112/3+ 81/4=57.58333333
(112/3+ 81/4
=(112*4)+(81*3)/12 (LCD=2*2*3*1=12)
=(448+243)/12
=691/12
=57.58333333
=57 7/12)
IF IT IS 11 2/3+ 8 1/4
=33+2/3 +32+1/4
=35/3+33/4
=(35*4+33*3)/12 LCD=12
=(140+99)/12
=239/12
=19.92
=19 11/12
2006-10-26 18:33:39
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answer #1
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answered by Anonymous
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In order to successfully add fractions with different denominators, the denominator of both fractions should be equal.
The common denominator is the LCM of the original denominators
LCM of 3 and 4 is 12
Therefore
112/3+81/4 = (112*4)/(3*4) + (81*3)/(4*3)
= 448/12 + 243/12
= (448 + 243)/12
= 691/12
= 57.5833333333333...
2006-10-28 06:49:44
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answer #2
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answered by Akilesh - Internet Undertaker 7
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112/3 + 81/4 = 448/12 + 243/12 = 691/12
2006-10-27 01:29:21
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answer #3
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answered by Pascal 7
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You fust find a common denominator, which is 4 x 3 = 12 in this case
(112 * 4) / (3 * 4) + (81 * 3) / (4 * 3)
448/12 + 243/12 = 691/12
This can be wrote as a mixed number 57 7/12
2006-10-27 01:29:19
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answer #4
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answered by Clayton A 2
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1. 112/3 + 81/4 Original problem
2. (112 x 4)/12 + (81 x 3)/12 Get a common denominator
3. 448/12 + 243/12 Add teh numerator
4. 691/12/4
2006-10-27 01:29:04
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answer #5
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answered by aftababhan 2
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1) Given 112/3 + 81/4
2) Taking common denominator
(112*4 + 81*3)/(3*4)
3) Simplifying
( 448 + 243 )/ (12)
= ( 691 )/ 12
= 57.583 (approx.)
2006-10-27 01:49:08
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answer #6
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answered by anjali 2
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LCD is 12
if the sum is 112/3+81/4 as it seems
448/12+243/12=691/12
=57 7/12
however if the sum is
11 2/3+8 1/4
35/3+33/4
LCD is again 12
=(140+99)/12
=239/12=19 11/12
2006-10-27 01:30:51
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answer #7
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answered by raj 7
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If that says what I think it says ( 112 over 3 plus 81 over 4) then the answer is 57 7/12 (57 and 7 twelths)
2006-10-27 01:28:58
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answer #8
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answered by Jeff 3
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112/3 + 81/4
Find Lowest Common Multple (12)
448/12 + 243/12
(691/12) --> Correct Answer
2006-10-27 01:30:13
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answer #9
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answered by Bentnalboy 3
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=(112 * 4)/(3 * 4) + (81 * 3) / (4 * 3)
=((112 * 4) + (81 * 3)) / 12
=(448 + 243) / 12
= 691/12
= 57.58
2006-10-27 01:57:37
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answer #10
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answered by calpal2001 4
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