Thread: How to include the right units?

Well I have a problem, about how when celsius increases, the fahrenheit will increase as well.
I have a graph, showing my celcius as the manipulative variable (on the x-axis), while the fahrenheit is the responding variable (y-axis)

So I used the slope equation, m=(change in y)/(change in x)
So I grabbed 2 points and did m=(212^oF-50^oF)/(100^oC-10^oC) = 162^oF/90^oC
I know that 162/90 = 1.80

But I have 2 different units in the final solution, fahrenhait (^oF) and celcius (^oC)
I always have to include the units so that when another person reads it, they would know what units I am talking about.

So can anyone explain to me which units I would use? I wasn't sure to convert one measurement to another, cause that would start messing things up maybe.

This means, when C increases by 1, F increases by 9/5. And yes, the units will be

Ok thanks!

So once I try to find points on the graph, like an x coordinate or y coordinate.
If I plug it into the linear equation: y=mx+b
and since m=1.80^oF/^oC

If I had to subtract it from Fahrenheit, how will the units become then?
Since if I subtract m, into y so I could get x, y is Fahrenheit, while m is Fahrenheit over Celsius.

Can anyone explain? Thanks

Originally Posted by Chaim

So once I try to find points on the graph, like an x coordinate or y coordinate.
If I plug it into the linear equation: y=mx+b
and since m=1.80^oF/^oC

If I had to subtract it from Fahrenheit, how will the units become then?
Since if I subtract m, into y so I could get x, y is Fahrenheit, while m is Fahrenheit over Celsius.

Can anyone explain?

subtract what ? why would you want to subtract the slope?

the equation is

... note that the equation makes the Celsius units cancel.

Thanks!
I saw where my mistakes were now!

So just wondering, if I was looking for an x coordinate, and I got a y coordinate, for example:
I know Y = 70^oF

So I put it into the linear equation
1. 70^oF=(9^oF/5^oC)x + 32^oF
2. 350^oF^oC = (9^oF)x + 32^oF Multiply both sides by 5^oC
3. 38.88^oC = x + 32^oF Divide both sides by 9^oF

So I got 2 numbers equaled to each other, but they're both different units.
Did I do something wrong or should I subtract 32 from both sides to find x?
Thanks

Originally Posted by Chaim

So just wondering, if I was looking for an x coordinate, and I got a y coordinate, for example:
I know Y = 70^oF ... then x would be in ^oC

So I put it into the linear equation
70^oF=(9^oF/5^oC)x + 32^oF ...

notice the degrees F cancel, leaving degrees Celsius ...

Originally Posted by Chaim

1. 70^oF=(9^oF/5^oC)x + 32^oF
2. 350^oF^oC = (9^oF)x + 32^oF Multiply both sides by 5^oC

The right-hand side should be (9^oF)x^oC + 160^oF^oC.