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anonymous
 one year ago
WILL MEDAL AND FAN!!!
Express answer in exact form.
Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.)
and is there any way to put the answer in this format?
https://suwannee.owschools.com/media/g_geo_2013/8/q255_x.gif
anonymous
 one year ago
WILL MEDAL AND FAN!!! Express answer in exact form. Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.) and is there any way to put the answer in this format? https://suwannee.owschools.com/media/g_geo_2013/8/q255_x.gif

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437507590402:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437507884957:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, but the internet here is running extremely slow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's taking me like five minutes to send a reply

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright I thin I understand how you got that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what about the second one?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1problems here too maybe it was this website

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1remember I did not see problem 1. I don't know what you have for the value of the area of the smaller segment shaded in my sketch

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh. It's A= (64/6)pi = 16 times sqrt3) inches squared

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1do you have a question for me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437569371363:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need to find the area of the part that's shaded in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you must first find the area of a) the circular pie shaped wedge b) the triangle then you subtract the two areas

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that how you find the first one im still trying to figure out the second one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I had to get help with the last one yesterday too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got that from this site if it helps http://www.mathopenref.com/segmentarea.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh. I'm gonna go ahead and write down the site then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But. I'm supposed to have the answer in this format here:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://suwannee.owschools.com/media/g_geo_2013/8/q254_x.gif

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I don't know how to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got you hold up ill see what i can do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, i appreciate it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A[16pi8]inches ^2 i think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how would you put it in that format

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A={(32over2)pi  ? } inches squared

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that what im still trying to figure out
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