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प्रश्न
The normal boiling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F. Find the value of F for 38°C
योग
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उत्तर
For 38°C To find F
(1) ⇒ C = `5/9("F" - 32^circ)`
`9/5"C"` = F – 32°
F = `9/5"C" + 32^circ`
= `9/5 xx 38^circ + 32`
= `342/5 + 32`
= `(342 + 160)/5`
= `502/5`
F = 100.4°
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अध्याय 6: Two Dimensional Analytical Geometry - Exercise 6.2 [पृष्ठ २६०]
