Further Mathematics Questions and Answers

The third quartile can be found in the \((\frac{3N}{4})^{th}\) position

= \((\frac{3 \times 100}{4})^{th}\) position

This is found in the class 14 - 16.

The upper class boundary = 16.5

Using the remainder theorem, the remainder when a polynomial \(ax^{2} + bx + c\) is divided by \((x - a)\) is equal to \(f(a)\).

\(2x^{3} + 3x^{2} + qx - 1\) divided by \((x + 2)\), the remainder = \(f(-2)\)

\(\implies f(-2) = f(1)\)

\(f(-2) = 2(-2^{3}) + 3(-2^{2}) + q(-2) - 1 = -16 + 12 - 2q - 1 = -5 - 2q\)

\(f(1) = 2(1^{3}) + 3(1^{2}) + q(1) - 1 = 2 + 3 + q - 1 = 4 + q\)

\(4 + q = -5 -2q \implies 4 + 5 = -2q - q = -3q\)

\(q = -3\)

The equation \(ax^{2} + bx + c\) is a perfect square if \(b^{2} = 4ac\).

\(x^{2} - x + p\)

\((-1)^{2} = 4(1)(p)\)

\(1 = 4p \implies p = \frac{1}{4}\)

No explanation has been provided for this answer.

\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)

\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)

\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)

\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)

= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)

= 8.66 N