75.0785 - 34.624 + 9.83 = 50.2845

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

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

x:x < 7 is read as x is such that x is less than 7.

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

Note: since X is a subset of the universal set (µ), members of X should be found in the universal set (µ)

y:y is a factor of 24 is read as y is such that y is a factor of 24.

Y = {1, 2, 3, 4, 6, 8}

Note: since Y is a subset of the universal set (µ), members of Y should be found in the universal set (µ)

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

X ∩ Y = {1, 2, 3, 4, 6}

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:

1. 16 = 4 x 4 = 4²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

2. 9 = 3 x 3 = 3²

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

3. 16 = 2 x 2 x 2 x 2 = 2⁴

Expressed this way in order to cancel the denominator 4 of the $\frac{-3}{4}$ power

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4^2}{3^2}\right)}^{\frac{-3}{2}}x{\left(2^4\right)}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

From the law of indices,

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

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{\mathrm{2\; x}\frac{-3}{2}}}{3^{\mathrm{2\; x}\frac{-3}{2}}}x{2}^{\mathrm{4\; x}\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:
1. The power 2 cancel the denominator 2 of the power $\frac{-3}{2}$

2. The power 4 cancel the denominator of 4 of the $\frac{-3}{4}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{4^{-3}}{3^{-3}}x{2}^{-3}]}^{\frac{1}{3}}$

$\frac{4^{-3}}{3^{-3}}$ = ${\left(\frac{4}{3}\right)}^{-3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{\left(\frac{4}{3}\right)}^{-3}x{2}^{-3}]}^{\frac{1}{3}}$

The $\frac{1}{3}$ affects each of the powers in the bracket.

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-3x\frac{1}{3}}x{2}^{-3x\frac{1}{3}}}^{}$

The power 3 cancels the denominator 3 in $\frac{1}{3}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${{\left(\frac{4}{3}\right)}^{-1}x{2}^{-1}}^{}$

Notes:

1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).

2. ${\left(\frac{4}{3}\right)}^{-1}$ = $\frac{3}{4}$

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

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

4. 2^-1 = $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{3}{4}$ x $\frac{1}{2}$

${[{\left(\frac{16}{9}\right)}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{\mathrm{3\; x\; 1}}{\mathrm{4\; x\; 2}}$ = $\frac{3}{8}$

The least is 9 dividing the number 1 time with a remainder of 7

4x = 9 x 1 + 7

4x = 9 + 7

4x = 16

Divide both sides by 4

x = $\frac{16}{4}$ = 4

Notes

1. log_bb = 1

log₁₀10 = 1

2. blog c = log c^b

2log₁₀3 = log₁₀3²

3. log b + log c = log b x c

1 + 2 log₁₀3 = log₁₀10 + log₁₀3²

1 + 2 log₁₀3 = log₁₀10 + log₁₀9

1 + 2 log₁₀3 = log₁₀10 x 9

1 + 2 log₁₀3 = log₁₀90

Change each of the bases to base 10.

101_two = 1 x 2² + 0 x 2¹ + 1 x 2⁰

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

101_two = 4 + 0 + 1

101_two = 5

12_y = 1 x y¹ + 2 x y⁰

12_y = y + 2

23_five = 2 x 5¹ + 3 x 5⁰

23_five = 10 + 3

23_five = 13

Substitute the values of each of the bases.

5 + (y + 2) = 13

5 + y + 2 = 13

y + 7 = 13

y = 13 - 7

y = 6

Total amount = Principal + Interest

Principal = x

Time = 5

Rate = 2

Simple Interest = $\frac{\mathrm{x\; x\; 5\; x\; 2}}{100}$ = $\frac{\mathrm{10x}}{100}$ = $\frac{\mathrm{x}}{10}$

Total amount = 550,000

Substitute the total amount and the interest into the total amount equation.

550000 = x + $\frac{\mathrm{x}}{10}$

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

550000 x 10 = 10 x x + x

5500000 = 10x + x

5500000 = 11x

Divide both sides by 11

x = $\frac{5500000}{11}$ = ₦ 500,000

Simplify the left hand side and compare with the right hand side to determine the values of x and y.

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$

Divide each of the expression at the numerator by the denominator $\sqrt{5}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}}{\sqrt{5}}$ + $\frac{\sqrt{5}}{\sqrt{5}}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}}{\sqrt{5}}$ + 1

Get rid of the denominator surd in $\frac{\sqrt{3}}{\sqrt{5}}$ by multiplying both the numerator and the denominator by $\sqrt{5}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}x\sqrt{5}}{\sqrt{5}x\sqrt{5}}$ + 1

Notes:

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

$\sqrt{5}$ x $\sqrt{5}$ = 5

2. $\sqrt{a}$ x $\sqrt{b}$ = $\sqrt{\mathrm{a\; x\; b}}$

$\sqrt{3}$ x $\sqrt{5}$ = $\sqrt{\mathrm{3\; x\; 5}}$ = $\sqrt{\mathrm{15}}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{\mathrm{3\; x\; 5}}}{5}$ + 1

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{15}}{5}$ + 1

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{1}{5}$ x $\sqrt{\mathrm{15}}$ + 1

Comparing with x + y$\sqrt{\mathrm{15}}$,

x = 1 and y = $\frac{1}{5}$

x + y = 1 + $\frac{1}{5}$

x + y = 1$\frac{1}{5}$

2(x² - y³)

When x = 3 and y = -1

2(x² - y³) = 2 x (3² - (-1)³)

(-1)³ = - 1 x -1 x -1 = 1 x -1 = -1

2(x² - y³) = 2 x (9 - (-1))

2(x² - y³) = 2 x (9 - -1)

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

2(x² - y³) = 2 x (9 + 1)

2(x² - y³) = 2 x (10)

2(x² - y³) = 2 x 10

2(x² - y³) = 20

3x - 2y = 10 ----------- eqn 1

x + 3y = 7 ------------ eqn 2

In order to get rid of x, multiply ----- eqn 2 by 3

3x + 9y = 21 ----------- eqn 3

Subtract ---- eqn 1 from ---- eqn 3

9y - - 2y = 11

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

9y + 2y = 11

11y = 11

Divide both sides by 11

y = $\frac{11}{11}$ = 1

Substitute y = 1 into ------- eqn 1.

3x - 2(1) = 10

3x - 2 = 10

3x = 10 + 2

3x = 12

Divide both sides by 3

x = $\frac{12}{3}$ = 4

Th nth term is given by a_n = ar^{n - 1}

Where a = first term, r = common ratio

a = 3

a₅ = 3 x r^{5 - 1}

a₅ = 3r⁴

But a₅ = 48

3r⁴ = 48

Divide both sides by 3

r⁴ = $\frac{48}{3}$

r⁴ = 16

16 = 2 x 2 x 2 x 2 = 2⁴

r⁴ = 2⁴

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

r = 2

$\frac{1}{3}$(5 - 3x) < $\frac{2}{5}$(3 - 7x)

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (3 and 5).

The L.C.M of 3 and 5 is 15

15 x $\frac{1}{3}$(5 - 3x) < 15 x $\frac{2}{5}$(3 - 7x)

5(5 - 3x) < 3 x 2(3 - 7x)

25 - 15x < 6(3 - 7x)

25 - 15x < 18 - 42x

- 15x + 42x < 18 - 25

27x < -7

Divide both sides by 27

x < $\frac{-7}{27}$

k = $\sqrt{\frac{\mathrm{m\; -\; y}}{\mathrm{m\; +\; 1}}}$

Square both sides to get rid of the square root.

k² = $\frac{\mathrm{m\; -\; y}}{\mathrm{m\; +\; 1}}$

Get rid of the fraction by multiplying both sides by the denominator m + 1

k² x (m + 1) = m - y

mk² + k² = m - y

y + k² = m - mk²

Factorize m out.

y + k² = m(1 - k²)

Divide both sides by (1 - k²)

m = $\frac{\mathrm{y}+{\mathrm{k}}^2}{1-{\mathrm{k}}^2}$

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

x is directly proportional to y

x ∝ y

x is inversely proportional to z

x ∝ $\frac{1}{\mathrm{z}}$

Combining the two, x ∝ $\frac{\mathrm{y}}{\mathrm{z}}$

Introduce a constant k to remove the proportionality constant.

x = k x $\frac{\mathrm{y}}{\mathrm{z}}$

Substitute the known values and solve for the constant k

x = 15
y = 10
z = 4

15 = k x $\frac{10}{4}$

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

4 x 15 = k x 10

60 = 10k

Divide both sides by 10

k = $\frac{60}{10}$ = 6

Substitute the value of k into the relation.

x = k x $\frac{\mathrm{y}}{\mathrm{z}}$

k = 6

x = 6 x $\frac{\mathrm{y}}{\mathrm{z}}$

x = $\frac{\mathrm{6\; x\; y}}{\mathrm{z}}$

x = $\frac{\mathrm{6y}}{\mathrm{z}}$

The sum of the distances covered by both buses should give 240 km

Let t that travelled by both when they are 240 km apart.

Distance covered by the first bus = 72 km/h x t

Distance covered by the second bus = 48 km/h x t

72t + 48t = 240

120t = 240

t = $\frac{240}{120}$ = 2 hours

Since they started the journey at 9.00 am, in 2 hours time, the time will be 11.00 a.m and both buses will be 240 km apart.

Total surface area = 2(7 x 5 + 7 x 4 + 5 x 4) cm²

Total surface area = 2(35 + 28 + 20) cm²

Total surface area = 2 x 83 cm²

Total surface area = 166 cm²

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

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

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}

The man is five times as old as his son means when you multiply the son's age by 5, you should get the man's age.

The son is y years now.

Man's age now = 5 x y = 5y

In four years time,

Son's age = y + 4

Man's age = 5y + 4

The product of their ages would be 340

Product means multiplication.

Man's age x Son's age in four years time = 340

(5y + 4) x (y + 4) = 340

5y(y + 4) + 4(y + 4) = 340

5y² + 20y + 4y + 16 = 340

5y² + 24y + 16 - 340 = 0

5y² + 24y + 16 - 340 = 0

5y² + 24y - 324 = 0

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{\mathrm{a\; x\; a}-\mathrm{b\; x\; b}-\mathrm{a\; x\; c}}{\mathrm{ab}}$

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{{\mathrm{a}}^2-{\mathrm{b}}^2-\mathrm{ac}}{\mathrm{ab}}$

The midpoint divides XY into two equal lengths, 3 cm each.

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

tan ∠ XZT = $\frac{\mathrm{XT}}{\mathrm{ZT}}$

XT = 3 cm but ZT is unknown.

You can use Pythagoream theorem to calculate ZT.

From pythagoras theorem

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

ZT² + XT² = XT²

ZT² + 3² = 6²

ZT² + 9 = 36

ZT² = 36 - 9

ZT² = 27

Take square root of both sides.

ZT = $\sqrt{\mathrm{27}}$

27 = 9 x 3

ZT = $\sqrt{\mathrm{9\; x\; 3}}$

But $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

ZT = $\sqrt{9}$ x $\sqrt{3}$

$\sqrt{9}$ = 3

ZT = 3 x $\sqrt{3}$ = 3$\sqrt{3}$

tan ∠ XZT = $\frac{3\sqrt{3}}{3}$

3 cancel out 3.

tan ∠ XZT = $\sqrt{3}$

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

Opposite side to θ = 13.6 m

Adjacent side to θ = 10 m

tan(θ) = $\frac{\mathrm{13.6\; m}}{\mathrm{10\; m}}$

Take tan inverse.

θ = tan^-1($\frac{\mathrm{13.6\; m}}{\mathrm{10\; m}}$)

θ = 53.6731740479° ≈ 53.67°

Selling Price = Cost Price + Profit

Selling Price of each = ₦ (x + y)

For the 10 tins, selling price = 10 x ₦ (x + y)

Amount received for the 10 tins = ₦ 10(x + y)

Base area = Length x Breadth

Let h = the height of the pyramid.

Volume = $\frac{1}{3}$ x 9 x 5 x h

Volume = 15h

But the volume = 105

15h = 105

Divide both sides by 15

h = $\frac{105}{15}$ = 7

The height is 7 cm

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

∠ MNP = ∠ NPQ = 2x = (3x - 50°) (Alternate angles)

2x = (3x - 50°)

2x = 3x - 50°

50° = 3x - 2x

50° = x

x = 50°

∠ NPQ = 2x = 2 x 50° = 100°

Alternatively,

∠ NPQ = 3x - 50°

∠ NPQ = 3 x 50° - 50°

∠ NPQ = 150° - 50°

∠ NPQ = 100°

Length of the arc = 1$\frac{19}{36}$ cm = $\frac{\mathrm{1\; x\; 36\; +\; 19}}{36}$ cm = $\frac{55}{36}$ cm

π = $\frac{22}{7}$

Radius = 3.5 cm

$\frac{55}{36}$ = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 3.5

$\frac{55}{36}$ = $\frac{\mathrm{\theta \; x\; 2\; x\; 22\; x\; 3.5}}{\mathrm{360\; x\; 7}}$

Cross multiply.

36 x θ x 2 x 22 x 3.5 = 55 x 360 x 7

Divide both sides by 36 x 2 x 22 x 3.5

θ = $\frac{\mathrm{55\; x\; 360\; x\; 7}}{\mathrm{36\; x\; 2\; x\; 22\; x\; 3.5}}$

θ = 25°

PTRS is a parallelogram.

Since line PU is parallel to line SR the height of PTRS is the same as the height of ∆SUR

Let h = the height of ∆SUR

Base of ∆SUR = 8 cm

Area of ∆SUR = 24 cm²

$\frac{1}{2}$ x 8 cm x h = 24 cm²

4 cm x h = 24 ²

Divide both sides by 4 cm

h = $\frac{\mathrm{24\; }{\mathrm{cm}}^2}{\mathrm{4\; cm}}$ = 6 cm

Height of the parallelogram = 6 cm

Base of the parallelogram = 8 cm

Area of the parallelogram = 8 cm x 6 cm = 48 cm²

OP = OQ (Both are radius)

Since two sides of ∆OPQ are the same (isosceles), the base angles are the same.

∠ OPQ = ∠ OQP = 48°

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

m = ∠ OPQ + ∠ OQP

m = 48° + 48° = 96°

360° represents the total population of the city.

Sum of all the angles should be 360°

Angle of sector for men = 360° - (120° + 169°)

Angle of sector for men = 71°

If 360° = 1,800,000

71° = $\frac{\mathrm{1800000\; x\; 71°}}{\mathrm{360°}}$ = 355000

Σ(x_i - µ)² = (15 - 21)² + (21 - 21)² + (17 - 21)² + (26 - 21)² + (18 - 21)² + (29 - 21)²

Σ(x_i - µ)² = 36 + 0 + 16 + 25 + 9 + 64

Σ(x_i - µ)² = 150

N = 6

σ = $\sqrt{\frac{150}{6}}$ = 5

Notes:

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

∠ QPS = 90°

Angles subtended by the same arc are the same.

From arc SP,

∠ PRS = ∠ PQS = 37° (Angles subtended by the same arc)

∠ PQS + ∠ QPS + ∠ PSQ = 180° (Sum of angles in a triangle)

37° + 90° + ∠ PSQ = 180°

127° + ∠ PSQ = 180°

∠ PSQ = 180° - 127°

∠ PSQ = 53°

Number of samples = 12

Number of green balls = 12 - (5 + 4)

Number of green balls = 12 - 9 = 3

P(Green) = $\frac{3}{12}$ = $\frac{1}{4}$

P(Both red) = P(first ball red) and P(second ball red)

And is multiplication (x)

P(Both red) = P(Red) x P(Red)

Number of samples = 12

Number of red balls = 5

P(Red) = $\frac{5}{12}$

P(Both red) = $\frac{5}{12}$ x $\frac{5}{12}$ = $\frac{\mathrm{5\; x\; 5}}{\mathrm{12\; x\; 12}}$ = $\frac{25}{144}$

m = $\frac{1}{2}$(x + y + z) ------- eqn 1

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

x + y + z = 180° - m

Substitute (x + y + z) = 180° - m into ----- eqn 1

m = $\frac{1}{2}$(180° - m)

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

2 x m = $\frac{1}{2}$(180° - m) x 2

2m = 180° - m

2m + m = 180°

3m = 180°

Divide both sides by 3

m = $\frac{\mathrm{180°}}{3}$ = 60°

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

Method I

Drawing a third parallel line.

∠ QUT = ∠ UTO = m° (Alternate angles)

∠ SVT = ∠ VTO = n° (Alternate angles)

∠ UTV = ∠ UTO + ∠ VTO = m° + n°

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

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

∠ UTV = m° + n° = 90°

Method II

Apply the alternate segment theorem.

∠ QUT = ∠ UVT = m° (alternate segment)

∠ SVT = ∠ TUV = n° (alternate segment)

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

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

m° + n° + 90° = 180° (Sum of angles in a triangle)

m° + n° = 180° - 90°

m° + n° = 90°

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

∠ LOM = 2 x ∠ LNM

∠ LNM = $\frac{\mathrm{\angle \; LOM}}{2}$ = $\frac{\mathrm{x}}{2}$

∠ LNM + ∠ NLM + ∠ LMN = 180° (Sum of angles in a triangle)

∠ LNM = $\frac{\mathrm{x}}{2}$

∠ NLM = 74°

∠ LMN = 39°

$\frac{\mathrm{x}}{2}$ + 74 + 39 = 180

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

x + 2 x 74 + 2 x 39 = 180 x 2

x + 148 + 78 = 360

x + 226 = 360

x = 360 - 226

x = 134

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

Upper quartile - Lower quartile (25th percentile) = Interquartile range

Lower quartile (25th percentile) = 16

Interquartile range = 7

Upper quartile - 16 = 7
Upper quartile = 7 + 16
Upper quartile = 23

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Intersection 1 = (2, 17 x 0.5) = (2, 8.5)

Note: Point = Number of small grids from 0 x value of small grid

Intersection 2 = (-15 x 0.1, 4 x 0.5) = (-1.5, 2)

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Number of small grids of a point on an axis = $\frac{\mathrm{Point\; value}}{\mathrm{Value\; of\; each\; small\; grid\; on\; that\; axis}}$

Number of small grids on the x-axis for -2.5 = $\frac{-2.5}{0.1}$ = -25 from 0 at the left

Number of small grids on the y axis of the corresponding -2.5 on the curve = 18 from 0

Each small grid = 0.5

Point = 18 x 0.5 = 9

If (x + a) is a factor, then when you substitute -a into the equation, you should get 0.

Since (x + 2) is a factor of x² + px - 10, substituting -2 into x² + px - 10 should give 0.

(-2)² + p(-2) - 10 = 0

4 - 2p - 10 = 0

-6 = 2p

Divide both sides by 2

p = $\frac{-6}{2}$ = -3

Number of samples = 32 + 26 + 26 + 28 + 44 + 36 = 192