How do you evaluate #10x^2y^-2# for #x=-1# and #y=-2#?

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Answer 1 · 2016-07-03T08:41:27

#2 1/2 -> 5/2#

Given:#" "10x^2y^(-2)#

Note that #y^(-2)# is the same as #1/y^2#

#(10x^2)/(y^2)#

By substitution

#color(brown)((10x^2)/(y^2))color(blue)(->(10xx(-1)^2)/((-2^2))#

'............................................................
There is a difference between # -y^2" and "y^2#

For #-y^2->-(-2)^2 = -(+4)=-4#
For #y^2-> (-2)^2=+4#
'.........................................................

#color(brown)((10x^2)/(y^2))color(blue)(->(10xx1)/((+4)) = 10/4#

Divide top and bottom by 2 giving:

#(10-:2)/(4-:2) = 5/2 = 2 1/2#