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Wow, I'm glad you guys-and girl-made this!

Do any of you know a way to overcome stage fright? I'm going to be dancing to Utada Hikaru's Simple and Clean (in Japanese), and I'm REALLY nervous! I could really use this advice.

Okay I have a question for you guys...

Since I got my learners permit, when ever I get behind the wheel of a car I either hit, or almost hit something. I've endanged the lives of everone who has ever been in the car with me, people nearby and in other cars too... I kept practiceing anyway until I almost killed a woman and baby. Now I'm kinda scared to drive but I have to. I'm 17 and am going into collage next year and the only way to get there is to drive. I don't know what to do!!!!! crying

All great advice. 3nodding

I'd like to add to it:

The first time I competed in All Region Choir... I was a train wreck. I kept dropping my music. I kept stumbling over my own feet. I kept fidgeting. And apparently I didn't look well, because people keep coming up to me and saying, "Are you okay? You're like....really pale."

The first thing to remember is to breathe. Long, deep, and at a steady rhythm. I do my relaxing breaths at a 3-count. Inhale-1-2-3, exhale-1-2-3, rinse, repeat.

Next, get your mind off it. Worrying never did anything except give you premature wrinkles. Find a book of jokes or something entertaining and lighthearted to entertain yourself with when you start worrying.

DonSnJ already said this, but it really needs to be repeated.

See the audience?

Forget about them. Never even look at them if you can. If you have to, look somewhere above them. Don't think about them. Think about yourself. Think about the music, and the movements.

As my old director used to say, "Leave your issues in the wings, get out there, and perform."

I know you'll do great! I wanna hear all about it when it's over.

Sociology? Awesome! can you give me some tips on socializing with a girl in my school? her name is Anna. (Sorry DonSnJ, but i'm not interested in becoming a pretty boy)

You should be nice to DonSnJ. Being a pretty boy could help you out in this situation.

I'm not exactly sure what you're looking for here. I've actually done research with the evolutionary theory of attractiveness, so if you want to know what a lady looks for I can post some of that work, but it is pretty dense.

If you're just interested in getting her to notice you, I can give you some advice.

Step 1: Saying 'hello' is half the battle. Chances are she's already noticed you. To her you may be that guy that never says hello. You can change that by simply saying hello.

Step 2: Stop rationalizing and ask her out! Seriously, the worst thing she'll say is 'no'. It's better to get shot down than to sit and wonder what might have been.

I know these don't seem like really big deals, but it's amazing how many people get caught up on just those two steps.

I'm short on time this morning, so I'll refer you to DonSnJ if you have further questions. Just send him a PM and he'll help you out. I don't think he wants to post some of his secrets.

She's my friend though. I'm pretty sure that that Changes everything (or at least almost everything).

It's been 6 months for you, about 4 and a half years for me XD

But I think I can help. I'll do each problem and then attempt to explain why, so that you hopefully learn something from the experience ^_^

1) |c-3| (less than or equal to) 4
Answer: -1 < c < 7 ----> The '<' symbols are 'less than or equal to'

The absolute value bars in this problem change the question a bit. When you are solving for a variable (c) when it's within absolute value bars, you have to realize something.

When a variable is within absolute value bars, there is a range of possible answers.

The reason behind this is the function of absolute value bars. You know that when a problem has an absolute value symbol in it that the whole term becomes positive, right? For example:

|2-3|

The abs. val. bars act as a set of parenthesis. You take care of everything inside of them first. This then becomes:

|-1|

The abs. val. bars then make everything within it positive. This then becomes:

1

So, let's look back at out original problem. We have |c-3| (less than or equal to) 4. The question is asking you, "What numbers can you put in for 'C' will give you an answer that is less than or equal to 4?"

Well, you know that 7 - 3 = 4. So, 7 is the highest possible answer, since you can't go above an answer is 4.

So far our answer will be: c (less than or equal to) 7.

But here is where the absolute value bars come into play. The abs. val. bars will turn any negative number into a positive one. It cancels out the negative sign. So, you should know that there is a negative answer for c that fits the answer. Now you have to ask yourself, "Hmm... What negative number fits in for 'c' that will give me an answer of 4?"

Try -1.

|c-3| less than or equal to 4
|-1 - 3| less than or equal to 4
|-4| less than or equal to 4
4 less than or equal to 4

Remember, the abs. val. symbol turns the number positive.

So, since your answer of "4" is less than or equal to 4, you have the lower range of your problem. Your final answer will be:

-1 < c < -7 , where the '>' are 'less than or equal to'.

This just means that 'c' needs to be any number between those values to satisfy the answer to this problem.

Hope that helps.

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2) Okay! All of my notes from before apply here, so let's go through the problem.

Remember: The problem does not change, even if you have a coefficient.

With that in mind, we have:

l2x+5l (less than or equal) to 9

Re-word the question if it will help you. It is asking: "What range of numbers will give us an answer that is less than or equal to 9?"

First, find the high end of this range. What positive number will givev you the answer of 9? An easy way to do this is to remove the abs. val. symbol and set it equal to 9. So, you get:

2x+5=9
2x=4
x=2

So, you know that 2 is the high range of your answer. So far, the answer is:

x (less than or equal to) 2

But remember the lower range! There is a negative number that will also give you an answer of 9. So, you have to ask yourself, "Hmm... What negative number will make the equation inside of the abs. val. symbols simplify down to -9?" (The absolute value bars will cancel out the negative sign, so you want to achieve the answer that is the negative of what you're looking for. You're trying to get '9', so you want to get '-9' inside the abs. val. bars. )In this case, it is -7:

|2x+5|
|2(-7)+5|
|-14+5|
|-9|
9

You now have the lower end of your range! Your full answer is:

-14 < x < 2 , where '<' is 'less than or equal to'.

Hope that helps.

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3) (radical)t = 2 (rad) 5

You have to solve for 't'. To do this, you know that 't' has to be on it's own. It can't be squared, it can't be under a radical, it can't be on top of a building performing a sonnet. 't' has to be on its own.

Since 't' is under a radical, you need to perform the opposite function of the radical to cancel it out to both sides of the equation. Remember, what you do to one side you have to do to the other.

And what function is the opposite of a radical? Taking the 'square' of something.

So, you need to square both sides of the equation:

([rad] t)^2 = (2 [rad] 5)^2

**Note: The '^' symbol is called a 'carrot', and it means 'to the ____ power'. It's used for square roots.

So, you know that the square and the radical cancel each other out. Then, you need to simplify the other side as well. You know that '2' and '(rad)5' are being multiplied, so you need to distribute the squared term through to both of them separately. You get:

t = 2^2 x ([rad]5)^2

Simplify:

t = 4 x 5

Then, simplify again:

t = 20

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4) x^7, all under a radical. We'll express this as: (rad)[x^7]

Okay! I'm guessing you have to simplify this.

The way you do this problem requires you to realize something: The 'x' is absolutely irrelevant until the very end. You are just dealing with the '7' and the radical.

Now, you need to break up 'x^7' as far as you can using the rules of exponents. In this case, you know that when you have an exponent multiplied to an exponent, you add the numbers together. So, you can break up x^7 into:

(x^2) x (x^2) x (x^2) x (x^1) ----> The 'x' outside of parenthesis are multiplication symbols.

Breaking up x^7 into those littler exponents makes this problem a lot easier to deal with. You'll see why now.

The next step is applying the radical to these little exponents. Oh my god! Look! Taking the radical of 'x^2' just leaves 'x'! How simple!

^_^

So, after doing this, you are left with:

(x) x (x) x (x) x (rad[x]) --> The 'x' outside of parenthesis are multiplication symbols.

Simplify it again:

(x^3) x (rad[x])

At this point, you've simplified the problem as far as you can go. So, your final answer should be left as:

(rad x) x^3

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5) [rad]2([rad]12 + [rad]8 )

Okay. Another simplification problem. When you're dealing with radicals, it's easy to think of it as any other normal problem. You go about them the same way, it's just an added symbol that you have to take into account.

The first step is to distribute the [rad]2 throughout the parenthesis, just like any non-radical problem:

([rad 2] x [rad 12]) + ([rad 2] x [rad 8])

I put the parenthesis to remind you that you have to deal with the multiplication side of things first, before addition.

Okay, so now you have this jumbled mess of numbers, symbols and math signs. What do you do?!? Well, you follow this like any other normal problem. If the radicals weren't there, you would multiply the numbers together, right? So, that's exactly what you do. Remember: When dealing with radicals, you need to simplify them first. So, take each one and simplify it as far as it can go:

[rad 2] --> Simplified as far as it can go.

[rad 12] --> 2 [rad 3]

[rad 8] --> 2 [rad 2]

So, first you need to multiply ([rad 2] x [rad 12]). For radicals, you can just multiply what is underneath them together to get a single digit, and then simplify it down:

[rad 2] x 2[rad 3] = 2 [rad 6]

**Note: [rad 6] can't be simplified, so you're all set.

Now you need to multiply ([rad 2] x [rad 8]):

[rad 2] x 2[rad 2] = 2 [rad 4]

Remember to simplify!

[rad 4] = 2

So, your answer for the second part reads:

2 [rad 4] = 2 x 2 = 4

Now, you need to add the resulting answers together to finish the problem. You had:

([rad 2] x [rad 12]) + ([rad 2] x [rad 8])

Which led to:

2[rad 6] + 4

Guess what? You're done! You can't simplify that anymore, so your final answer is:

2[rad 6] + 4

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I really hope that helped Sei. I wish you luck on your mathematical quests in the future!