We are asked to order the numbers $6\frac{1}{4}$, $\frac{13}{2}$, and $\sqrt{40}$ from least to greatest.

1. Problem Description

We are asked to order the numbers $6\frac{1}{4}$, $\frac{13}{2}$, and $\sqrt{40}$ from least to greatest.

2. Solution Steps

First, let's convert each number to a decimal or approximate decimal.

$6\frac{1}{4} = 6 + \frac{1}{4} = 6 + 0.25 = 6.25$

$\frac{13}{2} = 6.5$

To approximate $\sqrt{40}$, we can think about perfect squares near

4

0. $6^2 = 36$ and $7^2 = 49$.

Since 40 is between 36 and 49, $\sqrt{40}$ is between 6 and

7. Also, 40 is closer to 36 than 49, so $\sqrt{40}$ will be closer to 6 than to

7. We can say $\sqrt{40} \approx 6.3$. More accurately, $\sqrt{40} \approx 6.32$.

Alternatively, we can find $\sqrt{40}$ using a calculator, which results in $\sqrt{40} \approx 6.324555$.

Now we have $6.25$, $6.5$, and approximately $6.32$.

Ordering these from least to greatest, we get $6.25 < 6.32 < 6.5$.

Therefore, the order from least to greatest is $6\frac{1}{4}$, $\sqrt{40}$, and $\frac{13}{2}$.

3. Final Answer

$6\frac{1}{4}$, $\sqrt{40}$, $\frac{13}{2}$