Math, asked by nandashveta, 11 months ago

Draw a graph to represent the relationship between celcuus and Fahrenheit.
C=5/9 (F-32)
From the graph find the value of C when F = 176

Answers

Answered by Anonymous
7

In this step we have a look at the important conversion between the Celsius and Fahrenheit systems, illustrating the idea of a linear relation which is not a direct proportionality.

Celsius versus Fahrenheit

In most countries around the world, temperature is measured in the Celsius system, where water freezes at  0  and boils at  100 . Such countries include those in Europe, most of Asia and Africa, and Australia. However there are some countries, including notably the USA, where the Fahrenheit system is used, with water freezing at  32  and boiling at  212 . Canada uses a mix of Celsius and Fahrenheit.

So a natural question is: how do we convert from one temperature to another? One way is with a chart:

Celsius Fahrenheit

0 32

10 50

20 68

30 86

40 104

50 122

60 140

70 158

80 176

90 194

100 212

A more visual way is to plot corresponding points. Let’s denote the two variables by  F  and  C , standing (of course) for Fahrenheit and Celsius respectively.

A linear relationship

The value  F=32  goes with  C=0 , while the value  F=212  goes with the value  C=100 . So let’s make a graph.

Suppose we want primarily to convert from Celsius values to Fahrenheit. Then we would treat  C  as the horizontal variable, and  F  as the vertical variable. Here is then our  C−F  plane, with the two points that we know:For the line  F=mC+b , the  F -intercept is  b , which we can see is  32 . How about the slope of the line? It is

m=ΔFΔC=212−32100−0=95.

 

So

F=95C+32.

 

Converting temperatures

Let’s get some practice converting from one system to the other. You can use the formula, or possibly eyeball things from the graph of the line.

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Answered by Anonymous
6

Answer:

{\bold{\red{\huge{Your\:Answer}}}}

Step-by-step explanation:

The\:general\:formula - C/5 = (F-32)/9 = (K+273)/5\\Where,C - Temperature\:in\:Centigrade/Celsius scale\\F - Temperature\:in\:Fahrenheit\:scale\\K - Temperature\:in\:Kelvin\:scale

C=\frac{5}{9} F-\frac{160}{9}

We\:could\:also\:rewrite\:this\:as

C=\frac{5}{9}(F-32)

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