b) Given that y = a + b/x, make x the subject of the formula. c) Transpose l = a + (n−1)d for n.

$$x = \frac{b}{y - a}$$

The subject of the formula is $\boxed{x = \frac{b}{y - a}}$.

c) Transpose $l = a + (n-1)d$ for $n$.

$$n = \frac{l - a}{d} + 1$$

Alternatively, combine the terms on the right side.

$$n = \frac{l - a + d}{d}$$

The transposed formula is $\boxed{n = \frac{l - a + d}{d}}$.

QUESTION 7

a) In a right-angled triangle ABC, angle C = 90°, angle A = 38°, and the hypotenuse AB = 14 cm. Calculate the lengths of sides BC and AC.

$$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} \implies \sin 38^\circ = \frac{BC}{14}$$ $$BC = 14 \sin 38^\circ$$ $$BC = 14 \times 0.61566... \approx 8.6192...$$ $$BC \approx 8.62 \text{ cm}$$

$$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} \implies \cos 38^\circ = \frac{AC}{14}$$ $$AC = 14 \cos 38^\circ$$ $$AC = 14 \times 0.78801... \approx 11.0321...$$ $$AC \approx 11.03 \text{ cm}$$

The length of BC is approximately $\boxed{8.62 \text{ cm}}$ and the length of AC is approximately $\boxed{11.03 \text{ cm}}$.

b) Solve the equation $2\sin\theta - 1 = 0$ for $0^\circ \le \theta \le 360^\circ$.

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

In Quadrant I:

$$\theta_1 = \theta_R = 30^\circ$$

In Quadrant II:

$$\theta_2 = 180^\circ - \theta_R = 180^\circ - 30^\circ = 150^\circ$$

The solutions are $\boxed{\theta = 30^\circ \text{ or } \theta = 150^\circ}$.

c) Simplify $\log 50 + \log 4 - \log 2$, giving your answer as a single logarithm, then evaluate it.

$$\log(100) = 2$$

The single logarithm is $\boxed{\log(100)}$ and its value is $\boxed{2}$.

QUESTION 8

a) In triangle DEF, angle D = 75°, angle E = 65°, and side DE = 9 cm. Determine the length marked x in the figure below, where x = EF.

$$\frac{EF}{\sin D} = \frac{DE}{\sin F}$$ $$\frac{x}{\sin 75^\circ} = \frac{9}{\sin 40^\circ}$$

$$x = \frac{9 \times 0.965925...}{0.642787...} = \frac{8.69332...}{0.642787...} \approx 13.5244...$$ $$x \approx 13.52 \text{ cm}$$

The length of EF (x) is approximately $\boxed{13.52 \text{ cm}}$.

b) In triangle PQR, PQ = 15 cm, PR = 5 cm, and angle QPR = 75°. Calculate the length of QR.
We use the Cosine Rule: $a^2 = b^2 + c^2 - 2bc \cos A$. Here, we want to find QR (let's call it $p$), and we know $q=PR=5$, $r=PQ=15$, and the included angle $P=75^\circ$.

$$QR^2 = PR^2 + PQ^2 - 2(PR)(PQ) \cos(\angle QPR)$$ $$QR^2 = 5^2 + 15^2 - 2(5)(15) \cos(75^\circ)$$

$$QR^2 = 25 + 225 - 150 \cos(75^\circ)$$ $$QR^2 = 250 - 150 \cos(75^\circ)$$

$$QR^2 = 250 - 150(0.258819...)$$ $$QR^2 = 250 - 38.8228...$$ $$QR^2 = 211.1771...$$

$$QR = \sqrt{211.1771...} \approx 14.5320...$$ $$QR \approx 14.53 \text{ cm}$$

The length of QR is approximately $\boxed{14.53 \text{ cm}}$.