Answer:
4 hours
Step-by-step explanation:
The general formula for distance, rate and time is:
d = rt, where distance is the product of rate multiplied by time
Given that one train is moving at a rate of 70 mph and one is moving at a rate of 62 mph in the opposite direction, the combined distance over time is equal to 528 miles. Using the variable 't' for time:
70t + 62t = 528
Combine like terms: 132t = 528
Divide both sides by 132: 132t/132 = 528/132 or t = 4 hours
Answer:
It should be "one solution"
Step-by-step explanation:
After graphing the equations, the two lines only intersect at one point which makes it "one solution." Hope this helps.
Check the picture below.
so let's find the lengths of those two sides in red, since are the length and width of the rectangle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-3%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B-3-%28-6%29%5D%5E2%2B%5B6-3%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-3%2B6%29%5E2%2B%286-3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B9%2B9%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B18%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29~%5Chfill%20d%3D%5Csqrt%7B%5B-2-%28-6%29%5D%5E2%2B%5B-1-3%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B%28-2%2B6%29%5E2%2B%28-1-3%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B16%2B16%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B32%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20rectangle%7D%7D%7B%28%5Csqrt%7B18%7D%29%28%5Csqrt%7B32%7D%29%7D%5Cimplies%20%5Csqrt%7B18%5Ccdot%2032%7D%5Cimplies%20%5Csqrt%7B576%7D%5Cimplies%2024)
the area is 180 cm. to find the perimeter you would do 15+15+14+14 to get 54 cm. So for area you do (l×w) which is 15×14=180.
