To calculate the distance between two points, we can use a formula that is a variation Pythagorean Theorem. Look:
![\mathsf{d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bd%3D%5Csqrt%7B%28x_B-x_A%29%5E2%2B%28y_B-y_A%29%5E2%7D%7D)
"d" represents the distance and coordinates are expressed as follows: (x, y)
Let's go to the calculations.
![\mathsf{d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}\\\\ \mathsf{d=\sqrt{(5-2)^2+(9-3)^2}}\\\\ \mathsf{d=\sqrt{(3)^2+(6)^2}}\\\\ \mathsf{d=\sqrt{9+36}}\\\\ \mathsf{d=\sqrt{45}}\\\\ \mathsf{d=6.7082039324993690892275210061938...}\\\\ \underline{\mathsf{d\approxeq6.7}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bd%3D%5Csqrt%7B%28x_B-x_A%29%5E2%2B%28y_B-y_A%29%5E2%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bd%3D%5Csqrt%7B%285-2%29%5E2%2B%289-3%29%5E2%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bd%3D%5Csqrt%7B%283%29%5E2%2B%286%29%5E2%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bd%3D%5Csqrt%7B9%2B36%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bd%3D%5Csqrt%7B45%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bd%3D6.7082039324993690892275210061938...%7D%5C%5C%5C%5C%20%5Cunderline%7B%5Cmathsf%7Bd%5Capproxeq6.7%7D%7D)
The answer is 6.7 uc.
I’m pretty sure that’s true.
8z − 4 = 7z
move 8z to the other side
sign changes from +8z to -8z
8z-8z-4=7z-8z
-4= 7z-8z
-4= -z
mutiply both sides by -1 to get +z
(-4)(-1)= (-z)(-1)
Answer:
z= 4
yeah. I can't see none of you work so help ME please.
<span>y=x-4
y=-x+6
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Substitute x - 4 for y in </span>y=-x+6
x-4=-x+6<span>
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Add X on each side
</span><span>x-4+x</span>=-<span>x+6+x
</span>2x - 4 = 6
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Add 4 on each side
<span>2x-4+4</span>=<span>6+4
</span>2x = 10
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Divide by 2 on each side
2x ÷ 2 = 10 ÷ 2
x = 5
Now we have X
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To find Y we substitute 5 for x in y=<span>x-<span>4
</span></span>y = 5 - 4
y = 1
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Your answers are Y = 1 and X = 5