The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x
2
6
18
12
5
Correct answer is D
\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)
\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\)
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
The sides cannot be 0 hence x = 12.
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
\(\frac{x}{(x - 3)(x - 7)}\)
\(\frac{x}{(x + 3)(x - 7)}\)
\(\frac{x}{(x + 3)(x + 7)}\)
\(\frac{x}{(x - 3)(x + 7)}\)
Correct answer is C
\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
= \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\)
= \(\frac{x}{(x + 3)(x + 7)}\)
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
y = x2 + \(\frac{1}{\sqrt{x}}\)
y = x2 + \(\frac{1}{x}\)
y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
Correct answer is A
y = kx2 + \(\frac{c}{\sqrt{x}}\)
y = 2when x = 1
2 = k + \(\frac{c}{1}\)
k + c = 2
y = 6 when x = 4
6 = 16k + \(\frac{c}{2}\)
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = \(\frac{10}{31}\)
c = 2 - \(\frac{10}{31}\)
= \(\frac{62 - 10}{31}\)
= \(\frac{52}{31}\)
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
The value of (0.03)3 - (0.02)3 is
0.019
0.0019
0.00019
0.000019
0.000035
Correct answer is D
Using the method of difference of two cubes,
\(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\)
\((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\)
= \((0.01)(0.0009 + 0.0006 + 0.0004)\)
= \(0.01 \times 0.0019\)
= \(0.000019\)
Make T the subject of the equation \(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)
T = \(\frac{3av}{1 - v}\)
T = \(\frac{1 + v}{2a^2v^3}\)
T = \(\frac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2}\)
\(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
Correct answer is D
\(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)
\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
\(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
= \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)