anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@FEARLESS_JOCEY @ganeshie8

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anonymous
  • anonymous
@nincompoop @love10129151 @SolomonZelman @mathstudent55
anonymous
  • anonymous
please?
e.mccormick
  • e.mccormick
Well, first, did you chart the points they are taling about?
anonymous
  • anonymous
no? @e.mccormick
anonymous
  • anonymous
I am totally confuzzled
anonymous
  • anonymous
I graphed it, but idk where to go from there
e.mccormick
  • e.mccormick
Well, you need to chart out the points. it will be something like: \(\begin{matrix} t= & 1 & 2 & 3 & 4\\ H(t) & & & & \\ g(t) & & & & \end{matrix} \)
anonymous
  • anonymous
okayyy
anonymous
  • anonymous
dinner soon- can we speed this up?
e.mccormick
  • e.mccormick
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.
e.mccormick
  • e.mccormick
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.
anonymous
  • anonymous
h(t), t=2, h(t)=151
e.mccormick
  • e.mccormick
Yep.
anonymous
  • anonymous
g(t), t=2, g(t)=97.4
anonymous
  • anonymous
i just read all this and im sorry but i have no clue
anonymous
  • anonymous
g(t)=136.1, when t=3
anonymous
  • anonymous
h(t), when t=3, h(t)=131
anonymous
  • anonymous
good so far?
anonymous
  • anonymous
h(t)=79, when t=4
e.mccormick
  • e.mccormick
\(\begin{array}{|r|c|c|c|}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & \hline \\ \hline \end{array}\) Yah, fine.
e.mccormick
  • e.mccormick
\(\begin{array}{|r|c|c|c|}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & 79 \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & 174.8 \\ \hline \end{array}\)
e.mccormick
  • e.mccormick
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? "
e.mccormick
  • e.mccormick
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?
e.mccormick
  • e.mccormick
I seem to have lost you, so whenever you get back, look at this: https://www.desmos.com/calculator/dam9lkolsw

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