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

The significant figures in 0.00798516 are 7,9,8,5,1 and 6.

The third significant figure is 8.

If the significant figure next to the position of the significant figure you seek is 5 or more, you add 1 to the last significant figures you seek.

The next significant to the third (8) is 5, hence 1 is added to the 8, making 9.

Method I

A number raised to the power 2 (square) means the number multiplied by itself.

Note: 2 in base two is 10.

Method II

Change the binary (base two) to base 10 and simplify and change the result back to base two.

11_two = 1 x 2¹ + 1 x 2⁰

11_two = 1 x 2 + 1 x 1

Note: any number raised to the power 0 is 1.

11_two = 2 + 1 = 3

(11_two)² = 3² = 3 x 3 = 9

Change the 9 base 10 to base 2.

9 base 10 to base 2

$$\begin{array}{cc}\mathrm{}& 9\\ 2& \mathrm{4\; r\; 1}\\ 2& \mathrm{2\; r\; 0}\\ 2& \mathrm{1\; r\; 0}\\ 2& \mathrm{0\; r\; 1}\end{array}$$

You list the remainders from the bottom to the top.

9_ten = 1001_two

Before you can simplify a power in an equation, ensure that both the left and the right have the same base.

Express the 32 to base 2.

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

Solve: $2^{\sqrt{\mathrm{2x\; +\; 1}}}$ = 2⁵

Since the bases are the same, you can equate the powers.

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

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

2x + 1 = 5²

2x + 1 = 25

2x = 25 - 1

2x = 24

Divide both sides by 2.

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

24 = 8 x 3

8 = 2 x 2 x 2 = 2³

24 = 2³ x 3

log₁₀ ²⁴ = log₁₀ 2³ x 3

From law of logarithm,

1. log a^b = blog a

2. log a x b = log a + log b

log₁₀ 2³ x 3 = 3log₁₀ 2 x 3

3log₁₀ 2 x 3 = 3log₁₀ 2 + log₁₀ 3

But log₁₀ 2 = m and log₁₀ 3 = n

3log₁₀ 2 + log₁₀ 3 = 3 x m + n

3log₁₀ 2 + log₁₀ 3 = 3m + n

The rule of the sequence is n² + 1

Where n is the number of term.

When n = 1 → 1² + 1 = 1 + 1 = 2

When n = 2 → 2² + 1 = 4 + 1 = 5

When n = 3 → 3² + 1 = 9 + 1 = 10

When n = 4 → 4² + 1 = 16 + 1 = 17

When n = 5 → 5² + 1 = 25 + 1 = 26

P = {-3 < x < 1}, read as -3 is less than x and x is less than 1

P = {-2, -1, 0}

Q = {-1 < x <3}, read as -1 is less than x and x is less than 3.

Q = {0, 1, 2)

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

P ∩ Q = {0}

P ∩ Q = {-1 < x < 1}, read as -1 is less than x and x is less than 1 which is true for the 0. -1 is less than 0 and 0 is also less than 1.

6pq - 3rs - 3ps + 6qr = 6pq - 3rs - 3ps + 6qr

6pq - 3rs - 3ps + 6qr = 6pq + 6qr - 3rs - 3ps

6pq - 3rs - 3ps + 6qr = 6q(p + r) - 3s(r + p)

6pq - 3rs - 3ps + 6qr = 6q(p + r) - 3s(p + r)

6pq - 3rs - 3ps + 6qr = (6q - 3s)(p + r)

6pq - 3rs - 3ps + 6qr = 3(2q - s)(p + r)

6pq - 3rs - 3ps + 6qr = 3(p + r)(2q - s)

Change the mixed fractions to improper fractions.

2$\frac{1}{6}$ = $\frac{\mathrm{2\; x\; 6\; +\; 1}}{6}$ = $\frac{\mathrm{12\; +\; 1}}{6}$ = $\frac{13}{6}$

2$\frac{7}{12}$ = $\frac{\mathrm{2\; x\; 12\; +\; 7}}{12}$ = $\frac{\mathrm{24\; +\; 7}}{12}$ = $\frac{31}{12}$

3$\frac{1}{4}$ = $\frac{\mathrm{4\; x\; 3\; +\; 1}}{4}$ = $\frac{\mathrm{12\; +\; 1}}{4}$ = $\frac{13}{4}$

Let x the number to be subtracted.

Sum is addition.

$\frac{13}{6}$ + $\frac{31}{12}$ - x = $\frac{13}{4}$

$\frac{\mathrm{2\; x\; 13\; +\; 1\; x\; 31}}{12}$ - x = $\frac{13}{4}$

$\frac{\mathrm{26\; +\; 31}}{12}$ - x = $\frac{13}{4}$

$\frac{57}{12}$ - x = $\frac{13}{4}$

$\frac{57}{12}$ - $\frac{13}{4}$ = x

$\frac{\mathrm{1\; x\; 57\; -\; 3\; x\; 13}}{12}$ = x

$\frac{\mathrm{57\; -\; 39}}{12}$ = x

$\frac{18}{12}$ = x

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

x = $\frac{3}{2}$ = 1$\frac{1}{2}$

Mensah's age now = 5

Thrice means 3 times (3 x ).

Joyce's age now = 3 x Mensah's age

Joyce's age now = 3 x 5

Joyce's age now = 15

Let y years in the future that Joyce will be twice as old as Mensah.

In y years, Mensah will be (5 + y) years.

In y years, Joyce will be (15 + y) years.

In y years, Joyce will be twice as old as Mensah.

Twice means 2 times (2 x )

(15 + y) = 2 x (5 + y)

Now expand and solve for y

15 + y = 2 x 5 + 2 x y

15 + y = 10 + 2y

15 - 10 = 2y - y

5 = y

∴ Joyce will be twice as old as Mensah in 5 years time.

To solve for the x, make sure each of the numbers has the same base in order for you to equate the powers.

The lowest base is 2, hence express all the numbers to base 2.

16 = 2 x 2 x 2 x 2 = 2⁴

4 = 2 x 2 = 2²

8 = 2 x 2 x 2 = 2³

16 x 2^{(x + 1)} = 4^x x 8^{(1 − x)}

2⁴ x 2^{(x + 1)} = 2^{2 x x} x 2^{3 x (1 − x)}

2⁴ x 2^{(x + 1)} = 2^2x x 2^{3(1 − x)}

Apply the law of multiplication of indices, which states that when indices of the same base are multiplying, you can add the powers.

2^{4 + x + 1} = 2^{2x + 3(1 - x)}

Since the base at the left equals to the base at the right, you can equate the powers.

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

Expand the bracket.

x + 5 = 2x + 3 - 3x

x + 5 = 3 - x

x + x = 3 - 5

2x = -2

Divide both sides by 2.

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

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

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

Get rid of the denominator surds by multiplying the numerator and the denominator by the surd.

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ x $\frac{\sqrt{7}}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ x $\frac{\sqrt{\mathrm{21}}}{\sqrt{\mathrm{21}}}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14\sqrt{7}}{7}$ + $\frac{7\sqrt{\mathrm{21}}}{21}$

Every number is being divided by 1.

2$\sqrt{7}$ = $\frac{2\sqrt{7}}{1}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{2\sqrt{7}}{1}$ - $\frac{14\sqrt{7}}{7}$ + $\frac{7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{\mathrm{21\; x\; 2}\sqrt{7}-3x14\sqrt{7}+7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{42\sqrt{7}-42\sqrt{7}+7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{7\sqrt{\mathrm{21}}}{21}$

Note:

1. $42\sqrt{7}$ - $42\sqrt{7}$ = 0

2. 7 dividies itself 1 time and 21, 3 times

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{\sqrt{\mathrm{21}}}{3}$

4x + 2y = 16 ----- eqn 1

6x -2y = 4 ----- eqn 2

Get rid of y. eqn 1 + eqn 2

10x + 0 = 20

10x = 20

Divide both sides by 10

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

Substitute x = 2 into ----- eqn 1

4 x 2 + 2y = 16

8 + 2y = 16

2y = 16 - 8

2y = 8

Divide both sides by 2.

y = $\frac{8}{2}$ = 4

y - x = 4 - 2 = 2

Directly proportion → R ∝ L

Inversely proportional → R ∝ $\frac{1}{\mathrm{P}}$

When you join the two relations,

R ∝ $\frac{\mathrm{L}}{\mathrm{P}}$

Remove the proportionality and introduce a constant k.

R = k x $\frac{\mathrm{L}}{\mathrm{P}}$

Now use the given information to calculate for k.

R = 3
L = 9
P = 0.8

3 = k x $\frac{9}{0.8}$

Get rid of the fraction by multiplying both sides by the denominator 0.8.

3 x 0.8 = k x 9

Divide both sides by 9

k = $\frac{\mathrm{3\; x\; 0.8}}{9}$ = $\frac{0.8}{3}$

Substitute the value of k into the equation.

R = $\frac{0.8}{3}$ x $\frac{\mathrm{L}}{\mathrm{P}}$

R = $\frac{\mathrm{0.8L}}{\mathrm{3P}}$

When L = 15 and P = 1.8

R = $\frac{\mathrm{0.8\; x\; 15}}{\mathrm{3\; x\; 1.8}}$ = $\frac{12}{5.4}$ = 2.2222 ≈ 2.2

Right angle means 90°

Sum of angles of a regular polygon = (n - 2) x 180°

Number of sides (n) of the polygon = 4

Sum of angles of a regular polygon = (4 - 2) x 180°

Sum of angles of a regular polygon = 2 x 180°

Sum of angles of a regular polygon = 360°

∠ADC = ∠EDF = 70°

90° + 90° + t + 70° = 360°

t + 250° = 360°

t = 360° - 250°

t = 110°

n = k.

Sum of interior of regular polygon with k sides = (k - 2) x 180

But the sum is given as (3k - 10) x 90°

Note: right angle = 90°

(k - 2) x 180 = (3k - 10) x 90

Divide both sides by 90

(k - 2) x 2 = (3k - 10) x 1

2k - 4 = 3k - 10

10 - 4 = 3k - 2k

6 = k

∴ the polygon has 6 sides.

Number of sides of the polygon = 6

Exterior angle of the polygon = $\frac{\mathrm{360°}}{6}$ = 60°

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 kobo = 1 naira

38 kobo = $\frac{38}{100}$ naira = ₦ 0.38

0.38 x cost of engine = import duty

cost of engine = $\frac{\mathrm{import\; duty}}{0.38}$

cost of engine = $\frac{\mathrm{₦\; 22,800.00}}{\mathrm{₦\; 0.38}}$ = ₦ 60,000

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

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

Method I

Finding the angle x

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

Take sin inverse.

x = sin^-1($\frac{3}{5}$)

x = 36.9°

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

tan x = $\frac{\mathrm{|LM|}}{\mathrm{|MN|}}$

tan x = $\frac{6}{\mathrm{|MN|}}$

Multiply both sides by |MN|.

|MN| x tan x = 6

|MN| = $\frac{6}{\mathrm{tan\; x}}$

But x = 36.9°

|MN| = $\frac{6}{\mathrm{tan\; 36.9}}$ = 7.99 ≈ 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

Method II

Finding the other two sides.

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

The opposite side of x = |LM|

The hypotenuse side is |LN|

|LM| = 6

sin x = $\frac{6}{\mathrm{|LN|}}$

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

$\frac{6}{\mathrm{|LN|}}$ = $\frac{3}{5}$

Cross multiply.

3 x |LN| = 6 x 5

Divide both sides by 3

|LN| = $\frac{\mathrm{6\; x\; 5}}{3}$ = 10 cm

From pythagorean theorem,

|LM|² + |MN|² = |LN|²

|LN| = 10

|LM| = 6

6² + |MN|² = 10²

|MN|² = 10² - 6²

|MN|² = 64

Take the square root of both sides.

|MN| = $\sqrt{\mathrm{64}}$ = 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

All students offering Literature(L) also offer History(H) means Literature (L) is a subset of History(H)

Students offering History(H) do not offer Geography(G) means the two sets are disjoint.

If α and β are the roots, then

(x - α)(x - β) = 0

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

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

(x + 2q)(x - 5q) = 0

x(x - 5q) + 2q(x - 5q) = 0

x² - 5qx + 2qx - 10q² = 0

x² - 3qx - 10q² = 0

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 = - $\frac{4}{5}$

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

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

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{2x\; -\; 4\; -\; 3x\; +\; 9}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{2x\; -\; 3x\; +\; 9\; -\; 4}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

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

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

Compare $\frac{\mathrm{5\; -\; x}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$ to $\frac{\mathrm{p}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

The numerator p = 5 - x

12 cm = 6 cm + 6 cm (Diagonals meet at the center)

5 cm = 2.5 cm + 2.5 cm

From Pythagorean theorem,

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

c² = 2.5² + 6²

c² = 42.25

Take square root of both sides.

c = $\sqrt{\mathrm{42.25}}$

c = 6.5 cm

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

Perimeter of the rhombus = c + c + c + c

Perimeter of the rhombus = 4 x c

Perimeter of the rhombus = 4 x 6.5 cm

Perimeter of the rhombus = 26 cm

Isosceles triangles (triangle with two sides equal) have the base angles the same.

∠ YXZ = ∠ XZY

Let ∠ YXZ = ∠ XZY = x

x + x + 40° = 180° (Sum of angles in a triangle)

2x = 180° - 40°

2x = 140°

Divide both sides by 2.

x = $\frac{\mathrm{140°}}{2}$ = 70°

∠ YXZ = ∠ XZY = 70°

∠ XZY + ∠XZT = 180° (Sum of angles on a straight line)

70° + ∠XZT = 180°

∠XZT = 180° - 70°

∠XZT = 110°

The volume of both the cube and the cone formed are the same.

Volume of a cube = Length x Length x Length = L³

Height of the cube = Length of the cube = h

Volume of the cube = h³

Volume of a cone = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

h³ = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

The h at the right cancels one of the h at the left to produce h² at the left.

h² = $\frac{\mathrm{\pi }{\mathrm{r}}^2}{3}$

Multiply both sides by 3.

3 x h² = πr²

3h² = πr²

Divide both sides by π

r² = $\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}$

Take the square root of both sides.

r = $\sqrt{\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}}$

The square root of a square is the number itself.

$\sqrt{{\mathrm{h}}^2}$ = h

r = h x $\sqrt{\frac{3}{\mathrm{\pi }}}$

r = h$\sqrt{\frac{3}{\mathrm{\pi }}}$

Exterior angle of a regular polygon or not

Exterior angle of a regular polygon x n = 360°

n = $\frac{\mathrm{360°}}{\mathrm{Exterior\; angle\; of\; a\; regular\; polygon}}$

If n is whole number then the angle is an exterior angle.

66°

n = $\frac{\mathrm{360°}}{\mathrm{66°}}$ = 5.4545

Not an exterior angle of a regular polygon since n is not a whole number.

72°

n = $\frac{\mathrm{360°}}{\mathrm{72°}}$ = 5

Is an exterior angle of a regular polygon since n is a whole number.

24°

n = $\frac{\mathrm{360°}}{\mathrm{24°}}$ = 15

Is an exterior angle of a regular polygon since n is a whole number.

15°

n = $\frac{\mathrm{360°}}{\mathrm{15°}}$ = 24

Is an exterior angle of a regular polygon since n is a whole number.

Since two sides are equal, the right angle is isosceles and the base angles are the same.

Base angle = $\frac{\mathrm{180\; -\; 90}}{2}$ = 45°

Bearing starts from the north clockwise. If the angle is less than 100°, begin with a 0.

Note:

1.

Central angle is twice any inscribed angle subtended by the same arc.

∠ POR = 2 x ∠PQR = 2 x 72° = 144°

2.

∠ POT = ∠ OPS (Alternate angles).

∠ POT + ∠ POR = 180° (Angles on a straight line)

∠ POT + 144° = 180°

∠ POT = 180° - 144°

∠ POT = 36°

∠ POT = ∠ OPS = 36°

Area of region not covered by the trapezium = Area of the circle - Area of the trapezium

Area of a circle = πr².

Where r is the radius.

π = $\frac{22}{7}$

Radius = 7 cm

r² = r x r

Area of the circle = $\frac{22}{7}$ x 7 cm x 7 cm = 154 cm²

Area of a trapezium = $\frac{1}{2}$ x (a + b) x h

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

Area of the trapezium = $\frac{1}{2}$ x (10 cm + 21 cm) x 8 cm

Area of the trapezium = $\frac{1}{2}$ x 31 cm x 8 cm = 124 cm²

Area of region not covered by the trapezium = 154 cm² - 124 cm²

Area of region not covered by the trapezium = 30 cm²

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

Initial speed = X km/h

Time taken at the initial speed = 2 hours

Distance covered at the initial speed = X km/h x 2h

Distance covered at the initial speed = 2X km

Increased his speed by 2km/h means if you add 2 km/h to his current speed(initial speed), you will get the new speed.

New speed = (X + 2) km/h

Time taken at the new speed = 3 hours.

Distance covered at the new speed = (X + 2) km/h x 3h

Distance covered at the new speed = 3X km + 6 km

Total distance covered is 36 km

Distance covered at the initial speed + Distance covered at the new speed = 36 km

2X km + 3X km + 6 km = 36 km

2X + 3X + 6 = 36

5X = 36 - 6

5X = 30

Divide both sides by 5

X = $\frac{30}{5}$ = 6 km/h

b + 110 = 180 (Sum of angles on a straight line)

b = 180 - 110

b = 70°

a = b = 70° (Alternate angles)

a = y = 70° (Vertically opposite angles)

c = 35 (Vertically opposite angles)

c = d = 35 (Alternate angles)

x + d = 180 (Sum of angles on a straight line)

x + 35 = 180

x = 180 - 35

x = 145°

x + y = 70° + 145° = 215°

Apply the alternate segment theorem.

∠ QRN = 64°

The base angles of isosceles triangles (triangle with two sides equal) are the same.

∠ RQN = ∠ RNQ

Let ∠ RQN = ∠ RNQ = x

x + x + 64 = 180 (Sum of angles of a triangle)

2x = 180 - 64

2x = 116

Divide both sides by 2.

x = $\frac{116}{2}$ = 58°

∠ RQN = ∠ RNQ = 58°

From the the alternate segment theorem, ∠ RQN = t

∠ RQN = t = 58°

Arrange the numbers in ascending order.

6, 7, 7, 7, 8, 8, 8, 9, 9, 11

First Quartile (ungrouped data) = $\frac{\mathrm{n}}{4}$^th value

Where n is the number of items.

n = 10

First Quartile (ungrouped data) = $\frac{10}{4}$ = 2.5^th value

First Quartile = $\frac{\mathrm{7\; +\; 7}}{2}$ = 7.0

In cyclic quadrilateral, the opposite angles add up to 180°

x + (x + y) = 180

x + x + y = 180

2x + y = 180 ----- eqn 1

(2y - 30) + x = 180

2y - 30 + x = 180

x + 2y = 180 + 30

x + 2y = 210 ----- eqn 2

Get rid of y.

eqn 1 x 2

4x + 2y = 360 ----- eqn 3

eqn 3 - eqn 2

3x = 150

Divide both sides by 3

x = $\frac{150}{3}$ = 50°

A cone is produced from a sector of a circle.

Let l = the radius of the circle which produces the cone and L = the length of the arc of the sector.

Figure 1

Let h = height of the cone and r = base radius of the cone.

Figure 2

From Pythagorean theorem,

l² = h² + r²

l = $\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}$

h = 11 cm
r = 8 cm

l = $\sqrt{{11}^2+8^2}$

l = 13.6014705087 cm

Curved surface area = 8 cm x $\frac{22}{7}$ x 13.6014705087 cm

Curved surface area = 341.979829934 ≈ 341.98 cm²

Explantion to the formula curved surface area of a cone = r x π x l

The area of the sector of the circle which the cone is made from is the same as the curved surface area of the cone.

Area of a sector of a circle = $\frac{\mathrm{\theta }}{360}$ x π x r²

But the radius of the circle that produces the cone = l

Area of the sector the circle = $\frac{\mathrm{\theta }}{360}$ x π x l² ------------- eqn 1

The length of the arc of the circle becomes the circumference of the circular base of the cone.

Length of an arc of a circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x r

But the radius of the circle that produces the cone = l

Length of the arc of the circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x l

The length of the arc is the same as the circumference of the base of the cone.

Circumference of a circle = 2 x π x r

Circumference of the circular base of the cone = 2 x π x r

$\frac{\mathrm{\theta }}{360}$ x 2 x π x l = 2 x π x r

2 and π cancels each other.

$\frac{\mathrm{\theta }}{360}$ x l = r

Make $\frac{\mathrm{\theta }}{360}$ the subject.

$\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$

But from ---------- eqn 1,

Area of the sector of the circle = $\frac{\mathrm{\theta }}{360}$ x π x l²

Subtitute $\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$ into ------------ eqn 1

Area of the sector of the circle = $\frac{\mathrm{r}}{\mathrm{l}}$ x π x l²

The denominator, l cancels out one of the l in the l².

Area of the sector of the circle = r x π x l

∴ the curved surface area of the cone = r x π x l

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

SOH

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

The opposite side to x = 3 and the hypotenuse = 5

Draw the right angle and assign the values.

From pythagoras theorem

3² + b² = 5²

b² = 5² - 3²

b² = 25 - 9

b² = 16

Take the square root of both sides.

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

TOA

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

The opposite side to x is 3

The adjacent side to x is 4

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

CAH

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

The adjacent side to x is 4

The hypotenuse side to x is 5

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

tan x + 2 cos x = $\frac{3}{4}$ + 2 x $\frac{4}{5}$

tan x + 2 cos x = $\frac{3}{4}$ + $\frac{8}{5}$

tan x + 2 cos x = $\frac{\mathrm{5\; x\; 3\; +\; 4\; x\; 8}}{20}$

tan x + 2 cos x = $\frac{\mathrm{15\; +\; 32}}{20}$

tan x + 2 cos x = $\frac{47}{20}$ = 2$\frac{7}{20}$

1. A diameter subtends an angle of 90° at the circumference of the circle

In the diagram EC is a diameter

∠ EBC = 90° (subtended by the diameter)

∠ ABE + ∠ EBC = ∠ ABC

∠ ABE + 90° = 158°

∠ ABE = 158° - 90°

∠ ABE = 68°

2. Angles in the same segment are equal

∠ ABE = ∠ ADE = 68°(Angles on the same segment of chord AE)

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{6030}{37}$ = 162.97297 ≈ 163.0 (one decimal place)

For a midpoint between two points A(x₁, y₁) and B(x₂, y₂),

Midpoint x coordinate = $\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}$

Midpoint y coordinate = $\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}$

Calculating the midpoint of X(-8, -12) and Y(p, q)

Midpoint of XY x coordinate = $\frac{\mathrm{-8\; +\; p}}{2}$

Midpoint of XY y coordinate = $\frac{\mathrm{-12\; +\; q}}{2}$

Midpoint of XY = (-4, -2)

But Midpoint of XY x coordinate = -4

$\frac{\mathrm{-8\; +\; p}}{2}$ = -4

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

-8 + p = -4 x 2

-8 + p = -8

p = -8 + 8

p = 0

But Midpoint of XY y coordinate = -2

$\frac{\mathrm{-12\; +\; q}}{2}$ = -2

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

-12 + q = -2 x 2

-12 + q = -4

q = -4 + 12

q = 8

∴ the coordinates of Y = (0, 8)

Let n = number of adults sold to and m = number of children sold to.

Five hundred tickets were sold for a concert implies that when you add the number of adults and children, you should get 500

n + m = 500 ------- eqn 1

Total tickets sold = number of adults x cost for adult + number of children x cost for children.

Total tickets sold = n x 4.5 + m x 3

Total tickets sold = 4.5n + 3m

But total tickets sold = $1987.50

4.5n + 3m = 1987.50 ------ eqn 2

Get rid of m.

eqn 1 x 3

3 x n + 3 x m = 3 x 500

3n + 3m = 1500 ------ eqn 3

eqn 2 - eqn 3

1.5n = 487.5

Divide both sides by 1.5

n = $\frac{487.5}{1.5}$ = 325

∴ 325 tickets were sold to adults.

More than 18 km means the distance d is greater than 18 km and hence 18 km is less than the distance (18 < d)

Not more than 23 km means it can be 23 or less hence the less than or equal to sign is used (d ≤ 23)

When you combine the two, the inequality will be 18 < d ≤ 23 which is read as 18 is less than d and d is also less than or equal to 23.

The sum of all the angles is 360°

Angle of sector for orange = 360° - (100° + 80° + 60°)

Angle of sector for orange = 360° - 240°

Angle of sector for orange = 120°

The quantity for apple is known to be 60.

The angle of sector for apple is 80°

If 80° = 60

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

∴ there are 90 oranges.

Number of balls which are neither red nor blue = 40 - (10 + 12)

Number of balls which are neither red nor blue = 40 - 22

Number of balls which are neither red nor blue = 18

Number of samples = 40

P(Neither red nor blue) = $\frac{18}{40}$ = $\frac{9}{20}$

1. Calculate the total number of outcomes when a fair die is tossed twice: 6 × 6 = 36

At least 10 means 10 or more.

2. Find the favorable outcomes for a sum of at least 10: (4,6), (6,4) for 10; (5,5) for 10; (5,6), (6,5) for 11, (6,6) for 12

Number of favourable outcome is 6

P(Sum is at least 10) = $\frac{6}{36}$ = $\frac{1}{6}$

You can also draw a table for all the possible outcomes.

Number of samples = 6 x 6 = 36

Number of possible selection = 6

P(Sum is at least 10) = $\frac{6}{36}$ = $\frac{1}{6}$

The man's age now = Age in 8 years time - 8

The man's age now = (x + 10) - 8

The man's age now = x + 10 - 8

The man's age now = x + 2

2 years ago he was 63 years.

The man's age now = 63 + 2 = 65 years

But the man's age now = x + 2

x + 2 = 65

x = 65 - 2

x = 63

The general equation of a line is y = mx + c

Where m is the gradient or slope and c is the y-intercept.

Express the given equation to be in the form of the general equation of a line and compare to find m.

3x - 5y = 7

3x - 7 = 5y

Divide both sides by 5

$\frac{3}{5}$x - $\frac{7}{5}$ = y

Comparing the above with the general equation of a line, m = $\frac{3}{5}$ hence gradient (slope) = $\frac{3}{5}$

A function is undefined when the denominator is 0.

x + 1 = 0

x = 0 - 1

x = -1

Possible outcomes

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

Number of samples = 6 x 6 = 36