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to the nearest tenth of a unit, what is the distance on a coordinate pl…

Question

to the nearest tenth of a unit, what is the distance on a coordinate plane between the points (-6, -9) and (5, 7)?

units

Explanation:

Step1: Recall the distance formula

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \).
Here, \( x_1=-6 \), \( y_1 = - 9 \), \( x_2 = 5 \), \( y_2=7 \).

Step2: Calculate the differences in coordinates

First, find \( x_2 - x_1 \) and \( y_2 - y_1 \):
\( x_2 - x_1=5-(-6)=5 + 6=11 \)
\( y_2 - y_1=7-(-9)=7 + 9 = 16 \)

Step3: Substitute into the distance formula

Substitute these values into the formula:
\( d=\sqrt{(11)^2+(16)^2}=\sqrt{121 + 256}=\sqrt{377} \)

Step4: Approximate the square root

Calculate the square root of 377. \( \sqrt{377}\approx19.4 \) (to the nearest tenth)

Answer:

19.4