Express \(\frac{2}{x + 3} - \frac{1}{x - 2}\) as a simple fraction
\(\frac{x - 7}{x^2 + x - 6}\)
\(\frac{x - 1}{x^2 + x - 6}\)
\(\frac{x - 2}{x^2 + x - 6}\)
\(\frac{x - 27}{x^2 + x - 6}\)
Correct answer is A
\(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\)
= \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\)
= \(\frac{x -7}{x^2 + x - 6}\)
= \(\frac{x - 7}{x^2 + x - 6}\)
11
\(\frac{15}{2}\)
5
\(\frac{5}{2}\)
Correct answer is C
Let the number be y, subtract y from 2 i.e 2 - y
2 - y = 4 < \(\frac{1}{5}\) y,
2 - y = \(\frac{y}{5}\) - 4
2 - y + 4 = \(\frac{y}{5}\)
6 = \(\frac{y}{5}\) + y
6 = \(\frac{y + 5y}{5}\)
6 = \(\frac{6y}{5}\)
multiplying through by 5
6 * 5 = 6y
\(\frac{30}{6}\) = y
= 5
the bisector of the straight line joining P and M
an arc of a circle with PM as a chord
the bisector of angle PXM
a circle centre X and radius PM
Correct answer is B
No explanation has been provided for this answer.
The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k
zero
-2
-4
-8
Correct answer is D
y = x2 + 2x + k at point(2,0) x = 2, y = 0
0 = (2)2 + 2(20 + k)
0 = 4 + 4 + k
0 = 8 + k
k = -8
If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9
8
7
6
5
Correct answer is A
y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\)
6 = k\(\sqrt{3 + 1}\)
6 = k\(\sqrt{4}\)
6 = 2k
k = \(\frac{6}{2}\) = 3
y = \(\sqrt{(x + 1)}\)
9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)
\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)
3 = \(\sqrt{x + 1}\)(square both sides)
9 = x + 1
x = 9 - 1
x = 8