1/2 (1/3 - 1/p) =1/4
divide by 1/2 each side :
(1/3 - 1/p) = (1/4) / (1/2)
substract 1/3 each side :
-1/p = [ (1/4) / (1/2) ] - 1/3
-1/p = 1/6
cross product (numerator of first side by denominator of second side and vice vera ) :
p*1 = -1 *6
p = -6
Good luck !
2006-10-03 11:22:10
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answer #1
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answered by Vee A 1
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The correct answer is -6
1/2 (1/3-1/p) = 1/4
2*(1/2(1/3 - 1/p)) = (1/4)*2 (multiply both sides by 2)
1/3 - 1/p = 2/4
(1/3 - 1/p) -1/3 = (1/2)-1/3 (-1/3 from each side)
-1/p = 1/2 - 1/3
-1/p = 3/6 - 2/6
-1/p = 1/6 (multiply each side by -1)
1/p = -1/6 (cross multiply to solve for p)
p = -6
Hope this helps..
2006-10-03 18:33:59
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answer #2
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answered by sunshine 4
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1/2 * (1/3 - 1/p) = 1/4
2 * 1/2 * (1/3 - 1/p) = 2 * 1/4
1 * (1/3 - 1/p) = 2/4
1/3 - 1/p + 1/p - 2/4 = 2/4 - 2/4 + 1/p
1/3 - 2/4 = 1/p
4/12 - 6/12 = 1/p
(4-6)/12 =1/p
-2/12 = 1/p
-1/6 = 1/p
-6 = p
2006-10-03 18:16:28
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answer #3
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answered by spongeworthy_us 6
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1/2 (1/3 - 1/p) = 1/4 //multiply by 2
1/3-1/p=1/2 //add 1/p to both sides
1/3=1/p+1/2 //subtract 1/2 from both sides
1/3-1/2=1/p //find a common denominator
2/6-3/6=1/p //subtract
-1/6=1/p //cross-multiply
p=-6
you must have found 1/p instead of p itself
2006-10-03 18:20:19
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answer #4
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answered by Ramesh S 2
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since 1/2 * 1/2 = 1/4
and we have 1/2(1/3-1/p)=1/4
if A*X=B and A*Y=B then X=Y
then 1/3-1/p=1/2
-1/p=1/2-1/3=1/6
1/p= -1/6
therefore
p= -6
2006-10-03 18:34:34
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answer #5
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answered by washiko 2
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1/2(1/3-1/p) = 1/4
1/2(p-3)/3p = 1/4
1/6p(p-3) = 1/4
cross-multiply
4(p-3) = 6p
4p-12 = 6p
-12 = 2p
so, p= -6
2006-10-03 18:19:22
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answer #6
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answered by donnoMuch 2
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1/2(1/3-1/p)=1/4
1/3-1/p=1/2
1/p=1/3-1/2
1/p=2/6-3/6
1/p=-1/6
p=-6
2006-10-03 18:18:37
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answer #7
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answered by Anonymous
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1/2(1/3-1/p)=1/4
1/6-1/2p=1/4
-1/2p=1/4-1/6
-1/2p=-1/12
1/p=-1/6
p=-6 answer is -6
2006-10-03 18:14:43
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answer #8
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answered by lifescircle 5
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1/2(1/3-1/p)=1/4
1/3-1/p=1/2
1/p=1/3-1/2
1/p=2/6-3/6
1/p=-1/6
p=-6
2006-10-03 18:13:54
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answer #9
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answered by Katie 5
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i got that answer too but does that say 1 over p or 1p becasue that makes a big diffference?
2006-10-03 18:12:14
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answer #10
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answered by Chynadoll 2
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