3.8
3.0
2.8
2.3
Correct answer is A
\(\begin{array}{c|c} height(x) & frequency(f) & fx \\ \hline 2 & 2 & 4\\ 3 & 4 & 12\\ 4 & 5 & 20 \\ 5 & 3 & 15\\ 6 & 1 & 6\end{array}\)
mean\(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{57}{15}\)
= 3.8
If (x - a) is a factor pf bx - ax + x2, find the other factor.
(x + b)
(x - b)
(a + b)
(a - b)
Correct answer is A
No explanation has been provided for this answer.
Find the gradient of the line joining the points (2, -3) and 2, 5)
9
1
2
undefined
Correct answer is D
(2, -3)(2, 5)
gradient = \(\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2}\)
= \(\frac{5 + 3}{0}\)
= \(\frac{8}{0}\)
= undefined
F x varies inversely as y and y varies directly as Z, what is the relationship between x and z?
x \(\alpha\) z
x \(\alpha \frac{1}{z}\)
x \(\alpha z^2\)
x \(\alpha \frac{1}{z^2}\0
Correct answer is B
x \(\alpha \frac{1}{y}\)
y \(\alpha z\)
the relationship = x \(\alph \frac{1}{z}\)
Make u the subject of formula, E = \(\frac{m}{2g}\)(v2 - u2)
u = \(\sqrt{v^2 - \frac{2Eg}{m}}\)
u = \(\sqrt{\frac{v^2}{m} - \frac{2Eg}{4}}\)
u = \(\sqrt{v- \frac{2Eg}{m}}\)
u = \(\sqrt{\frac{2v^2Eg}{m}}\)
Correct answer is A
E = \(\frac{m}{2g}\)(v2 - u2)
multiply both sides by 2g
2Eg = 2g (\(\frac{M}{2g} (V^2 - U^2)\)
2Eg = m(V2 - U2)
2Eg - mV2 - mU2
mU2 = mV2 - 2Eg
divide both sides by m
\(\frac{mU^2}{m} = \frac{mV^2 - 2Eg}{m}\)
U2 = \(\frac{mV^2 - 2Eg}{m}\)
= \(\frac{mV^2}{m} - \frac{2Eg}{m}\)
U2 = V2 - \(\frac{2Eg}{m}\)
U = \(\sqrt{V^2 - \frac{2Eg}{m}}\)