From 21:50 to midnight = 2 h 10 min. From midnight to 05:18 = 5 h 18 min.

Total = 7 h 28 min

√5.76 = 2.4. 2.8³ = 21.952.

2.4 + 21.952 = 24.352

Collect m-terms: 4m−m=3m. Collect k-terms: 7k+3k=10k.

3m + 10k

The main horizontal strip of the net unfolds into the four side faces of the cuboid, wrapping all the way around the base — so its total width equals the perimeter of the 10×4 base:

a = 2(10+4) = 28 cm

The top flap folds up to become the top face (10×4), attached along one of the “length” (10 cm) sections of the strip, so its width matches the cuboid’s length:

b = 10 cm

The depth of that same top flap matches the cuboid’s width:

c = 4 cm

The main strip’s own height equals the cuboid’s height (5 cm), and the shaded Base flap (which becomes the bottom face) extends a further 4 cm below it, so the total vertical measurement on the right of the net is:

d = 5 + 4 = 9 cm

1. (a) (3÷20) × 360° = 54°
2. (b) P(not blue) = 17÷20 = 0.85 (17/20)

AB = B − A = (−3−7, 4−1) = (−10, 3)

As one variable increases, so does the other — this is a positive correlation

1. (a) n=1: 1−3=−2. n=2: 4−3=1. n=3: 9−3=6.
   −2, 1, 6
2. (b) Each term is 3 times the previous one (a geometric sequence with common ratio 3, starting at 1): nth term = 3ⁿ⁻¹

Substitute y=3: 3 = 2x−5 → 2x=8 → x=4

P = (4, 3)

Area of a trapezium = ½(sum of parallel sides) × height = ½(5.3+8.7) × 3.8

= ½(14) × 3.8 = 7 × 3.8 = 26.6 cm²

1¼ = &frac54;. Common denominator of 4 and 6 is 12: &frac54;=&frac{15}{12}, ⅚=&frac{10}{12}

&frac{15}{12} − &frac{10}{12} = 5/12

40 × 0.35 = 14

Since the bearing 107° is less than 180°, add 180° to find the back bearing: 107+180 = 287°

For the whole train to completely cross, the front must travel the bridge length plus the entire train length: 1750+480 = 2230 m

Speed = 55 km/h = 55 000÷3600 = 15.278 m/s

Time = 2230 ÷ 15.278 = 146 s (nearest second)

1. (a)(i) Comparing vertex (1,−1) on A with the corresponding vertex (−5,−1) on B: testing a reflection in a vertical line x=k, using the rule x′=2k−x: 2k−1=−5 gives k=−2. Checking the other vertices confirms this works throughout.
   Reflection in the line x = −2
2. (a)(ii) Triangle C is a smaller copy of A with the same angles, positioned in a different part of the grid — this is an enlargement. To find the exact scale factor and centre: measure the ratio of corresponding side lengths for the scale factor, then join at least two pairs of corresponding vertices with straight lines and extend them — the point where they cross is the centre of enlargement.
   Enlargement, scale factor and centre to be confirmed from the exact grid positions
3. (b) Using the rotation rule for 90° clockwise about a centre (a,b): a point (x,y) maps to (a+(y−b), b−(x−a)). Using centre (1,3) and A’s vertices (1,−1), (5,−1), (1,2):
   (1,−1) → (1+(−1−3), 3−(1−1)) = (−3, 3)
   (5,−1) → (1+(−1−3), 3−(5−1)) = (−3, −1)
   (1,2) → (1+(2−3), 3−(1−1)) = (0, 3)
   Plot these three points and join them to draw the rotated triangle.

𝒰 = {1,2,3,4,5,6,7,8,9,10}. P (even) = {2,4,6,8,10}. Q (multiples of 5) = {5,10}.

P∩Q (even AND multiple of 5) = {10}

P only = {2, 4, 6, 8}

Q only = {5}

Outside both circles = {1, 3, 7, 9}

Using the midpoint of each class:

Total = 3520+6875+8960+8330 = 27 685, over 200 people.

Mean = 27 685÷200 = 138 cm (3 s.f., exact value 138.425 cm)

Numeric HCF: 28=2²×7, 98=2×7². HCF(28,98) = 2×7 = 14

x-power: take the lower power, x³

14x³

Distance = area under the graph, split into three sections:

Rising (0 to 20s): ½ × 20 × 15 = 150 m

Constant (20 to 140s): 120 × 15 = 1800 m

Falling (140 to 190s): ½ × 50 × 15 = 375 m

Total = 150+1800+375 = 2325 m

Area = ½ × AB × CB × sin(B) = ½ × 4.9 × 5.6 × sin(23°)

= 13.72 × 0.39073 = 5.36 cm² (3 s.f.)

1. (a) The fifth root of 3 is 3^1/5: h = 0.2 (1/5)
2. (b) (4x³)³ = 4³ × x⁹ = 64x⁹

y = k÷(x+3)². Substituting x=5,y=0.375: (5+3)²=64, so 0.375=k÷64 → k=24

y = 24 / (x + 3)²

1. (a) The cosine curve starts at (0°,1), falls to (90°,0), continues down to (180°,−1), rises back to (270°,0), and returns to (360°,1) — a smooth wave shape.
2. (b) x = cos⁻¹(0.294) = 72.9°. Since cosine is also positive in the fourth quadrant, the second solution is 360−72.9 = 287.1°
   x = 72.9° or x = 287.1° (1 d.p.)

Expanding (x+b)² = x²+2bx+b². Comparing with x²−16x+a: matching the x-terms gives 2b=−16, so b=−8.

Then a must equal b² = (−8)² = 64 for the constant terms to match.

a = 64   b = −8

Total buttons = 2+5+6 = 13. It’s easier to find P(same colour) first, then subtract from 1.

P(both green) = (2/13)(1/12) = 2/156

P(both red) = (5/13)(4/12) = 20/156

P(both blue) = (6/13)(5/12) = 30/156

P(same colour) = (2+20+30)/156 = 52/156 = ⅓

P(different colours) = 1 − ⅓ = 2/3

Midpoint of AB = ((6+2)/2, (1+7)/2) = (4, 4)

Gradient of AB = (7−1)/(2−6) = 6/(−4) = −1.5

Perpendicular gradient = negative reciprocal of −1.5 = ⅔

Using the midpoint (4,4): y−4 = ⅔(x−4) → y = ⅔x − &frac83; + 4 = ⅔x + &frac43;

y = (2/3)x + 4/3