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Q4(1)    The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(c) = 9C/5 +32

Q4(2)    The function ‘t’ which maps temperature in degree Celsius into temperature in
       degree Fahrenheit is defined by t ( C ) = \frac{9 C }{5} + 32
        t (28) 

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Given function is 
t ( C ) = \frac{9 C }{5} + 32
Now,
t ( 28 ) = \frac{9 (28) }{5} + 32= \frac{252}{5}+ 32 = \frac{252+160}{5}= \frac{412}{5}
Therefore,
Value of t(28) is  \frac{412}{5}

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