plz help me with a math problem?

pupsikleva · #1 (**permalink**) 02-24-2008, 07:31 PM

Assume y varies inversely as x.
If x = 1/3 when y = 162, find x when y = 3.

=Tiffany · #2 (**permalink**) 02-24-2008, 07:39 PM

54
you do 162 dived by 3 equals 54 so
1/3 is 54
2/3 is 108
3/3 162

3408 · #3 (**permalink**) 02-24-2008, 07:39 PM

when y = 3 (it's been divided by 54), you do the same to x

1/3 divided by 54 is 1/162

x = 1/162

THIS ANSWERS RIGHT, EVERYONE ELSE IS APPARENTLY AS STUPID AS ****

clcclnc · #4 (**permalink**) 02-24-2008, 07:39 PM

Good luck with that.

kelsey93092 · #5 (**permalink**) 02-24-2008, 07:39 PM

well whats the equation to plug them into?

sweetz · #6 (**permalink**) 02-24-2008, 07:39 PM

Given y = k/x

x = 1/3, y = 162

x = ? when y = 3

when x = 1/3, y =162

162= k/0.333 or 3k/1

162 = 3k.
162/3 = k

k =54

y = 3

y = k/x i.e; 3 = 54/x

x = 54/3

x = 18

i hope its clear now...

for more problems of this type

http://www.glencoe.com/sec/math/algebra/algebra1/algebra1_03/study_guide/pdfs/alg1_pssg_G091.pdf

http://answers.yahoo.com/question/index?qid=20070405012616AAqPuup

D.W. · #7 (**permalink**) 02-24-2008, 07:40 PM

y/54 = 162/54 = 3
x = x + 54 = 1/3 * 54 = 18

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