Evaluate \(\int_{1}^{2} \frac{4}{x^{3}} \mathrm {d} x\)
\(-1\frac{1}{2}\)
\(-\frac{15}{16}\)
\(\frac{15}{16}\)
\(1\frac{1}{2}\)
Correct answer is D
\(\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x\)
\(\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}\)
= \(\frac{-2}{2^{2}} - \frac{-2}{1^{2}} = -\frac{1}{2} + 2 = 1\frac{1}{2}\)
70.6°
50.2°
39.8°
19.4°
Correct answer is A
\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)
\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)
\(m_{1} = \frac{3}{4}\)
\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)
\(m_{2} = -\frac{2}{3}\)
\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)
\(\tan \theta = \frac{17}{6}\)
\(\theta \approxeq 70.6°\)
If \(y = 2(2x + \sqrt{x})^{2}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
\(2\sqrt{x}(2x + \sqrt{2})\)
\(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)
\(4(2x + \sqrt{x})(2 + \sqrt{x})\)
\(8(2x + \sqrt{x})(2 + \sqrt{x})\)
Correct answer is B
\(y = 2(2x + \sqrt{x})^{2}\)
Let \(u = 2x + \sqrt{x}\)
\(y = 2u^{2}\)
\(\frac{\mathrm d y}{\mathrm d u} = 4u\)
\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
= \(4u(2 + \frac{1}{2\sqrt{x}}) \)
= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)
Which of the following items is found on he credit side of a sales Ledger Control Account?
Credit sales
Debtors cheques dishonoured
Interest on overdue account
Bad debts written off
Correct answer is C
No explanation has been provided for this answer.
Calculate, correct to one decimal place, the length of the line joining points X(3, 5) and Y(5, 1).
4.0
4.2
4.5
5.0
Correct answer is C
\(XY = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\) is the distance between a point \(X(x_{1}, y_{1})\) and \(Y(x_{2}, y_{2})\).
\(XY = \sqrt{(3 - 5)^{2} + (5 - 1)^{2}} = \sqrt{20}\)
= \(2\sqrt{5} = 4. 467 \approxeq 4.5\)