V1 - velocity first train
v2 - velocity second train
v2 > v1
v2 - v1 = 17 mph
We know, that:
![s_1+s_2=210 \ [miles] \\ \\ t_1=t_2=2h](https://tex.z-dn.net/?f=s_1%2Bs_2%3D210%20%5C%20%5Bmiles%5D%20%5C%5C%20%5C%5C%20t_1%3Dt_2%3D2h)
So:
NOw we've got simple system of equations:
![+\begin{cases} v_2-v_1=17 \\ v_2+v_1=105\end{cases} \\ \\ 2v_2=122 \qquad /:2 \\ \\ v_2=61 \qquad [mph] \\ \\ v_2-v_1=17 \\ \\ 61-v_1=17 \\ \\ v_1=44](https://tex.z-dn.net/?f=%2B%5Cbegin%7Bcases%7D%20v_2-v_1%3D17%20%5C%5C%20v_2%2Bv_1%3D105%5Cend%7Bcases%7D%20%5C%5C%20%5C%5C%202v_2%3D122%20%5Cqquad%20%2F%3A2%20%5C%5C%20%5C%5C%20v_2%3D61%20%5Cqquad%20%5Bmph%5D%20%5C%5C%20%5C%5C%20v_2-v_1%3D17%20%5C%5C%20%5C%5C%2061-v_1%3D17%20%5C%5C%20%5C%5C%20v_1%3D44)
Velocities of these trains are 61mph and 44mph
When two lines cross, there are four angles at the place where they cross.
Two angles that do NOT share a side are vertical angles. There are two pairs
of vertical angles where the lines cross.
Also, by the way, vertical angles are always equal to each other. It's often
very helpful to know that.
For this case we have that by definition, the equation of the line of the slope-intersection form is given by:
Where:
m: It's the slope
b: It is the cut-off point with the y axis
We have two points through which the line passes:
We found the slope:
Substituting we have:
Thus, the equation is of the form:
We substitute one of the points and find the cut-off point:
Finally, the equation is:
ANswer:
Answer:
425
Step-by-step explanation:
225/24 then multiply the answer by 40
