Can someone please explain how to simplify these, thanks so much.
(2b^4)^-1
(x^2y^-1)^-1
x^2/2x^3
When you have a term raised to a negative exponent, you take the reciprical of the term.
(2b^4)^-1 = 1/(2b^4)
(x^2y^-1)^-1 = 1/(x^2y^-1) = y/x^2 <-- notice that the y had a negative exponent and so could be brought back up the numerator
x^2/(2x^3) <-- in this case, we need to subtract the exponents since the "x"s are the same base. We get
1/2*x^(2 - 3) = 1/2*x^-1 = 1/(2x)
If that's not clear, say so and I'll try to explain it better.
Hello, Natalie!
(2b^4)^-1 .= .(2^-1)(b^4)^-1 .= .(2^-1)·(b^-4)1) .(2b^4)^-1
. . . . . . . . 1
Answer: .-------
. . . . . . .2b^4
(x^2·y^-1)^-1 .= .(x^2)^-1·(y^-1)^-1 .= .(x^-2)·(y)2) .(x^2·y^-1)^-1
. . . . . . . y
Answer: .---
. . . . . . .x²
. . . . . . . . x² . . . .13) .x²/2x³
We have: .---- .= .---
. . . . . . . .2x³ . . .2x