(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.
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(i) Write R in roster form.
(ii) Find: dom (R) and range (R)
(iii) Write R–1 in roster form
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(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.