A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ
\(\frac{3}{5}\)
\(\frac{2}{3}\)
\(\frac{3}{2}\)
\(\frac{5}{3}\)
Correct answer is C
Let: (x1, y1) = (1, 2)
(x2, y2) = (5, 8)
The gradient m of \(\bar{PQ}\) is given by
m = \(\frac{y_2 y_1}{x_2 - x_1}\)
= \(\frac{8 - 2}{5 - 1}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
3
5
6
8
Correct answer is A
Volume of pyramid = \(\frac{1}{3}\)lbh
90 = \(\frac{1}{3} \times x \times 6 \times 15\)
x = \(\frac{90 \times 33}{6 \times 15}\)
= 3
2.6cm
3.5cm
3.6cm
7.0cm
Correct answer is B
Curved surface area of cylinder = 2\(\pi\)rh
110 = 2 x \(\frac{22}{7}\) x r x 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
Factorize; (2x + 3y)2 - (x - 4y)2
(3x - y)(x + 7y)
(3x + y)(2x - 7y)
(3x + y)(x - 7y)
(3x - y)(2x + 7y)
Correct answer is A
(2x + 3y)2 - (x - 4y)2
= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)
= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)
= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2
= 3x2 + 20xy - 7y2
= 3x2 + 21xy - xy - 7y2
= 3x(x + 7y) - y(x + 7y)
= (3x - y)(x + 7y)
Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)
s = \(\frac{mrp}{nr + m^2}\)
s = \(\frac{nr + m^2}{mrp}\)
s = \(\frac{nrp}{mr + m^2}\)
s = \(\frac{nrp}{nr + m^2}\)
Correct answer is D
P = S + \(\frac{sm^2}{nr}\)
P = S(1 + \(\frac{m^2}{nr}\))
P = S(1 + \(\frac{nr + m^2}{nr}\))
nrp = S(nr + m2)
S = \(\frac{nrp}{nr + m^2}\)