Question

water boils at different temperatures at different elevations. the boiling temperature of water is 212°f at sea level (0 feet) but drops about 1.72°f for every 1,000 feet of elevation. write a formula for the boiling point at a given elevation. then solve the formula for the elevation when the boiling point for water is 190°f. determine a formula for the boiling point b, in degrees fahrenheit, at elevation h, in feet. (type an equation. simplify your answer. do not include the degree symbol in your answer.)

Explanation:

Step1: Identify the initial value and rate of change

The initial boiling - point at sea - level ($h = 0$) is $212$. The rate of change is $-\frac{1.72}{1000}$ (negative because the boiling - point drops as elevation increases).
Let $b$ be the boiling point and $h$ be the elevation. Using the slope - intercept form of a linear equation $y=mx + c$ (where $y = b$, $x = h$, $m=-\frac{1.72}{1000}$, and $c = 212$), we get $b=212-\frac{1.72h}{1000}$.

Step2: Solve for $h$ when $b = 190$

First, substitute $b = 190$ into the equation:
$190=212-\frac{1.72h}{1000}$.
Then, move the terms around to isolate $h$.
$\frac{1.72h}{1000}=212 - 190$.
$\frac{1.72h}{1000}=22$.
Multiply both sides by $1000$: $1.72h=22\times1000$.
$1.72h = 22000$.
Finally, divide both sides by $1.72$ to solve for $h$: $h=\frac{22000}{1.72}\approx12790.7$.

Answer:

The formula for the boiling point is $b = 212-\frac{1.72h}{1000}$. When the boiling point $b = 190$, the elevation $h\approx12790.7$ feet.