We have (x − 7)(x + 4) = 0.
When two things multiply to give 0, at least one of them must be 0.

So set each bracket equal to 0:
x − 7 = 0    →    x = 7
x + 4 = 0    →    x = −4

Check: put x = 7 → (0)(11) = 0 ✓
put x = −4 → (−11)(0) = 0 ✓

Look at both terms: 2x and 4xy.
Find what they have in common.

Numbers: the biggest number that goes into both 2 and 4 is 2.
Letters: both terms have x (but the second term also has y).
The common factor is: 2x.

Divide each term by 2x:
2x ÷ 2x = 1
4xy ÷ 2x = 2y

Put the common factor outside the brackets:
2x × (1 − 2y)

Identify the triangle. We have side BC = 3.5 m and side AB = 0.9 m, and we want angle BAC (at vertex A, between AB and AC).

The angle BAC is the angle opposite the side of length 0.9 m. We use the Sine Rule:
sin A / a = sin B / b
Or: sin(BAC) / 0.9 = sin(BCA) / 3.5

First find angle C:
sin C = 0.9 / 3.5 = 0.2571...
C = sin¹(0.2571) = 14.9°

Angles in a triangle add to 180°:
BAC = 180° − 90° − 14.9° = 75.1°

We have: 6n + 3 > 8n
The goal is to get n by itself on one side.

Subtract 6n from both sides:
6n + 3 − 6n > 8n − 6n
3 > 2n

Divide both sides by 2 (positive number, so inequality sign stays the same):
3 / 2 > n
1.5 > n

Write it as n < 1.5 (normal way round, n on the left).

"Similar" means the triangles have the same shape — all angles equal, sides in the same ratio.

Match up the sides. From the diagram:
AB corresponds to PQ,   BC corresponds to QR,   AC corresponds to PR.
So the ratio is: AB / PQ = BC / QR = AC / PR

We know:   AC = 5.2 cm,   PR = 12.4 cm,   BC = 21.7 cm.
Find the scale factor:
Scale factor = PR / AC = 12.4 / 5.2 = 2.3846...

Now apply the same scale factor to AB (which corresponds to PQ):
PQ = AB × 2.3846... = 9.1 cm

Let x = 0.4444... (the recurring decimal).

Multiply by 10 to move the decimal point one place:
10x = 4.4444...

Subtract the original equation from this:
10x − x = 4.444... − 0.444...
9x = 4

Divide by 9:
x = 4 / 9

We know two sides and the angle between them: 22.3 cm, 27.6 cm, and angle 25°.
When we have this information, we use the formula:
Area = ½ × side1 × side2 × sin(angle)

Plug in the numbers:
Area = ½ × 22.3 × 27.6 × sin(25°)

Calculate (using calculator):
sin(25°) = 0.4226...
Area = 0.5 × 22.3 × 27.6 × 0.4226
= 130.0...

For a 2×2 matrix A = [a b / c d], the determinant is written as det(A).
It is calculated by det(A) = ad − bc.
A matrix has an inverse only when det(A) ≠ 0.

Here A = [3 c / −2 8], so
det(A) = (3)(8) − (c)(−2) = 24 + 2c.
So the inverse exists provided 24 + 2c ≠ 0.

Use the inverse formula for a 2×2 matrix:
A⁻¹ = 1det(A)[d −b / −c a]
A⁻¹ = 124 + 2c[8 −c / 2 3]

To add fractions, we need a common denominator.
Find the LCM (lowest common multiple) of 12 and 20.
12 = 2² × 3,   20 = 2² × 5
LCM = 2² × 3 × 5 = 60

Convert each fraction to denominator 60:
1/12 = 5/60   (because 12 × 5 = 60, and 1 × 5 = 5)
7/20 = 21/60   (because 20 × 3 = 60, and 7 × 3 = 21)

Add the numerators:
5/60 + 21/60 = 26/60

Simplify: divide both numerator and denominator by 2:
26/60 = 13/30
13 and 30 have no common factor, so this is in simplest form.
As a mixed number: 13/30 (less than 1, so stays as an improper fraction)

The scale 1 : 20 000 means:
1 cm on the map = 20 000 cm in real life.
But we're dealing with area, not length. So we need to square the scale factor.

For area, the scale factor is squared:
Scale factor for area = 20 000² = 400 000 000

Multiply the map area by this squared scale factor:
Actual area = 1.6 × 400 000 000 = 640 000 000 cm²

Convert to square metres. Remember: 1 m = 100 cm, so:
1 m² = 100 × 100 = 10 000 cm²
Actual area = 640 000 000 / 10 000 = 64 000 m²

Formula for arc length:
Arc length = (angle / 360°) × (2 × π × radius)

Plug in the numbers:
AB = (38 / 360) × 2 × π × 25

Calculate step by step:
38 / 360 = 0.1055...
2 × π × 25 = 50π = 157.079...
AB = 0.1055... × 157.079... = 16.577...

Round to 4 significant figures:
16.577... → 16.58

"500 cm correct to the nearest cm" means the true length is between:
499.5 cm and 500.5 cm
We use the largest possible pole length to find the maximum number of rods.

"5.8 cm correct to the nearest mm" means each rod's true length is between:
5.75 cm and 5.85 cm
We use the shortest possible rod length to maximise the count.

To get the largest possible number of rods, divide the largest pole by the shortest rod:
500.5 / 5.75 = 87.043...

We can only have a whole number of rods. Take the integer part (floor):
87.043... → 87

Start dividing 2016 by small prime numbers:
2016 ÷ 2 = 1008
1008 ÷ 2 = 504
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63    (now we have 2&sup5;)

Continue with 63:
63 ÷ 3 = 21
21 ÷ 3 = 7    (now we have 3²)

And finally: 7 ÷ 7 = 1
Prime factors: 2, 2, 2, 2, 2, 3, 3, 7
Or: 2016 = 2&sup5; × 3² × 7

Standard form means: a × 10⊃n; where 1 ≤ a < 10.
Move the decimal point so only one non-zero digit is in front:
2016 = 2.016 × 10³

Part (a): x³y⁴ × x⁵y³
When multiplying powers with the same base: add the exponents.
x³ × x⁵ = x³⁺⁵ = x⁸
y⁴ × y³ = y⁴⁺³ = y⁷

Combine: x⁸y⁷

Part (b): (3p²m⁵)³
Apply the power to each factor inside the bracket:
(3p²m⁵)³ = 3³ × (p²)³ × (m⁵)³

Now simplify each power:
3³ = 27,   (p²)³ = p⁶,   (m⁵)³ = m¹⁵
So 27p⁶m¹⁵

We have all three sides (2.8, 3.6, 5.3 cm) and need to find angle p. p is the angle opposite the side of length 5.3 cm.
Use the Cosine Rule:
cos p = (b² + c² − a²) / (2bc)
where a is opposite p, and b, c are the other two sides.

Plug in: a = 5.3, b = 2.8, c = 3.6
cos p = (2.8² + 3.6² − 5.3²) / (2 × 2.8 × 3.6)

Calculate:
2.8² = 7.84,   3.6² = 12.96,   5.3² = 28.09
cos p = (7.84 + 12.96 − 28.09) / (2 × 2.8 × 3.6)
= −7.29 / 20.16
= −0.3615...

Find p using cos¹:
p = cos¹(−0.3615...) = 111.2°

For grouped data, we use the mid-point of each class interval as our best estimate of the average value in that group.
Mid-point = (lower bound + upper bound) / 2

Calculate each mid-point:
10<h≤20  →  mid = (10+20)/2 = 15
20<h≤40  →  mid = (20+40)/2 = 30
40<h≤50  →  mid = (40+50)/2 = 45
50<h≤60  →  mid = (50+60)/2 = 55
60<h≤90  →  mid = (60+90)/2 = 75

Multiply each mid-point by its frequency (fx) and add up:
7×15 = 105
15×30 = 450
27×45 = 1215
13×55 = 715
8×75 = 600
Σfx = 105+450+1215+715+600 = 3085

Divide total Σfx by total frequency:
Mean = 3085 / 70 = 44.07...
Rounded to 3 sf: 44.1 cm

This is a quadratic equation (has x²). Use the Quadratic Formula:
x = (−b ± √(b² − 4ac)) / 2a
Identify a, b, c from the equation:
a = 3,   b = −11,   c = 4

Calculate the discriminant (the bit inside the square root):
b² − 4ac = (−11)² − 4×3×4
= 121 − 48 = 73
73 > 0, so there are two real solutions.

Apply the formula:
x = (−(−11) ± √73) / (2×3)
= (11 ± 8.544...) / 6

Calculate both answers:
x&sub1; = (11 + 8.544) / 6 = 19.544 / 6 = 3.257... → 3.26
x&sub2; = (11 − 8.544) / 6 = 2.456 / 6 = 0.409... → 0.41

Look at the four boundary lines of the unshaded triangle. Each line gives one inequality.

Horizontal line at y = 2. The unshaded region is below this line.
y < 2

Vertical line at x = −2. The unshaded region is to the right of this line.
x ≥ −2

Slanted line going through (−2, 0) and (4, 3).
Slope = (3−0)/(4−(−2)) = 3/6 = 1/2.
Equation: y − 0 = (1/2)(x − (−2))
So: y = (1/2)x + 1.
The unshaded region is above this line.
y ≥ (1/2)x + 1

Slanted line going through (0, 3) and (4, 1).
Slope = (1−3)/(4−0) = −2/4 = −1/2.
Equation: y − 3 = (−1/2)(x − 0)
So: y = −(1/2)x + 3.
The unshaded region is below this line.
y ≤ −(1/2)x + 3

Part (a): The formula is an² + bn.
Replace n with 3:
3rd term = a(3)² + b(3) = 9a + 3b

Part (b): We have two equations.
3rd term = 21  →  9a + 3b = 21    ...(1)
6th term = 96  →  36a + 6b = 96    ...(2)

Solve by elimination. Multiply equation (1) by 2:
18a + 6b = 42    ...(3)

Subtract (3) from (2):
36a + 6b − (18a + 6b) = 96 − 42
18a = 54
a = 54 / 18 = 3

Substitute a = 3 back into equation (1):
9(3) + 3b = 21
27 + 3b = 21
3b = 21 − 27 = −6
b = −2

Check: 3rd term = 9(3) + 3(−2) = 27 − 6 = 21 ✓
6th term = 36(3) + 6(−2) = 108 − 12 = 96 ✓

Part (a): Probabilities on a branch always add to 1.
P(walks) + P(cycles) = 1
P(walks) = 1 − 1/3 = 2/3

Part (b): Fill in the tree diagram (shown above).
Cycles branch: P(late) = 1/8,   P(not late) = 7/8
Walks branch: P(late) = 3/8,   P(not late) = 5/8

Part (c)(i): P(cycles and late) = P(cycles) × P(late | cycles)
= 1/3 × 1/8 = 1/24

Part (c)(ii): P(not late) — add the two "not late" branches:
P(cycles & not late) = 1/3 × 7/8 = 7/24
P(walks & not late) = 2/3 × 5/8 = 10/24
P(not late) = 7/24 + 10/24 = 17/24