How do you evaluate #x^2y^0 # for x = -2 and y =5?

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Answer 1 · 2017-08-04T04:34:12

See a solution process below:

First, substitute #color(red)(-2)# for #color(red)(x)# and #color(blue)(5)# for #color(blue)(y)#:

#color(red)(x)^2color(blue)(y)^0# becomes:

#color(red)((-2))^2color(blue)(5)^0#

Next, use this rule of exponents to evaluate the #5# term:

#a^color(red)(0) = 1#

#color(red)((-2))^2color(blue)(5)^0 => color(red)((-2))^2 * 1 => color(red)((-2))^2#

Next, evaluate the #-2# term as:

#color(red)((-2))^2 => -2 xx -2 = 4#