The equation can be used to find the approach velocity v (in feet per second) of a pole-vaulter and where h represents the height a pole-vaulter can reach. Erwin and Elmer are pole-vaulters. In one of the events they participated, Erwin’s approach velocity was 40 feet per second while Elmer’s approach velocity was 35 per second. Who reached greater height? Justify your answer.

Missing part of question:

A pole-vaulter’s approach velocity v (in feet per second) and height reached h (in feet) are related by the following equation.

v=8√h

Erwin

Step-by-step explanation:

The equation is

v=8√h

If velocity is 40 feet per second,

40=8√h

√h=40/8

h=√5

height =2.2360ft

If velocity = 35 feet per second,

v=8√h

35=8√h

√h=35/8

√h= 4.375

height =2.09ft

Therefore Erwin reached greater height =2.2360ft

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

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!