Question
Use a calculator to solve the equation below. Use pencil and paper. Before computing anything, write down two perfect squares you know, one that is less than the given number and one that is greater than the given number. Then write down their square roots as two numbers you know w must be between.
wsquaredequals118
Question content area bottom
Part 1
walmost equals
enter your response here
(Round to the nearest tenth as needed. Use a comma to separate answers as needed.)
wsquaredequals118
Question content area bottom
Part 1
walmost equals
enter your response here
(Round to the nearest tenth as needed. Use a comma to separate answers as needed.)
Answers
GPT-4o mini
To solve the equation \( w^2 = 118 \), we first need to find two perfect squares: one that is less than 118 and one that is greater than 118.
1. **Identify perfect squares**:
- The perfect square less than 118 is \( 100 \) (since \( 10^2 = 100 \)).
- The perfect square greater than 118 is \( 121 \) (since \( 11^2 = 121 \)).
2. **Calculate the square roots**:
- The square root of \( 100 \) is \( 10 \) (i.e., \( \sqrt{100} = 10 \)).
- The square root of \( 121 \) is \( 11 \) (i.e., \( \sqrt{121} = 11 \)).
3. **Write down the range**:
- Since \( 10^2 < 118 < 11^2 \), we can conclude that \( w \) must be between \( 10 \) and \( 11 \).
Now, we can calculate the square root of \( 118 \) to find \( w \):
\[
w = \sqrt{118} \approx 10.863
\]
4. **Round to the nearest tenth**:
- Rounding \( 10.863 \) to the nearest tenth gives \( 10.9 \).
Thus, the final answer is:
\[
w \approx 10.9
\]
So, \( w \) is between \( 10 \) and \( 11 \), with the rounded result being \( 10.9 \).
1. **Identify perfect squares**:
- The perfect square less than 118 is \( 100 \) (since \( 10^2 = 100 \)).
- The perfect square greater than 118 is \( 121 \) (since \( 11^2 = 121 \)).
2. **Calculate the square roots**:
- The square root of \( 100 \) is \( 10 \) (i.e., \( \sqrt{100} = 10 \)).
- The square root of \( 121 \) is \( 11 \) (i.e., \( \sqrt{121} = 11 \)).
3. **Write down the range**:
- Since \( 10^2 < 118 < 11^2 \), we can conclude that \( w \) must be between \( 10 \) and \( 11 \).
Now, we can calculate the square root of \( 118 \) to find \( w \):
\[
w = \sqrt{118} \approx 10.863
\]
4. **Round to the nearest tenth**:
- Rounding \( 10.863 \) to the nearest tenth gives \( 10.9 \).
Thus, the final answer is:
\[
w \approx 10.9
\]
So, \( w \) is between \( 10 \) and \( 11 \), with the rounded result being \( 10.9 \).