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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

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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Two Dimensional Analytical Geometry - Exercise 6.2 [पृष्ठ २६०]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 6 Two Dimensional Analytical Geometry
Exercise 6.2 | Q 5. (iii) | पृष्ठ २६०
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