If \sin x = -\sin 70°, 0° < x < 360°, determine the two possible values of x.
110°, 250°
110°, 290°
200°, 250°
250°, 290°
Correct answer is D
The value of the sine of an angle is negative in the third and fourth quadrant. Hence options A and B are not the options.
\sin 250 = -\sin (250 - 180) = - \sin 70
\sin 290 = - \sin (360 - 290) = - \sin 70
If (x - 3) is a factor of 2x^{2} - 2x + p, find the value of constant p.
-12
-6
3
6
Correct answer is A
Using remainder theorem, since x - 3 is a factor, then
given 2x^{2} - 2x + p, f(3) = 0
2(3^{2}) - 2(3) + p = 0 \implies 18 - 6 = -p
p = -12
The roots of a quadratic equation are (3 - \sqrt{3}) and (3 + \sqrt{3}). Find its equation.
x^{2} - 6x - 9 = 0
x^{2} - 6x + 6 = 0
x^{2} + 6x - 9 = 0
x^{2} + 6x + 6 = 0
Correct answer is B
(x - \alpha)(x - \beta) = 0
(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0
(x^{2} - (3 - \sqrt{3})x - (3 + \sqrt{3})x + (9 + 3\sqrt{3} - 3\sqrt{3} - 3) = 0
x^{2} - 3x - x\sqrt{3} - 3x + x\sqrt{3} + 6 = 0
x^{2} - 6x + 6 = 0
\frac{560}{243}
\frac{841}{243}
\frac{1120}{243}
\frac{4481}{243}
Correct answer is C
^{10}C_{7 - 1} (2^{10 - 6}) (\frac{-1}{3})^{6}
\frac{10!}{(10 - 6)! 6!} \times 16 \times \frac{1}{243}
= 210 \times 16 \times \frac{1}{729}
= \frac{1120}{243}
Simplify \frac{\log_{5} 8}{\log_{5} \sqrt{8}}.
-2
\frac{-1}{2}
\frac{1}{2}
2
Correct answer is D
\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\log_{5} 8^{\frac{1}{2}}}
= \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8}
= \frac{1}{\frac{1}{2}}
= 2