Can someone help me out please? Thanks! Please explain 'cause I have more that I want to do on my own :)

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3 Answers
MysticSmile
Sep 18, 2017

#4sqrt5#

Explanation:

I can't really see the text so I just assumed it was #(4sqrt400)/(4sqrt5)#.

The first thing I will do is to cancel the #4# because #4/4=1# and the #4# is unnecessary so the equation would be #sqrt400/sqrt5#.
Let's focus on the top part which is #sqrt400#.

#sqrt400# is the same as #sqrt(2^4*5^2)# which is the same as #sqrt2^4sqrt5^2#.

#sqrt(2^4)# is the same as #2 4/2# and the same as #2^2#.

#sqrt# and #x^y# are opposites so #sqrt(5^2)# is equal to #5#.

So the numerator would be #2^2*5#. The whole equation would be #(2^2*5)/sqrt5# or #2^2sqrt5#. #2^2=4# so the equation would be #4sqrt5#. (Simplified)

Tony B
Sep 18, 2017

Assumption: the question reads as #root(4)(400)/root(4)(5)#

#2root(4)(5)#

Explanation:

Look for values raised to the power of 4

Tony B

This gives:

#root(4)(2^4xx5^2)/root(4)(5)#

#2xxroot(4)(5^2)/root(4)(5)#

#2xxroot(4)(5^2/5)#

#2root(4)(5)#

Jim G.
Sep 18, 2017

#2root(4)(5)#

Explanation:

#(root(4)(400))/(root(4)(5))=root(4)(400/5)=root(4)(80)#

#root(4)(80)=root(4)(16xx5)#

#color(white)(xxx)=root(4)(16)xxroot(4)(5)#

#color(white)(xxx)=2root(4)(5)#

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