The bearing of a point Y from point x is 150°. What is the bearing of X from Y?
330°
240°
150°
120°
30°
Correct answer is A
No explanation has been provided for this answer.
Simplify 27\(^{-\frac{1}{3}}\) \(\times\) 64\(^{-\frac{1}{3}}\) \(\times\) 4\(^{\frac{1}{3}}\)
48
12
\(\frac{4}{3}\)
\(\frac{1}{12}\)
\(\frac{1}{48}\)
Correct answer is D
No explanation has been provided for this answer.
Find the 19th term of the A.P. \(\frac{5}{6}\), \(\frac{8}{6}\), \(\frac{11}{6}\).................
7\(\frac{1}{2}\)
9
9\(\frac{1}{2}\)
9\(\frac{5}{6}\)
10
Correct answer is D
first term (a) = \(\frac{5}{6}\)
common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\)
A.P formula → T\(_n\) = a + (n - 1)d
T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\)
T\(_n\) = \(\frac{5}{6}\) + 9
→ 9\(\frac{5}{6}\)
11.60%
7.60%
15%
17.65%
Correct answer is C
Using these:
Loss = C.p - S.p
: Percentage Loss = ( Loss X 100) ÷ C.p
=15 %
If s = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)what does y equal?
\(\sqrt{\frac{(b^2 - a^2)}{(s^2 - x^2)}}\)
\(\sqrt{\frac{(b^2 - a^2)}{(s^2 \times 2)}}\)
\(\sqrt{\frac{x^2 - a^2 - b^2}{s}}\)
\(\frac{x^2 - a^2 - b^2}{s}\)
\(\sqrt{\frac{(b^2 x^2)}{(a^2 - s^2 x^2)}}\)
Correct answer is E
s = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)
s\(^2\) = \(\frac{a^2}{x^2} - \frac{b^2}{y^2}\)
\(\frac{b^2}{y^2}\) = \(\frac{a^2}{x^2}\) - s\(^2\)
\(\frac{b^2}{y^2}\) = \(\frac{a^2 - s^2 x^2}{x^2}\)
\(\frac{1}{y^2}\) = \((\frac{a^2 - s^2 x^2}{x^2}) \times \frac{1}{b^2}\)
\(\frac{1}{y^2}\) = \(\frac{a^2 - s^2 x^2}{b^2 x^2}\)
\(\therefore\) y\(^2\) = \(\frac{b^2 x^2}{a^2 - s^2 x^2}\)
y = \(\sqrt{\frac{b^2 x^2}{a^2 - s^2 x^2}}\)