Help me, please.

[Rosestar]

In certain deep parts of oceans, the pressure of sea water,

in pounds per square foot, at a depth of

feet below the surface, is given by the following equation:

If a scientific team uses special equipment to measures the pressure under water and finds it to be 562 pounds per square foot, at what depth is the team making their measurements?

Answer: The team is measuring at ____ feet below the surface.

 

[HallsofIvy]

In certain deep parts of oceans, the pressure of sea water,

in pounds per square foot, at a depth of

feet below the surface, is given by the following equation:

If a scientific team uses special equipment to measures the pressure under water and finds it to be 562 pounds per square foot, at what depth is the team making their measurements?

Answer: The team is measuring at ____ feet below the surface.

So P= 562 and putting that into the formula, \(\displaystyle 562= 16+ \frac{6d}{11}\). Can you solve that for d?

 

[Rosestar]

So P= 562 and putting that into the formula, \(\displaystyle 562= 16+ \frac{6d}{11}\). Can you solve that for d?

562=16+6d/11

562=22d/11

562/2=2d/2

281= d

 

[JeffM]

In certain deep parts of oceans, the pressure of sea water,

in pounds per square foot, at a depth of

feet below the surface, is given by the following equation:

If a scientific team uses special equipment to measures the pressure under water and finds it to be 562 pounds per square foot, at what depth is the team making their measurements?

Answer: The team is measuring at ____ feet below the surface.

As our name says, we do not give answers here, we give help so that YOU can get the answer. If we give you the answer, how will that help you on your test?

You are given a relationship between pressure and depth that lets you put in a value for depth and easily find the related pressure. The purpose of this exercise is to have you put the SAME relationship into a form so you can put in a value for pressure and easily find the related depth.

Think about this formula:

\(\displaystyle C = \frac{5}{9} * (F - 32),\) where C measures temperature in centigrade and F measures temperature in Fahrenheit.

You can turn this into a formula like:

\(\displaystyle F = 32 + \frac{9}{5}C.\)

How do you do that transformation?

\(\displaystyle C = \frac{5}{9} * (F - 32) \implies 9C = 5 * (F - 32) \implies 9C = 5F - 160 \implies 160 + 9C = 5F \implies F = \frac{1}{5} * 160 + \frac{1}{5} * 9C \implies 32 + \frac{9}{5}C.\)

So what is the first step in solving your problem?

 

[Rosestar]

So P= 562 and putting that into the formula, \(\displaystyle 562= 16+ \frac{6d}{11}\). Can you solve that for d?

562=16+6d/11

562= 22d/11

562/2d =2d/2d

281=d

 

[HallsofIvy]

562=16+6d/11

562= 22d/11

No, "16+ 6d/11" is NOT the same as "(16+ 6d)/11".

562/2d =2d/2d

281=d

If you don't like fractions, multiply each part of "562= 16+ 6d/11" by 11.

 

[guitarguy]

When solving an equation here are two things which might help:

When adding numbers they must have the same denominator.

You can add or subtract like quantities from both sides of an equation.