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anonymous
 5 years ago
Amy drives 45 miles to work. If she increases her average speed by 5 miles per hour, the time it takes to drive to work is decreased by 6 minutes. find her usual average speed.
anonymous
 5 years ago
Amy drives 45 miles to work. If she increases her average speed by 5 miles per hour, the time it takes to drive to work is decreased by 6 minutes. find her usual average speed.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.045 = rt usual 45 = (r+5)t if she does increase her speed Solve both for t 45/r = t 45 /(r+5) = t 45/(r + 5) + 1/10 = 45/r multiply each term by 10r(r+5) which is your common denominator 450r + r^2 + 5r = 450r + 2250 r^2 + 5r  2250 = 0 (r + 50)(r  45) = 0 r = 50 or r = 45 You throw out the negative number be it cannot be her speed. Original speed was 45 mph

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks but if her time is decreased by six, woudnt the one formula be 45/(r+5)=t6?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No because it says 6 minutes and everything else is in hours... so you need to keep everything in hours.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the formula d = rt time is always in hours.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok but still then 45/(r+5)=t1/60 ? or am I just being weird

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.06 minutes is 6 minutes out of 60 minutes which is 1 hour One minute would be 1/60 of an hour Six minutes would be 6/60 of an hour which reduces to 1/10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0....45/(r+5)=t1/10? sorry I'm really tired :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh haha wait nvm I see you added that in in the next step

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The way you have to look at it is two compare the two ways of driving. normal 45 = rt Going faster 45 = (r+5)t because we are comparing times between driving to work the two different ways we solve both sides for t. 45/r = t 45/(r + 5) = t Usually these problems say it took the same amount of time so you would put the left side of those equations equal to each other, but on this one it says that when she "sped up", it took her 6 minutes less to get to work. (which makes sense). The problem is "HOW DO WE get these equal to each other?" If "speeding" makes your time 6 minutes less than the other... then to make them equal you would add 6 to the "normal" driving. Ex. I am doing this in minutes because it is easier Normal Speeding 50m 44m 20m 14m 36m 30m So to make these "equal", you would add 6 minutes to the speeding side..
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