## anonymous 4 years ago Hi everyone! I'm working on some Algebra homework and I'm need some help w/a problem, please....

1. precal

2. anonymous

XD

3. anonymous

We're mathematicians (at least amateur :P), not psychics.

4. precal

and we are very patient

5. anonymous

Many people know that the weight of an object varies on different planets, but did you know that eh weight of an object on Earth also varies according to the elevation of the object? In particular, the weight of an object follows this equation: w=Cr^-2, where C is a constant and r is the distance that the object is from the center of the earth, and w is the weight of the object. (R=miles)(w=pounds) I figured out the equation on the 1st part that w=c/r^2: Now the ??? :) Use the value of C you found in the previous question to determine how much the object would weigh in New Orleans (2 feet below sea level) and Rocky Mountains (14,255 feet above sea level)

6. anonymous

Sorry for the delay

7. anonymous

On the previous question it asked me to determine if an object was 150 pounds at seal level to find the value of C that makes the equation true. This is what I came up with: C=wr^2 C=(150)(3963)^2 c=2,355,805,350

8. TuringTest

you have that r=3936ft, right?

9. TuringTest

r=3963 sorry

10. anonymous

yes

11. TuringTest

but the radius of the earth in feet is about 21,000,000ft so your number is in miles: r=3963mi

12. TuringTest

...which means C is proportional to miles, so we have to convert feet to miles r=3963mi is sea level 2 feet below:$2ft\times\frac{mi}{5280ft}\approx0.000379mi$14255 feet above:$14255ft\times\frac{mi}{5280ft}\approx2.7mi$so in new orleans$r=3963+0.000397$and in the Rocky Mountains$r=3963+2.7$so use those values of r to find the weight in your formula$W=\frac Cr$

13. anonymous

Great! I really appreciate your help!

14. TuringTest

*sorry, New Orleans should be$r=3963-0.000397$but I don't think it will affect the answer very much anyway