ⇒ the opposite side to x is 12 and the adjacent side to x is 5

Draw your right-angle triangle and get the third side and use the SOHCAHTOA to find the sin x and cos x and multiply them.

From Pythagorean theorem,

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

c² = 5² + 12²

c² = 25 + 144

c² = 169

Take the square root of both sides.

c = $\sqrt{\mathrm{169}}$ = 13

∴ the hypotenuse = 13

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

sin x = $\frac{5}{13}$

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

cos x = $\frac{12}{13}$

sin x cos x = $\frac{5}{13}$ x $\frac{12}{13}$

sin x cos x = $\frac{\mathrm{5\; x\; 12}}{\mathrm{13\; x\; 13}}$

sin x cos x = $\frac{60}{169}$

Negation, ~ symbol means not

Conjunction, ∧ symbol means and

Implication, ⇒ symbol means implies, if ..., then ...

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

The total perimeter before the sector was cut is the circumference of the circular plate.

Circumference of a circle = 2πr

The remaining perimeter will be the circumference - the length of the arc of the sector cut + the two radii.

Remaining perimeter = Circumference - length of cut arc + the two radii

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x circumference of the circle

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Angle = 150°

π = $\frac{22}{7}$

Radius = 14 cm

Length of cut arc = $\frac{\mathrm{150°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm

Length of cut arc = 36.666666667 ≈ 36.7 cm

Circumference = 2 x x $\frac{22}{7}$ x 14 cm

Circumference = 88 cm

Remaining perimeter = 88 cm - 36.7 cm + 14 cm + 14 cm = 79.3 cm

Alternatively

The total angle for the circle = 360°

Remaining angle for the sector left = 360° - 150°

Remaining angle for the sector left = 210°

Remaining perimeter = Length of arc left + the two radii

Length of arc left = $\frac{\mathrm{210°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm = 51.3 cm

Remaining perimeter = 51.3 cm + 14 cm + 14 cm

Remaining perimeter = 79.3 cm

Let the width of the gutter = w

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

tan 35° = $\frac{15}{\mathrm{w}}$

Multiply both sides by w

w x tan 35° = 15

Divide both sides by tan 35°

w = $\frac{15}{\mathrm{tan\; 35°}}$ = 21.42 cm

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

∠MNP = 90°

Sum of angles on a straight line is 180°

(5x + 18)° + ∠NPM = 180°

5x + 18 + ∠NPM = 180

∠NPM = 180 - 5x - 18

∠NPM = 162 - 5x

∠NPM + ∠MNP + ∠NMP = 180° (sum of angles in a triangle)

162 - 5x + 90 + 2x = 180

162 + 90 - 5x + 2x = 180

252 - 3x = 180

252 - 180 = 3x

72 = 3x

Divide both sides by 3

x = $\frac{72}{3}$ = 24°

Volume of a square base pyramid = $\frac{\mathrm{Area\; of\; base\; x\; height}}{3}$

Area of a square = Length x Length = L²

Area of the base = 324 cm²

L² = 324

Take square root of both sides.

L = $\sqrt{\mathrm{324}}$ = 18 cm

The height of a square base pyramid is thrice the length of a side of its base

Height = 3 x Length of its base

Length of base = 18 cm

Height = 3 x 18 cm = 54 cm

Volume of the square base pyramid = $\frac{\mathrm{324\; x\; 54}}{3}$ = 5,832 cm³

4^x = $\frac{1}{1024}$

Express both the left and the righ sides to be in the same base.

1024 = 4 x 4 x 4 x 4 x 4 = 4⁵

4^x = $\frac{1}{4^5}$

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

4^x = 4^-5

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

x = -5

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°

Method 1

Simplifying in base six

Solving for the addition

Note: 7 in base six is 11

Solving for the difference

Method II

Change the base six to base 10 and simplify and change the result back to base six

141_six = 1 x 6² + 4 x 6¹ + 1 x 6⁰

141_six = 36 + 24 + 1

141_six = 61

233_six = 2 x 6² + 3 x 6¹ + 3 x 6⁰

233_six = 72 + 18 + 3

233_six = 93

102_six = 1 x 6² + 0 x 6¹ + 2 x 6⁰

102_six = 36 + 0 + 2

102_six = 38

141_six + 233_six - 102_six = 61 + 93 - 38

141_six + 233_six - 102_six = 116

Change the base 10 to six

116 base 10 to base 6

$$\begin{array}{cc}\mathrm{}& 116\\ 6& \mathrm{19\; r\; 2}\\ 6& \mathrm{3\; r\; 1}\\ 6& \mathrm{0\; r\; 3}\end{array}$$

You list the remainders from the bottom to the top.

116_ten = 312_six

First year = $1,560

Second year = $1,560 + y

Third year = $1,560 + y + y

Fourth year = $1,560 + y + y + y

This is a linear sequence: $1,560,$1,560 + y, $1,560 + 2y $1,560 + 3y,...,$13,980

The value of the nth term of a linear sequence is given by U₁ + (n-1)d

Where U₁ = the first term
n = the nth term
d = the common difference

In the above story problem, the first term is $1,560

The common difference = y

We were given the 13th term which is $13,980

We can therefore express the 13th term using the above formula and solve for the value of y

$13980 = $1560 + (13-1) x y

$13980 - $1560 = 12y

$12420 = 12y

Divide both sides by 12

y = $\frac{\mathrm{\$12420}}{12}$ = $1035

Geometric Progression (G.P) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.

When each succeeding terms is divided by the preceding terms, the result (common ratio) is the same.

x ÷ $\frac{16}{9}$ = $\frac{1}{\mathrm{x}}$ = $\frac{\mathrm{y}}{1}$

Using $\frac{1}{\mathrm{x}}$ = $\frac{\mathrm{y}}{1}$

Cross multiply.

x x y = 1 x 1

xy = 1

Let r = the radius of the circle.

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

sin 75 = $\frac{6}{\mathrm{r}}$

Multiply both sides by r

r x sin 75 = 6

Divide both sides by sin 75

r = $\frac{6}{\mathrm{sin\; 75}}$ = 6.2 cm

At the point of intersection, the y and x values are the same.

3x + 2y = 4

2y = -3x + 4

Divide both sides by 2

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

y = $\frac{\mathrm{-3x}}{2}$ + 2 ---- eq. 1

y = 2x - 5 ---- eq. 2

eq. 1 = eq. 2

$\frac{\mathrm{-3x}}{2}$ + 2 = 2x - 5

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

2 x $\frac{\mathrm{-3x}}{2}$ + 2 x 2 = 2x x 2 - 5 x 2

-3x + 4 = 4x - 10

4 + 10 = 4x + 3x

14 = 7x

Divide both sides by 7

x = $\frac{14}{7}$ = 2

Substitute the value of x into --- eq. 2 and solve for the value of y.

y = 2x - 5 ---- eq. 2

y = 2 x 2 - 5

y = 4 - 5

y = -1

∴ P = (2, -1)

$\frac{\mathrm{t}}{\mathrm{s\; +\; u}}$ = $\frac{\mathrm{s}}{\mathrm{t\; -\; u}}$

Cross multiply

t x (t - u) = s x (s + u)

t² - tu = s² + su

t² - s² = tu + su

t² - s² = u(t + s)

Divide both sides by (t + s)

u = $\frac{{\mathrm{t}}^2-{\mathrm{s}}^2}{\mathrm{t\; +\; s}}$

From difference of two squares, a² - b² = (a + b)(a - b)

u = $\frac{\mathrm{(t\; +\; s)(t\; -\; s)}}{\mathrm{t\; +\; s}}$

t + s cancels out

u = t - s

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{150}{30}$ = 5

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.

Sum of number of students (frequencies) = 2 + 1 + 4 + 3 + 8 + 5 + 4 + 3

Sum of number of students (frequencies) = 30

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{30\; +\; 1}}{2}$ = $\frac{31}{2}$ = 15.5

∴ the median is the average of the 15^th and the 16^th position

The 15^th and the 16^th positions fall in the number of subjects = 5

Median = $\frac{\mathrm{5\; +\; 5}}{2}$ = $\frac{10}{2}$ = 5

Volume of a cylinder = πr²h

Where r = radius and h = height of the cylinder.

Volume of a rectangular shape = Length x Breadth x Height

Volume of the rectangular tank = 4 m x 8 m x 11 m

Since the volumes of the rectangular tank and cylindrical tank are the same,

Volume of the cylindrical tank = 4 m x 8 m x 11 m

πr²h = 4 m x 8 m x 11 m

h = 7 cm

π = $\frac{22}{7}$

$\frac{22}{7}$ x r² x 7 = 4 x 8 x 11

The 7s cancel each other

22 x r² = 4 x 8 x 11

Divide both sides by 22

r² = $\frac{\mathrm{4\; x\; 8\; x\; 11}}{22}$

r² = 16

Take square root of both sides

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

Each interior angle of a regular polygon = $\frac{\mathrm{(n-2)\; x\; 180°}}{\mathrm{n}}$

Where n = number of sides of the regular polygon.

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

Let x = sum of the ages of the second year class before the promotion examination
Let n = number of students in the second year class before the promotion examination

Mean = $\frac{\mathrm{x}}{\mathrm{n}}$

But the mean before the promotion examination = 18$\frac{2}{5}$

18$\frac{2}{5}$ = $\frac{\mathrm{5\; x\; 18\; +\; 2}}{5}$ = $\frac{92}{5}$

$\frac{92}{5}$ = $\frac{\mathrm{x}}{\mathrm{n}}$

Cross multiply

5 x x = 92 x n

Divide both sides by 5

x = $\frac{\mathrm{92n}}{5}$

3 students were repeated.

Sum of ages of the repeated students = 20 + 19 + 19
Sum of ages of the repeated students = 58

Sum of ages of the promoted students = Sum of ages before the promotion - Sum of ages of the repeated students
Sum of ages of the promoted students = x - 58

Number of promoted students = Number of students before the promotion - Number of students repeated
Number of promoted students = n - 3

Mean age of the promoted students = $\frac{\mathrm{x\; -\; 58}}{\mathrm{n\; -\; 3}}$

The new mean age of the class = 18$\frac{1}{3}$

18$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 18\; +\; 1}}{3}$ = $\frac{55}{3}$

$\frac{\mathrm{x\; -\; 58}}{\mathrm{n\; -\; 3}}$ = $\frac{55}{3}$

Cross multiply

3 x (x - 58) = 55 x (n - 3)

3x - 174 = 55n - 165

3x - 174 + 165 = 55n

3x - 9 = 55n

But x = $\frac{\mathrm{92n}}{5}$

3 x $\frac{\mathrm{92n}}{5}$ - 9 = 55n

Get rid of the fraction by multiplying both sides by 5

3 x 92n - 9 x 5 = 5 x 55n

276n - 45 = 275n

276n - 275n = 45

n = 45

∴ the number of students who were in the class before the promotion examination = 45