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