There are two scales of measuring the temperature, namely degree Fahrenheit `(.^(@)F)` and degree Celsius `(.^(@)C)`. The relation between the two scales is given by, `F=9/5C+32.`
(i) If the temperature is `0^(@) C`, what is the temperature in Fahrenheit?
(ii) If the temperature is `50^(a) C`, What is the temperature in Fahrenheit ?
(iii) If the temperature is `50^(@)C`, what is the temperature in Fahrenheit ?
(iii) If the temperature is `86^(@)F`, what is the temperature in Celsius ?
(iv) If the temperature is `0^(@)F`, what is the temperature in Celsius ?
(v) Find the numerical value of the temperature which is the same in both the scales.
(vi) Draw the graph of the linear equation `F= 9/5C + 32,` taking C along the x-axis and F along the y-axis.
(vii) Using the graph, fill in the blanks given below:
`-5^(@)C=(......)^(@)F and 14 ^ (@) F = (......) ^(@) C.`

Explanation

The given relation is `F=9/5 C+ 32." "` …(A)
Putting C= 0 in (A), we get
`F=(9/5 xx0)+32 = (0+32) = 32.`
`:. C=0 rArrF=32.`
Hence, `0^(@)C= 32^(@) F.`
(ii) Putting C= 50 in(A) , we get
`F=(9/5 xx 50) + 32 = (90 +32)=122.`
`:. C= 50 rArr F = 122.`
Hence, `50^(@) C= 122^(@)F.`
(iii) Putting F= 86 in(A) , we get
`9/5C+32 = 86 rArr 9/5 C = ( 86-32) = 54`
`rArr C = ( 54 xx 5/9) = 30.`
`:. F= 86 rArr C= 30.`
Hence, `86 ^(@)F = 30^(@)C.`
(iv) Putting F= 0 in (A) , we get
`9/5 C+ 32 = 0 rArr 9/5 C = -32`
`rArr C=(-32 xx 5/9)= -160/9 = -17.8.`
`:. F=0 rArr C = - 17.8.`
Hence, `0^(@)F = - 17. 8 ^(@) C.`
(v) Let C= F. Then, (A) becomes
`F=9/5 F + 32 rArr ( 9/5 - 1) F = -32`
`rArr 4/5 F = -32 rArr F = (-32 xx 5/4) = -40.`
`:. C = F= -40.`
Hence, `-40^(@)C = -40^(@) F.`
(vi) We have, `F=9/5 C + 32." "` ...(A)
Putting C= 0 in (A), we get
`F=(9/5 xx0) + 32 = ( 0+ 32) = 32.`
Putting C=10 in (A) , we get
`F=(9/5 xx 10) + 32 = ( 18+ 32) = 50.`
Putting C= 20 in (A), we get
`F= (9/5 xx 20 ) + 32 = ( 36 +32) = 68.`
Thus, we have the following table.
`{:(C,0,10,20),(F,32,50,68):}`
On a graph paper, we take C along the x-axis and the corresponding values of F along the y- axis.
We plot the points `A(0,32),B(10,50) and C(20,68)` on this graph paper.
Now, we join the points A, B, and B, C to get a line ABC, which we extend in both the direction to get the required graph of the given equation, as shown below.
(vii) On the x-axis, we take a point P at which C=-5.
From P, draw `PD bot X'OX`, meeting the line ABC at a point `D(-5,23).`
`:. -5^(@)C=23^(@)F.`
On the y-axis, take a point Q at which F= 14.
From Q draw Q||x-axis, meeting the line ABCD at the point `E(-10, 14).`
Hence, `14^(@)F = -10 ^(@) C.`