Correct Answer: C. 3y = -4x + 12

Explanation

The gradient of two parallel lines are equal.

m1 = m2

6i + 8j is parallel to →OP

m = \(\frac{y_2 - y_1}{x_2 - x_1}\)

gradient(m) of 6i + 8j

m = \(\frac{8}{6}\) = \(\frac{4}{3}\)

\(\frac{y}{x}\) = \(\frac{4}{3}\)

3y - 4x =0

|OP| = 80

|OP| = √(x\(^2\) + y\(^2\)) = 80

x\(^2\) + y\(^2\) = 80\(^2\)

x\(^2\) + y\(^2\) = 6400

\(y = \frac{4x}{3}\)

\(x^2 + ({\frac{4x}{3}})^2\) = 6400

\(x^2 + \frac{16x^2}{9}\) = 6400

multiply through by 9

9x\(^2\) + 16x2 = 57600

25x\(^2\) = 57600

x\(^2\) = \(\frac{57600}{25}\)

x\(^2\) = 2304

x = √2304

x = 48

\(y = \frac{4x}{3}\) → \(y = \frac{4*48}{3}\)

y = 64

→OP =xi + yj

→OP = 48i + 64j

8i - 6j is parallel to →OQ

gradient of 8i - 6j = \(\frac{-6}{8}\)

\(\frac{y}{x}\) = \(\frac{-6}{8}\)

y = \(\frac{-6x}{8}\)

\(x^2 + ({\frac{-6x}{8}})^2\) = 14400

\(x^2 + \frac{36x^2}{64}\) = 14400

multiply through by 64

64x\(^2\) + 36x\(^2\) = 921600

100x\(^2\) = 921600

divide both sides by 100

x\(^2\) = 9216

find the square root of both sides

x = √9216

x = 96

y = \(\frac{-6x}{8}\)

y = \(\frac{-6[96]}{8}\)

y = -72

→OQ = xi + yj

→OQ = 96i - 72j

→PQ = (96i - 72j) - (48i + 64j)

→PQ = (96-48)i + (-72-64)j

→PQ = 48i - 136j

|→PQ| = √(48\(^2\) + (-136)\(^2\))

|→PQ| = √(2304 + 18496)

|→PQ| = √20800

|→PQ| = 40√13

c = 40, k = 13

Correct Answer: D. 840N

Explanation

\(\text{Net force = Upward force - weight}\)

\(ma = F - mg \implies F = ma + mg \)

\(F = (80 \times 10) + (80 \times 0.5) = 800 + 40 = 840N\)

Correct Answer: A. (x - 2)

Explanation

Divide \((2x^{3} - 5x^{2} - x + 6)\) by \((2x^{2} - x - 3)\) to get the other factor. Be careful of sign conventions.

Correct Answer: (b) all it sides equal

Explanation

(a) \(3 \times 9^{1 + x} = 27^{-x}\)

\(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\)

\(3 \times 3^{2 + 2x} = 3^{-3x} \)

\(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\)

\(\implies 2x + 3 = - 3x\)

\(-5x = 3 \implies x = \frac{-3}{5}\)

(b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)

= \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\)

= \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\)

= \(\log_{10} \sqrt{10}\)

= \(\log_{10} 10^{\frac{1}{2}}\)

= \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)

Correct Answer: D. 3.4 cm

Correct Answer: D. 3y + x + 14 = 0

Explanation

\(3x - y + 11 = 0 \implies y = 3x + 11\)

\(Gradient = 3\)

Gradient of perpendicular line = \(\frac{-1}{3}\)

\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)

\(3(y + 5) = 1 - x\)

\(3y + x + 15 - 1 = 3y + x + 14 = 0\)

Correct Answer: A. 10 sin 5x cos 5x

Correct Answer: A. y = - \(\frac{5}{2}\)

Explanation

(0.25)\(^y\) = 32

2\(^{-2y}\) = 3\(^{2}\)

2 - 2y = 5

y = - \(\frac{5}{2}\)