Further Mathematics Questions and Answers

\(\frac{2 + 5+ (x + 2) + 7 + 9}{5} = 6 \implies 25 + x = 30\)

\(x = 5 \therefore x + 2 = 5 + 2 = 7\)

Arranging the numbers in ascending order: 2, 5, 7, 7, 9. 

Median = 7.0

\((a + 3x)^{6}\).

The coefficient of \(x^{2}\) is:

\(^{6}C_{4}(a)^{6 - 2} (3x)^{2} = \frac{6!}{(6 - 4)! 4!} (a^{4})(9x^{2})\)

\(15 \times a^{4} \times 9 = 135a^{4}\)

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

where (a, b) and r are the coordinates of the centre and radius respectively.

Given : \((a, b) = (-3, -8); r = 4\sqrt{6}\)

\((x - (-3))^{2} + (y - (-8))^{2} = (4\sqrt{6})^{2}\)

\(x^{2} + 6x + 9 + y^{2} + 16y + 64 = 96\)

\(x^{2} + y^{2} + 6x + 16y + 9 + 64 - 96 = 0\)

\(\implies x^{2} + y^{2} + 6x + 16y - 23 = 0\)

No explanation has been provided for this answer.

Let the original volume be V with radius r.

\(V = \frac{4}{3}\pi r^{3}\)

45% increased volume = 145%V. 

Let the %age increase in radius = m%r

\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)

\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)

\(1.45V = V(\frac{m}{100})^{3}\)

\(\implies 1.45 \times 10^{6} = m^{3}\)

\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)

\(\therefore \text{%age increase =}  113.2 - 100 = 13.2%\)

\(\approxeq 13%\)