Question

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
$\bf \: F = \bigg( \dfrac{9}{5} \bigg)C + 32$
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.​

Answer 1

$\large \sf \underline{Solution - }$

(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by :

$\sf : \implies\boxed{ \sf y = \bigg( \dfrac{9}{5} \bigg)x + 32 }$

For plotting the graph :

Putting x = 0,

$\sf \implies \: y = \bigg( \dfrac{9}{5} \bigg)0 + 32$

$\sf \implies \: y = 0 + 32 = 32$

$\sf : \implies \: Solution : \: A(0,32)$

Now,

Putting x = 5,

$\sf \implies \: y = \bigg( \dfrac{9}{5} \bigg)5 + 32$

$\sf \implies \: y = 9+ 32 = 41$

$\sf : \implies \: Solution : \: B (5,41)$

Now,

Putting x = 10,

$\sf \implies \: y = \bigg( \dfrac{9}{5} \bigg)10 + 32$

$\sf \implies \: y = 9 \times 2+ 32 = 18 + 32 = 50$

$\sf : \implies \: Solution : \: C (10,50)$

Hence, A(0,32) , B(5,41) , C(10,50) are the solution of the equation.

➢ Graph is in the attachment.

$\quad$

(ii) If the temperature is 30°C,

$\sf \implies \: F = \bigg( \dfrac{9}{5} \bigg)30 + 32$

$\sf \implies \: F = 9 \times 6 + 32 = 54 + 32 = 86^\circ F$

➢ Hence, if the temperature is 30°C, the temperature in fahrenheit is 86°F.

$\quad$

(iii) If the temperature is 95°F,

$\sf \implies \: 95= \bigg( \dfrac{9}{5} \bigg)C + 32$

$\sf \implies \: 95 - 32= \bigg( \dfrac{9}{5} \bigg)C$

$\sf \implies \: 6 3= \bigg( \dfrac{9}{5} \bigg)C$

$\sf \implies \: 6 3 \times \dfrac{5}{9} = C$

$\sf \implies \: 7 \times 5 = C$

$\sf \implies \: 35^\circ = C$

➢ Hence, if the temperature is 95°F, the temperature in celcius is 35°C.  

$\quad$

(iv) If the temperature is 0°C,

$\sf \implies \: F = \bigg( \dfrac{9}{5} \bigg)0 + 32$

$\sf \implies \: F = 0 + 32 = 32^\circ F$

$\quad$

If the temperature is 0°F,

$\sf \implies \: 0= \bigg( \dfrac{9}{5} \bigg)C + 32$

$\sf \implies \: - 32 = \bigg( \dfrac{9}{5} \bigg)C$

$\sf \implies \: - 32 \times \dfrac{5}{9} = C$

$\sf \implies \: \dfrac{ - 160}{9} = C$

$\sf \implies \: - 17.8^\circ = C$

➢ Hence, if the temperature is 0°C, the temperature in fahrenheit is 32°F and , if the temperature is  0°F, the temperature in celcius is - 17.8°C.

$\quad$

(v) Let x° be the same in both fahrenheit and celcius, then

$\sf \implies \sf x = \bigg( \dfrac{9}{5} \bigg)x + 32$

$\sf \implies \sf x-32 = \bigg( \dfrac{9}{5} \bigg)x$

$\sf \implies \sf (x-32)\times5 = 9x$

$\sf \implies \sf 5x-160 = 9x$

$\sf \implies \sf 4x=-160$

$\sf \implies \sf x= \dfrac{-160}{4}$

$\sf \implies \sf x= -40 ^{\circ}$

➢ Hence, -40° is the temperature which is numerically the same in both  fahrenheit and in celcius.

Answer 2

Answer:

(i) Since the equation between F and C is given, the graph can be drawn as shown above.

(ii) Given c=30, so F=

5

9

×30+32=86 F.

(iii) Given F=95, so C=

9

5

×(F−32)=35 C.

(iv) Given C=0, so F=32 and if F=0, we get C=−

9

160

.

(v) Put F=C=x in given equatin, we get

5

4x

=−32, from which we get x=−40

Yes, there is a temperature which is numerically the same in both fahrenheit and celsius, the numerical value is −40.

Step-by-step explanation:

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