kay was walking along the bridge above the river. she accidentally lost her grip on her camera when she was trying to get it out to photograph massive bird nests in the trees. the height of the camera above the water can be modeled by the equation
y = -16t^2 + 400, where t represents the seconds after the camera is dropped. how long will it take (in seconds) before kay sees the splash from the camera hitting the water.

The given equation in this case is:

 y = -16 t^2 + 400

We can see that the slope of the equation is negative (-16) which means that the position of Kay is on the positive y-axis , while the bottom of the river is in the origin. This is supported by taking t = 0, the initial y becomes y = 400. Therefore the position of Kay now is at (0, 400).

So to find for the time when the camera hits the water, then y must be zero (t, 0) so we have to find for t in this case.

When y = 0

0 = -16 t^2 + 400

16 t^2 = 400

t^2 = 25

t = 5 s

Therefore it takes 5 seconds for the camera to hit the water and for Kay to see a splash.

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!

uhhh idk u yolo?