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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Part 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
 one year ago
Best ResponseYou've already chosen the best response.0@FEARLESS_JOCEY @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop @love10129151 @SolomonZelman @mathstudent55

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Well, first, did you chart the points they are taling about?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am totally confuzzled

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I graphed it, but idk where to go from there

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.0dinner soon can we speed this up?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0So lets look at 1. You put it in each equation. \(\begin{array}{cccc}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t) & 139 & & & \\ \hline g(t) & 58.7 & & & \hline \\ \hline \end{array}\) And you just fill it in.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0All 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
 one year ago
Best ResponseYou've already chosen the best response.0g(t), t=2, g(t)=97.4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just read all this and im sorry but i have no clue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0g(t)=136.1, when t=3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0h(t), when t=3, h(t)=131

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{array}{rccc}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & \hline \\ \hline \end{array}\) Yah, fine.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{array}{rccc}\hline t= & 1 & 2 & 3 & 4\\ \hline H(t)= & 139 & 151 & 131 & 79 \\ \hline g(t) =& 58.7 & 97.4 & 136.1 & 174.8 \\ \hline \end{array}\)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0OK... 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
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0I seem to have lost you, so whenever you get back, look at this: https://www.desmos.com/calculator/dam9lkolsw
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