Schoolwork Help Thread

[kirby_superstar]

I feel like this is approaching the vicinity of being relevant...

I am terrible at math in french for some reason. You'd think it wouldn't be any different...and really it isn't. I'm confident actually that it was my non-francophone teacher. Her voice was monotonous and dull. Her classes made me hate Smartboards everywhere. And she had no business coaching the volleyball team while being so fat. Goodness, I hate that woman. She, and I mean this sincerely, stuffed her face while wading through her boring as ass classes. I used to be in AP math. AP!!!!! I passed the horrid strumpet's class...just barely. In fact, she single-handedly took my average and bent it over a desk and pounded it. Math is more engaging in English.

Also, logarithms aren't too difficult once you get the hang of them. Just gotta practise the problems and you'll catch on. I didn't learn much in my class--as I've mentioned. HOWEVER. I believe that logs make sense after you do a few applications with'em. Now for homework help.... wanna do my Physics write-up? :S

[Roz~]

I feel like this is approaching the vicinity of being relevant...

I am terrible at math in french for some reason. You'd think it wouldn't be any different...and really it isn't.

Also, logarithms aren't too difficult once you get the hang of them. Just gotta practise the problems and you'll catch on. I didn't learn much in my class--as I've mentioned. HOWEVER. I believe that logs make sense after you do a few applications with'em.

I suck at both math in French and English D:
As you mentioned, logs are not that hard. I loved doing that. T'was awsm.
Also dull teachers are the reason why I failed a couple of my math classes. Didn't feel like doing my homework ;~;

[kirby_superstar]

Ahahaha, well, I think it was because my teacher was so boring. Normally I'd catch on. But now I've joined the masses for whom math just isn't really intelligible.
I'm horrid with homework. Ahahaha, as such, I commend the originator of this thread. >.<

[Jub3r7]

The trouble I had with logarithms is that I wasn't really awake while we were learning it for the first time...

I was writing all the notes down, but that's all they were. None of it ever reached my head for a while.
After getting some help with it and actually paying attention, it wasn't too hard.

[SlowPokemon]

Exponent*

But yes, that is correct above.  That's actually one of the most simple log problems possible, so you must have not been paying attention in class :|

Umm well that's what I thought until I asked her about it and she said that we hadn't learned any logarithms yet and that I should just substitute random numbers in for R. -_- Wat. I got some smart kid in my class to explain the log to me.

[Winter]

Too bad I don't have a "classroom" to get help from.

[KefkaticFanatic]

sucks to your assmar

[SlowPokemon]

Reviving this topic because Jub said he was having math issues. Ask if you want, Jub.

[Jub3r7]

It's more of a not-getting-enough-sleep problem that prevents me from hearing anything my teacher says during fifth period.

So I just have to 1. get more sleep, 2. do the homework/study, and 3. be more organized so I can find my homework.
And if I run into anymore problems with number 2, I'll let you know.

[Jub3r7]

A company produces packets of soap powder labeled "Giant Size 32 Ounces." The actual weight of soap powder in a box has a normal distribution with a mean of 33 oz. and a standard deviation of 0.8 oz. What proportion of packets are underweight (i.e., weigh less than 32 oz.)?

If it helps, the answer is .106; I just need to see how to get there.

[Nebbles]

We just started on the unit circle in math... errr... anyone know any good tricks to help learn it?

[The Deku Trombonist]

Jub: Bah, the batteries in my graphics calculator have died. I don't remember exactly how to do it off the top of my head but there's some place where you can input the details of the distribution and find the probability that it's less than 32 (assuming you can use a graphics calculator, not that I know of any other way).

Nebbles: What bits of the unit circle are the problem? Is it converting from radians to degrees or something else?

[KefkaticFanatic]

Saria: lol idunno anything about Ben Franklin besides the kite thing

Nebbles: Memorize it.

[Cobraroll]

Nebbles:

The unit circle has its centre in origo (coordinates (0,0)). Its radius is 1.
The sine of an angle between the positive x-axis and a given spot on the line of the circle is given by the Y-coordinate of the spot. The cosine is given by the X-coordinate. Note that there will always be two angles that fit each sine and cosine value.

Could have gone on for longer, but school starts in half an hour. Later today, perhaps.

[Qew]

We never learned the unit circle like you did, but as far as I pieced together from various things I think I understand what you need to know, tell me if it's too much or too little.

Cobraroll explained the actual circle well, but I have a suspicion you're referring to the values on the unit circle rather than the concept itself.

Firstly radian measure. I won't go into the details of what it is because you may already know. The key thing to remember is pi is equal to 180 degrees. So 2pi = 360 degrees ; pi/2 = 90 ; pi/3 = 60 ; pi/6 = 30 and pi/4 = 45. Thus the formula really is if you have something in degrees, multiply by pi/180. If you have something in radians, multiply by 180/pi. An easy trick to remembering which is which is to look at the denominator, if what you have is in radians, multiply by the one with the pi on the bottom, vice versa for degrees

Now that that's settled, let's look at the unit circle.



Let P be a point on the unit circle of co-ordinates (x,y). Let P have an angle of θ with the x-axis. Therefore, by pythagoras, given the hypotenuse is the radius of the circle and equal to one

cosθ = x/1 = x
sinθ = y/1 = y
tanθ = y/x

Thus the co-ordinates for P can be written as (cosθ, sinθ).

Now the horrible part... how to memorise the values of the co-ordinates when θ is equal to various numbers.

Personally, since I wasn't taught this way, I'll teach you my way and how to convert it to the unit circle.

If θ is equal to pi/4, then the other angle has to be pi/4 because the angles in a triangle must add up to pi (180). Thus the triangle is isoceles because the two base angles are equal. Let one of the smaller sides equal one, the other side must also equal one since opposite sides of an isoceles triangle are equal. Now what about the hypotenuse? well by pythagoras the hypotenuse = square root of 1^2 + 1^2. The hypotenuse = square root of 2.



But the hypotenuse of the unit circle isn't the square root of 2, it's 1. Therefore we have to make the hypotenuse of this triangle 1. Look at the bottom left triangle. All the sides have been divided by the square root of 2. This is fine since I just let the two sides equal one, the ratio and pythagoras and everything still holds if I divide it by root 2. Except now we have the two co-ordinates that define the point for the unit circle.



Now we can use trigonometry! We found earlier that cosθ = x, sinθ = y, tanθ = y/x. If we substitute the values we have on the triangle

sin(pi/4) = 1/root 2 ; cos(pi/4) = 1/root 2 ; tan(pi/4) = 1/root 2 divided by 1/root 2 = 1

The other two values, pi/3 and pi/6 I have shown on the triangles below. For pi/6, you can just swap the values, as long as you know which side of the triangle represents x and which represents y you'll be fine.



Now, this is only for the first quadrant. For the other three you'll be pleased to know it's just these exact same triangles mirrored. For example, in the second quadrant, the x value will be negative, but the y value will be positive. So cosθ is negative, sinθ is positive, and tanθ is negative.

The trick to remembering the angles, I find, is to remember it's the same angle, but taken from the x axis. So for the third quadrant,


In that sense if we have an angle of θ in the first quadrant, the equivalent angles are:
Second Quadrant: pi - θ
Third Quadrant: pi + θ
Fourth Quadrant: 2pi - θ

Hope that helps.