12
8
9
16
Correct answer is C
\(p \propto \frac{1}{\sqrt{q}}\)
\(\implies p = \frac{k}{\sqrt{q}}\)
when p = 3, q = 16.
\(3 = \frac{k}{\sqrt{16}}\)
\(k = 3 \times 4 = 12\)
\(\therefore p = \frac{12}{\sqrt{q}}\)
when p = 4,
\(4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}\)
\(\sqrt{q} = 3 \implies q = 3^2 \)
\(q = 9\)
Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
\(4\frac{1}{2}\)
\(3\frac{1}{2}\)
\(7\frac{1}{2}\)
\(5\frac{1}{4}\)
Correct answer is C
\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)
= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)
= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)
= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)
= \(\frac{22}{3} - (-\frac{1}{6})\)
= \(\frac{22}{3} + \frac{1}{6}\)
= \(\frac{15}{2}\)
= \(7\frac{1}{2}\)
The nth term of a sequence is given by 2\(^{2n - 1}\). Find the sum of the first four terms.
74
32
42
170
Correct answer is D
\(T_n = 2^{2n - 1}\)
\(T_1 = 2^{2(1) - 1} \)
= 2
\(T_2 = 2^{2(2) - 1}\)
= 8
\(T_3 = 2^{2(3) - 1}\)
= 32
\(T_4 = 2^{2(4) - 1}\)
= 128
\(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\)
= 170
Solve the inequality: -7 \(\leq\) 9 - 8x < 16 - x
-1 \(\leq\) x \(\leq\) 2
-1 \(\leq\) x < 2
-1 < x < 2
-1 < x \(\leq\) 2
Correct answer is D
-7 \(\leq\) 9 - 8x < 16 - x
-7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x
-7 - 9 \(\leq\) -8x and -8x + x < 16 - 9
-16 \(\leq\) -8x and -7x < 7
\(\therefore\) x \(\leq\) 2 and -1 < x
-1 < x \(\leq\) 2.
Solve for x in \(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
\(x \leq 1\frac{1}{2}\)
\(x \leq \frac{21}{2}\)
\(x \geq \frac{21}{2}\)
\(x \geq 1\frac{1}{2}\)
Correct answer is B
\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
2(4x - 6) \(\leq\) 3(3 + 2x)
8x - 12 \(\leq\) 9 + 6x
8x - 6x \(\leq\) 9 + 12
2x \(\leq\) 21
\(x \leq \frac{21}{2}\)