How do you evaluate #3y(x+x^2-y)# given x=-2, y=-3, z=5?

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Answer 1 · 2016-10-30T19:27:00

#-45#

#3y(x+x^2-y)#

Well you would put in the given values for x and y.

So #-2# would go in for where each #x# is and

#-3# would go in for #y# making the expression:

# 3(-3)[-2+(-2)^2-(-3))] # I am not sure where z fits in but it may just be extra..

=#-9[-2+4+3]#

=#-9 xx 5#

=#-45#