Convert `30^(@) C` temperature into (a) Fahrenheit scale and (b) kelvin scale.
Here, `t_(c) = 30^(@)C`
(a) `(t_C)/(100) = (t_(F)-32)/(180)` or 30

Explanation

Here `l_(c)= 30^(@)`
`(l_(c))/(100) = (l_(F)) -( 32)/(180)` or, `(30)/(100) = (l_(F) - 32)/(180)` or, `l_(F) - 32 = (30 xx 180)/(100) = 54`
`l_(F) = 32 + 54 = 86 F`
`(l_(c))/(100) = (T_(K) - 273)/(100) ` or `T_(K) = l_(C) + 273 = 30 + 273 = 303K`