mutliply by 63:
63(2/9+ 5/7+ 13/3) = 14/63 + 45/63 + 273/53
= 332/ 63
2007-11-30 11:08:00
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answer #1
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answered by sayamiam 6
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I really don't understand the point behind this question, but here goes:
2/9 + 5/7 + 13/3 = 2/9 + 39/9 + 5/7 = 41/9 + 5/7 =
(41*7+9*5)/63 = (287+45)/63 = 332/63.
Now 332 = 166*2 = 83*4 while 63 = 9*7 = 3*3*7 have no common factors, so this is the reduced form.
And I did all that w/o a calculator;)
As for 2, the common denominator of 7/3 and 8/6 is 3, since 8/6 = 4/3. Then both expressed w/ that denominator are 7/3 (unchanged) and 4/3.
2007-11-30 19:12:35
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answer #2
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answered by Indicator Veritatis 2
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2/9+ 5/7+ 13/3?
2/9 + 5/7 + 13/3 = find common base, 9*7 = 63 so..
63/9 = 7
63/7 = 9
63/3 = 21
so 2/9 + 5/7 + 13/3=
2*7/9*7 + 5*9/7*9 + 13*21/ 3*21 =
14/ 63 + 45/ 63 + 273 / 63 =
(14+45+273) / 63 =
( 59 + 273 )63 =
332/63 = thats it, the simplest form
Answer: 332/63 or 5.27 round to two decimals
*still confused? write back
2007-11-30 19:13:34
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answer #3
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answered by john_lu66 4
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1,. 2/9 + 5/7 + 13/3
common denominator = 63
63/9 = 7 63/7= 9 63/3= 21
14/63 + 45/63 + 273/63 = 332/63
2. lcd= 12
because 4x3 = 12 and 6x2 = 12
7/3=28/12 8/6=16/12
2007-11-30 19:11:56
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answer #4
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answered by michael c 2
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1. You need to have a common denominater to add fractions..
the LCD of 9,7,and 3 is 63
2/9 (multiply numerator and denominator by 7)
2/9 = 14/63
5/7 (multiply numerator and denominator by 9)
5/7 = 45/63
13/3 (multiply numerator and denominator by 21)
13/3 = 273/63
now, that they all have the same denominator, you can add the numerators
(14 + 45 +273)/13
332/13
2. the LCD of 3 and 6 is 6
7/3 (multiply numerator and denominator by 2)
14/6
leave 8/6 as is
2007-11-30 19:12:09
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answer #5
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answered by albs 2
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1. 2/9 + 5/7 + 13/3
= (2*7 + 5*9 + 13*3*7)/9*7
= (14 + 45 + 273)/63
= 332/63
2007-11-30 19:15:59
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answer #6
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answered by sv 7
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2/9 + 5/7 + 13/3
LCD is 9*7 = 63
14/63 + 45/63 + 273/63
(14+45+273)/63
332/63
2007-11-30 19:13:09
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answer #7
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answered by Robert S 7
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1.332/9 or 36 and 8/9
2.14/6 and 8/6
2007-11-30 19:15:56
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answer #8
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answered by Anonymous
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