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Postby The Weird One on Mon Oct 22, 2007 11:42 am

I think I've got the following problem solved correctly, but I'ld like to be sure (this teacher grades hmwk for correctness)

(2/3)x-(5/2)y=2


standard form= 2x-5y=12

slope-intercept form= y=(2/5)x-(12/5)

x-intercept= (6,0)

y-intercept= (0,-[12/5])

slope= (2/5)

slope of a perpendicular line= -(5/2)


Thanks in advance and sorry for asking you to waste your time like this
-TWO
Last edited by The Weird One on Mon Oct 22, 2007 11:51 am, edited 1 time in total.
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Re: MATH HELP PLEASE!!!!!

Postby cena-rules on Mon Oct 22, 2007 11:48 am

The Weird One wrote:I think I've got the following problem solved correctly, but I'ld like to be sure (this teacher grades hmwk for correctness)

(2/3)x-(5/2)y=2


standard form= 2x-5y=12

slope-intercept form= y=(2/5)x-(12/5)

x-intercept= (6,0)

y-intercept= (0,-[12/5])

slope= (2/5)

slope of a perpendicular line= -(5/2)


Thanks in advance and sorry for asking you to waste your time like this
-TWO


I think you may have it wrong

if the brackets are 2/3 meaning 2 divided by 3 then it is 0.66x-2.5y=2
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Postby The Weird One on Mon Oct 22, 2007 11:48 am

:-k anybody?:-k

Edit: fastposted :oops:
Last edited by The Weird One on Mon Oct 22, 2007 11:51 am, edited 1 time in total.
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Re: MATH HELP PLEASE!!!!!

Postby cena-rules on Mon Oct 22, 2007 11:49 am

cena-rules wrote:
The Weird One wrote:I think I've got the following problem solved correctly, but I'ld like to be sure (this teacher grades hmwk for correctness)

(2/3)x-(5/2)y=2


standard form= 2x-5y=12

slope-intercept form= y=(2/5)x-(12/5)

x-intercept= (6,0)

y-intercept= (0,-[12/5])

slope= (2/5)

slope of a perpendicular line= -(5/2)


Thanks in advance and sorry for asking you to waste your time like this
-TWO


I think you may have it wrong

if the brackets are 2/3 meaning 2 divided by 3 then it is 0.66x-2.5y=2
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Re: MATH HELP PLEASE!!!!!

Postby The Weird One on Mon Oct 22, 2007 11:49 am

cena-rules wrote:
The Weird One wrote:I think I've got the following problem solved correctly, but I'ld like to be sure (this teacher grades hmwk for correctness)

(2/3)x-(5/2)y=2


standard form= 2x-5y=12

slope-intercept form= y=(2/5)x-(12/5)

x-intercept= (6,0)

y-intercept= (0,-[12/5])

slope= (2/5)

slope of a perpendicular line= -(5/2)


Thanks in advance and sorry for asking you to waste your time like this
-TWO


I think you may have it wrong

if the brackets are 2/3 meaning 2 divided by 3 then it is 0.66x-2.5y=2


the teacher hates decimals and makes us keep it in the form of improper fractions :wink:
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Re: MATH HELP PLEASE!!!!!

Postby cena-rules on Mon Oct 22, 2007 11:50 am

The Weird One wrote:
cena-rules wrote:
The Weird One wrote:I think I've got the following problem solved correctly, but I'ld like to be sure (this teacher grades hmwk for correctness)

(2/3)x-(5/2)y=2


standard form= 2x-5y=12

slope-intercept form= y=(2/5)x-(12/5)

x-intercept= (6,0)

y-intercept= (0,-[12/5])

slope= (2/5)

slope of a perpendicular line= -(5/2)


Thanks in advance and sorry for asking you to waste your time like this
-TWO


I think you may have it wrong

if the brackets are 2/3 meaning 2 divided by 3 then it is 0.66x-2.5y=2


the teacher hates decimals and makes us keep it in the form of improper fractions :wink:


I hate mproper fractions

Ill give your teacher a fight about that. Decimal also work and I cba to do it another way so thats my help
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Postby The Weird One on Mon Oct 22, 2007 11:51 am

thx. 8) 8)
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Postby Norse on Mon Oct 22, 2007 11:53 am

How did you come to that standard form from the initial equation?
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Postby cena-rules on Mon Oct 22, 2007 11:53 am

Norse wrote:How did you come to that standard form from the initial equation?


thats what threw me
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Postby Norse on Mon Oct 22, 2007 11:55 am

cena-rules wrote:
Norse wrote:How did you come to that standard form from the initial equation?


thats what threw me


I can see the logic in it, but I am pretty sure it aint correct
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Postby cena-rules on Mon Oct 22, 2007 11:56 am

Norse wrote:
cena-rules wrote:
Norse wrote:How did you come to that standard form from the initial equation?


thats what threw me


I can see the logic in it, but I am pretty sure it aint correct


because of the division in the brackets
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Postby The Weird One on Mon Oct 22, 2007 11:56 am

in order to eliminate the fractions, (which can't be in the first half of standard form, according to my teacher) I multiplied both sides by the denominaters.

k. . .thanks for the help, I gotta go the bell just rang 8)
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Postby cena-rules on Mon Oct 22, 2007 11:57 am

The Weird One wrote:in order to eliminate the fractions, (which can't be in the first half of standard form, according to my teacher) I multiplied both sides by the denominaters.

k. . .thanks for the help, I gotta go the bell just rang 8)


which makes 6x-10y=2
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Postby The Weird One on Mon Oct 22, 2007 4:47 pm

I screwed up. :oops: :oops: :oops:



What I did was:

Code: Select all
3[(2/3)x]{2[-(5/2y)]}=2*3*2



What I should've done was:

6[(2/3)x-(5/2)y]=2*6




-TWO
Last edited by The Weird One on Mon Oct 22, 2007 5:19 pm, edited 1 time in total.
sheepofdumb wrote:I'm not scum, just a threat to the town. There's a difference, thank you very much.
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Postby GreecePwns on Mon Oct 22, 2007 5:13 pm

The Weird One wrote:I screwed up. :oops: :oops: :oops:
What I did was:

3[(2/3)x]{2[-(5/2y)]}=2*3*2
Well, drawing a smiley will never get you a correct answer.
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Postby The Weird One on Mon Oct 22, 2007 5:19 pm

fixed it :wink:
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