anonymous
  • anonymous
Please see the attached file. I WILL MEDAL AND FAN, PLEASE HELP!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
http://i30.photobucket.com/albums/c324/JMLOVER2121/Number2.gif
anonymous
  • anonymous
Find the value of x in the figure below. Show all steps.
anonymous
  • anonymous
The larger arc is 158°

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anonymous
  • anonymous
I got an answer of 5.78°
anonymous
  • anonymous
is this correct?
anonymous
  • anonymous
Just a second.
anonymous
  • anonymous
tyt :)
anonymous
  • anonymous
I can give you a formula that will help you find the answer. (9x - 5) = 1/2 (larger arc - smaller arc) So 158-64 = 94 / (1/2) = 47 angle (9x - 5) = 47 so 47 - 5 = 42 42 / 9 = 4.6666666667 or you can round to 4.70 You can check it.
anonymous
  • anonymous
Oh sorry it's actually my bad. You take 47 and then you add 5 and then you divide by 9 and that's x
anonymous
  • anonymous
right, yeah, I was just going to say that would make my answer 5.78 right?
anonymous
  • anonymous
No, that would make your answer 6
anonymous
  • anonymous
52/9 = 5.78....unless they wanted me to round up to the nearest whole number...
anonymous
  • anonymous
ahhh sorry so yeah. You're original answer is correct. I don't know why but for some reason today I'm just being extra... well I feel dumber. Is the question multiple choice?
anonymous
  • anonymous
lol, it's fine no it's free response
anonymous
  • anonymous
this is my work: (9x - 5)º = (1/2)(158° - 64º) (9x - 5)º = (1/2)(94) (9x - 5)º = 47° 9x + (-5 + 5) = (47 + 5)° 9x = 52 9x/9 = 52/9 x = 5.78 x = 5.78
anonymous
  • anonymous
Well you can wait to see if anyone else will answer but I would say go with your gut. I'm not really sure how else to put it. Sorry if I wasn't of help.
anonymous
  • anonymous
Yup, that looks about right though.
anonymous
  • anonymous
It's fine, mostly I just wanted to make sure my equation was right thnx for your help! :D
anonymous
  • anonymous
Sure thing!

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