Whole 'lotta Work Ahead of Me...

So, this summer I've been given a review packet for my calculus class so that I may do it and *somehow* refresh my memory...except for the fact that they left me hanging out to dry on how to actually re-learn this stuff after two months of rotting my brain away with video-games and other such activities.

I realize that the rules say to limit a thread to one, or even two questions, but I hope I won't get two severly punished or, even better, an exception will be made for me.

Below I've got my work and the questions asked attached to pdfs (which I scanned myself).

__Important__

For some, I answered them and feel pretty confident (like #'s 6,7,8) however the REAL help begins at questions 12, 14(C), and 15 (B, C).

Now a question for the moderators/admins of the forums. Since I will probably have further questions about the packet (as I move on to the other questions) is it alright to make ANOTHER post in this thread for those questions and continue asking for help...or should I just start another thread?

Attachments deleted. Repost your questions with no more than two or three questions per thread.

Also give the threads titles descriptive of the nature of the uestions.

CB

Re: Whole 'lotta Work Ahead of Me...

Quote:

Originally Posted by

**UnstoppableBeast** Now a question for the moderators/admins of the forums. Since I will probably have further questions about the packet (as I move on to the other questions) is it alright to make ANOTHER post in this thread for those questions and continue asking for help...or should I just start another thread?

**Absolutely not!**

Always start a new thread for a new question.

Limit thread to at most three related questions.

Learn to use LaTeX.

Re: Whole 'lotta Work Ahead of Me...

12: Your 1st step is wrong. -- order of operations.

To start you could multiply by $\displaystyle \frac{2\sqrt{x+1}}{2\sqrt{x+1}}$. Don't forget the distributive law for the numerator.

14: Is $\displaystyle x^2 + 2$ less than 1 for any x ?

15: What are the definitions of **domain** and **range**?

Re: Whole 'lotta Work Ahead of Me...

Quote:

Originally Posted by

**UnstoppableBeast** So, this summer I've been given a review packet for my calculus class so that I may do it and *somehow* refresh my memory...except for the fact that they left me hanging out to dry on how to actually re-learn this stuff after two months of rotting my brain away with video-games and other such activities.

I realize that the rules say to limit a thread to one, or even two questions, but I hope I won't get two severly punished or, even better, an exception will be made for me.

Below I've got my work and the questions asked attached to pdfs (which I scanned myself).

__Important__

For some, I answered them and feel pretty confident (like #'s 6,7,8) however the REAL help begins at questions 12, 14(C), and 15 (B, C).

Now a question for the moderators/admins of the forums. Since I will probably have further questions about the packet (as I move on to the other questions) is it alright to make ANOTHER post in this thread for those questions and continue asking for help...or should I just start another thread?

6. I don't understand why you've taken out $\displaystyle \displaystyle x^2 + 2x$ as a common factor, as it is not a common factor.

The highest common factor is $\displaystyle \displaystyle 2x(x + 7)^3(x - 9)^3$, so

$\displaystyle \displaystyle \begin{align*}2x^2(x+7)^3(x-9)^4 + 4x(x - 7)^4(x-9)^3 &= 2x(x+7)^3(x-9)^3\left[x(x - 9) + 2(x - 7)\right] \\ &= 2x(x+7)^3(x-9)^3\left(x^2-9x+2x-14\right) \\ &= 2x(x+7)^3(x-9)^3\left(x^2-7x-14\right) \end{align*}$

7. You are correct that $\displaystyle \displaystyle \sin^4{x} - \cos^4{x} = (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x})$. You can factorise $\displaystyle \displaystyle \sin^2{x} - \cos^2{x}$ further using the difference of two squares again.

So

$\displaystyle \displaystyle \begin{align*} \sin^4{x} - \cos^4{x} &= (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x}) \\ &= 1(\sin^2{x} - \cos^2{x}) \\ &= \sin^2{x} - \cos^2{x} \\ &= (\sin{x} + \cos{x})(\sin{x} - \cos{x}) \end{align*}$

8. While you haven't done anything wrong mathematically, you would not get the marks as you were asked to factorise the equation first.

$\displaystyle \displaystyle \begin{align*} \sec{x}\tan^2{x} + \sec{x} &= \sec{x}(\tan^2{x} + 1) \\ &= \sec{x}\sec^2{x} \\ &= \sec^3{x} \end{align*}$

Re: Whole 'lotta Work Ahead of Me...

Quote:

Originally Posted by

**UnstoppableBeast** So, this summer I've been given a review packet for my calculus class so that I may do it and *somehow* refresh my memory...except for the fact that they left me hanging out to dry on how to actually re-learn this stuff after two months of rotting my brain away with video-games and other such activities.

I realize that the rules say to limit a thread to one, or even two questions, but I hope I won't get two severly punished or, even better, an exception will be made for me.

Below I've got my work and the questions asked attached to pdfs (which I scanned myself).

__Important__

For some, I answered them and feel pretty confident (like #'s 6,7,8) however the REAL help begins at questions 12, 14(C), and 15 (B, C).

Now a question for the moderators/admins of the forums. Since I will probably have further questions about the packet (as I move on to the other questions) is it alright to make ANOTHER post in this thread for those questions and continue asking for help...or should I just start another thread?

Please post according to the rules. (There are good reasons for these rules).

Thread closed.