mathdad
Full Member
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- Apr 24, 2015
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If a does not equal 0, and if n is a positive integer, the book goes on to define the following:
a^(-n) = 1/[a^(n)], if a is not 0.
Michael Sullivan makes this comment:
"Whenever you encounter a negative Exponent, think 'reciprocal.' "
Sample A
5^(-2) = 1/[5^(2)] = 1/25
Yes?
What about expressions?
Sample B
(x + 2)^(-2) = 1/[(x + 2)^2]
Can I leave the answer as 1/[(x + 2)^2]?
Must I simplify further?
(x + 2)^2 = (x + 2)(x + 2) = x^2 + + 4x + 4.
The answer then becomes 1/(x^2 + 4x + 4).
You say?
a^(-n) = 1/[a^(n)], if a is not 0.
Michael Sullivan makes this comment:
"Whenever you encounter a negative Exponent, think 'reciprocal.' "
Sample A
5^(-2) = 1/[5^(2)] = 1/25
Yes?
What about expressions?
Sample B
(x + 2)^(-2) = 1/[(x + 2)^2]
Can I leave the answer as 1/[(x + 2)^2]?
Must I simplify further?
(x + 2)^2 = (x + 2)(x + 2) = x^2 + + 4x + 4.
The answer then becomes 1/(x^2 + 4x + 4).
You say?
