Find the quadratic equation whose roots are \(\frac{1}{2}\) and -\(\frac{1}{3}\)
3x\(^2\) + x + 1 = 0
6x\(^2\) + x - 1 = 0
3x\(^2\) + x - 1 = 0
6x\(^2\) - x - 1 = 0
Correct answer is D
x = \(\frac{1}{2}\) and x = \(\frac{-1}{3}\)
(2x - 1) = 0 and (3x + 1) = 0
(2x - 1) (3x + 1) = 0
6x\(^2\) - x - 1 = 0
Make m the subject of the relation k = \(\frac{m - y}{m + 1}\)
m = \(\frac{y + k^2}{k^2 + 1}\)
m = \(\frac{y + k^2}{1 - k^2}\)
m = \(\frac{y - k^2}{k^2 + 1}\)
m = \(\frac{y - k^2}{1 - k^2}\)
Correct answer is B
k = \(\frac{m - y}{m + 1}\)
k\(^2\) = \(\frac{m - y}{m + 1}\)
k\(^2\)m + k\(^2\) = m - y
k\(^2\) + y = m - k\(^2\)m
\(\frac{k^2 + y}{1 - k^2}\) = m\(\frac{(1 - k^2)}{1 - k^2}\)
m = \(\frac{y + k^2}{1 - k^2}\)
Solve \(\frac{1}{3}\)(5 - 3x) < \(\frac{2}{5}\)(3 - 7x)
x > \(\frac{7}{22}\)
x < \(\frac{7}{22}\)
x > \(\frac{-7}{27}\)
x < \(\frac{-7}{27}\)
Correct answer is D
\(\frac{1}{3}\)(5 - 3x) < \(\frac{2}{5}\)(3 - 7x)
5(5 - 3x) < 6(3 - 7x)
25 - 15x < 18 - 42x
- 15x + 42x < 18 - 25
\(\frac{27x}{27}\) < \(\frac{-7}{27}\)
x < \(\frac{-7}{27}\)
The first term of a geometric progression (G.P) is 3 and the 5th term is 48. Find the common ratio.
2
4
8
16
Correct answer is A
T\(_5\) = ar\(^4\)
\(\frac{48}{3} = \frac{3r^4}{3}\)
16 = r\(^4\)
r = \(4\sqrt{16}\)
= 2
~ y \(\to\) ~ x
y \(\to\) ~ x
~ x \(\to\) ~ y
y \(\to\) x
Correct answer is C
No explanation has been provided for this answer.