Substitution and calculator brackets

(a) $1$

1. $4 + (-3) = 1$.

• B1 The correct answer, $1$.

(b) $18$

1. Put the values in, keeping $-3$ in a bracket.
   $$3 \times 4 - 2 \times (-3) = 12 - (-6) = 12 + 6 = 18$$

• M1 Writing $12$ and $-6$ (or $3 \times 4 - 2 \times (-3)$).
• A1 The correct answer, $18$.

(c) $9$

1. $b^2 = (-3)^2 = (-3) \times (-3) = 9$.

• B1 The correct answer, $9$.

(a) $-22$

1. $$3\times (-4)-2\times 5$$
2. Substitute the values: 3($-4$) $-$ 2(5).
3. Multiply first: $-12$ $-$ 10 = $-22.$

• M1 Establishing $3\times (-4)-2\times 5$ or an equivalent valid method.
• A1 Correct answer: $-22$

(a) £109

1. $C = 25 + 12 \times 7 = 25 + 84 = 109$.

• M1 Writing $25 + 12 \times 7$.
• A1 The correct answer, £109.

(b) $12$ days

1. $169 = 25 + 12d$, so $12d = 144$ and $d = 12$.

• M1 Subtracting 25 and dividing by 12 (in either form).
• A1 The correct answer, $12$ days.

(a) $-1.9$

1. $$v = 12.5 + (-3.2) \times 4.5 = 12.5 - 14.4 = -1.9$$

• M1 Working out $at = -14.4$.
• A1 The correct answer, $-1.9$.

(b) $7.2$

1. Square $v$ first, with the bracket.
   $$(-6)^2 = 36$$
2. $$E = 0.5 \times 0.4 \times 36 = 7.2$$

• M1 Finding $v^2 = 36$.
• A1 The correct answer, $7.2$.

(a) $24$

1. Square the whole negative input.
   $$(-2)^{2}$$
2. Subtract twice the product, keeping both signs.
   $$4-2\times (-2)\times 5$$
3. Therefore $24$.

• M1 Square the whole negative input.
• M1 Subtract twice the product, keeping both signs.
• A1 Correct answer: $24$

(b) $9$

1. Add the signed inputs before squaring.
   $$(-2+5)^{2}$$
2. Therefore $9$.

• M1 Add the signed inputs before squaring.
• A1 Correct answer: $9$

(a) $3$

1. Substitute with brackets around the negative acceleration.
   $$4\times 3+0.5\times (-2)\times 3^{2}$$
2. Therefore $3$.

• M1 Substitute with brackets around the negative acceleration.
• A1 Correct answer: $3$

(b) In $at^2$, only t is squared and the term is negative. In $(at)^2$, a is also squared, giving a positive result.

1. In $at^2$, only t is squared and the term is negative. In $(at)^2$, a is also squared, giving a positive result.

• C1 Correct conclusion with supporting reasoning: In $at^2$, only t is squared and the term is negative. In $(at)^2$, a is also squared, giving a positive result.

(a) $-4.5$ °C

1. Substitute h = 30, multiplying before subtracting.
   $$15-0.65\times 30$$
2. Therefore $-4.5$ °C.

• M1 Substitute h = 30, multiplying before subtracting.
• A1 Correct answer: $-4.5$ °C

(b) $40$

1. Substitute T = −11 into the formula.
   $$15-0.65h=-11$$
2. Rearrange and divide.
   $$26/0.65$$
3. Therefore $40$.

• P1 Substitute T = −11 into the formula.
• P1 Rearrange and divide.
• A1 Correct answer: $40$

(a) $-3$

1. $$s = 5 \times 3 + \tfrac{1}{2} \times (-4) \times 3^2 = 15 + (-2) \times 9 = 15 - 18 = -3$$

• M1 Finding $ut = 15$ and $\frac{1}{2}at^2 = -18$.
• A1 The correct answer, $-3$.

(b) $u = 4$

1. Substitute the known values.
   $$40 = 4u + \tfrac{1}{2} \times 3 \times 16 = 4u + 24$$
2. Solve.
   $$4u = 16 \Rightarrow u = 4$$

• P1 Substituting to get $40 = 4u + 24$.
• P1 Rearranging to $4u = 16$.
• A1 The correct answer, $u = 4$.