Answer:
The height of rocket is 102.7 meter.
Step-by-step explanation:
Given : Brynn and Denise launch their rockets at the same time.
The height of Brynn’s rocket, in meters, is given by the function [tex]f(x)=-4.9x^2+75x[/tex] , where x is the number of seconds after the launch.
The height of Denise’s rocket, in meters, is given by the function
[tex]g(x)=-4.9x^2+50x+38[/tex], where x is the number of seconds after the launch.
There is a moment when the rockets are at the same height.
To find : The height
Solution :
When the rockets have same height
So, [tex]f(x)=g(x)[/tex]
[tex]-4.9x^2+75x=-4.9x^2+50x+38[/tex]
[tex]-4.9x^2+75x+4.9x^2-50x=38[/tex]
[tex]25x=38[/tex]
[tex]x=\frac{38}{25}[/tex]
[tex]x=1.52[/tex]
Now, we put x value in any of the function to find height.
[tex]f(x)=-4.9x^2+75x[/tex] , x=1.52
[tex]f(x)=-4.9(1.52)^2+75(1.52)[/tex]
[tex]f(x)=-11.32096+114[/tex]
[tex]f(x)=102.67[/tex]
Nearest tenth = 102.7
Therefore, The height of rocket is 102.7 meter.
Answer: 102.7 meters
Step-by-step explanation:
Given: Brynn and Denise launch their rockets at the same time.
The height of Brynn’s rocket, in meters, is given by the function [tex]f(x)=-4.9x^2+75x[/tex] , where x is the number of seconds after the launch.
The height of Denise’s rocket, in meters, is given by the function
[tex]g(x)=-4.9x^2+50x+38[/tex], where x is the number of seconds after the launch.
The moment when both rockets are on same height, then [tex]f(x)=g(x)[/tex]
[tex]\Rightarrow-4.9x^2+75x=-4.9x^2+50x+38\\\\\Rightarrow\ 75x=50x+38\\\\\Rightarrow\ 25x=38\\\\\Rightarrow\ x=1.52[/tex]
It means at 1.52 seconds the rockets are at the same height.
To calculate the height substitute, the value of x in any of the function.
[tex]f(1.52)=-4.9(1.52)^2+75(1.52)=102.67904\approx102.7[/tex] meters
