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

1 Answer

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

Related Questions

Let R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.

Class 11 Maths Relations and Functions View Answer

Let R = {(x, x2) : x is a prime number less than 10}.

(i) Write R in roster form.

(ii) Find dom (R) and range (R).

Class 11 Maths Relations and Functions View Answer