Correct Answer: A. 5.5m

Explanation

From S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);

Distance traveled in 2 seconds

S = 12[2] + \(\frac{5}{2[2]^2}\) - 2\(^3\).

= 24 + 10 - 8 = 26m

Distance traveled in 3 seconds

S = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)

= 36 + 22.5 - 27 = 31.5m

Distance traveled in the 3rd second

= 31.5m - 26m

: S = 5.5m

Correct Answer: A. 0, 1

Correct Answer: C. 8

Explanation

\(104_x = 68\\

1 \times x^2 + 0 \times x + 4 \times x^0 = 68\\

x^2 = 68 - 4; x^2 = 64\\

x = \sqrt{64}=8\)

Correct Answer: A. 8i + j

Explanation

\(\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ}\)

\(\overrightarrow{RQ} = -\overrightarrow{QR} = - (3i + 2j) = -3i - 2j\)

\(\overrightarrow{SQ} = (-5i + 3j) - 3i - 2j = -8i + j\)

Explanation

(i)

< NPQ = 180\(^o\) - 50\(^o\) = 130\(^o\)

< MNP = 130\(^o\) - 90\(^o\) = 40\(^o\)

< NMP = 40\(^o\) (Isolceles triangle)

< MNP = 180\(^o\) - (40\(^o\) + 40\(^o\))

= 180\(^o\) - 80\(^o\) = 100\(^o\)

(ii) < OPQ = 90\(^o\)(50\(^o\) - 40\(^o\)) = 80\(^o\)

< OPQ = 80\(^o\) (Isoseles triangle)

< POQ = 180\(^o\) (80\(^o\) + 80\(^o\))

180\(^o\) - 160\(^o\) = 20\(^o\)

(b) 8y + 4x = 24

\(\frac{8y}{8} = \frac{-4y}{8} + \frac{24}{8}\)

y = \(\frac{-1}{2}x + 6\)

m = \(\frac{-1}{2}\)

The equation will give

\(\frac{y - 12}{x + 8} = \frac{-1}{2}\)

2y - 24 = x - 8

2y = - x - 8 + 24

2y = -x + 16

Correct Answer: D. N288

Correct Answer: B. 116₇

Correct Answer: B. 7 − \(\frac{q}{4}\)

Explanation

(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two.

I'll be using the latter method.

\(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\)

= \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\)

\(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \)

= \(4 + 0 + 1 = 5_{ten}\)

\(\frac{255}{5} = 51_{ten}\)

= \(110011_{two}\)

(b)

(i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\)

= \(11cm\)

\(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\)

(ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\)

= \(33 cm^{2}\)

Correct Answer: b. N240,000