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anonymous

  • 5 years ago

help please in 1920 the record for a certain race was 45.9 seconds. In 1960 it was 44.7 sec. let R(t) = the record in the race and t=the number of years since 1920. a. find a linear function that fits the data b. use the function in (a) to predict the record in 2003 and in 2006 c. find the year when the record will be 43.26 sec. round to nearest hundredth

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  1. dumbcow
    • 5 years ago
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    use 2 points (t,R) (0,45.9) and (40,44.7) slope = (44.7-45.9)/(40-0) = -1.2/40 = -.03 y_intercept is when t=0: (0,45.9) R(t) = -.03t + 45.9

  2. anonymous
    • 5 years ago
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    hi dumbcow what is the ans to the other two questions b and c

  3. dumbcow
    • 5 years ago
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    use R(t) and substitute in t=83 and t=86 for part c 43.26 = -.03t +45.9 solve for t

  4. anonymous
    • 5 years ago
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    what year will the record be 43.26 sec

  5. dumbcow
    • 5 years ago
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    get t by itself 1) subtract 45.9 on both sides 2) divide by -.03 on both sides

  6. anonymous
    • 5 years ago
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    I'm very confuse now

  7. dumbcow
    • 5 years ago
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    do you know how to solve linear equations? like 4x -2 = 10 solve for x

  8. anonymous
    • 5 years ago
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    no i do not

  9. dumbcow
    • 5 years ago
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    well that would help you with this question and class im guessing... t=88 this means 88 years after 1920 =2008

  10. anonymous
    • 5 years ago
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    that helped me thank you very much

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