Question

100 POINTS!!! PLEASE HELP ASAP EXPERTS! PLEASE I'M BEGGING!!

Answer 1

Answer:

A)  

$\begin{array}{|c|c|c|}\cline{1-3} t & H(t) & g(t)\\\cline{1-3} 1 & 60 & 20.4\\\cline{1-3} 2 & 76 & 30.8\\\cline{1-3} 3 & 60 & 41.2\\\cline{1-3} 4 & 12 & 51.6\\\cline{1-3} \end{array}$

Between 3 and 4 seconds.

B)  The baseballs will collide between 3 and 4 seconds.

Step-by-step explanation:

Given functions:

$H(t)=-16t^2+64t+12$

$g(t)=10+10.4t$

Part A

Substitute the values of t = 1, 2, 3 and 4 into the two functions:

$H(1)=-16(1)^2+64(1)+12=60$

$H(2)=-16(2)^2+64(2)+12=76$

$H(3)=-16(3)^2+64(3)+12=60$

$H(4)=-16(4)^2+64(4)+12=12$

$g(1)=10+10.4(1)=20.4$

$g(2)=10+10.4(2)=30.8$

$g(3)=10+10.4(3)=41.2$

$g(4)=10+10.4(4)=51.6$

Create a table with the found values:

$\begin{array}{|c|c|c|}\cline{1-3} t & H(t) & g(t)\\\cline{1-3} 1 & 60 & 20.4\\\cline{1-3} 2 & 76 & 30.8\\\cline{1-3} 3 & 60 & 41.2\\\cline{1-3} 4 & 12 & 51.6\\\cline{1-3} \end{array}$

The solution to H(t) = g(t) is between 3 and 4 seconds as:

When t = 3, H(t) > g(t)

When t = 4, H(t) < g(t)

To prove this, equate the equations and solve for t:

$\implies -16t^2+64t+12=10+10.4t$

$\implies -16t^2+53.6t+2=0$

Using the Quadratic Formula to solve for t:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$\implies t=\dfrac{-53.6 \pm \sqrt{53.6^2-4(-16)(2)} }{2(-16)}$

$\implies t=\dfrac{53.6 \pm \sqrt{3000.96}}{32}$

$\implies t=3.39, -0.04\:\:(\sf 2\:d.p.)$

As time is positive, t = 3.39 s (which is between 3 and 4 seconds).

Part B

When the two baseballs are at the same height they will collide.

Therefore, the baseballs will collide between 3 and 4 seconds (when t = 3.39 s).

Answer 2

The solution to H(t) = g(t) is located in the function between 1 and 2.

How to illustrate the function?

H(t) = −16t2 + 64t + 12

g(t) = 10 + 10.4t,

t.  H(t)   g(t)

0   12.    10

1.   60.  20.4

2.  76.   30.8

The ball that follows function H(t) increase its height, reaches a maximum and, then, decreases its height. The ball that follows function g(t) increases its height all the time. In the beginning, the ball that follows function H(t) increases its height faster than the other ball, but after it reaches its maximum height, its height starts to decrease, giving the opportunity to the other ball to reach it.

Learn more about functions on:

brainly.com/question/16134033

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