Five friends go out to the football field, and decide to throw tide pods onto the field, while standing at different places in the bleachers. each throw is modeled by functions below, where the height of the throw is a function of how much time it is in the air: alastor: h= -6t^2 + 36t + 96. bael: h= -8t^2 + 32t + 40. corson: h= -4t^2 + 16t + 48. danjal: h= -6t^2 + 54. eblis: h= -4t^2 + 32t + 80. in a thorough explanation and description of the methods you are using, and sharing and relevant computations and work, describe 1. how far up in the bleachers they started, 2. the maximum height their tide pod reached on the air, and 3. how long it took for it to hit the football field. also share whose went highest, and whose took longest to land.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: