Question
The function, \( p(d)=1+\frac{d}{33} \), gives the pressure, in atmospheres (atm), at a depth \( d \) in the sea ( \( d \) is in feet). Note that \( p(0)=1 \) atm, \( p(33)=2 \), and so on. Find the pressure at 60 feet. The pressure at 60 feet is \( \square \) atm. (Type an integer or a simplified fraction.)
Ask by Barnett Fleming. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The pressure at 60 feet is \( \frac{31}{11} \) atm.
Solution
To find the pressure at a depth of 60 feet using the given function:
\[
p(d) = 1 + \frac{d}{33}
\]
Substitute \( d = 60 \) feet into the function:
\[
p(60) = 1 + \frac{60}{33}
\]
Simplify the fraction:
\[
\frac{60}{33} = \frac{20}{11} \quad \text{(dividing numerator and denominator by 3)}
\]
Now, add the fractions:
\[
p(60) = 1 + \frac{20}{11} = \frac{11}{11} + \frac{20}{11} = \frac{31}{11}
\]
So, the pressure at 60 feet is:
\[
\boxed{\dfrac{31}{11}}
\]
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
To find the pressure at 60 feet using the function \( p(d) = 1 + \frac{d}{33} \), we'll substitute \( d = 60 \): \[ p(60) = 1 + \frac{60}{33} \] First, simplify \( \frac{60}{33} \): \[ \frac{60}{33} = \frac{20}{11} \quad (dividing\, numerator\, and\, denominator\, by\, 3) \] Now plug this back into the equation: \[ p(60) = 1 + \frac{20}{11} = \frac{11}{11} + \frac{20}{11} = \frac{31}{11} \] Thus, the pressure at 60 feet is \( \frac{31}{11} \) atm.
