Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. it decreased from 1400 to 1000 problems in only 25 minutes. how long would it take until only 500 problems remained?

Step-by-step explanation:

Well it is simple.If he was able to solve 400 problems in just 25 ' then how long would it take him to solve 100(1/4 of 400)?It would take him 6.25' to solve 100 problems(1/4 of 25).So if he had to do another 500 (because 1000 -500=500) it would take him 31.25' (5*6.25) to complete them.If you have any further questions please contact me.

Yours sincerely,

Manos

4x and 6y

im not sure though