The function ‘ t ’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by . Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t (C) = 212

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Solution

The given function is,

t( C )= 9C 5 +32

(i)

Substitute C=0 in the given function to calculate the value of t( 0 ).

t( 0 )= 9( 0 ) 5 +32 =32

(ii)

Substitute C=28 in the given function to calculate the value of t( 28 ).

t( 28 )= 9( 28 ) 5 +32 = 252 5 +32 = 252+160 5 = 412 5

(iii)

Substitute C=−10 in the given function to calculate the value of t( −10 ).

t( -10 )= 9( −10 ) 5 +32 = −90 5 +32 =−18+32 =14

(iv)

Substitute t( C )=212 in the given function to calculate the value of C.

212= 9C 5 +32 9C 5 =212−32 9C=5( 180 ) C= 5( 180 ) 9

Therefore, C=100.

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Similar questions

The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.

Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212

The function t which maps temperature in degree Celsius to temperature in degree Fahrenheit is defined by $t (C) = \frac{9 C}{5} + 32.$ Find

(i) t(0) (ii) t(28) (iii) t(-10)

(iv) The value of C when t(C) = 212.

t

t (C) = \frac{9 C}{5} + 32^{o}

t (0^{o})

t (28^{o})

t (- 10^{o})

C

t (C) = 212^{o} F

t

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

(i) t (0) (i i) t (28) (i i i) t (- 10)

(i v)

C,

t (C) = 212

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