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anonymous

  • one year ago

The function H(t) = −16t^2 + 60t + 95 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 20 + 38.7t, where g(t) is the height, in feet, of the object from the ground at time t seconds.

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  1. anonymous
    • one year ago
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    Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points) Part B: Explain what the solution from Part A means in the context of the problem. (4 points) @e.mccormick @welshfella @ali2x2

  2. anonymous
    • one year ago
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    @dan815

  3. anonymous
    • one year ago
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    @FEARLESS_JOCEY @ganeshie8

  4. anonymous
    • one year ago
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    @nincompoop @love10129151 @SolomonZelman @mathstudent55

  5. anonymous
    • one year ago
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    please?

  6. e.mccormick
    • one year ago
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    Well, first, did you chart the points they are taling about?

  7. anonymous
    • one year ago
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    no? @e.mccormick

  8. anonymous
    • one year ago
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    I am totally confuzzled

  9. anonymous
    • one year ago
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    I graphed it, but idk where to go from there

  10. e.mccormick
    • one year ago
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    Well, you need to chart out the points. it will be something like: \(\begin{matrix} t= & 1 & 2 & 3 & 4\\ H(t) & & & & \\ g(t) & & & & \end{matrix} \)

  11. anonymous
    • one year ago
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    okayyy

  12. anonymous
    • one year ago
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    dinner soon- can we speed this up?

  13. e.mccormick
    • one year ago
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    So lets look at 1. You put it in each equation. \(\begin{array}{|c|c|c|c|}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t) & 139 & & & \\ \hline g(t) & 58.7 & & & \hline \\ \hline \end{array}\) And you just fill it in.

  14. e.mccormick
    • one year ago
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    All it is saying is that at t=1, H(t) = 139 and g(t) = 58.7. You just plug the number in to each equation to get these answers.

  15. anonymous
    • one year ago
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    h(t), t=2, h(t)=151

  16. e.mccormick
    • one year ago
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    Yep.

  17. anonymous
    • one year ago
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    g(t), t=2, g(t)=97.4

  18. anonymous
    • one year ago
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    i just read all this and im sorry but i have no clue

  19. anonymous
    • one year ago
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    g(t)=136.1, when t=3

  20. anonymous
    • one year ago
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    h(t), when t=3, h(t)=131

  21. anonymous
    • one year ago
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    good so far?

  22. anonymous
    • one year ago
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    h(t)=79, when t=4

  23. e.mccormick
    • one year ago
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    \(\begin{array}{|r|c|c|c|}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & \hline \\ \hline \end{array}\) Yah, fine.

  24. e.mccormick
    • one year ago
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    \(\begin{array}{|r|c|c|c|}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & 79 \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & 174.8 \\ \hline \end{array}\)

  25. e.mccormick
    • one year ago
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    OK... so, there is the chart. So the question about the chart is " Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? "

  26. e.mccormick
    • one year ago
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    If you look at the chart and the equations, you know that H(t) is a curve and g(t) is a line that intersect somewhere. Now H(t) obviously goes up from 1 to 2, but some time between 1 and 3 it peaked and is going down. With all that info, can you see where they might cross?

  27. e.mccormick
    • one year ago
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    I seem to have lost you, so whenever you get back, look at this: https://www.desmos.com/calculator/dam9lkolsw

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