Correct Answer: B. 0.0084N

Explanation

\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(Mass = 28g = 0.028kg\)

\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)

\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)

Correct Answer: A. 5.7\(ms^{-1}\)

Explanation

\(s = ut + \frac{1}{2}at^{2}\)

\(250 = 6u + \frac{1}{2}(12)(6^{2})\)

\(250 = 6u + 216 \implies 6u = 250 - 216 = 34\)

\(u = \frac{34}{6} \approxeq 5.7ms^{-1}\)

Correct Answer: D. 9.0 cm

Explanation

Length of arc (in radians) = \(r \theta\)

\(10.8 = 1.2r\)

\(r = \frac{10.8}{1.2} = 9.0 cm\)

Correct Answer: A. \(\frac{5\pi}{12}\)

Explanation

\(180° = \pi rads\)

\(1° = \frac{\pi}{180}\)

\(\therefore 75° = \frac{\pi}{180} \times 75 \)

= \(\frac{5\pi}{12}\)

Correct Answer: D. 1

Explanation

P(white or red) = P(white) + P(red)

= \(\frac{3}{4} + \frac{1}{4}\)

= 1

Explanation

(a) a=8;l=52;n=?

\(S_n=\frac{n}{2}(a+l)=286\)

\(=\frac{n}{2}(-8+52)=286\)

\(=\frac{n}{2}(44)=286\)

=22n=286

\(∴n=\frac{286}{22}=13\)

(b) l = a + (n - 1)d = 52

= -8 + (13 - 1)d = 52

= -8 + 12d = 52

= 12d = 52 + 8

= 12d = 60

\(∴d=\frac{60}{12}=5\)

Correct Answer: C. y - x - 6=0

Explanation

From y= 2x- 5x 10 ; dy/dx= 4x-5

But at (1, 7); m\(_1\) = (dy/dx)\(_{1,7}\)

= 4(1) - 5 = -1

Using m\(_1\)m\(_2\),= -1; m\(_2\) = 1.

Gradient m\(_2\), of normal at (1,7) is 1.

Using y - y\(_1\) = m(x - x\(_1\))

y - 7 = 1(x - 1) ;

y - 7 = x - 1

y - x - 6 = 0

Explanation

(a) \(-3 \ast 2 = \frac{(-3)^{2} - (2)^{2}}{2(-3)(2)}\)

= \(\frac{9 - 4}{-12}\)

= \(-\frac{5}{12}\).

(b) The operation \(\ast\) is associative if given a, b and c which are real numbers, then

\((a \ast b) \ast c = a \ast (b \ast c)\)

Let a = -3 ; b = 2 ; c = 1.

\((-3 \ast 2) \ast 1 = (-\frac{5}{12}) \ast 1\)

= \(\frac{(-\frac{5}{12})^{2} - 1^{2}}{2(-\frac{5}{12})(1)}\)

= \(\frac{\frac{25}{144} - 1}{-\frac{5}{6}}\)

= \(\frac{-\frac{119}{144}}{\frac{-5}{6}}\)

= \(\frac{119}{120}\)

\(-3 \ast (2 \ast 1) \)

\(2 \ast 1 = \frac{2^{2} - 1^{2}}{2(2)(1)}\)

= \(\frac{3}{4}\)

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

= \(\frac{9 - \frac{9}{16}}{-\frac{9}{2}}\)

= \(-\frac{15}{8}\)

\((-3 \ast 2) \ast 1 \neq -3 \ast (2 \ast 1)\)

The operation \(\ast\) is not associative.

Correct Answer: B. 2

Explanation

\(Gradient = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}\)

Line passing through \((0, 0)\) & \((1\frac{1}{4}, 2\frac{1}{2})\),

\(Gradient = \frac{2\frac{1}{2} - 0}{1\frac{1}{4} - 0} = \frac{\frac{5}{2}}{\frac{5}{4}} \)

= 2.

Correct Answer: C. 29.5

Explanation

£f = 4 + 12 + 16 + 6 +2 = 40

Upper quartile class

= 3/4 of 40 = 30th position

Counting from the distribution table to the 30th position,

the upper class boundary is 29.5