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}

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

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 $\frac{1}{2}$

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

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

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) = $\frac{6}{36}$ = $\frac{1}{6}$

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

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.

∠ QFG + ∠ EFG = 180°

105° + ∠ EFG = 180°

∠ EFG = 180° - 105°

∠ EFG = 75°

∠ FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)

50° + m + 75° = 180°

m + 125° = 180°

m = 180° - 125°

m = 55°

The angle (θ) substended are the same.

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

Area = 231 cm²

$\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 231

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator 360.

360 x $\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 360 x 231

θ x π x radius x radius = 360 x 231

Divide both sides by π x radius x radius

θ = $\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ --- eqn 1

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

Length = 66 cm

$\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 66

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator, 360.

360 x $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 360 x 66

θ x 2 x π x radius = 360 x 66

Divide both sides by 2 x π x radius

θ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$ --- eqn 2

--- eqn 1 = --- eqn 2

$\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$

Cross multiply.

360 x 231 x 2 x π x radius = 360 x 66 x π x radius x radius

Divide both sides by 360 x π x radius to cancel out.

$\frac{\mathrm{360\; x\; 231\; x\; 2\; x\; \pi \; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66\; x\; \pi \; x\; radius\; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$

231 x 2 = 66 x radius

Divide both sides by 66

Radius = $\frac{\mathrm{231\; x\; 2}}{66}$ = 7 cm

Concept

(a + b)² = a² + 2 x a x b + b² = a² + 2ab + b²
(a - b)² = a²-2 x a x b + b² = a² -2ab + b²

Alternatively
(a + b)² = (a + b)x(a + b) [Expand and simplify]
(a -b)² = (a - b)x(a - b) [Expand and simplify]

(x - 2)² = x² - 2 x x x 2 + 2² = x² -4x + 4
(x + 1)² = x² + 2 x x x 1 + 1² = x² + 2x + 1

(x - 2)² + 3 = (x + 1)² - 6
x² - 4x + 4 + 3 = x² + 2x + 1 - 6
x² - 4x + 7 = x² + 2x - 5
x² - x² - 4x - 2x = -5 - 7
-6x = -12

Divide both sides by -6

x = $\frac{-12}{-6}$ = 2

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°

The area of the trapezium is the summation of the area of the triangle and the rectangle.

Area of triangle = $\frac{1}{2}$ x base x height

Area of a rectangle = Length x Breadth

Base of the triangle = 24 - 16 = 8

Height of the triangle = x
Breadth of rectangle = x

300 = $\frac{1}{2}$ x 8 x x + 16 x x

2 divides itself once and 8, 4 times.

300 = 4x + 16x
300 = 20x

Divide both sides by 20

x = $\frac{300}{20}$ = 15

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43°

∠ OBC = ∠ OCB = y = 43°

∠ OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)

∠ OBC = ∠ OCB = 43°

43° + 43° + ∠ BOC = 180°

86° + ∠ BOC = 180°

∠ BOC = 180° - 86°

∠ BOC = 94°

x = $\frac{\mathrm{BOC}}{2}$ = $\frac{\mathrm{94°}}{2}$ = 47°

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

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

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

Y₂ = 6
Y₁ = 4
X₂ = -2
X₁ = 1

Gradient = $\frac{\mathrm{6\; -\; 4}}{\mathrm{-2\; -\; 1}}$ = $\frac{2}{-3}$ = -$\frac{2}{3}$

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.

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$

Express each surds to be in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{28}}$ = $\sqrt{\mathrm{4\; x\; 7}}$

$\sqrt{\mathrm{50}}$ = $\sqrt{\mathrm{25\; x\; 2}}$

$\sqrt{\mathrm{72}}$ = $\sqrt{\mathrm{36\; x\; 2}}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{4\; x\; 7}}$ - 3$\sqrt{\mathrm{25\; x\; 2}}$ + $\sqrt{\mathrm{36\; x\; 2}}$

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

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x $\sqrt{4}$ x $\sqrt{7}$ - 3 x $\sqrt{\mathrm{25}}$ x $\sqrt{2}$ + $\sqrt{\mathrm{36}}$ x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x 2 x $\sqrt{7}$ - 3 x 5 x $\sqrt{2}$ + 6 x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 15$\sqrt{2}$ + 6$\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 9$\sqrt{2}$

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

9 = 3 x 3 = 3²
27 = 3 x 3 x 3 = 3³

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

$\frac{1}{27}$ = $\frac{1}{3^3}$ = 3^-3

3^{2(y + 1)} = 3^{-3(y - 2)}

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

2(y + 1) = -3(y - 2)
2y + 2 = -3y + 6

Note: - x - = +
-3 x -2 = + 6

2y + 2 = -3y + 6
2y + 3y = 6 - 2
5y = 4

Divide both sides by 5

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

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

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)

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

Angle of sector of orange = 360° - (100° + 60° + 120°)

Angle of sector of orange = 360° - 280°

Angle of sector of orange = 80°

Angle of sector of apple = 120°

Number of oranges = 60

If 80° = 60

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

There are 90 apples

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

Expand the brackets

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

Note: - x - = +

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

The Least Common Multiples (L.C.M) of the denominators (2,3 and 6) is 6

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

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

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

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

$\frac{1}{2}$(x-1)-$\frac{1}{3}$($\frac{1}{2}$x-1) = $\frac{1}{6}$(2x - 1)

Distance = $\sqrt{{\left(\mathrm{8\; -\; 3}\right)}^2+{\left(\mathrm{10\; --\; 2}\right)}^2}$

Note: - - = +

Distance = $\sqrt{{\left(5\right)}^2+{\left(\mathrm{10\; +\; 2}\right)}^2}$

Distance = $\sqrt{5^2+{12}^2}$

Distance = $\sqrt{\mathrm{25\; +\; 144}}$

Distance = $\sqrt{169}$ = 13 units

∠ QRS = ∠ PQT = 40° (Corresponding angles)

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

Isosceles triangle has base angles equal.

∠ RPT = ∠ PQT = 40° (Base angles equal)

cos (3x + 28°) = sin (2x + 48°)

Use the trigonometric values to change either cos to sin or sin to cos and equate the angles to solve for x.

Method I

Changing the cos to sin.

cos (3x + 28°) = sin (90° - [3x + 28°])

cos (3x + 28°) = sin (2x + 48°)

sin (90° - [3x + 28°]) = sin (2x + 48°)

Since both sides are sin, the angles are the same.

90° - (3x + 28°) = 2x + 48°

90° - 3x - 28° = 2x + 48°

90° - 28° - 48° = 2x + 3x

90° - 76° = 5x

14° = 5x

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°

Method II

Changing the sin to cos.

sin (2x + 48°) = cos(90° - [2x + 48°])

cos (3x + 28°) = sin (2x + 48°)

cos (3x + 28°) = cos(90° - [2x + 48°])

Since both sides are cos, the angles are the same.

3x + 28° = 90° - (2x + 48°)

3x + 28° = 90° - 2x - 48°

3x + 2x = 90° - 48° - 28°

5x = 14°

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°

lb = $\frac{1}{2}$(a + b)h

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

lb x 2 = h(a + b)

2lb = ah + bh

2lb - bh = ah

Factorize b out.

b(2l - h) = ah

Divide both sides by (2l - h)

b = $\frac{\mathrm{ah}}{\mathrm{2l\; -\; h}}$

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

∠ EPG = 2 x ∠ EFG

∠ EPG = 2 x 142^o

∠ EPG = 284^o

∠ EPG + x = 360^o (Sum of angles of a circle)

284^o + x = 360^o

x = 360^o - 284^o

x = 76^o

Sum of interior angles of a quadrilateral is 360^o

Note: sum of interior angles = (n - 2) x 180^o, where n is the number of sides.

63^o + 142^o + ∠ FEP + x = 360^o

63^o + 142^o + ∠ FEP + 76^o = 360^o

∠ FEP + 281^o = 360^o

∠ FEP = 360^o - 281^o

∠ FEP = 79^o

Sum of interior angles = 1260°

(n - 2) x 180° = 1260°

(n - 2) x 180 = 1260

Divide both sides by 180

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

n - 2 = 7

n = 7 + 2

n = 9

Sum of exterior angles of a polygon = 360°

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

10x + 80 = 360

10x = 360 - 80

10x = 280

Divide both sides by 10

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

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

Numbers in a die rolled once = {1,2,3,4,5,6}

Numbers less than 3 = {1,2}

Number of possible selection = 2
Number of samples = 6

Probability of less than 3 = $\frac{2}{6}$ = $\frac{1}{3}$

(4x² + 3x - 1)(3x + 1) = (4x² + 3x - 1)(3x + 1)

(4x² + 3x - 1)(3x + 1) = 3x(4x² + 3x - 1) + 1(4x² + 3x - 1)

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² - 3x + 4x² + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² + 4x² - 3x + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 13x² + 0x - 1

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

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

$\sqrt{a}$ = $a^{\frac{1}{2}}$

$\sqrt{\mathrm{64}}$ = ${\mathrm{64}}^{\frac{1}{2}}$

64 = 2⁶

$\sqrt{\mathrm{64}}$ = $2^{\mathrm{6\; x}\frac{1}{2}}$ = 2³

2a = 2³

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

a = 3

$\frac{\mathrm{b}}{\mathrm{a}}$ = 3

b = 3 x a

b = 3 x 3

b = 9

a² + b² = 3² + 9²

a² + b² = 9 + 81

a² + b² = 90

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°

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 = $\frac{1}{2}$

E = $\frac{\mathrm{k}{\mathrm{x}}^2}{\mathrm{2y}}$ + z

Multiply both sides the denominator 2y

E x 2y = $\frac{\mathrm{k}{\mathrm{x}}^2}{\mathrm{2y}}$ x 2y + z x 2y

2Ey = kx² + 2yz

kx² = 2Ey - 2yz

Divide both sides by k.

x² = $\frac{\mathrm{2Ey\; -\; 2yz}}{\mathrm{k}}$

Factorize 2y out.

x² = $\frac{\mathrm{2y(E\; -\; z)}}{\mathrm{k}}$

Take square root of both sides.

x = $\sqrt{\frac{\mathrm{2y(E\; -\; z)}}{\mathrm{k}}}$

Since 3 can go into 6 and 9, split -6pqr into -3pqr and -3pqr

p²q² - 6pqr + 9r² = p²q² - 3pqr - 3pqr + 9r²

p²q² - 6pqr + 9r² = pq(pq - 3r) - 3r(pq - 3r)

p²q² - 6pqr + 9r² = (pq - 3r)(pq - 3r)

Since pq - 3r = pq - 3r, the multiplication makes it square

p²q² - 6pqr + 9r² = (pq - 3r)²

The equation above is similar to the general factorization rule of (a - b)²

(a - b)² = a² - 2ab + b²

Thus the first term square then the sign (either + or -) and 2 times the first and second terms + the second term square

Similarly (a + b)² = a² + 2ab + b²

The first term for the equation, p²q² - 6pqr + 9r² is pq and the second term is 3r and the sign is -

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

Pick a point on the line of the relation: y = d - x and substitute into the relation to find d.

y = d - x

Using the point (1, 0)

x = 1 and y = 0

0 = d - 1

1 = d

d = 1

OR

Using the point (0, 1)

x = 0, y = 1

1 = d - 0

1 = d

d = 1

OR

Using the point (-2, 3)

x = -2 and y = 3

3 = d - (-2)

Note: -- or -(-) = +

3 = d + 2

3 - 2 = d

1 = d

d = 1

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

x² + 2x + 1 = 0

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

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

x² + x + x + 1 = 0

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

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

When x + 1 = 0, x = -1

Let y = area of the field

First day area cleared = 40% of y

First day area cleared = $\frac{40}{100}$ x y = 0.4y

Remaining uncleared area = y - 0.4y

Remaining uncleared area = 0.6y

Second day area cleared = 60% of remaining

Second day area cleared = $\frac{60}{100}$ x 0.6y

Second day area cleared = 0.36y

Remaining area after the second day = 0.6y - 0.36y

Remaining area after the second day = 0.24y

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Percentage of the land remaining at the end of the second day = $\frac{\mathrm{0.24y}}{\mathrm{y}}$ x 100%

Note: the ys cancel each other.

Percentage of the land remaining at the end of the second day = $\frac{\mathrm{0.24\; x\; 1}}{1}$ x 100% = 24%

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

Mean = $\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$

The mean is not greater than 5 means the mean can be less than 5 or equal to 5.

$\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$ ≤ 5

$\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5

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

4 x $\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5 x 4

14 + 2x ≤ 20

2x ≤ 20 - 14

2x ≤ 6

Divide both sides by 2.

x ≤ $\frac{6}{2}$

x ≤ 3

4x² - 16x + 15 = 0

Notes:

1. a x c = 4 x 15 = 60

2. Split b (-16) such that two numbers when you sum you will get b (-16) and when you multiply you will get ac (60). The two numbers are -10 and -6

3. -16x = -10x - 6x

4x² -10x - 6x + 15 = 0

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

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

When 2x - 5 = 0, x = $\frac{5}{2}$ = 2$\frac{1}{2}$

When 2x - 3 = 0, x = $\frac{3}{2}$ = 1$\frac{1}{2}$

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

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

Let x = the length of the ladder.

From Pythagorean theorem,

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

x² = 6² + 8²

x² = 36 + 64

x² = 100

Take square root of both sides.

x = $\sqrt{\mathrm{100}}$ = 10

The ladder is 10 m long.