Find the value of the angle marked x in the diagram above
60\(^0\)
45\(^0\)
90\(^0\)
30\(^0\)
Correct answer is A
\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\)
\(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\)
\(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\)
\(\implies cos Q = \frac{64 + 25 - 49}{80}\)
\(\implies cos Q = \frac{40}{80} = 0.5\)
\(\implies Q = cos^{-1} (0.5) = 60^0\)
\(\therefore x = 60^0\)
\(^{-1}/_3\)
2
\(^{4}/_3\)
\(^{2}/_9\)
Correct answer is A
\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)
\(T_n = ar^n - 1\)
∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)
\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)
= 2a = 3(1 - r)
= 2a = 3 - 3r
∴ a = \(\frac{3 - 3r}{2}\)
Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)
= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)
= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)
= 3(3r - 3r\(^2\)) = -4
= 9r - 9r\(^2\) = -4
= 9r\(^2\) - 9r - 4 = 0
= 9r\(^2\) - 12r + 3r - 4 = 0
= 3r(3r - 4) + 1(3r - 4) = 0
= (3r - 4)(3r + 1) = 0
∴ r = \(\frac{4}{3} or - \frac{1}{3}\)
For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.
∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)
15 cm
19 cm
13 cm
21 cm
Correct answer is C
Let the length of the longer side = \(x\) cm
∴ The length of the shorter side = (\(x\) - 6) cm
If we increase each side's length by 2 cm, it becomes
(\(x\) + 2) cm and (\(x\) - 4) cm respectively
Area of a rectangle = L x B
\(A_1 = x(x - 6) = x^2 - 6x\)
\(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\)
\(A_1 + 68 = A_2\) (Given)
⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\)
⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68
⇒ -4\(x\) = -76
⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm
∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm
Evaluate \(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} \times \frac{1}{4}\)
- \(\frac{3}{40}\)
\(\frac{3}{40}\)
\(\frac{7}{40}\)
-\(\frac{263}{40}\)
Correct answer is C
\(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} x \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{3}{4} ÷ \frac{5}{12}) \times \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\)
⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\)
⇒ \(\frac{5}{8} - \frac{9}{20}\)
∴ \(\frac{7}{40}\)
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