the formula p = 15 + \\frac{5d}{11} describes the pressure of sea water, p, in pounds per square foot, at a…

the formula p = 15 + \\frac{5d}{11} describes the pressure of sea water, p, in pounds per square foot, at a depth of d feet below the surface. if a diver was subjected to a pressure of 40 pounds per square foot, to what depth did she descend? at what depth is the pressure 162 pounds per square foot?\n(a) 55 ft; 323.4 ft\n(b) 321.4ft; 321.4 ft\n(c) 57 ft; 325.4 ft\n(d) 353.4 ft; 353.4 ft

the formula p = 15 + \\frac{5d}{11} describes the pressure of sea water, p, in pounds per square foot, at a depth of d feet below the surface. if a diver was subjected to a pressure of 40 pounds per square foot, to what depth did she descend? at what depth is the pressure 162 pounds per square foot?\n(a) 55 ft; 323.4 ft\n(b) 321.4ft; 321.4 ft\n(c) 57 ft; 325.4 ft\n(d) 353.4 ft; 353.4 ft

Answer

Explanation:

Step1: Set up equation for first - pressure value

Given $p = 15+\frac{5d}{11}$, when $p = 40$, we have the equation $40=15+\frac{5d}{11}$.

Step2: Isolate the term with $d$

Subtract 15 from both sides: $40 - 15=\frac{5d}{11}$, so $25=\frac{5d}{11}$.

Step3: Solve for $d$

Multiply both sides by 11: $25\times11 = 5d$, then $275 = 5d$. Divide both sides by 5, $d=\frac{275}{5}=55$.

Step4: Set up equation for second - pressure value

When $p = 162$, we have the equation $162=15+\frac{5d}{11}$.

Step5: Isolate the term with $d$

Subtract 15 from both sides: $162 - 15=\frac{5d}{11}$, so $147=\frac{5d}{11}$.

Step6: Solve for $d$

Multiply both sides by 11: $147\times11 = 5d$, then $1617 = 5d$. Divide both sides by 5, $d=\frac{1617}{5}=323.4$.

Answer:

A. 55 ft; 323.4 ft