## Micheellexox Group Title Algebra help pleasee one year ago one year ago

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1. Micheellexox

2. dumbcow

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 $\sqrt{x^{7}} = x^{7/2} = x^{3} x^{1/2} = x^{3} \sqrt{x}$

3. Jhannybean

$\frac{ 7q^4 }{{\sqrt{q ^{15}}} }$ 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 $\frac{ 7 }{ q ^{7/2} }$

4. Jhannybean

Hope that's not too complicated! :(

5. dumbcow

never leave a square root in denominator either... $\frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{x}$

6. Micheellexox

Im confused... Lol. This whole thing is confusing. I am just trying to finish up this last question. /:

7. Jhannybean

What part is confusing?

8. Jhannybean

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.

9. mahdi2020

$\frac{ 7q ^{4} }{ \sqrt{q ^{8}}*\sqrt{q ^{7}} }$ = $\frac{ 7 }{ \sqrt{q ^{7}} }= \frac{ 7 }{ q ^{3}*\sqrt{q} }$ $ans*\frac{ \sqrt{q} }{ \sqrt{q} }$ =$\frac{ 7\sqrt{q} }{ q ^{3}*q }=\frac{ 7\sqrt{q} }{ q ^{4} }$

10. mahdi2020

q^4 = sqrt(q*8)