[mattimeo_boyd]

Given that the relationship between the two temperature scales is linear, use the facts that water freezes at 0 degrees C and 32 degrees F and boils at 100 degrees C and 212 degrees F to DERIVE a formula to express Fahrenheit temperature as a function of Celsius temperature. Show the derivation of the formula.

1. 40 degrees C=____________

2. -5 degrees C=____________

3. 15 degrees C=____________

[ROLCAM]

Celsius 100 degrees= fahrentite 180degrees.
Or C 5 degrees = F 9 degrees.

One starts at 0 and the other starts at 32. We need to adjust for this.

1. 40 degrees C= 40* 9/5 = 72+32 = 104 F

2. -5 degrees C= -5*9/5 = -9 +32
=23 F

3. 15 degrees C= 15*9/5 =27 +32
= 59 F

[mattimeo_boyd]

f(c) = 5
and
f(c) = y

??

[ebaines]

You should approach this by deriving a linear equation that expresses Degrees F as a function of Degrees C. Since you are told that this is a linear equation, the general form is like this:

F = mC + b

where "m" is the slope of the line and "b" is the "F" intercept (the value of F when C = 0). You are told that C = 0 when F = 32, and C = 100 when F = 212, so you have two equations in two unknowns:

(1) 212 = m x 100 + b
(2) 32 = m x 0 + b

You must now solve for the values of m and b. Can you take it from here?

[mattimeo_boyd]

Thanks guys I think I figured it out now.

212-100m=b
32=b

212-100m=32
-100m=-180

m=9/5

[husenbekele]

Using by function degree fahreneit to degree celsius

[ebaines]

C = (5/9) x (F-32)

[Mikhail2202]

F = A + BxC
to find constants A and B.
compose two equetions for boilt point and freezed point:
220 = A + Bx100
32 = A + Bx0
Decision of this two equestion system:
A = 32
220 = 32 + Bx100
220 -32 = Bx100
B = (220 - 32)/100 = 1.88
Then
F = 32 + 1.88xC
C = (F - 32)/1.88

[ebaines]

Mikhail2202: If you use the correct value for boiling point of water (212 degrees Fahrenheit) you will find that:

F = 32 + 1.8C
C = (F-32)/1.8

[Pain_11]

Transform the function so that Celsius is expressed in terms of fahrenheit