To estimate [tex]\(\sqrt{50}\)[/tex] to the hundredths place, let's follow a detailed, step-by-step solution:
### 1. Estimate between two whole numbers:
We need to find two whole numbers such that [tex]\(\sqrt{50}\)[/tex] lies between them.
[tex]\[
\begin{aligned}
&7^2 = 49 \quad \text{(since \(49 < 50\))} \\
&8^2 = 64 \quad \text{(since \(64 > 50\))}
\end{aligned}
\][/tex]
So, [tex]\(\sqrt{50}\)[/tex] is between 7 and 8.
### 2. Estimate further to the tenths place:
Now, we refine our estimate to figure out more precisely where [tex]\(\sqrt{50}\)[/tex] lies by looking at tenths within 7 and 8. Starting from 7.0:
[tex]\[
\begin{aligned}
&7.0^2 = 49.0 \quad \text{(since \(49.0 < 50\))} \\
&7.1^2 = 50.41 \quad \text{(since \(50.41 > 50\))}
\end{aligned}
\][/tex]
So, [tex]\(\sqrt{50}\)[/tex] is between 7.0 and 7.1.
### 3. Estimate further to the hundredths place:
Now, we look more closely between 7.0 and 7.1 to refine our estimate to the hundredths place. We need to test values between 7.0 and 7.1, increasing in steps of 0.01:
[tex]\[
\begin{aligned}
&7.00^2 = 49.00 \quad \text{(since \(49.00 < 50\))} \\
&7.01^2 = 49.1401 \quad \text{(since \(49.1401 < 50\))} \\
&7.02^2 = 49.2804 \quad \text{(since \(49.2804 < 50\))} \\
&7.03^2 = 49.4209 \quad \text{(since \(49.4209 < 50\))} \\
&7.04^2 = 49.5616 \quad \text{(since \(49.5616 < 50\))} \\
&7.05^2 = 49.7025 \quad \text{(since \(49.7025 < 50\))} \\
&7.06^2 = 49.8436 \quad \text{(since \(49.8436 < 50\))} \\
&7.07^2 = 49.9849 \quad \text{(since \(49.9849 < 50\))} \\
&7.08^2 = 50.1264 \quad \text{(since \(50.1264 > 50\))}
\end{aligned}
\][/tex]
So, [tex]\(\sqrt{50}\)[/tex] is between 7.07 and 7.08.
Thus, [tex]\(\sqrt{50}\)[/tex] is approximately [tex]\(\boxed{7.07 \ \text{and} \ 7.08}\)[/tex].
