Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.