Correct Answer: A. N3,900.00

Explanation

A = P + I

p + \(\frac{PRT}{100}\)

A = N3000 (1 + \(\frac{6 \times 5}{100}\))

= N3000(1 + 0.3)

A = N3000 (1.3)

= N3,900.00

Correct Answer: D. 24cm

Correct Answer: B. \(x^{2} + y^{2} + 6x + 16y - 23 = 0\)

Explanation

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

where (a, b) and r are the coordinates of the centre and radius respectively.

Given : \((a, b) = (-3, -8); r = 4\sqrt{6}\)

\((x - (-3))^{2} + (y - (-8))^{2} = (4\sqrt{6})^{2}\)

\(x^{2} + 6x + 9 + y^{2} + 16y + 64 = 96\)

\(x^{2} + y^{2} + 6x + 16y + 9 + 64 - 96 = 0\)

\(\implies x^{2} + y^{2} + 6x + 16y - 23 = 0\)

Explanation

(a) In \(\Delta ACB, < CAB = 180° - (90° + 61°)\)

= \(180° - 151°\)

= \(29°\)

In \(\Delta ADB, < DAB = (29 + x)°\)

\(\cos (29 + x) = \frac{3}{5} = 0.6\)

\((29 + x)° = \cos^{-1} (0.6)\)

\((29 + x)° = 53.13°\)

\(x = 53.13° - 29°\)

= \(24.13°\)

\(\approxeq 24.1°\) (1 decimal place).

(b) (i) \(DB = 2DN = 2NB\)

\(OA = OB = OC = OD = 4cm\)

\(AB = BC = CD = DA = 2cm\)

\(DB^{2} = BC^{2} + DC^{2}\)

= \(2^{2} + 2^{2}\)

= \(\sqrt{8}\)

= \(2\sqrt{2} cm\)

\(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\)

In \(\Delta DON\),

\(OD^{2} = ON^{2} + DN^{2}\)

\(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\)

\(h^{2} = 14 \implies h = \sqrt{14}\)

(ii) \(V = \frac{1}{3} \times \base area \times \perp distance\)

\(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\)

= \(\frac{4\sqrt{14}}{3}\)

= \(4.99 cm^{3}\)

(a) In \(\Delta ACB, < CAB = 180° - (90° + 61°)\)

= \(180° - 151°\)

= \(29°\)

In \(\Delta ADB, < DAB = (29 + x)°\)

\(\cos (29 + x) = \frac{3}{5} = 0.6\)

\((29 + x)° = \cos^{-1} (0.6)\)

\((29 + x)° = 53.13°\)

\(x = 53.13° - 29°\)

= \(24.13°\)

\(\approxeq 24.1°\) (1 decimal place).

(b) (i) \(DB = 2DN = 2NB\)

\(OA = OB = OC = OD = 4cm\)

\(AB = BC = CD = DA = 2cm\)

\(DB^{2} = BC^{2} + DC^{2}\)

= \(2^{2} + 2^{2}\)

= \(\sqrt{8}\)

= \(2\sqrt{2} cm\)

\(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\)

In \(\Delta DON\),

\(OD^{2} = ON^{2} + DN^{2}\)

\(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\)

\(h^{2} = 14 \implies h = \sqrt{14}\)

(ii) \(V = \frac{1}{3} \times \base area \times \perp distance\)

\(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\)

= \(\frac{4\sqrt{14}}{3}\)

= \(4.99 cm^{3}\)

Correct Answer: B. 10%

Explanation

15 = (\(\frac{6,900 - C.P \times 100}{C.P}\))

15 C.P = 690000 - C.P 100

C.P = \(\frac{690000}{115}\)

C.P = N6,000

%profit = \(\frac{6,600 - 6,000}{6,000}\) x 100

= \(\frac{600}{6,000}\) x 100

= 10%

Correct Answer: B. ϕ

Correct Answer: D. 20.352

Correct Answer: C. (1,4)

Explanation

\(f(x) = 5x^{2} - 4x + 3\)

\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)

When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)

The coordinates are (1,4)

Correct Answer: B. 90cm²

Explanation

V = L x B x H

V = \(2x \times x \times x = 2x^3\)

= 54cm3

i.e. 2x3 = 54

x3 = \(\frac{54}{2}\) = 27

x = 3\(\sqrt{27}\)

x = 3

L = 2x i.e. 2 x 3 = 6cm

B = x i.e. x = 3cm

H = x i.e. x = 3cm

but A = 2(LB + LH + BH) = 2(6 x 3) + (6 x 3) + (3 x 3) = 2(18 + 18 + 9) = 2(45)

= 90cm²

Correct Answer: B. 6