Given two points
and
The distance between them is >>>
![D=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}](https://tex.z-dn.net/?f=D%3D%5Csqrt%5B%5D%7B%28y_2-y_1%29%5E2%2B%28x_2-x_1%29%5E2%7D)
The points given are (Sqrt(20), Sqrt(50)) and (Sqrt(125), Sqrt(8)), so their distance is >>>
![\begin{gathered} D=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ D=\sqrt[]{(\sqrt[]{8}-\sqrt[]{50})^2+(\sqrt[]{125}-\sqrt[]{20})^2} \\ D=\sqrt[]{(\sqrt8)^2-2(\sqrt[]{8})(\sqrt[]{50})+(\sqrt[]{50})^2^{}+(\sqrt[]{125})^2-2(\sqrt[]{125})(\sqrt[]{20})+(\sqrt[]{20})^2} \\ D=\sqrt[]{8-2(2\sqrt[]{2})(5\sqrt[]{2})+50+125-2(5\sqrt[]{5})(2\sqrt[]{5})+20} \\ D=\sqrt[]{8-40+50+125-100+20} \\ D=\sqrt[]{63} \\ D=3\sqrt[]{7} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D%5Csqrt%5B%5D%7B%28y_2-y_1%29%5E2%2B%28x_2-x_1%29%5E2%7D%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B%28%5Csqrt%5B%5D%7B8%7D-%5Csqrt%5B%5D%7B50%7D%29%5E2%2B%28%5Csqrt%5B%5D%7B125%7D-%5Csqrt%5B%5D%7B20%7D%29%5E2%7D%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B%28%5Csqrt8%29%5E2-2%28%5Csqrt%5B%5D%7B8%7D%29%28%5Csqrt%5B%5D%7B50%7D%29%2B%28%5Csqrt%5B%5D%7B50%7D%29%5E2%5E%7B%7D%2B%28%5Csqrt%5B%5D%7B125%7D%29%5E2-2%28%5Csqrt%5B%5D%7B125%7D%29%28%5Csqrt%5B%5D%7B20%7D%29%2B%28%5Csqrt%5B%5D%7B20%7D%29%5E2%7D%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B8-2%282%5Csqrt%5B%5D%7B2%7D%29%285%5Csqrt%5B%5D%7B2%7D%29%2B50%2B125-2%285%5Csqrt%5B%5D%7B5%7D%29%282%5Csqrt%5B%5D%7B5%7D%29%2B20%7D%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B8-40%2B50%2B125-100%2B20%7D%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B63%7D%20%5C%5C%20D%3D3%5Csqrt%5B%5D%7B7%7D%20%5Cend%7Bgathered%7D)
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The midpoint formula is >>>
Plugging in the points, we have >>>
![\begin{gathered} M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ M=(\frac{\sqrt[]{20}+\sqrt[]{125}}{2},\frac{\sqrt[]{50}+\sqrt[]{8}}{2}) \\ M=(\frac{2\sqrt5+5\sqrt[]{5}}{2},\frac{5\sqrt[]{2}+2\sqrt[]{2}}{2}) \\ M=(\frac{7\sqrt[]{5}}{2},\frac{7\sqrt[]{2}}{2}) \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20M%3D%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cfrac%7By_1%2By_2%7D%7B2%7D%29%20%5C%5C%20M%3D%28%5Cfrac%7B%5Csqrt%5B%5D%7B20%7D%2B%5Csqrt%5B%5D%7B125%7D%7D%7B2%7D%2C%5Cfrac%7B%5Csqrt%5B%5D%7B50%7D%2B%5Csqrt%5B%5D%7B8%7D%7D%7B2%7D%29%20%5C%5C%20M%3D%28%5Cfrac%7B2%5Csqrt5%2B5%5Csqrt%5B%5D%7B5%7D%7D%7B2%7D%2C%5Cfrac%7B5%5Csqrt%5B%5D%7B2%7D%2B2%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%29%20%5C%5C%20M%3D%28%5Cfrac%7B7%5Csqrt%5B%5D%7B5%7D%7D%7B2%7D%2C%5Cfrac%7B7%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%29%20%5Cend%7Bgathered%7D)
Answer:
Step-by-step explanation:
Whatever you do to the left you must do to the right, he multiplied 12 but forgot to multiply the -4 on the left and the + 1 on the right by the number 12
Answer: 20.833mph.
Step-by-step explanation:
Using the motion Formula,
speed = distance/time.
This means:
distance = speed * time.
To find the value of "s" in the question, we have to calculate the distance both trains travelled in terms of "s" and sum it to give us 300miles.
For train A,
Speed = S miles per hour
time taken to reach current position= 7hours.
Hence, the distance travelled by train A
Distance A = 7s.
For train B,
Speed= (S + 10)miles per hour.
time taken to reach current position= 5hours (since train B left two hours after the departure of train A)
Distance = speed * time
Distance B = (S + 10) * 5
Distance B = 5s + 50.
Hence, adding both distance to sum up to 300miles, it becomes:
300 = 7s + 5s + 50
300-50 = 12s
250 = 12s
s = 20.833mph.
Answer:
No. There would be less then half left
Step-by-step explanation:
