Hi. I need help on a couple of algebra problems. Can you please explain step by step how to do these?
Thanks.
�ã(3x (whole thing has radical sign)
----
2y^3
^3(�ã2x^4y^4)
------------ (the whole thing is cubed
9x
7 answers
The squiggly A was originally a radical sign, but I guess it got changed.
I will assume I am looking at
√(3x)/(2y^3) and
( √(2x^4 y^4) )/(9x)
are they multiplied or what ?
explain please
√(3x)/(2y^3) and
( √(2x^4 y^4) )/(9x)
are they multiplied or what ?
explain please
No. These two are both separate problems. The book asks to simplify each problem. And for the 2nd problem the index is a four.
Still not too clear if the √ governs the whole thing in the first
Actually there is very little that can be done in terms of simplification.
If it is
√( 3x/(2y^3) ) remember that √ is equivalent to having an exponent of 1/2
so all you get is
√(3/2)x^(1/2) / y^(3/2) which is certainly not any more simpified
for the second I will read it as:
( √2x^4 y^4/(9x) )^3
= 2√2x^12 y^12/(729x^3
= (2√2/729)x^9 y^12
I think it was more "simplified" at the start.
Actually there is very little that can be done in terms of simplification.
If it is
√( 3x/(2y^3) ) remember that √ is equivalent to having an exponent of 1/2
so all you get is
√(3/2)x^(1/2) / y^(3/2) which is certainly not any more simpified
for the second I will read it as:
( √2x^4 y^4/(9x) )^3
= 2√2x^12 y^12/(729x^3
= (2√2/729)x^9 y^12
I think it was more "simplified" at the start.
Well for the first one, the answer s supposed to be 1�ã6xy/2y^2.
The second problem's answer is xy/3 times the cubed root of 6y
The second problem's answer is xy/3 times the cubed root of 6y
Can you see how important brackets are when typing on this forum?
I can't see any way to obtain those answers the way the questions were typed.
I can't see any way to obtain those answers the way the questions were typed.
That's alright then. Thank you very much.