1. ## Couple questions

Can you guys help me workout these problems for my HW due tonight - sooner the better please

1. 1.431^9t = 3

2. At sea level, the weight of the atmosphereexerts a pressure of 14.7 pounds per square inch, commonly reffered to as 1 atmoshphere of pressure. As an object descends in water, the pressure P ad depth d are linearly related. In sal water, the pressure at a depth of 33 ft is 2 atms, or 29.4 pounds per square inch.

A) What is the correct linear model?
P = ?d + ?
B) What is the pressure at a depth of 50 ft
C) What is the depth at which the pressure is 5 atms?

2. ## Re: Couple questions

It is against MHF policy to either help or solve problems for class credit. Please do not do this again.

-Dan

3. ## Re: Couple questions

Originally Posted by topsquark
It is against MHF policy to either help or solve problems for class credit. Please do not do this again.

-Dan
Actually i read the policy and all the numbers above are completely scewed... I am giving random numbers and just need help with the formula.

4. ## Re: Couple questions

All right. Let's play the Socrates game.

Show us what you've been able to do on each problem, starting with the first one. But don't expect any direct answers...

-Dan

5. ## Re: Couple questions

Originally Posted by topsquark
All right. Let's play the Socrates game.

Show us what you've been able to do on each problem, starting with the first one. But don't expect any direct answers...

-Dan
Ok soo... I was given an example to go off of for the class on number 1... it was this: 6^2x-x^2 = 6^-8... So i took away 6 on either side... I was left with 2x-x^2=-8... Then i did +8 to make the equation equal to zero... Then i got 2x-x^+8=0.. Then I got my answer of x = -2 and x = 4... I get that... But i dont get what you do when you have unequal numbers on either end...

The second one... I am not sure whether its P = .445d + 14.7 OR P = .445d + 29.4

Trust me... I am not a moron to post on here the actual questions... And no reason to give me a warning as I know what I am doing..

6. ## Re: Couple questions

Originally Posted by Bosanac
Ok soo... I was given an example to go off of for the class on number 1... it was this: 6^2x-x^2 = 6^-8... So i took away 6 on either side... I was left with 2x-x^2=-8... Then i did +8 to make the equation equal to zero... Then i got 2x-x^+8=0.. Then I got my answer of x = -2 and x = 4... I get that... But i dont get what you do when you have unequal numbers on either end...
If I'm correct, what you are saying you've done is this:
$6^2 \cdot x - x^2 = 6^{-8}$

This looks a tad funny, but let's go with what it looks like and continue:
Your next step says: "So i took away 6 on either side" which would give:
$6^2 \cdot x - x^2 - 6 = 6^{-8} - 6$

And your next line is incorrect.

What I would do is to expand the original, giving
$6^2 \cdot x - x^2 = 6^{-8}$

$36x - x^2 = \frac{1}{6^8}$

$36x - x^2 = \frac{1}{1679616}$

Not the result you seem to be working with. Let's validate this stage before we do any more on it.

Originally Posted by Bosanac
Trust me... I am not a moron to post on here the actual questions... And no reason to give me a warning as I know what I am doing..
I never start with the assumption that anyone is a moron. But we are not a homework solving help site. We absolutely cannot tolerate doing someone's homework and the loss of reputation that this site has been able to garner. That would be the end of any credibility of this site. I am reeeealy stretching it by even making comments about your work. I will allow indirect help on these problems, but only this once.

-Dan