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Rudiy27
3 years ago
9

What is the distance between (-6, 2) and (8, 10) on a coordinate grid?

Mathematics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

The answer is 2(the square root of 65 )

Step-by-step explanation:

Plug ( -6, 2) and (8, 10) into the distance formula and solve.

Distance = the square root of ((x2 - x1)^2 + (y2 - y1)^2))

Distance = the square root of ((8 - -6)^2 + (10 - 2)^2))

Distance = the square root of (14^2 + 8^2)

Distance = the square root of (196 + 64)

Distance = the square root of 260 which can be simplified to 2(the square root of 64)

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Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi
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This problem is represented in the Figure below. So, we can find the components of each vector as follows:


\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m


\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m


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