Question 198043
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a) Find a linear function that gives celsius degrees in terms of fahrenheit.


Let us appoint the following:
<font color=blue>Celsius = x</font>, where{{{system(x[1]=0(freezing),x[2]=100(boiling))}}}
Also,
<font color=blue>Fahrenheit = y</font>, where{{{system(y[1]=32(freezing),y[2]=212(boiling))}}}


Via Point-Slope Form:
{{{m=(y[2]-y[1])/(x[2]-x[1])=(212-32)/(100-0)=180/100}}}, divide both by 20:
{{{red(m=9/5)}}}


Then, via Slope-Intercept Form,
{{{y=mx+b}}}
Thru point (x1,y1)= (0,32):


{{{32=(9/5)(0)+b}}}
{{{b=32}}}


It follows ---> {{{red(y=(9/5)x+32)}}}, Answer ---> {{{highlight(F=(9/5)C+32)}}}


b) What celsius degrees temperature corresponds to 98.6 degrees fahrenheit?


Susbtituting to our Eqn in (a):
{{{F=(9/5)C+32}}} ---> {{{F-32=(9/5)C}}} ---> {{{(F-32)/(9/5)=(cross(9/5))(C))/cross(9/5)}}}
{{{C=(F-32)(5/9)}}}
{{{C=(98.6-32)(5/9)=66.6(5/9)}}}
{{{red(C=37^o)}}}, Answer (body temp.)


c. At what temperature do the two scales give the same reading.


When Fahrenheit=Celsius ----> y=x


Going back to eqn in (a):
{{{y = (9/5)x+32}}} ---> {{{x=(9/5)x+32}}}
{{{-32=(9/5)x-x}}}
{{{-32=(9x-5x)/5}}}
{{{-32=4x/5}}} ----> {{{(-32)(5)=4x}}} ----> {{{-160=4x}}} ----> {{{(cross(-160)-40)/cross(4)=cross(4)x/cross(4)}}}
{{{red(x=-40)}}}, Answer


Thank you,
Jojo</font>