Let's analyze Timothy's steps in detail to see if he correctly evaluated the expression given \( x = 3 \) and \( y = -4 \).
The expression to evaluate is: \(\left(\frac{1}{3}\right) x^{-1} y^2\).
Step-by-step solution:
1. Substitute \( x = 3 \) and \( y = -4 \) into the expression:
[tex]\[
\left(\frac{1}{3}\right) x^{-1} y^2
= \left(\frac{1}{3}\right) 3^{-1} (-4)^2
\][/tex]
2. Evaluate the powers:
- \( 3^{-1} = \frac{1}{3} \)
- \( (-4)^2 = 16 \)
So the expression becomes:
[tex]\[
\left(\frac{1}{3}\right) \left(\frac{1}{3}\right) \cdot 16
\][/tex]
3. Simplify the fractions and multiply:
[tex]\[
\left(\frac{1}{3}\right) \left(\frac{1}{3}\right) \cdot 16
= \frac{1}{3} \cdot \frac{1}{3} \cdot 16
= \frac{1}{9} \cdot 16
= \frac{16}{9}
\][/tex]
4. Convert the fraction to a decimal:
[tex]\[
\frac{16}{9} \approx 1.7777777777777777
\][/tex]
So, the result of the evaluated expression is approximately \( 1.7777777777777777 \).
### Analysis of Timothy's Steps:
Let's consider each of Timothy's steps:
1. \(\left(\frac{1}{3}\right) x^{-1} y^2\)
2. \(\left(\frac{1}{3}\right) 3^{-1}(-4)^2\)
3. \(\left(\frac{1}{3}\right)\left(\frac{1}{3^1}\right)(-4)^2\)
These steps are correct so far.
4. \(\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)(-16)\)
Here, Timothy made a mistake. Evaluating \((-4)^2\) correctly should yield \(16\), not \(-16\).
Let's check the next suggestion in the question:
- No, his value of \( (-4)^2\) should be positive because an even exponent indicates a positive value.
This suggestion is correct. An even exponent makes the value positive, therefore:
Timothy's calculations were incorrect because the value of [tex]\( (-4)^2\)[/tex] should be positive 16.
