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anonymous
 one year ago
Really simple problem, so can someone help me out! I appreciate it!
I had to measure the radii, circumferences, and areas of some circles. I put the data into a table and was asked to graph the radius (cm) to Circumference (cm). I did so and then it asked me to find the experimental equation. I know that the accepted is A=2pir, so what is the experimental? Thanks a lot!
anonymous
 one year ago
Really simple problem, so can someone help me out! I appreciate it! I had to measure the radii, circumferences, and areas of some circles. I put the data into a table and was asked to graph the radius (cm) to Circumference (cm). I did so and then it asked me to find the experimental equation. I know that the accepted is A=2pir, so what is the experimental? Thanks a lot!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or how do I find the experimental I should say.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero @Abhisar @dakotahducharme @zepdrix @Luigi0210 @Compassionate @nincompoop @freckles @midhun.madhu1987 @adrynicoleb

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Suppose on y you graph circumference and on x you graph radius then we will get a straight line of equation \[y=mx+c\] Since this straight line passes through origin, c=0 we have \[y=mx\] \[2\pi.r=mr\] Where m is the slope using this you can find the slop as 2pi, I'm not sure if this is what they are asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0They want an equation with another number that is similar, as I am also asked to find percent error.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after I find the experimental equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hm, I can't understand what they want, lol sorry :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but is there a way to calculate something like pi that can replace it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with the specific data I found

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well if you divide the circumference by the diameter, you will get pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's the definition of pi itself

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, I think I get what I need to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do that to find pi and see how close it is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, thanks for the help!
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