An exponential sequence (G.P.) is given by 8√2, 16√2, 32√2, ... . Find the nth term of the sequence
8√2n
2(n+2)√2
√2(n+3)
8n√2
Correct answer is B
8√2, 16√2, 32√2, ..
a=8√2;r=T2T1=16√28√2=2
Tn=arn−1
Tn=8√2×2n−1
Tn=23×2n−1×√2
Tn=23+n−1×√2
∴
Solve 6 sin 2θ tan θ = 4, where 0º < θ < 90º
18.43º
30.00º
35.26º
19.47º
Correct answer is C
6 sin 2θ tan θ = 4, where 0º < θ < 90º
sin 2θ = 2sin θ cos θ and tanθ = \frac{sinθ}{cosθ}
= 6 x 2sin θ cos θ x \frac{sin θ}{cos θ} = 4
= sin^2 θ = 4
= sin^2 θ = \frac{4}{12}=\frac{1}{3}
=sin θ = \frac{\sqrt1}{3}=\frac{1}{\sqrt3}
= θ = sin^{-1}(\frac{1}{\sqrt3})
∴ θ = 35.26º
Given that r = (10 N , 200º) and n = (16 N , 020º), find (3r - 2n).
(62 N , 240º)
(62 N , 200º)
(62 N , 280º)
(62 N , 020º)
Correct answer is D
r = (10 N, 200º) and n = (16 N, 020º)
In rectangular form:
r = 10cos 200ºi + 10sin 200ºj = -9.397i - 3.420j
n = 16cos 20ºi + 16sin 20ºj = 15.035i + 5.472j
3r = -28.191i - 10.260j
2n = 30.070i + 10.945j
3r - 2n = (-28.191i - 10.260j) - (30.070i + 10.945j)
3r - 2n = -58.261i - 21.205j
|3r - 2n| = √((-58.261)^2 + (-21.205)^2) = 62 N
tan θ =\frac{-21.205}{-58.261} = 0.3640
θ = tan^{-1} (0.3640) = 20^o
∴ (62 N , 020º)
Simplify: \frac{log √27 - log √8}{log 3 - log 2}
\frac{3}{2}
-\frac{1}{4}
-\frac{3}{2}
\frac{1}{4}
Correct answer is A
\frac{log √27 - log √8}{log 3 - log 2}
= \frac{log √3^3 - log √2^3}{log 3 - log 2}
= \frac{log3^{3/2} - log2^{3/2}}{log3}
=\frac{^3/_2(log 3 - log 2)}{log 3 - log 2}
\therefore\frac{3}{2}
The table shows the mark obtained by students in a test.
Marks | 1 | 2 | 3 | 4 | 5 |
Frequency | 2 | k | 1 | 1 | 2 |
4
1
2
3
Correct answer is B
x̄ =\frac{∑fx}{∑f}= 3
=\frac{(1 \times 2)+(2 \times k)+(3 \times 1)+(4 \times 1)+(5 \times 2)}{2 + k + 1 + 1 + 2}= 3
=\frac{2 + 2k + 3 + 4 + 10}{6 + k} = 3
=\frac{19 + 2k}{6 + k} = 3
=\frac{19 + 2k}{6 + k} = \frac{3}{1}
=19+2k=3(6+k)
=19+2k=18+3k
=2k-3k=18-19
=-k=-1
∴k=1