Let's say her speed was x miles/hour during the first 3 miles runThen, time = distance/speedt1 = 3/x eq1 In the next 4 miles run, her speed = x-1 miles/hourTime taken:t2 = 4/(x-1) eq2 Now, total time:t1 + t2 = 1 3/5 hourssubstitute t1 and t2 from eqs. 1 and 2 3/x + 4/(x-1) = 1 3/5=> 3/x + 4/(x-1) = 8/5
=> 3(x-1) + 4x = 8x(x-1)/5=> 35x - 15 = 8x2 - 8x=> 8x2 - 43x + 15 = 0=> (8x-3)*(x-5) = 0=> x = 3/8 or 5 miles/hourx can not be 3/8 miles/hour because in that case, the speed during 4 miles run would be 3/8-1 = negative numberi.e. speed during 3 miles segment = 5 miles/hourand speed during 4 miles segment = 5-1 = 4 miles/hour
Whats the question tell me and i'll try to help
It would be 13 because you are adding 5 and 8
Lets say you have an addition problem
3 + 7 = ?
To figure that you can just add in your head since its easy.
3 + 7 = 10
You can work backwords to check to see if your awnser is right.
When you work backwards you do the opposite of what you are doing, so if your adding the opposite would be subtracting, and if your multiplying than the opposite would be dividing.
So you could work it out like this:
(Regular⬇️)
3 + 7 = 10
(Backwards⬇️)
10 - 7 = 3
And if you get the same numbers that were in the original awnser than you are correct! Hope this helps!!!!
