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12w^4/21w^6 (simplifying rational functions)
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would it be 4w^4 / 7w^6
You've correctly simplified the numbers, but the variables can cancel out too.
x^a * x^b = x^(a+b). For example, 2^3 * 2^4 = (2*2*2) * (2*2*2*2) = 2^7 since there's seven 2's. Another example: 2^5 / 2^3 = (2*2*2*2*2) / (2*2*2) = 2*2 = 2^2 since three of the 2's in the numerator cancel with three of the twos in the denominator. So you have w^4 / w^6 = (w*w*w*w) / (w*w*w*w*w*w) = 1/ (w*w) = 1/w^2.
would it be 4w^2/7w^3?
@TurtleMuffin
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Not quite. You have 4 w's in the top, and 6 w's in the bottom. 4 of the w's cancel, and you're left with only 2 w's in the bottom. So the answer is 4 / (7w^2).
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