Conquer Club

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

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

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 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

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

cena-rules wrote:

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

thats what threw me

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

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

Code: Select all

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


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

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