How do you divide #5sqrt(16y^4) + 7sqrt(2y)#?
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"Find the sum of the integers between 2 and 100 which are divisible by 3 ?"
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#color(red)(5sqrt(16y^4) + 7sqrt(2y)=20y^2+7sqrt(2)sqrt(y)#
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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.
#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)#
