Find the value of the angle marked x in the diagram above
600
450
900
300
Correct answer is A
PR2=PQ2+RQ2−2(PQ)(RQ)cosQ
⟹cosQ=PQ2+RQ2−PR22(PQ)(RQ)
⟹cosQ=82+52−722×8×5
⟹cosQ=64+25−4980
⟹cosQ=4080=0.5
⟹Q=cos−1(0.5)=600
∴
^{-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
⇒ -4x = -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}