Correct Answer: D. p=q^r

Explanation

If \(log_q p\) = r convert from logarithm to indices

P = q\(^r\)

Correct Answer: D. legitimacy

Explanation

Legitimacy is a political concept which refers to the acceptance and recognition by the citizens of a country the right of their rulers to govern. Power is transformed into legal authority by legitimizing it.

Correct Answer: D. -3

Explanation

x + 3 = 0 => x = -3

Correct Answer: C. afforestation

Explanation

An environmental hazard is a substance, a state or an event which has the potential to threaten the surrounding natural environment and adversely affect people's health, including pollution and natural disasters such as storms and earthquakes.

Correct Answer: C. Witr

Correct Answer: B. Stock exchange

Correct Answer: C. 64

Explanation

Third quartile Q3 = \(\frac{3}{4}\)Nth

Alternatively:

Split the score into equal quarters

44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75.

Third quartile Q3 = [63 + 65] / 2 = 64

Explanation

(a) \(y = 6 - x - x^{2}\)

When x = 0, y = 6.

When y = 0, \(x^{2} + x - 6 = 0\)

\(x^{2} + 3x - 2x - 6 = 0 \implies x(x + 3) - 2(x + 3) = 0\)

\(x = -3 ; x = 2\)

Graph passes through (0, 6), (-3, 0) (2, 0).

Turning point : \(\frac{\mathrm d y}{\mathrm d x} = - 1 - 2x\)

\(-1 - 2x = 0 \implies x = \frac{-1}{2}\)

When \(x = \frac{-1}{2}\), \(y = 6 - (-\frac{1}{2}) - (\frac{-1}{2})^{2} = 6\frac{1}{4}\).

Turning point : \((-\frac{1}{2}, 6\frac{1}{4})\).

\(y = 3x^{2} - 2x + 3\)

When x = 0, y = 3.

\(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 < 0\)

Graph does not cut x - axis.

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\)

At turning point, \(6x - 2 = 0 \implies x = \frac{2}{6} = \frac{1}{3}\)

When \(x = \frac{1}{3}, y = 3(\frac{1}{3})^{2} - 2(\frac{1}{3}) + 3 = 2\frac{2}{3}\)

Turning point : \((\frac{1}{3}, 2\frac{2}{3})\).

(b) For the x- coordinates, we use the equation.

\(3x^{2} - 2x + 3 = 6 - x - x^{2}\)

\(4x^{2} - x - 3 = 0 \implies 4x^{2} - 4x + 3x - 3 = 0\)

\(4x(x - 1) + 3(x - 1) = 0 \implies (4x + 3)(x - 1) = 0\)

\(x = -\frac{3}{4} ; 1\)

(c) Area of shaded portion :

\(\int_{-\frac{3}{4}} ^{1} [(6 - x - x^{2}) - (3x^{2} - 2x + 3)] \mathrm {d} x\)

= \(\int_{-\frac{3}{4}} ^{1} (3 + x - 4x^{2}) \mathrm {d} x\)

= \([3x + \frac{x^{2}}{2} - \frac{4}{3}x^{3}]|_{-\frac{3}{4}} ^{1}\)

= \((3(1) + \frac{1}{2} - \frac{4}{3}) - (3(-\frac{3}{4}) + \frac{(\frac{-3}{4})^{2}}{2} - \frac{4}{3} (-\frac{3}{4})^{3}\)

= \(\frac{13}{6} - (-\frac{45}{32})\)

= \(\frac{343}{96} sq. units\)

Correct Answer: B. vertical section of the soil

Correct Answer: B. Reduce water loss