Asked by Anonymous
How am I supposed to solve 2x^0y^-3 when x = 7 and y = -4?
Also, what steps must I take to solve x^10/d^-3?
Also, what steps must I take to solve x^10/d^-3?
Answers
Answered by
Steve
x^0 = 1
So, plug in y=-4 and you have
2(1)(-4)^-3 = 2/-64 = -1/32
If you have values for x and d, just plug them in.
Remember that odd powers of negative numbers are negative, and even powers of any number are positive.
So, plug in y=-4 and you have
2(1)(-4)^-3 = 2/-64 = -1/32
If you have values for x and d, just plug them in.
Remember that odd powers of negative numbers are negative, and even powers of any number are positive.