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anonymous
 4 years ago
solve the equations!
1.553t=0.1
1.553t= 0.1
anonymous
 4 years ago
solve the equations! 1.553t=0.1 1.553t= 0.1

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i hate decimals multiply by 10 and solve \[15530t=1\] for \(t\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[530t=14\] \[t=\frac{530}{14}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry if this is off subject but are you home schooled because I think I had this problem before

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, my alg teacher said there is a question like this on the alg test for tomorrow, and I need help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the actual full question is this: Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate of 53 miles per hour. The distance of the car from the checkpoint is given by d = 1.5 – 53t. At what times is the car 0.1 miles from the checkpoint? Calculate your answer in seconds. but I got it down to some equations. But im still super confused. Do you think you could help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I honestly didnt understand this question when I had it, but I think you can copy and paste this on google and yahoo may have it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I found this, it may help The car drives 53 miles per hour = 0.014722 miles per second 1 hour = 3600 seconds 0.1 miles takes 0.1 / 0.014722 = 6.7926 second
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