Explanation

\(p \propto q^2\)

\(q \propto \frac{1}{\sqrt{r}\)

\(p = kq^2\)

\(q = \frac{c}{\sqrt{r}}\)

where c and k are constants.

\(q^2 = \frac{c^2}{r}\)

\(p = \frac{kc^2}{r}\)

If k and c are constants, then kc\(^2\) is also a constant, say z.

\(p = \frac{z}{r}\)

p varies inversely as r.

Correct Answer: A. 30°

Explanation

Exterior angle = 180\(^o\) - 168\(^o\)

Number of sides = \(\frac{360^o}{12}\)

= 30\(^o\)

Correct Answer: D. 4.62 x 10^-2

Explanation

4.62 x 10-2

Correct Answer: A. 2

Explanation

\(log_{10}5 + log_{10}20\\

log_{10}[5\times 20]\\

log_{10}100\\

log_{10}10^2\\

2log_{10}10=2\)

Explanation

(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)

\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)

\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

Put \(x=1\)

\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)

⇒25=25A

=A=\(\frac{25}{25}=1\)

Put \(x=-\frac{3}{2}\)

\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)

⇒15=-2.5C

\(=C=-\frac{15}{2.5}=-6\)

Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)

\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)

\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)

\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)

By comparing the coefficient of \(x, 8=12A+B+c\)

=8=12(1)+B-6

=8=12+B-6

=8=6+B

=8-6=B

=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)

(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)

Where "a" and "b" are the coordinate of the center and "r" is the radius

2πr = 6π (given)

∴ r = 3 units

=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)

= \((x + 2)^2 + (y - 5)^2 = 9\)

= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)

= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)

∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)

Correct Answer: C. 5

Explanation

\(5+x\rightarrow 500\\

3 \rightarrow 150\\

∴ 3 \times 500 = 150 \times (5+x)\\

1500 = 750 + 150x\\

x=\frac{750}{150}=5\)

Correct Answer: C. N560.00

Correct Answer: B. 2

Explanation

Using elimination method:

4x+2y=16 * 6 --> 24x +12y=96 ... eqn iii

6x-2y=4 * 4 --> 24x - 8y = 16 ... eqn iv

Subtract eqn iv from iii

20y = 80

y = 4

Subst. y for 4 in 4x + 2y = 16

--> 4x + 2(4) = 16

--> 4x = 16 - 8

--> 4x = 8

--> x = 2

: The value of (y-x) is 4 - 2 = 2

Explanation

\(\stackrel\frown{AOB} = \stackrel\frown{BOC} = \stackrel\frown{COD} = \stackrel\frown{DOE} = 60°\)

In \(\Delta AOB\), Area of sector = \(\frac{60}{360} \times \frac{22}{7} \times 7^{2}\)

= \(25.67 cm^{2}\)

Area of triangle AOB = \(\frac{1}{2} \times \7 \times 7 \times \sin 60\)

= \(24.5 \times 0.866\)

= \(21.22 cm^{2}\)

Area of the 4 sectors = \(4 \times 25.67 = 102.68 cm^{2}\)

Area of the 4 triangles = \(4 \times 21.22 = 84.88 cm^{2}\)

Area of the shaded portion = \((102.68 - 84.88)cm^{2}\)

= \(17.80 cm^{2}\)

\(\approxeq 18 cm^{2}\)

Correct Answer: D. 2x + \(\frac{1}{x^2}\)