Correct Answer: C. 12.06

Correct Answer: C. -2

Correct Answer: C. 11/4, 8/3

Explanation

y = x\(^2\)(1 + x)\(^{\frac{3}{2}}\)

Let u = x\(^2\)

v = (1 + x)\(^{\frac{3}{2}}\)

\(\frac{du}{dx} 2x\)

\(\frac{dv}{dx} = \frac{3}{2}\)(1 + x)\(^\frac{1}{2}\)

\(\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}\)

= (1 + x)\(^{\frac{3}{2}}\)(2x) + x\(^2\)\(\frac{3}{2}(1 + x)^{\frac{1}{2}}\)

= 2x(1 + x)\(^{\frac{3}{2}}\) + \(\frac{3x^2(1 + x)^{\frac{1}{2}}}{2}\)

(b)

2y - x = 3

when x = 0, y = \(\frac{3}{2}\)

when y = 0, x = -3

h = -3, k = \(\frac{3}{2}\)

Using

(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)

(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\) = r\(^2\)

r(2, 3), x = 2, y = 3

(2 + 3)\(^2\) + (3 + \(\frac{3}{2}\))\(^2\) = r\(^2\)

\(\frac{109}{4}\) = r\(^2\)

The equation will give

(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\)

= \(\frac{109}{4}\)

Explanation

(a) \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\)

\(\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8}) = \frac{3}{4}(\frac{27}{8} + \frac{13}{8})\)

= \(\frac{3}{4}(\frac{40}{8})\)

= \(\frac{15}{4}\)

\(2\frac{1}{8} - 1\frac{1}{2} = \frac{17}{8} - \frac{3}{2}\)

= \(\frac{17}{8} - \frac{12}{8}\)

= \(\frac{5}{8}\)

\(\therefore \frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}} = \frac{15}{4} \div \frac{5}{8}\)

= \(\frac{15}{4} \times \frac{8}{5}\)

= \(6\)

(b) \(\log_{10} 2 = 0.3010 ; \log_{10} 3 = 0.4771\)

\(\log_{10} 1.125 = \log_{10}(\frac{1125}{1000})\) (Dividing through with 125)

= \(\log_{10} (\frac{9}{8})\)

= \(\log_{10} 9 - \log_{10} 8\)

\(\log_{10} 9 = \log_{10} 3^{2} = 2\log_{10} 3 = 2 \times 0.4771 = 0.9542\)

\(\log_{10} 8 = \log_{10} 2^{3} = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030\)

\(\therefore \log_{10} (\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\)

\(\approxeq 0.051\) ( 2 significant figures)

Explanation

(a) \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)

\(4\frac{2}{9} - 1\frac{13}{15} = \frac{38}{9} - \frac{28}{15}\)

= \(\frac{190 - 84}{45}\)

= \(\frac{106}{45}\)

\(2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3} = \frac{11}{5} + (\frac{4}{7} \times \frac{7}{3})\)

= \(\frac{11}{5} + \frac{4}{3}\)

= \(\frac{33 + 20}{15}\)

= \(\frac{53}{15}\)

\(\therefore \frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}} = \frac{106}{45} \div \frac{53}{15}\)

= \(\frac{106}{45} \times \frac{15}{53}\)

= \(\frac{2}{3}\)

(b) \(\frac{7\sqrt{5}}{\sqrt{7}}\)

= \(\frac{7\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\)

= \(\frac{7\sqrt{35}}{7}\)

= \(\sqrt{35}\)

Correct Answer: A. K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

Explanation

T = \(\sqrt{\frac{TK - H}{K - H}}\)

Taking the square of both sides, give

T² = \(\frac{TK - H}{K - H}\)

T²(K - H) = TK - H

T²K - T²H = TK - H

T²K - TK = T²H - H

K(T² - T) = H(T² - 1)

K = \(\frac{H(T^2 - 1)}{T^2 - T}\)

Correct Answer: A. -\(\frac{1}{3}\)

Explanation

a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)

= 7 - 10

= \(\frac{-3}{15}\)

d = - \(\frac{-1}{5}\)

T6 = a + 5d

= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)

= \(\frac{2}{3}\) - 1

= \(\frac{2 - 3}{3}\)

= \(\frac{-1}{3}\)

Correct Answer: D. -1

Explanation

tan 315

Correct Answer: C. 4⁵/₆