Answer:
the rate of each train is 139 km/h and 159 km/h.
Step-by-step explanation:
let
v = the rate of the slower train
v + 20 = the rate of the faster train
d = 1490 km the distance that the two train will travel
t = 5 hr the time of the travel
since speed = distance/time => distance = speed*time
v*t + (v+20)t = d
5v + 5(v + 20) = 1490
5v+5(v+20)=1490
Step 1: Simplify both sides of the equation.
5v+5(v+20)=1490
5v+(5)(v)+(5)(20)=1490(Distribute)
5v+5v+100=1490
(5v+5v)+(100)=1490(Combine Like Terms)
10v+100=1490
10v+100=1490
Step 2: Subtract 100 from both sides.
10v+100−100=1490−100
10v=1390
Step 3: Divide both sides by 10.
10v/10 = 1390/10
v = 139km/h
Next
we add 139 to 20
139 + 20 = 159 km/h
the rate of each train is 139 km/h and 159 km/h.
we know all it's doing is adding 6 over again to each term to get the next one, so then
now for the explicit one
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=7\\ d=6 \end{cases} \\\\\\ a_n=7+(n-1)6\implies a_n=7+6n-6\implies \stackrel{\textit{Explicit Formula}}{\stackrel{f(n)}{a_n}=6n+1} \\\\\\ therefore\qquad \qquad f(10)=6(10)+1\implies f(10)=61](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a_1%3D7%5C%5C%20d%3D6%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20a_n%3D7%2B%28n-1%296%5Cimplies%20a_n%3D7%2B6n-6%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BExplicit%20Formula%7D%7D%7B%5Cstackrel%7Bf%28n%29%7D%7Ba_n%7D%3D6n%2B1%7D%20%5C%5C%5C%5C%5C%5C%20therefore%5Cqquad%20%5Cqquad%20f%2810%29%3D6%2810%29%2B1%5Cimplies%20f%2810%29%3D61)
The answer would be -4.5/5. You would have to plot the number in between -2 and - 1 4/5. Hope this helps!
