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Area of lengthwise 2 walls = \(2( L\times H ) = 2 \times 7 \times x = 14xcm^2\) Area of breadthwise 2 walls = \(2( B \times H) = 2 \times 5 \times x = 10xcm^2\) T.S.A of the room( walls only) = \(14xcm^2 + 10xcm^2 = 96cm^2\) x = \(\frac{ 96}{14} = 4cm^2\)

The expression is not defines when \(y^2\) - y - 6 = 0 => \(y^2\) - 3y + 2y - 6 = 0 => (y-3)(y+2) = 0 => y -3 = 0 or y + 2 = 0 => y = 3 or -2

The two points of contact of the curve with the x-axis are x = 1 and x = 4. The roots of the equation are x = 1 and x = 4

The factor for the curve are (x-1)(x-4). The product off the curves gives \(=x^2-5x+4=y\)

∠QPT = ∠QTP = 70o. Since ∠QTR = 140o ∠PRS = TSR = 1/2(180-40) = 70o

Area of a sector \(=\frac{\theta}{360}\times \pi r^2\\ \left(\frac{144}{360}\times \pi \times \frac{3.5}{1} \times 3.5\right)cm^2 = 4.9\pi cm^2\)

Volume of rectangular packet = (16 x 12 x 6)cm3 = 1152cm3. Volume of cube of sugar = 2 x 2 x 2 = 8cm ∴ The number of cubes f sugar \(=\frac{1152}{8}=144\)

y \(\propto x^2\), y = 1\(\frac{1}{4}\), x = 4 y = \(\frac{\text{k}}{x^2}\) k = y x\(^2\) = \(\frac{5}{4}\) x 4\(^2\) = 20 y = \(\frac{20}{x^2}\) = \(\frac{20}{x^2}\) = \(\frac{20}{(\frac{1}{2})^2}\) = \(\frac{20}{\frac{1}{4}}\) = 20 x 4 = 80

\(\% increase = \frac{Increase}{initial\hspace{1mm} salary}\times 100\%\) Increase = N3,200 - N2,700 = N500 Initial \(\hspace{1mm} salary = N2,700.00 = \frac{N500}{N2,700}\times 100\% = 18.5\%\)

Reflex + ∠OML + ∠ONL + ∠NLM = 360 280o + 10o + 15o + x = 360o => 305o + xo = 360o x = 360o - 305o = 55o

Since PQRS is a quadrilateral with equal sides, PQR cannot be an equilateral triangle unless R coincides with S, which is not the case in a quadrilateral.

FROM the diagram above 100 + a = 180 (sum of angles on a straight line) a = 180 - 100 = 80º Z = 80 + 40 = 120º (exterior angles of a triangle = sum of the two opp. interior angles) Z = x = 120º (corresponding angles are equal).

The bearing of R from S = 180 + 44 = 224º, because -Following the clockwise movement at 90º: -NORTH -EAST, EAST-SOUTH, SOUTH - WEST & WEST - NORTH - It is observed that R is situated between SOUTH - WEST poles on S

Area of a rhombus = \(\frac{pq}{2}\) where p and q are the two diagonals of the rhombus.

No expanded working has been attached for this question yet. The verified answer is Option B.