Correct Answer: D. 30

Explanation

54 x 40 = 35 x 10 + 60F

160 = 350 + 60F

F = \(\frac{1810}{60}\)

= 30N

Correct Answer: C. \(x\le -2 and x\ge1\)

Correct Answer: D. \(y = 3x^{2} - 2x + 4\)

Explanation

The criterion for the quadratic curve to intersect the x- axis is \(b^{2} > 4ac\).

Explanation

(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)

\(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\)

= \(\frac{45 + 28}{36}\)

= \(\frac{73}{36}\)

\(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\)

= \(\frac{13}{9} - \frac{3}{8}\)

= \(\frac{104 - 27}{72}\)

= \(\frac{77}{72}\)

\(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\)

= \(\frac{73}{36} \times \frac{72}{77}\)

= \(\frac{146}{77}\)

= \(1\frac{69}{77}\)

(b) \(\sin x = \frac{2}{3}\)

In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\)

\(\therefore |BC|^{2} = 9 - 4 = 5\)

\(|BC| = \sqrt{5}\)

\(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\)

\(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\)

= \(\frac{6 - 5}{3\sqrt{5}}\)

= \(\frac{1}{3\sqrt{5}}\)

Rationalizing, we have

\(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)

Correct Answer: D. 440m

Explanation

radius = \(\frac{diameter}{2}\) = \(\frac{1.4}{2}\)

= 0.7m

circumference = 2\(\pi r = 2 \times \frac{22}{7} \times 0.7m = 4.4m\)

distance covered = 4.4m x 100 = 440m

Explanation

(a) \(T_{1} = a = -8\)

\(\frac{T_{7}}{T_{9}} = \frac{5}{8}\)

\(\frac{a + 6d}{a + 8d} = \frac{5}{8} \implies \frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\)

\(-8 + 6d = \frac{5}{8}(-8 + 8d) \implies -8 + 6d = -5 + 5d\)

\(6d - 5d = -5 + 8\)

\(d = 3\)

(b) Let :

the cost price \(CP_{1}\) of a basket of pawpaw be w;

the cost price \(CP_{2}\) of a basket of mangoes be m.

Then \(30w + 100m = N2,450 ..... (1)\)

Also, let :

The selling price of the baskets of pawpaw be \(SP_{1}\);

the selling price of the baskets of mangoes be \(SP_{2}\).

Then \(SP_{1} + SP_{2} = N(2,450 + 855) = N3,305 ...... (2)\)

%age profit on the sale of pawpaw = \(\frac{SP_{1} - CP_{1}}{CP_{1}} \times 100%\)

i.e. \(40% = \frac{SP_{1} - 30w}{30w} \times 100%\)

\(\implies 40 \times 3w = (SP_{1} - 30w) \times 10\)

\(12w = SP_{1} - 30w\)

\(SP_{1} = 12w + 30w = 42w\)

%age profit on the sale of mangoes = \(\frac{SP_{2} - CP_{2}}{CP_{2}} \times 100%\)

i.e. \(30% = \frac{SP_{2} - 100m}{100m} \times 100%\)

\(\implies 30m = SP_{2} - 100m\)

\(SP_{2} = 30m + 100m = 130m\)

But \(SP_{1} + SP_{2} = N3,305\)

\(\implies 42w + 130m = 3305 .... (2)\)

From (1), \(m = \frac{2450 - 30w}{100}\)

\(m = \frac{245 - 3w}{10}\)

Substitute \(m = \frac{245 - 3w}{10}\) in (3)

\(42w + 130(\frac{245 - 3w}{10}) = 3305\)

\(42w + 3,185 - 39w = 3305\)

\(3w = 3305 - 3185 = 120\)

\(w = N40.00\)

(i) Hence, the cost price of a basket of pawpaw = N40.00.

(ii) From (2),

\(SP_{1} + SP_{2} = 3305\)

\(42w + 130m = 3305 \implies 42 \times 40 + 130m = 3305\)

\(1680 + 130m = 3305\)

\(130m = SP_{2} = 3305 - 1680\)

= \(N1,625.00\)

Hence, the selling point of 100 baskets of mangoes = N1,625.00.

Explanation

(b)

(c)(i) \(2x^{2} - 9x = 4\)

\(2x^{2} - 9x - 1 = 4 - 1 = 3\)

At y = 3, x = 0.2 and 4.3.

(ii) Gradient of \(y = 2x^{2} - 9x - 1\) at x = 3

= \(\frac{BC}{AB} = \frac{9.5}{4.3}\)

= \(2.2\)

Correct Answer: A. 4

Explanation

\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\)

\(lim_{x\to2} \frac {x + 6}{x}\)

\(\frac {2 + 6}{2} = \frac {8}{2}\) = 4

Correct Answer: A. q = \(\frac{rt^2}{(p - r^3)}\)

Explanation

t = √pq/r - r\(^2\)q

multiply both sides by the L.C.M, r

r\(^2\) = pq - qr\(^3\)

collect like terms on the RHS

q(p - r3) = rt\(^2\)

q = \(\frac{rt^2}{(p - r^3)}\)

Correct Answer: B. \(x< -5\) or \(x > \frac{3}{2}\)

Explanation

\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)

\(x(2x - 3) + 5(2x - 3) > 0\)

\((x + 5)(2x - 3) > 0\)

For their product to be positive, they are either both +ve or -ve.

\(x + 5 > 0 \implies x > -5\)

\(2x - 3 > 0 \implies 2x > 3\)

\(x > \frac{3}{2}\)

Check:

\(x > -5: x = -3\)

\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)

\(\therefore x < -5\)

\(x > \frac{3}{2}: x = 2\)

\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)