Given that tan x = 5/12, what is the value of sin x + cos x ?
5/13
7/13
12/13
17/13
5/12
Correct answer is D
\(\tan x = \frac{opp}{adj} = \frac{5}{12}\)
\(Hyp^2 = opp^2 + adj^2\)
\(Hyp^2 = 5^2 + 12^2\)
= \(25 + 144 = 169\)
\(Hyp = \sqrt{169} = 13\)
\(\sin x = \frac{5}{13}; \cos x = \frac{12}{13}\)
\(\sin x + \cos x = \frac{5}{13} + \frac{12}{13}\)
= \(\frac{17}{13}\)
20km
24km
25km
31km
84km
Correct answer is C
No explanation has been provided for this answer.
If sin x = cos 50o, then x equals
40o
45o
50o
90o
130o
Correct answer is A
No explanation has been provided for this answer.
120km
165km
195km
(150 + 45√3)km
240km
Correct answer is C
No explanation has been provided for this answer.
State the fifth and seventh terms of the sequence \(-2, -3, -4\frac{1}{2}, ...\)
\(-\frac{81}{8}, -\frac{729}{32}\)
\(\frac{8}{81}, \frac{72}{39}\)
\(\frac{27}{729}, \frac{718}{39}\)
\(-\frac{27}{16}, -\frac{79}{81}\)
\(-\frac{21}{8}, \frac{32}{618}\)
Correct answer is A
\(-2, -3, -4\frac{1}{2}, ...\)
This is a G.P with r = 1\(\frac{1}{2}\).
\(T_{n} = ar^{n - 1}\) (terms of a G.P)
\(T_{5} = (-2)(\frac{3}{2})^{5 - 1}\)
= \(-2 \times \frac{81}{16}\)
= \(-\frac{81}{8}\)
\(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\)
= \(-2 \times \frac{729}{64}\)
= \(-\frac{729}{32}\)