Correct Answer: D. median

Explanation

(a) Width = x m; Length = 2x m; Height = (x - 4) m.

Total surface area = 2Lh + WL + 2Wh

= \(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\)

= \(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\)

= \(8x^{2} - 24x = 1440\)

= \(x^{2} - 3x - 180 = 0\)

= \((x - 15)(x + 12) = 0\)

\(x = 15m\)

Length = 2(15m) = 30m.

(b) \(I = \frac{PRT}{100}\)

\(I_{P} + I_{Q} = I_{T}\)

\(\frac{3000 \times x \times 1}{100} + \frac{2000 \times y \times 1}{100} = 640\)

\(30x + 20y = 640 \implies 3x + 2y = 64 .... (1)\)

\(\frac{2000 \times x \times 1}{100} + \frac{3000 \times y \times 1}{100} = 660\)

\(20x + 30y = 660 \implies 2x + 3y = 66 .... (2)\)

To eliminate x , multiply (1) by 2 and (2) by 3.

\(6x + 4y = 128 ... (1)\)

\(6x + 9y = 198 .... (2)\)

(2) - (1) : \(9y - 4y = 198 - 128 \implies 5y = 70\)

\(y = 14\)

\(2x + 3y = 66 \implies 2x + 3(14) = 66\)

\(2x + 42 = 66 \implies 2x = 24\)

\(x = 12\)

\(\therefore x = 12 ; y = 14\).

Correct Answer: d) 1/2

Correct Answer: A. 5 or \( \frac{3}{2} \)

Explanation

The fraction is undefined when the denominator is equal to zero

\(2x^2 - 13x + 15 = 0\\

2x^2 - 3x - 10x + 15\\

x(2x-3)-5(2x-3) = 0\\

(2x-3)(x-5)=0\\

x = \frac{3}{2} or x = 5\)

Explanation

(a) Mean \(\bar{x} = \frac{\sum x}{n} = 8\)

\(\frac{1 + 2 + x + 11 + y + 14}{6} = 8\)

\(28 + x + y = 48 \implies x + y = 20 .... (1) \)

The middle numbers are x and 11 therefore, \(\frac{x + 11}{2} = 9\)

\(x + 11 = 18 \implies x = 18 - 11 = 7\)

Putting x = 7 in (1),

\(7 + y = 20 \implies y = 20 - 7 = 13\)

(x, y) = (7, 13).

(b) \(\frac{LM}{LP} = \frac{3}{4}\)

\(\frac{\text{Area of LMN}}{\text{Area of LPQ}} = \frac{3^{2}}{4^{2}}\)

Area of \(\Delta\) LPQ = \(\frac{16}{9} \times 18 = 32 cm^{2}\)

\(\therefore \text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\)

Explanation

\(3x^{2} + 5x + 1 = 0\)

\(x^{2} + \frac{5}{3}x + \frac{1}{3} = 0\)

\(\alpha \beta = \frac{c}{a}\)

= \(\frac{1}{3}\)

\(\alpha + \beta = \frac{-b}{a}\)

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

\((\alpha + \beta)^{3} = \alpha^{3} + 3\alpha^{2} \beta + 3\alpha \beta^{2} + \beta^{3}\)

\(\therefore \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha^{2} \beta - 3\alpha \beta^{2}\)

= \((\alpha + \beta)^{3} - 3\alpha \beta (\alpha + \beta)\)

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

= \((-\frac{125}{27}) + \frac{5}{3}\)

\(\therefore (\alpha^{3} + \beta^{3}) = -\frac{80}{27}\)

\(27(\alpha^{3} + \beta^{3}) = 27(-\frac{80}{27})\)

= -80.

Correct Answer: B. 63°

Correct Answer: D. which is equidistant from the points X and Y

Correct Answer: A. {x: -1 ? x ? 3}

Explanation

Let x\(^2\) - 2x - 3 = 0;

(x+1)(x-3) = 0

The zeros of the function are x = -1 and 3

The solution set is { x: -1 ≤ x ≤ 3}

Correct Answer: A. 2.5

Explanation

\(\begin{array}{c|c} x & d = x - \bar{x} & |d| \\ \hline 10 & -4.5 & 4.5 \\ 12 & -2.5 & 2.5\\ 14 & -0.5 & 0.5\\ 15 & 0.5 & 0.5\\ 17 & 2.5 & 2.5\\ 19 & 4.5 & 4.5\\ \hline \sum x = 87 & & \sum|d| = 15\end{array}\)

\(\bar{x} = \frac{\sum x}{n}\)

= \(\frac{87}{6}\)

= 14.5