Explanation

(a) ∠XYZ = \(\frac{1}{2}\) (146) = 73º

∠OZX = 34º

∠OZY = 73 - 34

∠OYZ = ∠OZY = 39

(b) x = y - 7

42 + 38 + 57 + x + x + y + 2x - 15 + 3x - y = 360

122°+7x = 360° 7x 238°

x = 34

: 34 = y - 7

34 + 7 = y

41 = y

Explanation

(a)(i) ∠PSQ = ∠QPS

∠PQS = \(\frac{2}{8}\) x 180°

=45°

∠SPQ = ∠PSQ

= \(\frac{3}{8}\) x 180 = 67.5°

∠SQR = ∠SPR= 26°

∠PQR = ∠PQS + ∠SQR

= 45°+ 26°

= 71°

(ii) ∠RPQ = ∠SPQ - ∠SPR

= 67.5°- 26°

= 41.5°

(iii) ∠PRQ = 180° - 41.5° - 71°

= 67.5°

(b) √29 = √{ (7 - 5)\(^2\) + (3-x)\(^2\)}

√29 = √{ 2\(^2\) + (3 - x)\(^2\)}

29 = 4 + (3 - x)\(^2\)

25 = (3 - x)\(^2\)

x\(^2\) -6x - 16

(x + 2)(x - 8) = 0

x = -2 or 8

Correct Answer: B. \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)

Explanation

\(a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\); \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)

\(\implies 2 \times a = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) and \(\frac{1}{4} \times b = \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\)

\(\therefore 2a - \frac{1}{4}b = \begin{pmatrix} 4 - \frac{-1}{4} \\ 6 - 1 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)

Correct Answer: B. 17°

Explanation

The straight line MN = 180 = 2(x+30^o) + 86^o =>

180 = 2x + 60^o + 86^o => 180^o = 2x + 146^o => 2x = 180 - 46 = 34^o => x = 34/2 = 17^o

Correct Answer: D. \(\frac{y}{4x+x-3}\)

Explanation

2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

4y \(\frac{dy}{dx}\) + 0 = 3 \(\frac{dy}{dx}\) - x \(\frac{dy}{dx}\) + y

{4x+x-3} \(\frac{dy}{dx}\) = y

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

Explanation

(a) The formula for the first principles method of calculating derivatives is

\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\)

\(f(x) = (2x + 3)^{2} = 4x^{2} + 12x + 9\)

\(f(x + \Delta x) = 4(x + \Delta x)^{2} + 12(x + \Delta x) + 9\)

= \(4x^{2} + 8x \Delta x + 4 (\Delta x)^{2} + 12x + 12\Delta x + 9\)

\(\lim \limits_{\Delta x \to 0} \frac{(4x^{2} + 8x \Delta x + 4(\Delta x)^{2} + 12x + 12\Delta x + 9) - (4x^{2} + 12x + 9)}{\Delta x}\)

= \(\lim \limits_{\Delta x \to 0} \frac{4(\Delta x)^{2} + (8x + 12)\Delta x}{\Delta x}\)

= \(\lim \limits_{\Delta x \to 0} 4 \Delta x + (8x + 12)\)

\(\frac{\mathrm d y}{\mathrm d x} = 8x + 12\)

(b) \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)

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

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

= \(\int_{1} ^{2} (x - 1 + 2x^{2}) \mathrm {d} x\)

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

= \((\frac{2^{2}}{2} - 2 - \frac{2}{2}) - (\frac{1^{2}}{2} - 1 - \frac{2}{1})\)

= \((2 - 2 - 1) - (\frac{1}{2} - 1 - \frac{2}{1})\)

= \(-1 - (-\frac{5}{2}) = \frac{3}{2}\)

Correct Answer: B. -2

Explanation

\(\frac{64}{27} = (\frac{3}{4})^{t-1}\)

\((\frac{3}{4})^t = \frac{64}{27} \times \frac{3}{4} = \frac{16}{9}\)

\((\frac{3}{4})^t = (\frac{9}{16})^{-1}\)

\((\frac{3}{4})^t = (\frac{3}{4})^{-2}\)

t = -2

Correct Answer: A. 7.5

Explanation

Since it's in increasing order, let's have some scenarios. Let's say a,b are both 5, then that means e,f at can be 5,15 with f maxing out its value up to 10,10 with e maxing its value. That leaves c,d must be 5 with f maxing out its value and c,d being in a range of 5?10 with e maxing its value and not violating the set. Now, lets say a,b are 0,10respectively, this leaves e,f only being 10 and c,d being 10 as well. Let's find the range of c,d. If c,d were both 5, that means the median of them is 5. If c,d are both 10, that means the median of them is 10. 10 is highest median of both c,d and 5 is the lowest median of c,d. We need to find a number in the answer choices that fit this description. Answer is 7.5.

Correct Answer: C. 38cm³

Explanation

Volume of a cone = \(\frac{1}{3} \pi r^2h\)

h² = 5² = 3²

= 25 - 9 = 16

h = \(\sqrt{16}\)

h = 4cm

v = \(\frac{1}{3} \times \frac{22}{7} \times 3^2 \times 4\)

\(\frac{1}{3} \times \frac{22}{7} \times 9 \times 4\)

= \(\frac{22 \times 3 \times 4}{7}\)

= 37.7cm³

= 38cm³

Explanation

he will be late to office = 2/5

he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

\(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\)

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

\(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\)

p = 0.0467

(c) be late to office = 2/5

\(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\)

p = 0.0041