anonymous
  • anonymous
In circle O, CD = 44, OM = 20, ON = 19,CD l OM, EF l ON (The figure is not drawn to scale.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
a.Find the radius. If your answer is not an integer, express it in radical form. b.Find FN. If your answer is not an integer, express it in radical form. c.Find EF. Express it as a decimal rounded to the nearest tenth.
anonymous
  • anonymous
@jim_thompson5910

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jim_thompson5910
  • jim_thompson5910
|dw:1360903051401:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1360903107022:dw|
anonymous
  • anonymous
i hate geometry..lol
jim_thompson5910
  • jim_thompson5910
|dw:1360903151843:dw| solve for x to find the radius a^2 + b^2 = c^2 20^2 + 22^2 = x^2
anonymous
  • anonymous
is this the answer
jim_thompson5910
  • jim_thompson5910
tell me what you get for x
anonymous
  • anonymous
42?
anonymous
  • anonymous
oops 22
jim_thompson5910
  • jim_thompson5910
20^2 + 22^2 = x^2 400 + 484 = x^2 x^2 = 884 x = ???
anonymous
  • anonymous
52
anonymous
  • anonymous
i don't know
jim_thompson5910
  • jim_thompson5910
x^2 = 884 x = sqrt(884) x = 2*sqrt(221)
jim_thompson5910
  • jim_thompson5910
so thats your radius
jim_thompson5910
  • jim_thompson5910
now you can use this radius to find EF and FN
anonymous
  • anonymous
202 + 222
jim_thompson5910
  • jim_thompson5910
solve for y y^2 + 19^2 = (2*sqrt(221))^2 y^2 + 381 = 884 keep going
anonymous
  • anonymous
What do you do after you solve for y? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I let y = EN so to find EF, you need to find 2y this is because EF = 2*EN EF = 2y
jim_thompson5910
  • jim_thompson5910
also, FN = EN, so once you get EN, you have found FN
anonymous
  • anonymous
y^2 + 361=884?? y^2=523 right?
jim_thompson5910
  • jim_thompson5910
oh somehow I had 381...hmm lol yeah it's 361
jim_thompson5910
  • jim_thompson5910
y^2=523 y=sqrt(523)
anonymous
  • anonymous
How did you get 884 up there?
jim_thompson5910
  • jim_thompson5910
(2*sqrt(221))^2 = 4*221 = 884
anonymous
  • anonymous
alright so y=29.73?
jim_thompson5910
  • jim_thompson5910
leave it in radical form
jim_thompson5910
  • jim_thompson5910
unless it asks about the approximation and rounding it
anonymous
  • anonymous
to leave it in radical form do I just leave it as y=sqrt(523)?
jim_thompson5910
  • jim_thompson5910
yes square roots are considered radicals
anonymous
  • anonymous
Solving for y solves for both FN and EF..?
jim_thompson5910
  • jim_thompson5910
FN yes since FN = EN
jim_thompson5910
  • jim_thompson5910
but you have to double it to find EF
anonymous
  • anonymous
So the one we just solved (y) is FN right? and I add FN and EN together.. but how do I do that since its in radical form?
jim_thompson5910
  • jim_thompson5910
or you can double FN to get EF
anonymous
  • anonymous
1,046? right
anonymous
  • anonymous
is EF
jim_thompson5910
  • jim_thompson5910
no EF = 2*sqrt(523)
anonymous
  • anonymous
How do I double Fn to get Ef
jim_thompson5910
  • jim_thompson5910
I just did it
anonymous
  • anonymous
-___- if 2*sqrt(523) is EF then half of that is Fn and En right..
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
Alright lol sorry I was confused on which was which. How do I find out what half of the radical is? just divide 523 by 2?
jim_thompson5910
  • jim_thompson5910
it's just 2*sqrt(523) and sqrt(523)
anonymous
  • anonymous
EF=2*sqrt(523) and EN and FN = sqrt(523) .. wow haha I feel stupid. Thank you so much though!
jim_thompson5910
  • jim_thompson5910
it's fine, yw
anonymous
  • anonymous
to find 2*sqrt(523) as a decimal rounded to the nearest tenth I multiply 523*2 and then take the square root of that number right?
jim_thompson5910
  • jim_thompson5910
you first take the square root of 523, then you double that result
jim_thompson5910
  • jim_thompson5910
then you round to whatever they want
anonymous
  • anonymous
so the answer would be 45.74?
jim_thompson5910
  • jim_thompson5910
yep 2*sqrt(523) = 45.74 roughly
anonymous
  • anonymous
Thank you again
jim_thompson5910
  • jim_thompson5910
sure thing

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