Correct Answer: A. (4, 2)

Explanation

The point of intersection for the two lines exists at the point where the two lines are equal to each other. Make anyone of the variables the subject of the formula and equate the two lines to each other and solve for the coordinates of the point of intersection.

Given lines \(2y + 3x - 16 = 0\) and \(7y - 2x - 6 = 0\) , making x the subject of the formula:

Line 1 : \(2y + 3x - 16 = 0 \implies 3x = 16 - 2y\)

\(\therefore x = \frac{16}{3} - \frac{2}{3}y\)

Line 2 : \(7y - 2x - 6 = 0 \implies -2x = 6 - 7y\)

\(\therefore x = \frac{7}{2}y - 3\)

Equating them together and solving, we have:

\(\frac{16}{3} - \frac{2}{3}y = \frac{7}{2}y - 3 \implies \frac{16}{3} + 3 = \frac{7}{2}y + \frac{2}{3}y\)

\(\frac{25}{3} = \frac{25}{6}y \therefore y = 2\)

Putting y = 2 in the equation \(3x = 16 - 2y\), we have

\(3x = 16 - 2(2) = 16 - 4 = 12 \implies x = 4\)

The coordinate of P is (4, 2).

Correct Answer: D. \(\frac{15}{16}\)

Explanation

P(at least one head) = 1 - P(4 tails)

Let \(p = \frac{1}{2}\) = probability of head and \(q = \frac{1}{2}\) = probability of tail.

\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)

P(4 tails) = \(q^{4} = (\frac{1}{2})^{4} = \frac{1}{16}\)

P(at least one head) = \(1 - \frac{1}{16} = \frac{15}{16}\)

Correct Answer: D. 1/2 (Upper Quartile - Lower Quartile)

Explanation

"Semi" means "half"

Interquartile range = Upper quartile - Lower quartile

Correct Answer: C. \(\frac{-12i}{13} + \frac{5j}{13}\)

Explanation

\(\hat{n} = \frac{\overrightarrow{n}}{|n|}\)

\(\hat{n} = \frac{-12i + 5j}{\sqrt{(-12)^{2} + (5)^{2}}}\)

= \(\frac{-12i}{13} + \frac{5j}{13}\)

Explanation

\(\frac{x^3+8}{x+2}\).

x\(^3\) + 8 = x\(^3\) + 23

recall that

x\(^3\) + y\(^3\) = (x+y)(x\(^2\) - xy + y\(^2\))

x = x, y = 2

x\(^3\) + 8 = (x+2)(x\(^3\) - 2(x) + 22)

\(\frac{x^3+8}{x+2} = \frac{(x+2)(x^3 - 2(x) + 22)}{x+2}\) → x\(^2\) - 2x + 4

x = -2

(-2)\(^2\) - 2(2) + 4 = 4 - 4 + 4

= 4

Correct Answer: A. 70 kg ms\(^{-1}\)

Explanation

Change in momentum

= F x t = 35N x 2

= 70kg ms\(^{-1}\)

Correct Answer: A. 324

Explanation

\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)

Let the power on x be t, so that the power on \(x^{-2}\) = 9 - t

\((x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3\)

\(3t = 3 + 18 = 21 \therefore t = 7\)

To obtain the coefficient of \(x^{3}\), we have

\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)

= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)

Correct Answer: B. \(x^{2} - 4x + 5\)

Explanation

\(f(x) = x^{2} + 1\) and \(g(x) = x - 2\)

\(f o g = f(g(x)) = f(x - 2) = (x - 2)^{2} + 1 \)

= \(x^{2} - 4x + 4 + 1 = x^{2} - 4x + 5\)

Explanation

\(^9C_x = 4[^7C_{x - 1}]\)

\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)

\(={9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)

\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)

Cancel out the common terms

\(=\frac{9 \times 8}{x(9 - x)}=\frac{4}{1}\)

\(=x(9-x)=9*8\)

\(=9x-x^2=72\)

\(=x^2-9x+72=0\)

\(=x^2-6x-3x+18=0\)

\(=x(x-6)-3(x-6)=0\)

\(=(x-6)(x-3)=0\)

\(\therefore x\) = 6 or 3

Explanation

(a) \(\int_{1} ^{4} \frac{x(3x -2)}{2\sqrt{x}} \mathrm {d} x\)

= \(\int_{1} ^{4} \frac{1}{2} x(x^{-\frac{1}{2}})(3x - 2) \mathrm {d} x\)

= \(\frac{1}{2} \int_{1} ^{4} (3x^{\frac{3}{2}} - 2x^{\frac{1}{2}}) \mathrm {d} x\)

= \(\frac{1}{2} [\frac{3x^{\frac{5}{2}}}{\frac{5}{2}} - \frac{2x^{\frac{3}{2}}}{\frac{3}{2}}]_{1} ^{4}\)

= \(\frac{1}{2} [\frac{6x^{\frac{5}{2}}}{5} - \frac{4x^{\frac{3}{2}}}{3}]_{1} ^{4}\)

= \(\frac{1}{2} [\frac{6(4^\frac{5}{2})}{5} - \frac{4(4^\frac{3}{2})}{3}] - \frac{1}{2} [\frac{6(1^\frac{5}{2})}{5} - \frac{4(1^\frac{3}{2})}{3}]\)

= \(\frac{1}{2} [\frac{6 \times 32}{5} - \frac{4 \times 8}{3}] - \frac{1}{2} [\frac{6}{5} - \frac{4}{3}]\)

= \(\frac{1}{2}[38.4 - 10.67] - \frac{1}{2} [1.2 - 1.33]\)

= \(\frac{1}{2} [27.73 + 0.13]\)

= \(13.93\)

(b) \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\)

\(x^{2} + y^{2} - 4x + \frac{5}{2}y = 5\)

\(x^{2} - 4x + [\frac{1}{2} (-4)]^{2} + y^{2} + \frac{5}{2}y + [\frac{1}{2} (\frac{5}{2})]^{2} = 5 + 4 + \frac{25}{16} = \frac{169}{16}\)

\((x - 2)^{2} + (y + \frac{5}{2})^{2} = \frac{169}{16}\)

(i) coordinates of the centre = \((2, -\frac{5}{4})\)

(ii) Radius = \(\sqrt{\frac{169}{16}} = \frac{13}{4}\)