Why is 3^-2 not a negative number?

The rule you use is x^(-k) = 1/(x^k)

In this case x = 3 and k = 2, so,

3^(-2) = 1/(3^2) = 1/9

One way to think of it, is to consider the equation below

3^2*3^x = 1

We can solve for x like this

3^2*3^x = 1

3^(2+x) = 1

3^(2+x) = 3^0

2+x = 0 the bases are equal, so the exponents must be equal

x = -2

meaning that

3^2*3^x = 1

updates to

3^2*3^(-2) = 1

Now isolate the 3^(-2)

3^2*3^(-2) = 1

3^(-2) = 1/(3^2)

3^(-2) = 1/9

angle addition postulate

2x=56

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles.

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