$\text{Let}\ A(x_1,\ y_1)\ \text{and}\ B(x_2,\ y_2).\ \text{Then the distance between}\ A\ \text{and}\ B\ \text{is:}\\\\|AB|=\sqrt{(x_2-x_1)^2+(y_2-y1)^2}\\\\Example:\\\\A(2,\ 5),\ B(-1,\ 1)\\\\|AB|=\sqrt{(-1-2)^2+(1-5)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$

7 0

3 years ago

Read 2 more answers

If your navigator tells you your position is 0.7 nautical miles from the West Point light, how far is that in feet? And how far

Nataly_w [17]

Answer:

that would be 4253.28 feet

that would be 1296.4 m

Step-by-step explanation:

I think that's the answer and I hope it helps :)

8 0

2 years ago

Solve by substitution 3x+y=3 and 7x+2y=1

balu736 [363]

3x + y = 3

7x + 2y = 1

First isolate one of the variables (x or y) in one of the equations.

Isolate "y" in the first equation(because it is the easiest to isolate) and substitute it into the second equation.

3x + y = 3 Subtract 3x on both sides

3x - 3x + y = 3 - 3x

y = 3 - 3x

7x + 2y = 1

7x + 2(3 - 3x) = 1 [since y = 3 - 3x, you can substitute (3-3x) for "y"]

Multiply/distribute 2 into (3 - 3x)

7x + (3(2) - 3x(2)) = 1

7x + 6 - 6x = 1

x + 6 = 1 Subtract 6 on both sides

x = -5

Now that you know "x", substitute it into one of the equations (I will do both)

3x + y = 3

3(-5) + y = 3 [since x = -5, you can plug in -5 for "x"]

-15 + y = 3 Add 15 on both sides

y = 18

7x + 2y = 1

7(-5) + 2y = 1

-35 + 2y = 1 Add 35 on both sides

2y = 36 Divide 2 on both sides

y = 18

x = -5, y = 18 or (-5, 18)

5 0

2 years ago