Correct Answer: C. 13cm

Explanation

perimeter = 2(l + b) = 32

l + b = \(\frac{32}{2}\)

l + b = 16

b = 16 - 1.......(1)

Area = l + b = 39

lb = 39 .....(2)

put (1) into (2)

l(16 - 1) = 39

16l - l² = 39

l² = 13l - 3l + 39 = 0

l(l - 13) - 3(1 - 13) = 0

(l - 3)(l + 130 = 0

l - 3 = 0 or l - 13 = 0

l = 3cm or l = 13cm; The length in 13cm

Correct Answer: C. 297cm²

Explanation

T.S.A of s closed cylinder = \(2\pi r(r+h)\\

=\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right)=\frac{22}{1}\left(\frac{27}{2}\right)=27 \times 11=297cm^2\)

Correct Answer: D. \(9\frac{9}{16}\)

Explanation

\(T_{n} = ar^{n-1}\) (for an exponential sequence)

\(T_{1} = 36 = a\)

\(T_{2} = ar = 36r = p\)

\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)

\(T_{4} = ar^{3} = 36r^{3} = q\)

\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)

\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)

\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)

\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)

Correct Answer: D. 11

Correct Answer: A. (2a-3b) (2a+3b)

Explanation

(2a-3b) (2a+3b)

4a\(^2\) + 6ab - 6ab - 9b\(^2\)

4a\(^2\) - 9b\(^2\)

Correct Answer: D. 5, 4

Correct Answer: A. \(\frac{1}{9}\)

Explanation

\(y \propto \frac{1}{\sqrt{x}}\)

\(y = \frac{k}{\sqrt{x}}\)

When x = 16, y = 2.

\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)

\(k = 8\)

\(y = \frac{8}{\sqrt{x}}\)

When y = 24,

\(24 = \frac{8}{\sqrt{x}}\)

\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)

\(\therefore x = (\frac{1}{3})^2\)

\(x = \frac{1}{9}\)

Correct Answer: B. \(\frac{1}{x^2}\)

Explanation

(a) 7 - \(\frac{6}{x} = 3 - 2x\)

Simplifying; \(x^2 + 2x - 2 = 0\)

x = 1 or x = -3

Substituting for x; y = 3 - 2(1) = 3 - 2 = 1 or y = 3 - 2(-3) = 3 + 6 = 9

The coordinate of the two points are (x y) = (1, 1), (-3, 9)

(b) (\(\frac{1 - 3}{2}, \frac{1 + 9}{2}\)) = (-1, 5)

The gradient of the point of intersection ; \(\frac{9 - 1}{-3 -1} = \frac{8}{-4}\) = -2

The gradient of the perpendicular bisector; \(\frac{1}{2}\)

Thus, the equation of the perpendicular bisector; y - 5 = \(\frac{1}{2}\) (x + 1)

Therefore, 2y - x - 11 = 0

Correct Answer: D. {-1, 5, 13, 49}

Explanation

\(P = {x : \text{x is a factor of 6}} \implies P = {1, 2, 3, 6}\)

\(g(x) = x^{2} + 3x - 5\)

\(g(1) = 1^{2} + 3(1) - 5 = 1 + 3 - 5 = -1\)

\(g(2) = 2^{2} + 3(2) - 5 = 4 + 6 - 5 = 5\)

\(g(3) = 3^{2} + 3(3) - 5 = 9 + 9 - 5 = 13\)

\(g(6) = 6^{2} + 3(6) - 5 = 36 + 18 - 5 = 49\)

\(\therefore Range(g(x)) = {-1, 5, 13, 49}\)