If \((x - 3)\) is a factor of \(2x^{3} + 3x^{2} - 17x - 30\), find the remaining factors.
(2x - 5)(x - 2)
(2x - 5)(x + 2)
(2x + 5)(x - 2)
(2x + 5)(x + 2)
Correct answer is D
Divide \(2x^{3} + 3x^{2} - 17x - 30\) by \((x - 3)\). You get \(2x^{2} + 9x + 10\).
Factorizing, we have \(2x^{2} + 9x + 10 = 2x^{2} + 4x + 5x + 10\)
\(2x(x + 2) + 5(x + 2)\)
= \((2x + 5)(x + 2)\)
\(\sqrt{3}\)
\(\frac{3\sqrt{2}}{4}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{\sqrt{2}}{2}\)
Correct answer is C
\(a ♦ b = \frac{ab}{4}\)
\(\sqrt{2} ♦ \sqrt{6} = \frac{\sqrt{2} \times \sqrt{6}}{4} = \frac{\sqrt{12}}{4}\)
= \(\frac{2\sqrt{3}}{4} \)
= \(\frac{\sqrt{3}}{2}\)
Simplify \(\sqrt{(\frac{-1}{64})^{\frac{-2}{3}}}\)
-4
\(-\frac{1}{4}\)
\(\frac{1}{8}\)
4
Correct answer is D
\((\frac{-1}{64})^{\frac{-2}{3}} = -64^{\frac{2}{3}}\)
\((-4^{3})^{\frac{2}{3}} = -4^{2} = 16\)
\(\therefore \sqrt{(\frac{-1}{64})^{\frac{-2}{3}} = \sqrt{16} = 4\)
Solve the inequality \(2x^{2} + 5x - 3 \geq 0\).
\(x \leq -3\) or \(x \geq \frac{1}{2}\)
\(x < -\frac{1}{2}\) or \(x \geq 3\)
\(-3 \leq x \leq \frac{1}{2}\)
\(-\frac{1}{2} \leq x \leq 3\)
Correct answer is A
No explanation has been provided for this answer.
\(P = {x : 1 \leq x \leq 6}\) and \(Q = {x : 2 < x < 9}\) where \(x \in R\), find \(P \cap Q\).
\({x : 2 \leq x \leq 6}\)
\({x : 2 \leq x < 6}\)
\({x : 2 < x < 6}\)
\({x : 2 < x \leq 6}\)
Correct answer is D
No explanation has been provided for this answer.