If the angles of a quadrilateral are (3y + 10)°, (2y + 30)°, (y + 20)° and 4y°. Find the value of y.
66°
12°
30°
42°
Correct answer is C
Sum of angles in a quadrilateral = 360°
\(\therefore\) (3y + 10) + (2y + 30) + (y + 20) + 4y = 360
10y + 60 = 360 \(\implies\) 10y = 300
y = 30°
Find \(\frac{\mathrm d y}{\mathrm d x}\) if \(y = \cos x\).
\(\sin x\)
\(- \sin x\)
\(\tan x\)
\(- \tan x\)
Correct answer is B
If \(y = \cos x\)
\(\frac{\mathrm d y}{\mathrm d x}\) =—\(\sin x\)
52 cm
43 cm
40 cm
15 cm
Correct answer is B
\(\tan 16 = \frac{x}{150}\)
\(x = 150 \tan 16\)
= \(43 cm\)
30\(\pi\) cm\(^2\)
60\(\pi\) cm\(^2\)
120\(\pi\) cm\(^2\)
150\(\pi\) cm\(^2\)
Correct answer is B
Area of sector = \(\frac{\theta}{360°} \times \pi r^{2}\)
= \(\frac{150}{360} \times \pi \times 12^{2}\)
= 60\(\pi\) cm\(^{2}\)
If \(N = \begin{pmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{pmatrix}\), find \(|N|\).
65
23
17
91
Correct answer is C
\(\begin{vmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{vmatrix}\)
= \(3(-3 - (-10)) - 5(6 - 10) + (-4)(12 - 6)\)
= \(21 + 20 - 24\)
= 17
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