The atmospheric pressure P in pounds per square inch (psi) is given by

P = 14.7 e−0.21 a
where a is the altitude above sea level (in miles). If a city has an atmospheric pressure of 12.31 psi, what is its altitude? (Recall that 1 mi = 5,280 ft. Round your answer to the nearest foot.)

14.7 e^(-.21a) = 12.31

e^(-.21a) = 0.837
-.21a = ln 0.837 = -0.177
a = 0.845 mi = 4462 ft

Why did the city go to therapy? Because it couldn't handle the pressure!

To solve this problem, we'll substitute the given atmospheric pressure of 12.31 psi into the equation P = 14.7 e^(-0.21a). Now, let’s solve for a:

12.31 = 14.7 * e^(-0.21a)

Divide both sides by 14.7:

12.31/14.7 = e^(-0.21a)

0.837= e^(-0.21a)

To remove the exponent, take the natural logarithm of both sides:

ln(0.837) = ln(e^(-0.21a))

Now, recall that ln(e^x) = x:

ln(0.837) = -0.21a

Solve for a:

a = ln(0.837) / -0.21

Using a calculator, we find:

a ≈ 2.376

So, the altitude above sea level is approximately 2.376 miles.

To convert this to feet, multiply by 5280:

2.376 miles * 5280 feet/mile ≈ 12,541 feet

Therefore, the city's altitude is approximately 12,541 feet above sea level. Keep your head high, the air might be thin up there!

To find the altitude of the city given an atmospheric pressure of 12.31 psi, we can rearrange the equation:

P = 14.7 e^(-0.21a)

Substituting the given pressure:

12.31 = 14.7 e^(-0.21a)

Now, we need to solve for a, the altitude above sea level.

Divide both sides of the equation by 14.7:

12.31/14.7 = e^(-0.21a)

0.837 = e^(-0.21a)

Take the natural logarithm of both sides to remove the exponential term:

ln(0.837) = ln(e^(-0.21a))

Using the property of logarithms (ln(e^x) = x):

ln(0.837) = -0.21a

Now, divide both sides by -0.21:

a = ln(0.837) / -0.21

Using a calculator:

a ≈ -3.47

To convert the altitude from miles to feet, multiply by 5280:

altitude ≈ -3.47 * 5280 ≈ -18,285.6 ft

Rounding to the nearest foot, the altitude of the city is approximately -18,286 ft above sea level.

To find the altitude of a city given its atmospheric pressure, we need to rearrange the equation:

P = 14.7e^(-0.21a)

In this equation, P represents atmospheric pressure in psi, and a represents altitude in miles.

We are given that the atmospheric pressure is 12.31 psi, so we can substitute this value into the equation:

12.31 = 14.7e^(-0.21a)

To solve for a, we need to isolate it on one side of the equation. First, let's divide both sides of the equation by 14.7:

12.31 / 14.7 = e^(-0.21a)

Next, we can take the natural logarithm (ln) of both sides of the equation:

ln(12.31 / 14.7) = ln(e^(-0.21a))

Using the logarithmic property ln(e^x) = x, we can simplify the equation:

ln(12.31 / 14.7) = -0.21a

Now, divide both sides by -0.21:

a = ln(12.31 / 14.7) / -0.21

Using a calculator, evaluate the right side of the equation:

a ≈ ln(12.31 / 14.7) / -0.21 ≈ -3.9954

Since altitude cannot be negative, we discard the negative value and only consider the positive value.

Therefore, the altitude of the city is approximately 3.9954 miles above sea level.

To convert this to feet, multiply by 5280:

3.9954 miles * 5280 ft/mile ≈ 21,097.47 ft

Rounding this to the nearest foot, we get approximately 21,097 feet.

Therefore, the altitude of the city is approximately 21,097 feet above sea level.