The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by

$$t(C)=\dfrac{9C}{5}+32$$

Find

(i) t (0)

(ii) t (28)

(iii) t (–10)

(iv) The value of C, when t(C) = 212

Asked by Abhisek | 11 months ago |  101

##### Solution :-

(i) We have our given function as, $$t(C)=\dfrac{9C}{5}+32$$

Thus, to find the values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t(0)  $$=\dfrac{9\times 0}{5}+32=32$$

(ii) Thus, to find the values of the function we just have to put the values in the given function and simplify it.

So, we get now,

$$t(28)=\dfrac{9\times 28}{5}+32$$

$$=\dfrac{252+160}{5}$$

$$=\dfrac{412}{5}$$

= 82.4

(iii) Thus, to find the values of the function we just have to put the values in the given function and simplify it.

We are now getting,

$$t(-10)=\dfrac{9\times (-10)}{5}+32$$

= -18+32

= 14

(iv) We have our given function as, $$t(C)=\dfrac{9C}{5}+32$$

For this problem, we are given that, t(C) = 212

So, it can be written as,$$\dfrac{9C}{5}+32=212$$

Simplifying further,

$$\dfrac{9C}{5}=212-32$$

$$\dfrac{9C}{5}=180$$

C = $$\dfrac{900}{9}=100$$

Thus, it can be said that, for t(C) = 212, the value of t is 100

Answered by Pragya Singh | 11 months ago

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