typical temperatures\n212°f water boils\n98.6°f body temperature\n70°f room temperature\n32°f freezing point…

typical temperatures\n212°f water boils\n98.6°f body temperature\n70°f room temperature\n32°f freezing point of water\n0°f cold day\n-40°f very cold day\nthe basic unit of temperature in the metric system that is used internationally is the degree celsius. (°c).\nuseful conversion formulas:\n°c to °f multiply by 9, then divide by 5, then add 32\n°f to °c deduct 32, then multiply by 5, then divide by 9 water freezes at ____ °c\nwater boils at ____ °c\nnormal human body temperature is ____ °c.

typical temperatures\n212°f water boils\n98.6°f body temperature\n70°f room temperature\n32°f freezing point of water\n0°f cold day\n-40°f very cold day\nthe basic unit of temperature in the metric system that is used internationally is the degree celsius. (°c).\nuseful conversion formulas:\n°c to °f multiply by 9, then divide by 5, then add 32\n°f to °c deduct 32, then multiply by 5, then divide by 9 water freezes at ____ °c\nwater boils at ____ °c\nnormal human body temperature is ____ °c.

Answer

Explanation:

Step1: Convert freezing - point of water

Given the formula for converting $^{\circ}F$ to $^{\circ}C$: $C=\frac{(F - 32)\times5}{9}$. For the freezing - point of water, $F = 32$. Substitute $F = 32$ into the formula: $C=\frac{(32 - 32)\times5}{9}=0$.

Step2: Convert boiling - point of water

For the boiling - point of water, $F = 212$. Substitute into the formula $C=\frac{(F - 32)\times5}{9}=\frac{(212 - 32)\times5}{9}=\frac{180\times5}{9}=100$.

Step3: Convert body temperature

For body temperature, $F = 98.6$. Substitute into the formula $C=\frac{(F - 32)\times5}{9}=\frac{(98.6 - 32)\times5}{9}=\frac{66.6\times5}{9}=37$.

Answer:

Water freezes at 0 $^{\circ}C$. Water boils at 100 $^{\circ}C$. Normal human body temperature is 37 $^{\circ}C$.