−log34
−log32
3log32
3log34
Correct answer is C
2log38−3log32=log382−log323
= log3(648)
= log38=log323
= 3log32
Calculate, correct to the nearest degree, the angle between the vectors (131) and (14)
58°
72°
74°
87°
Correct answer is B
a.b=|a||b|cosθ
(131).(14)=13×1+1×4=13+4=17
17=(√132+12)(√12+42)cosθ
17=(√170)(√17)cosθ
cosθ=1717√10=√1010=0.3162
\theta = \cos^{-1} 0.3162 = 72°
The position vectors of A and B are (2i + j) and (-i + 4j) respectively; find |AB|.
3\sqrt{2}
\sqrt{34}
\sqrt{34}
9\sqrt{2}
Correct answer is A
No explanation has been provided for this answer.
1
\frac{1}{2}
\frac{5}{12}
\frac{1}{12}
Correct answer is C
p(P) = \frac{1}{2}, p(P') = \frac{1}{2}
p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}
p(R) = \frac{3}{4}, p(R') = \frac{1}{4}
p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')
= (\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})
= \frac{1}{24} + \frac{6}{24} + \frac{3}{24}
= \frac{5}{12}