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^{o}F-50^{o}F)/(100^{o}C-10^{o}C) = **162**^{o}F/90^{o}C

I know that 162/90 = 1.80

But I have 2 different units in the final solution, fahrenhait (^{o}F) and celcius (^{o}C)

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.

Re: How to include the right units?

$\displaystyle F = \frac{9}{5}C + 32$

This means, when C increases by 1, F increases by 9/5. And yes, the units will be $\displaystyle \frac{^{\circ}F}{^{\circ}C}$

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=1.80^{o}F/^{o}C

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 :)

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=1.80^{o}F/^{o}C

**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 $\displaystyle y^\circ F = \frac{9^\circ F}{5^\circ C} \cdot (x^\circ C) + 32^\circ F$

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

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 = 70^{o}F

So I put it into the linear equation

1. 70^{o}F=(9^{o}F/5^{o}C)x + 32^{o}F

2. 350^{o}F^{o}C = (9^{o}F)x + 32^{o}F **Multiply both sides by 5**^{o}C

3. 38.88^{o}C = x + 32^{o}F **Divide both sides by 9**^{o}F

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

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 = 70^{o}F ... then x would be in ^{o}C

So I put it into the linear equation

70^{o}F=(9^{o}F/5^{o}C)x + 32^{o}F ...

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

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

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

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

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

$\displaystyle 21.1^\circ C \approx x$

Re: How to include the right units?

Quote:

Originally Posted by

**Chaim** 1. 70^{o}F=(9^{o}F/5^{o}C)x + 32^{o}F

2. 350^{o}F^{o}C = (9^{o}F)x + 32^{o}F **Multiply both sides by 5**^{o}C

The right-hand side should be (9^{o}F)x^{o}C + 160^{o}F^{o}C.