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A graph shows four quadrants with the x-axis and y-axis both ranging from negative 5 to positive 5 in increments of 1. Unlabeled data points marked are at left parenthesis negative 2 comma negative 1 right parenthesis, and at left parenthesis 2 comma 4 right parenthesis. A straight line joins the two lines with a positive slope.
Apply the Pythagorean Theorem to find the length between the two points. Round your answer to the nearest hundredth, if necessary.
(1 point)
A graph shows four quadrants with the x-axis and y-axis both ranging from negative 5 to positive 5 in increments of 1. Unlabeled data points marked are at left parenthesis negative 2 comma negative 1 right parenthesis, and at left parenthesis 2 comma 4 right parenthesis. A straight line joins the two lines with a positive slope.
Apply the Pythagorean Theorem to find the length between the two points. Round your answer to the nearest hundredth, if necessary.
(1 point)
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Answered by
GPT-4o mini
To find the length between the two points \((-2, -1)\) and \((2, 4)\), we can use the Pythagorean Theorem. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
First, we identify our points:
- \(x_1 = -2\), \(y_1 = -1\)
- \(x_2 = 2\), \(y_2 = 4\)
Now, we calculate the differences in the coordinates:
\[
x_2 - x_1 = 2 - (-2) = 2 + 2 = 4
\]
\[
y_2 - y_1 = 4 - (-1) = 4 + 1 = 5
\]
Now we can substitute these values into the distance formula:
\[
d = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}
\]
Finally, we can approximate \(\sqrt{41}\).
Calculating \(\sqrt{41} \approx 6.403124237\).
Rounding to the nearest hundredth, the distance \(d\) is approximately:
\[
\boxed{6.40}
\]
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
First, we identify our points:
- \(x_1 = -2\), \(y_1 = -1\)
- \(x_2 = 2\), \(y_2 = 4\)
Now, we calculate the differences in the coordinates:
\[
x_2 - x_1 = 2 - (-2) = 2 + 2 = 4
\]
\[
y_2 - y_1 = 4 - (-1) = 4 + 1 = 5
\]
Now we can substitute these values into the distance formula:
\[
d = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}
\]
Finally, we can approximate \(\sqrt{41}\).
Calculating \(\sqrt{41} \approx 6.403124237\).
Rounding to the nearest hundredth, the distance \(d\) is approximately:
\[
\boxed{6.40}
\]
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