MATH SOLVE

4 months ago

Q:
# PLEASE HELP!!!!!!A reflecting telescope is purchased by a library for a new astronomy program. The telescope has a horizontal parabolic frame and contains two mirrors, 3 inches apart from each other- the first in the base and the second at the focal point. The astronomy teacher would like to attach a digital monitoring system on the edge of the telescope, creating a straight line distance above the focal point. Assuming that the telescope is placed on the edge of the roof in such a way that it is parallel to the ground, the position of the monitoring system, in relationship to the distance between the base and the focal point is modeled by the equation, x =[tex] \frac{1}{12} [/tex] y2 (x = the distance between the base and the focal point; y= the height of the monitoring system).Rewrite the model so that the height of the digital monitoring system is a function of the distance between the base and the focal point of the telescope. How high above the focal point is the digital monitoring system attached to the telescope? Include your function and the height, rounded to the nearest tenth of an inch, in your final answer.

Accepted Solution

A:

Hello! To rewrite the model so that the height of the digital monitoring system is a function of the distance between the base and the focal point of the telescope, we simply need to alter the equation such that the other side would be left with y alone.

[tex]x= \frac{1}{12} y^{2} [/tex]

[tex]12x=y^{2}[/tex]

[tex] \sqrt{12x}=y [/tex]

[tex]f(x)=2 \sqrt{3x} [/tex]

We have replaced y with f(x) to show that it is a function.

For the second part we just use the fact that the two mirrors are 3 inches apart. From the definition of the variables we know that this is just the value of x. We plug this into the function to know how high above the focal point the digital monitoring system is attached.

[tex]f(3)=2 \sqrt{3(3)}=2\sqrt{9}=2(3)=6[/tex]

ANSWER: The function is given by [tex]f(x)=2 \sqrt{3x} [/tex] and the digital monitoring system is 6.0 inches above the focal point.

[tex]x= \frac{1}{12} y^{2} [/tex]

[tex]12x=y^{2}[/tex]

[tex] \sqrt{12x}=y [/tex]

[tex]f(x)=2 \sqrt{3x} [/tex]

We have replaced y with f(x) to show that it is a function.

For the second part we just use the fact that the two mirrors are 3 inches apart. From the definition of the variables we know that this is just the value of x. We plug this into the function to know how high above the focal point the digital monitoring system is attached.

[tex]f(3)=2 \sqrt{3(3)}=2\sqrt{9}=2(3)=6[/tex]

ANSWER: The function is given by [tex]f(x)=2 \sqrt{3x} [/tex] and the digital monitoring system is 6.0 inches above the focal point.