# Lots of Pre-Calculus review problems (HELP!)

• August 18th 2009, 03:44 PM
Jph93
Lots of Pre-Calculus review problems (HELP!)
Hey so school's starting against soon and obviously I've procrastinated on my summer work.

If any of you would be so kind as to maybe choose a problem and help me with it I would be greatly appreciative. Of course, I want to actually understand the problems, so I don't want just answers otherwise I would just go copy one of my friends.

Hopefully this can be helpful to others going into Calc BC as well.

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• August 18th 2009, 04:19 PM
pickslides
Pls do not post such a large document.

Post each problem and your attempts to solve it in a separate post.

Answers to Q1 & 2 look correct.
• August 18th 2009, 04:19 PM
artvandalay11
For number 6, $\log_b y=\frac{\log_k y}{\log_k y}$ for any value of k

And the natural logarithm is the log functionn with base e, so $\log_b y=\frac{\ln y}{\ln b}$ and now you should be able to do that problem

For number 8, call the ? y, and now the perimeter of your fence is x+x+y=200 and the enlcosed area (A)=x*y=5500 sq ft, so you have two equations and two unknowns and you should be able to solve that

(you're not going to get much more help on this forum until the users see some more effort on the rest of the problems)
• August 18th 2009, 04:37 PM
Jph93
okay I'll try some more problems in a bit but I don't really know how to do math on the computer y'know?

EDIT:

Okay for problem #7 plugged all those numbers into the x y chart that you get when you choose "stat" on your calculator than went to calculate linear regression and got

y=ax+b
a= -2.257
b= 22341.75

but that makes almost zero sense to me. Help here?
• August 18th 2009, 05:32 PM
pickslides
Quote:

Originally Posted by Jph93
Okay for problem #7 plugged all those numbers into the x y chart that you get when you choose "stat" on your calculator than went to calculate linear regression and got

y=ax+b
a= -2.257
b= 22341.75

but that makes almost zero sense to me. Help here?

This cannot be correct, the values for salary in your table are increasing over time therefore your value for a (or gradient value) should be positive.
• August 18th 2009, 05:34 PM
Jph93
Quote:

Originally Posted by artvandalay11
For number 6, $\log_b y=\frac{\log_k y}{\log_k y}$ for any value of k

And the natural logarithm is the log functionn with base e, so $\log_b y=\frac{\ln y}{\ln b}$ and now you should be able to do that problem

I don't really get this.

so $\log_5 (x+3)=\frac{\log_9 (x+3)}{\log_9 (x+3)}=1?$ What?

For the second part you're saying $\log_5(x+3)=\frac{ln(x+3)}{ln(5)}$ ?

That would make the answer what?
• August 18th 2009, 05:42 PM
Jph93
Quote:

Originally Posted by pickslides
This cannot be correct, the values for salary in your table are increasing over time therefore your value for a (or gradient value) should be positive.

Stupid mistake. How does a=256.5085714 and b=15326.47 look?
• August 18th 2009, 07:29 PM
mr fantastic
A post like this abuses the whole spirit of MHF. I suggest you ask your teacher for solutions or answers for checking purposes. Thread closed.