How to include the right units?

• Aug 9th 2012, 08:51 AM
Chaim
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=(212oF-50oF)/(100oC-10oC) = 162oF/90oC
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.
• Aug 9th 2012, 10:15 AM
richard1234
Re: How to include the right units?
$F = \frac{9}{5}C + 32$

This means, when C increases by 1, F increases by 9/5. And yes, the units will be $\frac{^{\circ}F}{^{\circ}C}$
• Aug 13th 2012, 07:35 AM
Chaim
Re: How to include the right units?
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=1.80oF/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 :)
• Aug 13th 2012, 08:40 AM
skeeter
Re: How to include the right units?
Quote:

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=1.80oF/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 $y^\circ F = \frac{9^\circ F}{5^\circ C} \cdot (x^\circ C) + 32^\circ F$

... note that the equation makes the Celsius units cancel.
• Aug 14th 2012, 09:04 AM
Chaim
Re: How to include the right units?
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 = 70oF

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

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
• Aug 14th 2012, 10:14 AM
skeeter
Re: How to include the right units?
Quote:

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 = 70oF ... then x would be in oC

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

$70^\circ F = \frac{9^\circ F}{5^\circ C} \cdot x + 32^\circ F$

$70^\circ F - 32^\circ F = \frac{9^\circ F}{5^\circ C} \cdot x$

$38^\circ F = \frac{9^\circ F}{5^\circ C} \cdot x$

$38^\circ F \cdot \frac{5^\circ C}{9^\circ F} = x$

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

$21.1^\circ C \approx x$
• Aug 14th 2012, 10:16 AM
emakarov
Re: How to include the right units?
Quote:

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

The right-hand side should be (9oF)xoC + 160oFoC.