If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24
\(\frac{1}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
Correct answer is A
\(y \propto \frac{1}{\sqrt{x}}\)
\(y = \frac{k}{\sqrt{x}}\)
When x = 16, y = 2.
\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)
\(k = 8\)
\(y = \frac{8}{\sqrt{x}}\)
When y = 24,
\(24 = \frac{8}{\sqrt{x}}\)
\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)
\(\therefore x = (\frac{1}{3})^2\)
\(x = \frac{1}{9}\)
Subtract (-y + 3x + 5z) from (4y - x - 2z)
5y - 4x - 7z
3y + 2x + 3z
-5y + 4x + 7z
2x - 5y + 3z
Correct answer is A
(4y - x - 2z) - (-y + 3x + 5z)
= 4y + y - x - 3x - 2z - 5z
= 5y - 4x - 7z
y = -1
y = 1
y = 3
y = -3
Correct answer is D
Let the straight line PQ = y
2x\(^2\) - 5x - 1 - y = 2x\(^2\) - 5x + 2
y = 2x\(^2\) - 5x - 1 - 2x\(^2\) + 5x - 2
y = -3
which of the following is not quadratic expression?
\(y = 2x^2 - 5x\)
\(y=x(x-5)\)
\(y = x^2 - 5\)
y = 5(x-1)
Correct answer is D
A quadratic equation is an equation of the second order. The highest power in the equation is 2.
If \(2x^2 + kx - 14 = (x+2)(2x-7)\), find the value of K
-3
5
9
11
Correct answer is A
\(2x^2 + kx - 14 = (x+2)(2x-7)\\
∴2x^2 + kx - 14 = 2x^2 - 3x - 14\)
equating coefficient K = -3