At the surface of the ocean, the water pressure is the same as the air pressure above the water, about 15 lblin^2, below the surface the water pressure increases by about 4.54 lblin^2 for every 10ft of descent. F(x) which expresses the water pressure in pounds per square inch as a function of the depth in inches below the ocean surfaces. F(x) = _ At what depth is the pressure 80 lblin^2? Include the units in the answer: __

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04-11-2022
Mathematics

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At the surface of the ocean, the water pressure is the same as the air pressure above the water, about 15 lblin^2, below the surface the water pressure increases by about 4.54 lblin^2 for every 10ft of descent. F(x) which expresses the water pressure in pounds per square inch as a function of the depth in inches below the ocean surfaces. F(x) = _ At what depth is the pressure 80 lblin^2? Include the units in the answer: __

Sagot :

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LizaM697469 LizaM697469 · Answer 1 · 2022-11-04T07:57:37-04:00

The pressure above the water is 15 lb/in^2 and for every 10 ft, the pressure rises 4.54 lb/in^2

The equation would be:

[tex]f(x)=4.54\frac{lb}{IN^2}\cdot10ft\cdot x+15\frac{lb}{IN^2}[/tex]

If x = 0, the pressure in the surface is 15 lb/in^2, for each 10 ft of x, the pressure rises 4.54 lb/in^2

Now to solve when the pressure is 80 lb/in^2, we subtitute in the equation:

[tex]80\frac{lb}{IN^2}=4.54\frac{lb}{IN^2}\cdot10ft\cdot x+15\frac{lb}{IN^2}[/tex]

And solve for x:

[tex]\begin{gathered} (80-15)\frac{lb}{IN^2}=4.54\frac{lb}{\text{IN}^2}\cdot10ft\cdot x \\ \frac{65\frac{lb}{IN^2}}{45\frac{lb}{IN^2}\cdot10ft}=x \\ \frac{1.4317}{ft}=x \\ \end{gathered}[/tex]

Then the answer to question 2 is 1.43ft

This is basically a linear equation. The slope is the 4.54lb/in^2 for each 10ft. Then we need to adjust the 10ft of x, to reprensents each step of 10 ft for x. And we need to add the 15lb/in^2 for the surface pressure. With all this, we can construct the function of the pressure dependant of the deepness x.

Sagot :

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