SOLUTION: To simplify the expression {{{ sqrt (x^2*y^2) * 3 sqrt (x^5*y^4) * 5 sqrt (x^5) }}} into the form x^r*y^s we first rewrite each term in fractional powers: {{ sqrt (x^2*y^2) = x^1*

Algebra ->  Square-cubic-other-roots -> SOLUTION: To simplify the expression {{{ sqrt (x^2*y^2) * 3 sqrt (x^5*y^4) * 5 sqrt (x^5) }}} into the form x^r*y^s we first rewrite each term in fractional powers: {{ sqrt (x^2*y^2) = x^1*      Log On


   

Question 155266: To simplify the expression +sqrt+%28x%5E2%2Ay%5E2%29+%2A+3+sqrt+%28x%5E5%2Ay%5E4%29+%2A+5+sqrt+%28x%5E5%29+ into the form x^r*y^s we first rewrite each term in fractional powers:
{{ sqrt (x^2*y^2) = x^1*y^1 }},
{{ 3 sqrt (x^5*y^4) = x^(5/3)*y^(4/3) }}
and {{ 5 sqrt (x^5) = x^1 }}.
Combining all the powers we get
+sqrt+%28x%5E2%2Ay%5E2%29+%2A+3+sqrt+%28x%5E5%2Ay%5E4%29+%2A+5+sqrt+%28x%5E5%29+ = x^__*y^__.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
IF THE 3 & 5 ARE POWER ROOTS THEN:
(X^1Y^1)(X^5/3Y^4/3)(X^5/5)=(X^1Y^1)(X^5/3Y^4/3)(X^1)=X^11/3Y^7/3 ANSWER.