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OpenStudy (anonymous):

You drive 200 miles to city at an average speed 10 miles per hour faster than your usual average speed. If you completed the trip in 1 hour less than usual. What is your usual driving speed, in miles per hour?

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OpenStudy (anonymous):

d=(r)(t)

OpenStudy (anonymous):

please help!!!

OpenStudy (anonymous):

d=rt 200=(s+10)(t-1)

OpenStudy (anonymous):

d=rt 200=rt

OpenStudy (anonymous):

200= rt-r+10t-10 200=200-r+10t-10 200=190-r+10t

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OpenStudy (anonymous):

200-190=-r+10t 10=-r+10t

OpenStudy (anonymous):

r=10t-10

OpenStudy (anonymous):

200= (r+10)(t-1) 200=(10t-10+10)(t-1) we have substituted r

OpenStudy (anonymous):

200=(10t)(t-1) 200=10t^2-10t 0=10t^2-10t-200 which we can factorise

OpenStudy (anonymous):

0=10(t^2-t-20) 0=t^2-t-20 0=(t-5)(t+4) t=5 t=-4

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OpenStudy (anonymous):

since time cannot be negative we take 5hrs

OpenStudy (anonymous):

now we find the speed =distance/time =200/5= 40mil/hr

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