Check the picture below.
notice the alternate interior angles in the picture.
![\bf tan(68^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{x}}\implies x=\cfrac{1.5}{tan(68^o)}\implies x\approx 0.61 \\\\[-0.35em] ~\dotfill\\\\ tan(56^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{w}}\implies w=\cfrac{1.5}{tan(56^o)}\implies w\approx 1.01 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{width of the crater}}{x+w\implies 1.62}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20tan%2868%5Eo%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B1.5%7D%7D%7B%5Cstackrel%7Badjacent%7D%7Bx%7D%7D%5Cimplies%20x%3D%5Ccfrac%7B1.5%7D%7Btan%2868%5Eo%29%7D%5Cimplies%20x%5Capprox%200.61%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20tan%2856%5Eo%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B1.5%7D%7D%7B%5Cstackrel%7Badjacent%7D%7Bw%7D%7D%5Cimplies%20w%3D%5Ccfrac%7B1.5%7D%7Btan%2856%5Eo%29%7D%5Cimplies%20w%5Capprox%201.01%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bwidth%20of%20the%20crater%7D%7D%7Bx%2Bw%5Cimplies%201.62%7D~%5Chfill)
Answer:
y equals 10
Step-by-step explanation:
5x5= 25 and x equals 25 so 2×5=10 so y=10
Answer:
This route is 112 miles long.
Step-by-step explanation:
Let's say that the passeneger meant to get off at Station A but instead got off at Station B. We are told that the distance from Boston to Station A is 1/3 of the train's total route. Let's say that the train's total route is x miles long. Now we can say that the distance from Boston to Station A is 1/3x.
We are told that the distance from Station A to Station B is 14 miles.
We are also told that the distance from Station B to the end of the route is 13/24 of the whole route. Since we already decided that the whole route is x miles long, we can say that the distance from Station B to the end route is 13/24x.
Add up all of these distances to get the total distance of the trains route:
13/24x+14+1/3x=x.
14+13/24x+8/24x=x
14+21/24x=x
14=3/24x
14=1/8x
112=x
Hope this helps!
The first is in the form y=Mx+b where m is the slope and b is the y-intercept. Start at 600 on the y-axis and graph with a slope of six. The second is in the same form but the y-intercept is 0, so start at the origin and graph with a slope of 8. The last is in the form y=b+mx, so start at 1300 and graph with a slope of 3. Remember, slope is rise/run or (change in y)/(change in x). Since the domain and range start at 0 these graphs will only be in the first quadrant with an x limit of 650 and a y limit of 1500.
4x +1 so I got it like this 5x +3-x-2