Further Mathematics Questions and Answers

No explanation has been provided for this answer.

\(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9}) = \lim\limits_{x \to 3} (\frac{x^{3} - 3x^{2} + 4x^{2} - 12x}{(x - 3)(x + 3)}\)

\(\lim\limits_{x \to 3} (\frac{(x^{2} + 4x)(x - 3)}{(x - 3)(x + 3)} = \lim\limits_{x \to 3} (\frac{x^{2} + 4x}{x + 3})\)

=\(\frac{3^{2} + 4(3)}{3 + 3} = \frac{21}{6} = \frac{7}{2}\)

Distance between two points (a, b) and (c, d) = \(\sqrt{(d - b)^{2} + (c - a)^{2}}

Distance between (2, 5) and (5, 9) = \(\sqrt{(9-5)^{2} + (5-2)^{2}}\)

= \(\sqrt{16 + 9} = \sqrt{25} = 5 units\)

Maximum height (H) = \(\frac{u^{2}}{2g}\)

= \(\frac{15^{2}}{2 \times 10} = \frac{225}{20}\)

= \(11.25m\)

For collinear points (points on the same line), the slopes are equal for any 2 points on the line.

Given (-1, t - 1), (t, t - 3), (t - 6, 3), 

\(slope = \frac{(t-3) - (t-1)}{t - (-1)} = \frac{3 - (t-3)}{(t-6) - t} = \frac{3 - (t-1)}{(t-6) - (-1)}\)

Taking any two of the equations above, solve for t.

\(\frac{t - 3 - t + 1}{t + 1} = \frac{6 -t}{-6}\)

\(12 = (6 - t)(t + 1)\)

\(-t^{2} + 5t + 6 - 12 = 0 \implies t^{2} - 5t + 6 = 0\)

Solving, we have t = 2 and 3.