Question

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by:  \(t(C)=\frac {9C}{5}+32\)
Find:

1. t(0) 
2. t(28) 
3. t(-10) 
4. The value of C, when t(C) = 212

Answer 1

The given function is \(t(C)=\frac {9C}{5}+32\)
Therefore,
(i) \(t(0)\) = \(\frac {9\times 0}{5}\)\(+32\) = \(0+32\)

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(ii) \(t(28)\) = \(\frac {9\times 28}{5}+32\) = \(\frac {252+160}{5}\) = \(\frac {412}{5}\)

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(iii) \(t(-10)\) = \(\frac {9\times (-10)}{5}+32\) = \(9\times (-2)+32\) = \(-18+32\) = 14

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(iv) It is given that t(C) = 212

\(212 = \frac {9C}{5} +32\)

\(\frac {9C}{5}=212 - 32\)

\(\frac {9C}{5}=180\)

\(9C=180\times 5\)

\(C = \frac {180 \times 5}{9}\)

\(C = 100\)

Thus the value of t when t(C) = 212 is 100.