Last updated at Dec. 13, 2024 by Teachoo
Ex 2.3, 4 The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by . t(C) = 9/5 C+ 32 . Find (i) t (0) (ii) t (28) (iii) t (-10) Given t(C) = 9/5 C + 32 Putting C = 0, t(0) = (9 )/5 × (0) + 32 = 0 + 32 = 32 Putting C = 28, t(C) = 9/5 C + 32 t(28) = (9 )/5 × (28) + 32 = (252 )/5 + 32 = (252 + 32 × 5)/5 = (252 + 160)/5 = 𝟒𝟏𝟐/𝟓 Putting C = – 10, t(C) = 9/5 C + 32 t(−10) = (9 )/5 × (–10) + 32 = 9 × (–2) + 32 = –18 + 32 = 14 Ex 2.3, 4 The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by . t(C) = 9/5 C+ 32 . Find (iv) The value of C, when t(C) = 212 Given t(C) = 9/5 C+ 32 Putting t(C) = 212 212 = 9/5 C+ 32 212 – 32 = 9/5 C 9/5 C = 212 – 32 9/5 C = 180 C = (180 × 5)/9 C = 100
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo