Correct Answer: C. 14m

Explanation

\(a = \frac{v - u}{t} = \frac{9 - 5}{2} = 2 ms^{-2}\)

\(s = ut + \frac{1}{2}at^{2} = 5(2) + \frac{1}{2}(2 \times 2^{2})\)

= \(14m\)

Correct Answer: B. \(\sqrt\frac{3T - M}{K}\)

Correct Answer: A. 3

Correct Answer: D. log y = 3 log 2 + \(\frac{1}{2}\) log m

Explanation

y = 8\(\sqrt{m}\); log y = log 8\(\sqrt{m}\)

log y = log 8 + log \(\sqrt{m}\)

log y = log 2³ + log m^{\(\frac{1}{2}\)}

log y = 3 log 2 + \(\frac{1}{2}\) log m

Correct Answer: C. \(\frac{5\sqrt{2}}{2}\)

Explanation

\(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} = \frac{\sqrt{3} + 4\sqrt{3}}{\sqrt{6}}\)

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

\(\frac{5\sqrt{18}{6} = \frac{15\sqrt{2}}{6}\)

= \(\frac{5\sqrt{2}}{2}\)

Explanation

\(|AB| = \sqrt{(-1 - 3)^{2} + (2 - 5)^{2}}\)

= \(\sqrt{16 + 9} = \sqrt{25}\)

= 5 units.

\(|AC| = \sqrt{(4 - 1)^{2} + (8 - 2)^{2}}\)

= \(\sqrt{25 + 36}\)

= \(\sqrt{61}\) units.

\(|BC| = \sqrt{(4 - 3)^{2} + (8 - 5)^{2}}\)

= \(\sqrt{1 + 9} \)

= \(\sqrt{10}\) units.

From the cosine rule,

\((\sqrt{61})^{2} = 5^{2} + (\sqrt{10})^{2} - 2(5)(\sqrt{10}) \cos <PBQ\)

\(\cos < PBQ = \frac{25 + 10 - 61}{2(5)(\sqrt{10})}\)

\(\cos < PBQ = \frac{-26}{10\sqrt{10}} = -0.4747\)

\(< PBQ = 118°, \beta = < PBQ = 118°\)

\(< BPS = \alpha = 180° - 118° = 62°\)

The magnitude of the resultant

\(|R|^{2} = 5^{2} + (3\sqrt{10})^{2} - 2(5)(3\sqrt{10}) \cos 62\)

= \(25 + 90 - (30\sqrt{10})(0.4695)\)

= \(115 - 44.538\)

= \(70.462\)

\(|R| = \sqrt{70.462} = 8.39N\)

Let \(\delta\) be the angle the resultant makes with the 5N force.

\(PS^{2} = BS^{2} + BP^{2} - 2(BS)(BP) \cos PBS\)

\((3\sqrt{10})^{2} = (\sqrt{70.462})^{2} + 5^{2} - 2(8.39)(5) \cos \delta\)

\(90 = 70.462 + 25 - 83.9 \cos \delta\)

\(\cos \delta = \frac{70.462 + 25 - 90}{83.9}\)

\(\cos \delta = 0.0651\)

\(\delta = \cos^{-1} (0.0651) = 86.27°\)

\(\approxeq 86°\).

The resultant makes an angle of 86° with the 5N force.

(b)(i) \(u = 0 ms^{-1} ; v = 20 ms^{-1} ; s = 8 m ; a = ?\)

\(v^{2} = u^{2} + 2as \)

\(20^{2} = 0^{2} + 2(8)a \implies 400 = 16a\)

\( a = 25 ms^{-2}\)

(ii) \(t = ?\)

\(s = ut + \frac{1}{2} at^{2}\)

\(40 = 0 + \frac{1}{2} (25)(t^{2})\)

\(80 = 25t^{2}\)

\(t^{2} = \frac{80}{25} = 3.2\)

\(t = \sqrt{3.2} = 1.79 s\)

It takes 1.79s to cover a distance of 40m.

Correct Answer: B. 495

Explanation

(a) 3, x, y, 18.

Note : When given the consecutive terms of an A.P to be \(a, b, c, d\), then

\(\frac{a + c}{2} = b ; \frac{b + d}{2} = c\).

\(\therefore \frac{3 + y}{2} = x \implies 3 + y = 2x .... (1)\)

Also, \(\frac{x + 18}{2} = y \implies x + 18 = 2y ..... (2)\)

From (2), x = 2y - 18. Putting that in (1), we have

\(3 + y = 2(2y - 18) \implies 3 + y = 4y - 36\)

\(3 + 36 = 4y - y \implies 3y = 39\)

\(y = 13\)

\(x = 2y - 18\)

\(x = 2(13) - 18 = 26 - 18 = 8\)

\((x, y) = (8, 13)\).

(b)(i) G.P

\(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3}\)

\(T_{2} + T_{3} = 6T_{4}\)

\(ar + ar^{2} = 6ar^{3}\)

\(6ar^{3} - ar^{2} - ar = 0\)

\(6ar^{3} - 3ar^{2} + 2ar^{2} - ar = 0\)

\(3ar^{2}(2r - 1) + ar(2r - 1) = 0\)

\((3ar^{2} + ar)(2r - 1) = 0\)

\(ar(3r + 1)(2r - 1) = 0\)

\(\implies 3r + 1 = 0 ; 2r - 1 = 0\)

\(r = \frac{-1}{3} ; r = \frac{1}{2}\)

(ii) Since r is positive, then \(r = \frac{1}{2}\).

\(T_{2} = ar = 8 \implies \frac{a}{2} = 8\)

\(a = 16\)

\(\therefore \text{The first 6 terms} = 16, 8, 4, 2, 1, \frac{1}{2}\)

Correct Answer: C. \(\frac{m-3}{2}\ge10\)

Explanation

Total number of mangoes = m

Amina ate 3 mangoes \(\implies\) Remainder = m - 3

Shared equally with Uche \(\implies \frac{m - 3}{2}\)

\(\frac{m - 3}{2} \geq 10\)

Correct Answer: D. 6