approximate √48 to the tenth place

1 answer

To approximate \( \sqrt{48} \) to the tenth place, we can first simplify it:

\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]

Next, we calculate \( \sqrt{3} \), which is approximately 1.732. Therefore,

\[ \sqrt{48} \approx 4 \times 1.732 = 6.928 \]

Now, rounding 6.928 to the tenth place gives us:

\[ \sqrt{48} \approx 6.9 \quad \text{(to the tenth place)} \]

For more precision, if we compute \( \sqrt{48} \) directly or use a calculator, we find:

\[ \sqrt{48} \approx 6.928203230275509 \]

Thus, rounded to the tenth place, the value is still:

\[ \sqrt{48} \approx 6.9 \]