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iwanttogotostanford

  • one year ago

HELP WITH PRACTICE ACT MATH QUESTION: the braking distance, y feet, for Damon's car to come to a complete stop is modeled by y=3(x^2+10x)/40, where x is the speed of the car in miles per hour. According to this model, which of the following is the maximum speed, in miles per hour, Damon can be driving so that the braking distance is less than or equal to 150 feet?

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  1. iwanttogotostanford
    • one year ago
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    @Vocaloid

  2. tkhunny
    • one year ago
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    Have you considered solving: 150=3(x^2+10x)/40

  3. iwanttogotostanford
    • one year ago
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    no, i am not really trying to solve the problem as much as to know the steps TO solve it!:-) I'm practicing for the ACT

  4. iwanttogotostanford
    • one year ago
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    for the final step of the equation would it be 6,000+33x^3?

  5. tkhunny
    • one year ago
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    Well, then that is what you will have to do. You should break out your best Quadratic Equation solver that you have stored in your head.

  6. iwanttogotostanford
    • one year ago
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    i am just learning algebra 2 this year, math is my weakest subject

  7. iwanttogotostanford
    • one year ago
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    could you help me with the steps of solving that equation please?

  8. iwanttogotostanford
    • one year ago
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    because when i do it all i get 6000+3x^2+30x and i do not know how to solve that. Could you please explain

  9. tkhunny
    • one year ago
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    I might do this: 150=3(x^2+10x)/40 50=(x^2+10x)/40 2000=(x^2+10x) x^2 + 10x - 2000 = 0 10^2 - 4(1)(-2000) = 100 + 8000 = 8100 sqrt(9100) = 90 Thus, (-10 + 90)/2 = 80/2 = 40 and we are done. No need to worry about (-10 - 90)/2 because that is negative and doesn't concern us in this problem.

  10. iwanttogotostanford
    • one year ago
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    thank you so much! my ACT scores thank you too:-)))) @tkhunny

  11. iwanttogotostanford
    • one year ago
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    @tkhunny but on the second step how did you get 50?

  12. tkhunny
    • one year ago
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    Look carefully. Division by 3.

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