How do you divide #5sqrt(16y^4) + 7sqrt(2y)#?

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Answer 1 · 2018-08-06T03:59:13

#" "#
#color(red)(5sqrt(16y^4) + 7sqrt(2y)=20y^2+7sqrt(2)sqrt(y)#

#" "#
Given: #color(red)(5sqrt(16y^4) + 7sqrt(2y)#

My understanding of the question: Simplify the radical expression.

#rArr (5)(sqrt(16))sqrt(y^4) + 7sqrt(2)sqrt(y)#

#rArr 5*4*y^2+7sqrt(2)sqrt(y)#

#rArr 20y^2+7sqrt(2)sqrt(y)#

You can stop here.

If you so wish, you can simplify in a different way.

Consider the step:

#rArr (5)(sqrt(16))sqrt(y^4) + 7sqrt(2)sqrt(y)#

#rArr (5)(sqrt(16))sqrt(y^3)sqrt(y) + 7sqrt(2)sqrt(y)#

#rArr 5*4*sqrt(y^3)sqrt(y)+7sqrt(2) sqrt(y)#

#rArr sqrt(y)[20sqrt(y^3)+7sqrt(2)]#

Hope this helps.

Answer 2 · 2018-08-13T12:27:57

#color(magenta)(=> (40/7) (y)^(3/2)#

I am taking the sum as

#(5 * sqrt (16y^4)) / (7 sqrt (2y))#

as it has been asked to divide.

#=> (5 * sqrt (2^4y^4)) / ( 7 sqrt (2y)#

#=> (5/7) * sqrt (2^4y^4)/(sqrt 2y)#

#=> (5/7) * sqrt (2^3y^3)#

#=> (40/7) (y)^(3/2)#