Question

when you dive below the surface of the water, the pressure increases. you can feel this in your ears when you dive to the bottom of a deep swimming pool or spring. the function that describes this relationship is ( p(d)=14.5+\frac{29}{66}d ), where ( d ) is the depth in feet below the surface of the water and ( p ) is the pressure in pounds per square inch (psi).
what is the pressure at the surface of the water? select the correct units from the second box.

at what depth is the pressure exactly double the pressure at the surface of the water?select the correct units from the second box.

what is the pressure at a depth of 99 feet? use function notation in the first box and give the answer in the second box.select the correct units from the third box.

what is the pressure at a depth of 132 feet? use function notation in the first box and give the answer in the second box.select the correct units from the third box.

Explanation:

First Question: Pressure at the surface of the water

Step1: Substitute \(d = 0\) into the function

When \(d = 0\) (surface of the water), we use the formula \(P(d)=14.5+\frac{29}{66}d\).
Substituting \(d = 0\) gives \(P(0)=14.5+\frac{29}{66}\times0\).

Step2: Simplify the expression

Since \(\frac{29}{66}\times0 = 0\), then \(P(0)=14.5\) pounds per square inch (PSI).

Second Question: Depth when pressure is double the surface pressure

Step1: Find double - the - surface pressure

The surface pressure \(P(0)=14.5\) PSI. Double - the - surface pressure is \(P = 2\times14.5=29\) PSI.

Step2: Set up the equation and solve for \(d\)

We set \(P(d)=29\), so \(29 = 14.5+\frac{29}{66}d\).
First, subtract \(14.5\) from both sides: \(29 - 14.5=\frac{29}{66}d\).
\(14.5=\frac{29}{66}d\).
Then, solve for \(d\) by multiplying both sides by \(\frac{66}{29}\): \(d=\frac{14.5\times66}{29}\).
Since \(14.5=\frac{29}{2}\), then \(d=\frac{\frac{29}{2}\times66}{29}\).
The \(29\) cancels out, and \(d = 33\) feet.

Third Question: Pressure at \(d = 99\) feet

Step1: Substitute \(d = 99\) into the function

Using \(P(d)=14.5+\frac{29}{66}d\), when \(d = 99\), we have \(P(99)=14.5+\frac{29}{66}\times99\).

Step2: Simplify the expression

\(\frac{29}{66}\times99=\frac{29\times99}{66}=\frac{29\times3}{2}=\frac{87}{2}=43.5\).
Then \(P(99)=14.5 + 43.5=58\) PSI.

Fourth Question: Pressure at \(d = 132\) feet

Step1: Substitute \(d = 132\) into the function

Using \(P(d)=14.5+\frac{29}{66}d\), when \(d = 132\), we have \(P(132)=14.5+\frac{29}{66}\times132\).

Step2: Simplify the expression

\(\frac{29}{66}\times132=29\times2 = 58\).
Then \(P(132)=14.5+58 = 72.5\) PSI.

Answer:

• First question: \(14.5\) pounds per square inch (PSI)
• Second question: \(33\) feet
• Third question: \(P(99)=58\) pounds per square inch (PSI)
• Fourth question: \(P(132)=72.5\) pounds per square inch (PSI)