Looking to "Hire" Algebra Tutors!

I'm taking Algebra 2 for the first time, after just barely passing Algebra 1 and Geometry. In both of those, it really helped to have YWS people (particularly CL, Grif, Snoink and Brad,) who knew a lot about the subject. I'd like to do that again this year, only now I'll pay my tutors in points!

Rates: (lol)

25 points if you help me for fifteen minutes.
100 for half an hour.
+ tip.

Or, if you don't have time, 10 points for every problem you help me with. If these are too low, we can talk.

Here's what I'm doing:

- Find the area or a trapezoid with bases 4 cm and 11 cm and a height of 7 cm if A=1/2(b1+b2)h.

-Solve the inequality and graph their solutions on a number line: 3x+7<22

-2x-4+3(x-2)=0

And that's just from the very, very beginning of my book. *is afraid* I'm taking this correspondence, so I don't have a teacher to ask. If you're interested in helping, just post! It would be an easy way to earn points, or points for your user group, (e.g. CCF or SPEW.)

Just a warning; I have a really hard time understanding this stuff. You may have to explain it twelve zillion times.

Thanks in advance!

Hey, Jane! I may be able to help here and there, problems depending. If I'm online, feel free to just holler my way. And I'll probably be talking algebra next semester, too (since I refuse to ever take calculus again if I can avoid it) of some kind, so I'll be up to snuff (at least in theory).

I took Algebra II last year, so I should be pretty good at it! Providing I remember anything... I was in honors, so we did learn a whole lot. Now, you seem to be doing a bit of geometry to start, which... I detested, but can do, so I can help you out with that.


I'm open to help! I suppose just find me in the chat (or if you see I am on you can PM me and I will go into the chat to help you) Or, you do have my MSN ^_~ I can't always promise I know what I'm doing though!

I'd also like to volunteer. I may be doing made-up math right now (aka calculus), but I understood the problems you posted, and I'll do my best to help you out. Feel free to PM or email me anytime. I may not be online, but when I get the message (and you know I'm online pretty much every night) I'll work on it.

You'll do fine! We'll help you through it. No worries.

Jenna! I'm also around to [s]put in my two cents[/s] help [I've gone through Algebra 2 and PreCalc] so feel free to message me if needed.

I'm not too bad at maths so feel free to ask me any questions you have and I'll try to explain it to you.

Hi Jenna!

I took Algebra 3/4 last year (the same thing as Algebra 2... 'cept it goes by a different name here. We base the numbers off semesters) so I'll be happy to help if you like! And, if you feel like it, you can ask Tassen Spellbinder about my teaching skills. I've helped him with loads of his maths.

Also, if no one is available, visit this site: www.purplemath.com , which is a huge help and is the reason that I got through 3/4 last year.

By the way, the problems you posted are currently what we're covering in my PreCalc class, so it's fresh in my mind.

I can help, too! I'm taking Algebra 2 this summer/fall. I'm about halfway through. I've actually got my book out right now. (And I'm supposed to be working on it. Gotta love homeschooling. ) Anyway, I'm pretty good at it and I certainly understood the questions you posted, so feel free to PM me if you ever want my help.

I'd be able to help if I was still in school, I'd have to go over the stuff. It starts again in about a week so If you need math help in the future let me know, I'm a math wiz.

Jenna,


I'm more than willing.


Best!
Brad

I absolutely love algebra. In class, I'd write my own problems to have other solve. Go ahead and get ahold of me somehow, even though I'm the 20th person to say I'd help, and I have a very slim chance to being a huge help, with everyone else having pre-calc, and me only up through trig.

Okay, I tutored a girl last year in algebra so I'll try and help you out. You seem to have gotten alot of replies, but no one has posted something. I'm just going to put it here and someone can build on it is they want.

I'll take this problem:

-2x-4+3(x-2)=0

Step 1:Change your negatives.
This is a way to keep your head clear. So change -4 to +-4, and (x-2) to (x+-2). Now you know where the negatives are.

We now have: -2x+ -4+3(x+ -2)=0.

2:Distribute
What -2x+ -4+3(x+ -2)=0 really means is -2x+ -4+(x+ -2) + (x+ -2)+(x+ -2). See, there are really 3 "(x+ -2)"s. So, if you add all the (x+ -2)'s together you get 3x+ -6. Which is the same as distributing.

Distributing means you take the 3 and distribute it to everything inside the (parentheses). So, we get 3 times x plus 3 times -2. We get 3x+ -6. See it?

We now have: -2x+ -4+3x+ -6=0.

3:Add like terms.
Look at the problem and see what terms are alike. -2x and 3x are alike because they both have an "x", and -4 and -6 are alike because they are both by themselves, without an "x" (or other variable).

So, add -2x and 3x. You get 1x. Add -4 to -6, and you get -10.

We now have: 1x + -10=0.

4: Solve for "x"
When we solve for "x", you must have x=something, and you must have x by itself. In 1x + -10=0, we have x=something, but x is not on that side of the equation all by itself. So, we must get rid of the -10.

Do -10 + 10 and you get zero. But you must remember the rule:whatever you do to one side of the equation, you must do to the other. So, you also have to add 10 to the other side.

It looks like this: 1x + -10 +10=0+ 10. Do the math and we get 1x=10. 1x is really just "x" so,

we now have: x=10.

Step 5: Check your Answer

Yeah, I know you thought it was over, but it's not. EVeryone should check their work, especially those who struggle.

Take your solution, x=10, and plug it into the original problem. So we get this: -2(10)+ -4+3(10+ -2)=0.

Break it down into pieces.
-2 x 10= -20
Distribute 3(10+ -2) to get 30+ -6.

So, we get -20+ -4 +30 + -6=0.
Add like terms: -30+30=0. 0=0. This is true, so our solution works.

Ta da! You just solved that problem!

That should be right, if anyone finds it wrong, just say so. Jennafina, hope that helps. If you need clarification, just ask.

Well, I'll follow in the steps of alleycat13, and since both of these are quick, I'll do both for you.

Jennafina wrote:- Find the area or a trapezoid with bases 4 cm and 11 cm and a height of 7 cm if A=1/2(b1+b2)h.

Okay, so your bases are 4 and 11. Let's let 4 stand for "b1" and 11 stand for "b2." Also, 7 will stand for "h."

So, step by step breakdown:

1. Plug in your numbers!
Alright, so now we just got all the variables set for numbers, so we stick the numbers into where the variables were.

So: A = ½*(4+11)*7 (the asterisks mean multiply, and you can do the ½ by holding alt and pressing 171 on the number pad.)

PEMDAS
Well, in case you don't know what that means, it's the order of operations. Parenthesis, exponents, multiply, divide, add, subtract. When you have a problem like this, you want to do things in that order.

We have (4+11) in parenthesis, so we add that together first, which gives us 15. So the new equation looks like this:

A = ½*(15)*7

Now we have multiple numbers to multiply together. When that happens, go left to right. So, if you like, you can move the parenthesis just to keep everything straight.

(½*15)*7

Then you multiply! 15 is not an even number, so it doesn't halve cleanly. So, when we simplify the parenthesis, we get 15/2.

The last step is this:

15/2*7/1
Multiply straight across- 15*7 and 2*1

So you have 105/2. That does not simplify, so your answer is 105/2!

Solve the inequality and graph their solutions on a number line: 3x+7<22

Alright, this one's a lot like the one Alleycat explained. You want to isolate the X.

So, to do that, we do PEMDAS... backwards! Don't freak, I'll explain.

1. Subtract 7 from both sides
You're doing this because you want to move the seven to the other side of the less than sign. And to do that, you have to make it negative- something you need to do to both sides. So!

3x<15

2. Divide by 3
Same thing as before, only since the 3 is multiplying the X, you want to divide it. Anything divided by itself is one, which cancels it.

x<5

3. Interval notation and graphing
To graph, it makes life easier to look at it in "interval notation." In this case, the interval is (-∞, 5). And I'll tell you why.

Negative infinity (the little sideways means that the graph, when you put x<5 on a number line, will go on forever to the left, in the negative direction. A positive infinity will go on forever to the right. Why are we bringing infinity in? Because the inequality does not state anything that X has to be greater than- only that it cannot meet or exceed 5. So any solution to the left of 5 will work. 4 is less than 5, 0 is less than 5, -94968986983946 is less than 5. Doesn't stop.

So, your graph will look like this:

<-------------------------°
<-------------------------5----->
Where the bottom line is a number line and the top line is the graph.


If you have any questions or if I spoke complete gibberish, let me know.

w00t! Thank you. I just got a fresh set of these problems, so expect to be bugged about them soon!

Hi 'Fina, love! I'm in Calculus this year, so I finally understand Algebra II (everything catches up to me at least a year later XD). If you want my help just shoot me a PM or an e-mail. That would be the quickest way to get a hold of me.

Looks like you've got a lot of great tutors here!