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Distance and the Pythagorean Theorem Practice Complete this assessment to review what you've learned. It will not count toward...Question
Distance and the Pythagorean Theorem Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
5 of 55 of 5 Items
Question
Find the distance between (−8, −3)
and (−2, −5)
. Round your answer to the nearest hundredth, if necessary.(1 point)
units
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the distance between the points \((-8, -3)\) and \((-2, -5)\), you can use the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1) = (-8, -3)\) and \((x_2, y_2) = (-2, -5)\).
Now we can plug in the values:
1. Calculate the differences:
- \(x_2 - x_1 = -2 - (-8) = -2 + 8 = 6\)
- \(y_2 - y_1 = -5 - (-3) = -5 + 3 = -2\)
2. Substitute these values into the distance formula:
\[
d = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}
\]
3. Simplify the square root:
\[
\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}
\]
4. Now, calculate the approximate value of \(2\sqrt{10}\):
\[
\sqrt{10} \approx 3.162 \quad \Rightarrow \quad 2\sqrt{10} \approx 2 \times 3.162 \approx 6.324
\]
Finally, rounding \(6.324\) to the nearest hundredth gives \(6.32\).
Thus, the distance between the points \((-8, -3)\) and \((-2, -5)\) is approximately \(\boxed{6.32}\) units.
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1) = (-8, -3)\) and \((x_2, y_2) = (-2, -5)\).
Now we can plug in the values:
1. Calculate the differences:
- \(x_2 - x_1 = -2 - (-8) = -2 + 8 = 6\)
- \(y_2 - y_1 = -5 - (-3) = -5 + 3 = -2\)
2. Substitute these values into the distance formula:
\[
d = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}
\]
3. Simplify the square root:
\[
\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}
\]
4. Now, calculate the approximate value of \(2\sqrt{10}\):
\[
\sqrt{10} \approx 3.162 \quad \Rightarrow \quad 2\sqrt{10} \approx 2 \times 3.162 \approx 6.324
\]
Finally, rounding \(6.324\) to the nearest hundredth gives \(6.32\).
Thus, the distance between the points \((-8, -3)\) and \((-2, -5)\) is approximately \(\boxed{6.32}\) units.
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