Question Library

Fountain is pronounced /ˈfaʊntɪn/ Mountain is pronounced /ˈmaʊntɪn/

The idiomatic expression to have a bee in one's bonnet means to be preoccupied or obsessed with an idea, to talk about it continually, or to have an excessive, fixation-like passion for something. Option (C) captures this meaning.

This sentence uses the Third Conditional to express an unreal past condition and its hypothetical past result. The standard grammatical formula is: If+past perfect (had known),⋯+would/wouldn’t+have+past participle (have given)

Emphatic stress placed on the capitalized word MEETING corrects an erroneous location/event while confirming the rest of the sentence. Question (D) asks if they came to the party, to which the direct corrective reply is: "No, they came with me to the MEETING."

To applaud someone's actions, decisions, or motives in a non-literal context means to praise, express strong approval of, or commend them.

To applaud someone's actions, decisions, or motives in a non-literal context means to praise, express strong approval of, or commend them.

When a quadratic expression is a perfect square, its discriminant (\(b^2 - 4ac\)) must equal zero. Set \(b^2 - 4ac = 0\) 4x\(^2\) - 8x + k Let what must be added be k Comparing \(b^2 - 4ac = 0\) with 4x\(^2\) - 8x + k a = 4, b = -8, c = k \(b^2\) = 4ac (-8)\(^2\) = 4 . 4 x k 64 = 16k k = \(\frac{64}{16}\) k = 4

There are 3s Therefore, mode = 3

Range = maximum value - minimum value Range = 10 - 2 = 8

m\(_1\) = \(\frac{y_2 - y_1}{x_2 - x_1}\) The given points are -1, 3 and 3, 1 m\(_1\) = \(\frac{1 - 3}{3 - (-1)}\) = \(\frac{-2}{4}\) = - \(\frac{1}{2}\) The perpendicular gradient(m_2\) = m\(_1\) x m\(_2\) = -1 = - \(\frac{1}{2}\) x m\(_2\) = -1 m\(_2\) = -1 x - \(\frac{2}{1}\) = 2

\(\frac{3}{5}\)x + 3 > \(\frac{2}{3}\)x - 4 Multiply through by the LCM of 5 and 3, which are the denominators of the two fractions. 15 X \(\frac{3}{5}\)x + (3 X 15) > 15 X \(\frac{2}{3}\)x - (4 X 15) 3 X 3x + 45 > 5 X 2x - 60 9x + 45 > 10x - 60 Collect like terms 9x - 10x > -60 - 45 - x > - 105 Divide through by -1 \(\frac{- x}{ - 1}\) > \(\frac{-105}{-1}\) x < 105

To rationalise \(\frac{\sqrt{2}}{3 + \sqrt{2}}\), multiply both the numerator and denominator of the fraction by the conjugate of the denominator, that is, 3 - \(\sqrt{2}\) \(\frac{\sqrt{2}}{3 + \sqrt{2}}\) x \(\frac{3 - \sqrt{2}}{3 - \sqrt{2}}\) = \(\frac{3\sqrt{2} - 2}{9 - 2}\) = \(\frac{3\sqrt{2} - 2}{7}\)

The gradient of the curve is given by the derivative \(\frac{dy}{dx}\) Given that y = x\(^3\) + 3x\(^2\) − 9x + 5 \(\frac{dy}{dx}\) = 3x\(^2\) + 6x - 9 = 0 3x\(^2\) + 6x - 9 = 0 Divide through by 3 x\(^2\) + 2x - 3 = 0 Factorise x\(^2\) + 3x - x - 3 = 0 x(x + 3) - 1(x + 3) = 0 (x - 1)(x + 3) = 0 x = 1 Or x = -3 Therefore, x = -3 and x = -1

\(\begin{vmatrix} 3 & 1 & 2 \\ -1 & 4 & 1 \\ 2 & 1 & 3 \end{vmatrix} = 3\begin{vmatrix} 4 & 1 \\ 1 & 3 \end{vmatrix} - 1\begin{vmatrix} -1 & 1 \\ 2 & 3 \end{vmatrix} + 2\begin{vmatrix} -1 & 4 \\ 2 & 1 \end{vmatrix}\) = 3(12 - 1) - 1(-3 - 2) + 2(-1 - 8) = 3(11) - 1(-5) + 2(-9) = 33 + 5 - 18 = 20.

The locus of points equidistant from two fixed points P and Q is the perpendicular bisector of the line segment PQ. The midpoint M of PQ is found using the midpoint formula: M = \(\left(\frac{10 + 8}{2},\ \frac{4 + 6}{2} \right)\) = (9, 5). Thus, the perpendicular bisector of PQ passes through the point (9, 5).