Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
I suppose you meant z > 0, or meant to use x's in place of z'sā¦
If z > 0, then
[tex]\dfrac{20\sqrt{z^6}}{\sqrt{16z^7}} = \dfrac{20\sqrt{\left(z^3\right)^2}}{\sqrt{4^2\left(z^3\right)^2z}} = \dfrac{20\left|z^3\right|}{4\left|z^3\right|\sqrt{z}} = \boxed{\dfrac{5}{\sqrt{z}}}[/tex]
because
[tex]\sqrt{z^2} = |z|[/tex]
and if z is positive, we have |z| = z, and we can eliminate common factors of z in the fraction,
[tex]\dfrac{\left|z^3\right|}{\left|z^3\right|} = \dfrac{z^3}{z^3} = 1[/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.