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lyzzy7

  • 3 years ago

solve the equations! 1.5-53t=0.1 1.5-53t= -0.1

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  1. anonymous
    • 3 years ago
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    i hate decimals multiply by 10 and solve \[15-530t=1\] for \(t\)

  2. anonymous
    • 3 years ago
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    \[530t=14\] \[t=\frac{530}{14}\]

  3. aussy123
    • 3 years ago
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    sorry if this is off subject but are you home schooled because I think I had this problem before

  4. lyzzy7
    • 3 years ago
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    Yeah, my alg teacher said there is a question like this on the alg test for tomorrow, and I need help.

  5. aussy123
    • 3 years ago
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    oh I knew it!

  6. lyzzy7
    • 3 years ago
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    the 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?

  7. aussy123
    • 3 years ago
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    I 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

  8. aussy123
    • 3 years ago
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    I 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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