Answer :
So that the ratio is defined:
* The denominator can not be zero
* Being an integer index pair, the filing must be greater or equal to zero.
It is concluded that:
x ≠ 0 ∧ 2x ≥ 0
x ≥0
.:. x > 0
R/ alternative d) x>0
Jeizon1L :)
* The denominator can not be zero
* Being an integer index pair, the filing must be greater or equal to zero.
It is concluded that:
x ≠ 0 ∧ 2x ≥ 0
x ≥0
.:. x > 0
R/ alternative d) x>0
Jeizon1L :)
[tex] \sqrt{\frac{8x^{2}}{2x}} = \frac{\sqrt{8x^{2}}}{\sqrt{2x}} = \frac{\sqrt{4 * x^{2} * 2}}{\sqrt{2x}} = \frac{ \sqrt{4}\sqrt{x^{2}} \sqrt{2}}{\sqrt{2x}} = \frac{2x\sqrt{2}}{\sqrt{2x}} = \frac{2x}{\sqrt{x}} = 2\sqrt{x} [/tex]
The answer is D, x > 0.
The answer is D, x > 0.
