If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y.
-2
\(-\frac{1}{2}\).
\(\frac{1}{2}\)
2
Correct answer is D
\(2^{-2(2-y)}-x=2^{0}; -4 + 2y = 0\\
2y = 4; y = 2\)
In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS
35o
70o
100o
125o
Correct answer is A
∠PSQ = 90° (angle in a semi-circle)
(when SR is joined to SP)
∠SPQ = 180 – (90+35) = 180 – 125 = 55°
∠QRS + ∠SPQ = 180° (opposite angles in a cyclic quad is supplementary)
∠QRS = 180° - 55°
= 125°
Given that x + y = 7 and 3x-y = 5, evaluate \(\frac{y}{2}-3\).
-1
1
3
4
Correct answer is A
\(x + y = 7 ------ I\\
3x – y = 5 ------ II\)
Add I and II;
\(4x = 12 => x = \frac{12}{4} = 3\\
Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\\
\frac{4}{2} – 3 => 2-3 = -1\)
8√3cm2
12√3cm2
16√2cm2
16√3cm2
Correct answer is D
|AB| = 4cm
|BC| = 8cm
In right-angled BAC;
8\(^2\) = 4\(^2\) + |AC|\(^2\)
|AC|\(^2\) = 8\(^2\) - 4\(^2\)
|AC\\(^2\) = 64 - 16 → 48
|AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\)
The area of rectangle = L x B
= |AB| x |AC|
= \((4 \times 4\sqrt{3}cm^2\)
=\(16\sqrt{3}cm^2\)
62.35m
20.78m
18.00m
10.39m
Correct answer is B
Where H is the height of the tower H = ?
\(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\\
H = 36 \times 0.5774 = 20.79\)
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