Certainly! Let's break down the expression [tex]\(\left(2^{-3} \times 3^{-2}\right)+7^{-1}\)[/tex] step by step to find the solution.
### Step 1: Evaluate [tex]\(2^{-3}\)[/tex]
First, we'll calculate [tex]\(2^{-3}\)[/tex]:
[tex]\[2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125\][/tex]
### Step 2: Evaluate [tex]\(3^{-2}\)[/tex]
Next, we calculate [tex]\(3^{-2}\)[/tex]:
[tex]\[3^{-2} = \frac{1}{3^2} = \frac{1}{9} \approx 0.1111\][/tex]
### Step 3: Multiply [tex]\(2^{-3}\)[/tex] and [tex]\(3^{-2}\)[/tex]
Now, we multiply the results from Step 1 and Step 2:
[tex]\[0.125 \times 0.1111 \approx 0.0139\][/tex]
### Step 4: Evaluate [tex]\(7^{-1}\)[/tex]
Next, we evaluate [tex]\(7^{-1}\)[/tex]:
[tex]\[7^{-1} = \frac{1}{7} \approx 0.1429\][/tex]
### Step 5: Add the products from Step 3 and Step 4
Finally, we add the product from Step 3 to the result from Step 4:
[tex]\[0.0139 + 0.1429 \approx 0.1568\][/tex]
So, the final result of the expression [tex]\(\left(2^{-3} \times 3^{-2}\right) + 7^{-1}\)[/tex] is approximately:
[tex]\[
0.1568
\][/tex]
