Kayden is a stunt driver. one time, during a gig where she escaped from a building about to she drove at a constant speed to get to the safe zone that was 160160 meters away. after 33 seconds of driving, she was 8585 meters away from the safe zone. let d(t)d(t) denote the distance to the safe zone dd (measured in meters) as a function of time tt (measured in seconds). write the function's formula.

The required function formula is,

D(t) = 160 - 25 t

Step-by-step explanation:

Given,

The total distance of her from the safe zone = 160 meters,

And, after 3 seconds she is 85 meters far away from the safe zone,

Thus, the total distance she covered in 3 seconds = 160 - 85 = 75 meters,

Since, she drove at a constant speed,

Also,

Thus, the distance she will cover in t seconds = 25 t

( Because, Distance = Speed × Time )

Hence, the total distance of her from the safe zone after t seconds, D(t) = 160 - 25t

Which is the required function formula.

Kayden drives at a constant speed. The function D(t) can be represented as .

Given information:

Kayden is a stunt driver.

Total distance of safe zone is 160 m.

She drove with a constant speed.

After 3 seconds of driving, she was 85 meters away from the safe zone.

So, the distance traveled in 3 seconds will be 160-85=75 m.

Now, the speed of the driver will be,

The distance traveled in t seconds will be 25t.

So, the equation representing the given condition will be,

Therefore, the function D(t) can be represented as .

For more details, refer to the link:

link

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!