The function H(t) = −16t2 + 75t + 25 shows the height H(t), in feet, of a baseball after t seconds. A second baseball moves in the air along a path represented by g(t) = 5 + 5.2t, where g(t) is the height, in feet, of the object from the ground at time t seconds. Part A: Create a table using integers 2 through 5 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)
@dude
|dw:1526598319337:dw| In order to solve for the y value, substitute in the x values (if you don't have a graph)
ok but how do i do this
how do i solve for x or y if neither of them are in the equation
@dude
x is 2, 3, 4, 5
ok
For example: \(H(2) = −16(2)^2 + 75(2) + 25\) and \(g(2) = 5 + 5.2(2)\)
This is only for when x=2
im still so confused
if x = 2,3,4,5 how do you know which one is going in the equation and which ones wont
i dont get this
It asks you in part A `Create a table using integers 2 through 5 for the 2 functions.` "Integers 2 through 5"
so what you gave is the answer for part A?
Part of it yes, we still have to solve for when x is 3, 4 and 5
but we didnt solve when it was 2 did we
Im sorry but im so lost
\(H(2) = −16(2)^2 + 75(2) + 25\\ -16\times 4+150+25\\ 64+150+25\\ H(2)=239\) and \(g(2) = 5 + 5.2(2)\\ 5+5.2\times 2\\ 5+10.4\\ g(2)=15.4\)
and do the same for 3 4 and 5
Right
ok ill try
h(3)=-16(3)^2+75(3)+25 h(3)=-16 x 3^2+75 x 3 +25 h(3)=-3^2 x 16+225+25 h(3)=250-3^2x16 3^2x16=144 h(3) 250-144 h(3) = 106 ^CORRECT?
That is right
h(4)=-16(4)^2+75(4)+25 h(4)=-16x4^2+75x4+25 h(4)=-4^2x16+300+25 h(4)=325-4^2x16 4^2x16=256 h(4)=325-256 H(4)=69
^CORRECT?
That is also correct
h(5)=-16(5)^2+75(5)+25 h(5)=-16x5^2+75x5+25 h(5)=-5^2x16+375+25 h(5)=400-5^2x 16 5x16=400 h(5)= 400-400= h(5)=0
^CORRECT?
Whats next
oh i forgot g
g(2) = 5 + 5.22 Add 10.22 Divide by both sides by 2 g(2)=5.11
^CORRECT
?
g(3) = 5 + 5.23 Add 10.23 Divide both sides by 3 g(3)=3.41 ^CORRECT?
g(4) = 5 + 5.24 Add 10.24 Divide both sides by 4 g(4)=2.56 ^CORRECT?
h(5) is correct, re check g(2)
g(5) = 5 + 5.25 Add 10.25 Divide both sides by 5 g(5)=2.05
You're supposed to multiply 5.2 and the x variable first for g(t) I showed an example for g(2)
There is no division required for g(t)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude and \(g(2) = 5 + 5.2(2)\\ 5+5.2\times 2\\ 5+10.4\\ g(2)=15.4\) \(\color{#0cbb34}{\text{End of Quote}}\)
That is the example for g(2)
im confused now
i just looked it up https://www.symbolab.com/solver/algebra-calculator/g%5Cleft(2%5Cright)%3D5%2B5.22
That is assuming that g is a variable within the function, it is not important to the problem you only have to type whats after the g(#)
omg im sooooooo confused can you please explain more
i dont get thissssss
g(2) is just y when x equals 2 It doesn't mean anything that can help us in this problem You can completely ignore that when solving https://www.symbolab.com/solver/algebra-calculator/5%2B5.2%5Cleft(2%5Cright)
can you just help me find the answer for this i seriously am confused and i dont got how were going to answer this its 8PM im tired and i dont understand a word your saying
g(3) = 5 + 5.2*3 Multiply 5.2 and 3 \(\bf you~can~completely~ignore~the~g(3)for now\) 5+15.6 =20.6 That is g(3)
For g(4): g(4) = 5 + 5.2*4 Multiply 5.2 and 4 \(\bf you~can~completely~ignore~the~g(4)for~now\) 5+20.8 \(\bf =25.8\) This is g(4)
For g(5): g(5)=5+5.2*5 Multiply 5.2 and 5 \(\bf you~can~completely~ignore~the~g(5)for~now\) 5+26 \(\bf 31\) This is g(5)
so g(2)=15.4 g(3)=20.6 g(4)=25.8 g(5)=31 and h(2)=239 h(3)=106 h(4)=69 h(5)=0 Then what
Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? how do we get 2 seconds to solution to = g(t)
|dw:1526601848984:dw|
ok
question is how do i put that in writing
The question asked for a table, so I gave you one
ok nvm ill figure it out ok so what does that make the answer
Between 4 and 5 seconds because we see that they both might have been at the same height and would've crossed each other
and that graph dosnt help it makes it confusing
ohhh ok so if thats the answer to part a lets work on part b
Explain what the solution from Part A means in the context of the problem.
So you see when x is 4, they haven't crossed each other? Now when you look at x=5 the line is already past the other equation They had intersected in between those points which was labeled `intersection!`
passed*
ok
So this represents that shortly after 4 seconds of being thrown both baseballs would have reach the same height
^answer for part B
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