Algebra help pleasee

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Algebra help pleasee

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
the square root is same as exponent of (1/2) ... so divide the exponent in half to nearest integer , then leave the remainder inside square root \[\sqrt{x^{7}} = x^{7/2} = x^{3} x^{1/2} = x^{3} \sqrt{x}\]
\[\frac{ 7q^4 }{{\sqrt{q ^{15}}} }\] so now we're focusing on just q.\[\frac{ q^4 }{ q ^{\frac{ 15 }{ 2 }} }\]So now we have \[q ^{(15/2)-4}= q ^{(15/2)-(8/2)}= q ^{7/2}\] wecan also rewrite q under a radical. \[\sqrt{q ^{7}}\] putting it altogether, we would have \[\frac{ 7 }{ \sqrt{q ^{7}}}\] or you can write this as \[\frac{ 7 }{ q ^{7/2} }\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hope that's not too complicated! :(
never leave a square root in denominator either... \[\frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{x}\]
Im confused... Lol. This whole thing is confusing. I am just trying to finish up this last question. /:
What part is confusing?
Oh, in correlation to what @dumbcow was referring to, \[\frac{ 7 }{ \sqrt{q ^{7}} }*\frac{ \sqrt{q ^{7}} }{ \sqrt{q ^{7}} }=\frac{ 7\sqrt{q ^{7}} }{ q ^{7} }\] As he said,you can't have a radical in the denominator.
\[\frac{ 7q ^{4} }{ \sqrt{q ^{8}}*\sqrt{q ^{7}} }\] = \[\frac{ 7 }{ \sqrt{q ^{7}} }= \frac{ 7 }{ q ^{3}*\sqrt{q} }\] \[ans*\frac{ \sqrt{q} }{ \sqrt{q} }\] =\[\frac{ 7\sqrt{q} }{ q ^{3}*q }=\frac{ 7\sqrt{q} }{ q ^{4} }\]
q^4 = sqrt(q*8)

Not the answer you are looking for?

Search for more explanations.

Ask your own question