\(\frac{48}{65}\)
\(\frac{13}{15}\)
\(\frac{-33}{65}\)
\(\frac{-48}{65}\)
Correct answer is C
From Pythagoras' theorem
when sin x = \(\frac{12}{13}\), cos x = \(\frac{5}{13}\)
Using cos (x + y) = cosx cosy - sinxsiny
\(\frac{5}{13}\) * \(\frac{3}{5}\) - \(\frac{12}{13}\) * \(\frac{4}{5}\)
= \(\frac{-33}{65}\)
If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)
-32
-16
16
32
Correct answer is D
( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\)
compared with: 1 - 8x + 24x\(^2\) - 32x\(^3\) + 16\(^4\)
q = 24 and p = -8
(q - p) = 24 - [-8] = 32
Given that F\(^1\)(x) = x\(^3\)√x, find f(x)
\( \frac{2x^{9/2}}{9} + c \)
\( 2x^{9/2} + c \)
\( \frac{2x^{5/2}}{5} + c \)
\( x^4 + c \)
Correct answer is A
F1 (x) = x\(^3\) √x = x\(^{7/2}\)
F(x) = \(\frac{x^{7/2 +1}}{7/2 + 1}\) + c
= \(\frac{x^{9/2}}{9/2}\)
= \(\frac{2x^{9/2}}{9}\) + c
5.5m
14.5m
26.0m
30.0m
Correct answer is A
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
418.5m
56.0m
31.5m
30.0m
Correct answer is C
S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
ds/dt = 12 + 5t - 3t\(^2\)
At max height ds/dt = 0
i.e 12 + 5t - 3t\(^2\)
(3t + 4)(t -3) = 0;
t = -4/3 or 3
Hmax = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 45/2 - 27
= 31.5m