If [tex]$b=3$[/tex], then [tex]$b^{-2}$[/tex] is equivalent to:

A. [tex]-6[/tex]
B. [tex]\frac{1}{9}[/tex]
C. [tex]9[/tex]
D. [tex]-9[/tex]



Answer :

To solve the given problem, let's follow the steps to understand how to compute [tex]\( b^{-2} \)[/tex] where [tex]\( b = 3 \)[/tex].

1. Define the problem: We need to find the value of [tex]\( b^{-2} \)[/tex] when [tex]\( b \)[/tex] is equal to 3.

2. Interpret the exponent: The notation [tex]\( b^{-2} \)[/tex] indicates that we are dealing with a negative exponent. A negative exponent means we are taking the reciprocal of the base raised to the corresponding positive exponent. So, [tex]\( b^{-2} \)[/tex] is equivalent to [tex]\( \frac{1}{b^2} \)[/tex].

3. Substitute the value of [tex]\( b \)[/tex]: Plug the given value of [tex]\( b \)[/tex] (which is 3) into the expression [tex]\( \frac{1}{b^2} \)[/tex].

[tex]\[ b^{-2} = \frac{1}{3^2} \][/tex]

4. Calculate the square of the base: Compute the square of 3, which is:

[tex]\[ 3^2 = 9 \][/tex]

5. Find the reciprocal: Now, take the reciprocal of 9:

[tex]\[ \frac{1}{9} \][/tex]

6. Conclusion: The value of [tex]\( b^{-2} \)[/tex] when [tex]\( b = 3 \)[/tex] is [tex]\( \frac{1}{9} \)[/tex].

Therefore, the correct answer from the given options is [tex]\( \boxed{\frac{1}{9}} \)[/tex].