**(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 | 1 year agoLet 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.

Let A = (1, 2, 3} and B = {4} How many relations can be defined from A to B.

Let R = {(x, x^{2}) : x is a prime number less than 10}.

**(i) **Write R in roster form.

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

If A = {5} and B = {5, 6}, write down all possible subsets of A × B.