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प्रश्न
The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by `t(C) = "9C"/5 + 32`
Find
(i) t(0)
(ii) t(28)
(iii) t(–10)
(iv) The value of C, when t(C) = 212.
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उत्तर
`(9"C")/5`+32
- t(0) = `9/5xx` 0 + 32 = 32
- t(28) = `(9xx28)/5+ 32 = 252/5+ 32`= `(252+160)/5`
= `412/5` - t(-10) = `9/5xx`(-10) + 32
= - 18 + 32 = 14 - t(C) = 212
∴ 212 = `9/5xx` C + 32
0r `9/5xx` C = 212 - 32 = 180
C = `(180xx5)/9`=100
Thus, the value of t, when t(C) = 212, is 100.
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