x > -\(\frac{1}{6}\)
x > 0
0 < x < 6
0 < x <\(\frac{1}{6}\)
Correct answer is A
\(\frac{1}{3x}\) + \(\frac{1}{2}\)x = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4(2 + 3x) > 6x = 12x\(^2\) - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x - 1 > 0
= x < 0, x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x - 1 > 0 = x > 0
x > \(\frac{1}{6}\)
Combining solutions in cases (1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
Simplify \(\frac{(√12-√3)}{(√12+√3)}\) ...
Use the graph of the curve y = f(x)to solve the inequality f(x) \(\leq\) 0...
If sin θ = \(\frac{\sqrt{3}}{2}\) and 0 < θ < 180°, what is cos θ ?...
For what value of x is the expression x - 5/x2 - 2x - 3 not defined?...
Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx) ...