4.19cm2
8.38cm2
10.5cm2
20.9cm2
Correct answer is A
Given q2 = 25 - r2.....(1)
from pythagora's theorem
P2 = q2 + r2.......(2)
put (1) into (2)
p2 = 25 - p2 + x2
p2 = 25
p = \(\sqrt{25}\)
= 5
Simplify: \(\frac{2x^2 - 5x - 12}{4x^2 - 9}\)
\(\frac{x + 4}{2x + 3}\)
\(\frac{x + 4}{2x - 3}\)
\(\frac{x - 4}{2x + 3}\)
\(\frac{x - 4}{2x - 3}\)
Correct answer is D
\(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) = \(\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}\)
= \(\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}\)
= \(\frac{x - 4}{2x - 3}\)
If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r
\(\frac{9}{25} pr^2\)
\(\frac{9}{25} p^2r\)
\(\frac{25}{9} p^2r\)
\(\frac{25}{9} pr^2\)
Correct answer is C
p = \(\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}\)
= (\(\frac{5}{3}p\))2
= \(\frac{q}{r}\)
= \(\frac{25p^2}{9} = \frac{q}{r}\)
q = \(\frac{25}{9} p^2 r\)
If 4y is 9 greater than the sum of y and 3x, by how much is y greater than x?
3
6
9
12
Correct answer is A
4y - 9 > y + 3x; 4y - y > 3x + 9
3y > 3(x + 3); y = > \(\frac{3(x + 3)}{3}\)
y > x + 3; y - 3 > x
y is greater than x by 3
(a - y)(5y - 3a)
(y - a)(5y - 3a)
(y - a)(5y + 3a)
(y + a)(5y - 3a)
Correct answer is D
5y2 + 2ay - 3a2 = 5y2 + 5ay - 3a2
= 5y(y + a) - 3a(y + a)
= (y + a)(5y - 3a)