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Post Re: Homework Thread
In Lord of the Flies, who would you say are Simon's friends?

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Thu Apr 21, 2011 9:50 am
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Post Re: Homework Thread
Ralph and Piggy definitely, but everyone Liked Simon. They killed him accidentally because they thought he was the monster.

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Thu Apr 21, 2011 11:15 am
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Post Re: Homework Thread
Hey everyone! I've got a US history project I'm doing and I need a little help.

First off, I'm having a hard time finding any information about the President William McKinley assassination from reliable websites. Does anyone know of any websites I can use that have a .org .edu or .gov domain? No wikipedia, please. And yes, I did check the external links on Wiki and found no help.

Also, (and this is the main question) I'm having a hard time categorizing some of Teddy Roosevelt's reforms. Basically, I'm doing a PowerPoint presentation about how his reforms changed America socially, economically, and politically.

Before you read everything, can someone please explain to me what exactly constitutes as a social, economic, or political reform? It looks to me as if most of these can fit into more than one category.

So, if anyone is willing to look at this and correct me for the category, this is what I've got as reforms:

~Any environmental conservation is a social reform?

~Improving any living conditions (whether they be around the world or in the US) [i.e. safer food laws, safer working conditions, etc.] is a social reform?

~ Breaking up big business and monopolization is economic? Or political?

~ Anything with the military deal is a political reform?

~Foreign affairs are political reforms?

~ Is peace-keeping negotiations with other countries considered a political or social reform?

~ The building of the Panama Canal AND laws that regulate roads and railroads... I really have no idea what those would be classified as.

~Would establishing a new federal Department of Labor and Commerce be considered a political or social reform?




If anyone wants to talk about this more with me or want some exact situations (so they know what I'm talking about) feel free to ask here or PM me. [Please? I'm a little more stuck than I appear.]

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Tue Jul 12, 2011 8:39 pm
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Post Re: Homework Thread
Well this seems like an interesting thread, might as well revive it before it gets killed.
So I'm stuck with this math homework about systems of linear equations. I just don't understand how to do them T_T, here's an example:

x+1/y=1/3 (1)
x/y+1=1/4 (2)

Can someone please explain to me how to find x and y? I would really appreaciate it! Thank you.

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Sat Jul 23, 2011 1:40 pm
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Post Re: Homework Thread
To find the values represented by the letters 'x' and 'y,' apply the inverse order of operations.

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Sat Jul 23, 2011 2:06 pm
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Post Re: Homework Thread
yehoshua wrote:
Well this seems like an interesting thread, might as well revive it before it gets killed. So I'm stuck with this math homework about systems of linear equations. I just don't understand how to do them T_T, here's an example: x+1/y=1/3 (1) x/y+1=1/4 (2) Can someone please explain to me how to find x and y? I would really appreaciate it! Thank you.


Yeho, I'll be more than willing to help you with this if you send me a PM. I'm on my phone right now; I'll be on my laptop later. Give me... five hours-ish? Or I can PM you if our time zones are too incompatible.

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Sat Jul 23, 2011 2:59 pm
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Post Re: Homework Thread
Are you sure you copied that down correctly? That equation was giving me problems, and even WolframAlpha as well (graph is not a point as a solution set should be, but a line).

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Sat Jul 23, 2011 3:41 pm
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Post Re: Homework Thread
*edited this post out since it's solved*


Last edited by CY_Law on Sun Jul 24, 2011 12:36 am, edited 1 time in total.



Sat Jul 23, 2011 3:43 pm
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Post Re: Homework Thread
I echo the sentiments of those above me.
I can graph those lines by solving for y (but it gives you an equation not a number).
I can then use those equations to get a number solution for x (but it's not a pretty number; it's actually two numbers).
Thus I don't think I can give you a nice x = ?, y = ? solution.

My deduction from all of this is you either copied the problems wrong, or you do not in fact want a nice x = ?, y = ? solution.

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Sat Jul 23, 2011 6:45 pm
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Post Re: Homework Thread
Let's look at the equations, guys.

We have
(x+1)/y = 1/3

and

(x) / (y+1) = 1/4






For a system of equations, we have to isolate the y value. So, let's do this for each equation.

(x+1)/y = 1/3

Step 1) Multiply each slide of the equation by y. This will cross out the y on the left side of the equation.

(x+1) = (1/3)*y

Step 2) We need to get rid of the fraction on the right side of the equation. So, the inverse of (1/3) is 3. So, multiply each side by 3.

3(x+1)= y

Step 3) Because that x is attached to something by means of an addition or subtraction sign, we have to distribute that 3 to each part of the left hand side.

3x+3=y




Good! Now, to the second equation.




(x) / (y+1) = 1/4

For a system of equations, we have to isolate the y value. So, let's multiply each side of the equation by (y+1). This will eliminate the (y+1) on the left side.

x = (1/4)(y+1)

As in the first equation, we have to distribute that (1/4) to both the y and the +1, because the variable is attached to a number by an addition or subtraction sign.

NOTE: I am now going to use .25 to represent 1/4 so it's easier to understand.
x= .25y + .25

Now, x= anything is not a rational answer. So, again, we have to isolate the y value. But first, we need to take care of that +.25 on the right side of the equation. Let's subtract .25 from each side of the equation.

So,
x- .25 = .25y

Now, the inverse of .25 (or 1/4) is 4! So, let's multiply each side of the equation by 4, so we can finally fully isolate the y value.

4(x- .25) = y

Now, as before, a variable is attached to a number by an addition or subtraction sign, so we must distribute that 4.

4x-1=y




Nice! Now we have:

3x+3=y (from the first equation)
4x-1=y



Now, I can teach you both ways to solve this if you'd like. However, graphing is the easiest. So graph both equations at the same time. Your answer is where ever the two lines intersect. To see this more clearly, you might have to zoom in to the spot. I use a TI-84 graphing calculator, so once I've zoomed in on the spot, I hit 2nd -> CALC -> intersect. Get on the first line (to where it looks like it is intersecting), hit enter, get on the second line (to where it looks like it is intersecting), hit enter.

And your answer will appear on the screen.

x=4
y=15

Is this okay, or do you need to know how to do it without graphing?

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Sat Jul 23, 2011 7:57 pm
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Post Re: Homework Thread
Ooooooooh, there are parentheses there?!?!?!?!? :shock:
That makes a WHOLE lot more sense! :lol:

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Sat Jul 23, 2011 8:27 pm
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Post Re: Homework Thread
Without the parentheses the problem is unsolvable. Make sure you post properly written equations if you want people to solve them quickly! Order of operations is big deal!

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Sat Jul 23, 2011 8:29 pm
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Post Re: Homework Thread
You do not actually need these parenthesis, if you know how to read fractions when unformatted on a computer. When typing out math on a computer, it's much easier to just use parenthesis to make sure people understand you better. I only added them in for y'all's benefit.

EDIT: D'oh, never mind, zeek. I looked it over, and yes, it could be easy to get it confused without parenthesis. I just inferred that it was simple math (systems of equations usually are), so I didn't over think it.

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Last edited by Beagle on Sat Jul 23, 2011 8:38 pm, edited 1 time in total.



Sat Jul 23, 2011 8:33 pm
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Post Re: Homework Thread
And we thank you greatly Beagle :D (I'm used to calculus, so we have to deal with whatever the teacher throws our way :? )
I actually find the non-graphing way easier, cause I'm not as familiar with my graphing calculator (plus it's out of batteries currently :lol: )
so, here we go...

(3x + 3 = y)
-(4x - 1 = y)
(-x + 4 = 0y)
So x = 4
Then plug x into the first equation: 3(4) + 3 = y = 15

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Sat Jul 23, 2011 8:38 pm
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Post Re: Homework Thread
Obbl wrote:
And we thank you greatly Beagle :D
I actually find the non-graphing way easier, cause I'm not as familiar with my graphing calculator (plus it's out of batteries currently :lol: )
so, here we go...

(3x + 3 = y)
-(4x - 1 = y)
(-x + 4 = 0y)
So x = 4
Then plug x into the first equation: 3(4) + 3 = y = 15


That sounds like a good way, but if I'm not mistaken, there is a risk that you could run into complications with that method. There is another way to teach it, but I can't remember right at this very moment on how to do it. I think it has something to do with Completing The Square and factoring.

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Sat Jul 23, 2011 8:42 pm
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Post Re: Homework Thread
Completing the square is for solving quadratic equations. I don't see its application here. :|
Yes, this way gets a bit complicated when you have one term that doesn't conveniently go to zero when you subtract, but in that case you can multiply by some factor on either or both equations to make it work :lol:
This way may only work for linear equations though, so don't go extending its use ;)

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Sat Jul 23, 2011 8:45 pm
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Post Re: Homework Thread
Here's a helpful tip for everyone with a graphing calculator. You can get your TI-83/TI-84 to solve algebra. You need to go to Math, then at the bottom to Solve. Format the equation you want to solve so that it equals 0, and type it into your calculator (the field should look like 0= INPUT). There should be a few fields as to where the calculator should start it's search, basically a guess. Make any guess, but guess both far above and far below if you think the equation has two solutions.

Sometimes it's faster to work it out yourself, but use this if the equation is messy.

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Sat Jul 23, 2011 8:49 pm
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Post Re: Homework Thread
@Obbl

Yeah, I'm running into some weird numbers when I try to apply completing the square here.

Uhh.... that still doesn't sound right to me.... Either I'm so far out of basic Algebra (calculus in the fall for me) that it doesn't matter, or I'm having a severe brain fart right now. Probably both. :lol:

@zeekgenateer- That is an awesome tip! I'll definitely look into trying that! =) If you're setting the equation to 0, is that how you can get factors out of a quadratic equation without completing the square and factoring?

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Sat Jul 23, 2011 8:54 pm
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Post Re: Homework Thread
Hmm, I tried solving the problem just as it is, or like so:

x+(1/y)=1/3 -> (1)
(x/y)+1=1/4 -> (2)

Solving (1),
x+(1/y)=1/3
(1/y) = (1/3)-x
(1/y) = (1-3x)/3
Multiplying across,
3=y(1-3x)
y(1-3x)=3
y=3/(1-3x)

Solving (2),
(x/y)+1=1/4
(x/y)=-(3/4)
Multiplying across,
4x=-3y
-3y=4x
y=-(4/3)x

And then, I forgot what else to do.... :?
And this might be wrong, but I guess I would just post it here...
I missed learning calculus, maybe I could take it up again =3

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Sat Jul 23, 2011 9:07 pm
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Post Re: Homework Thread
Thanks a lot everyone, and Beagle, I am doing it without graphing. I knew this was a complicated system but I didn't think it was THAT complicated, the ones I already know how to solve are simpler such as
x+y=1
x-y-6
but I still don't understand so well how to do it when they have division/fractions like
y/0.6=x
2x+2y=80

I'm totally lost with these kinds of things T_T

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Sat Jul 23, 2011 9:32 pm
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Post Re: Homework Thread
yehoshua wrote:
Thanks a lot everyone, and Beagle, I am doing it without graphing. I knew this was a complicated system but I didn't think it was THAT complicated, the ones I already know how to solve are simpler such as
x+y=1
x-y-6
but I still don't understand so well how to do it when they have division/fractions like
y/0.6=x
2x+2y=80

I'm totally lost with these kinds of things T_T


y/0.6 = x, hence

y = 0.6 (x).

0.6 = 3/5, so y = 3x/5, or 5y = 3x.

So what you have is:

3x - 5y = 0
2x + 2y = 80

or

3x - 5y = 0
x + y = 40.

Multiply the bottom equation by 5 and you get:

3x - 5y = 0
5(x + y = 40)

3x - 5y = 0
5x + 5y = 200.

Add them together and you get:

8x = 200.

x = 25.

If 5y = 3x, then 5y = 3(25) = 75.

If 5y = 75, then y = 15.

x = 25, y = 15.

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Last edited by ChewyChewy on Sat Jul 23, 2011 10:00 pm, edited 1 time in total.



Sat Jul 23, 2011 9:39 pm
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Post Re: Homework Thread
In regards to fractions, the best way to deal with anything in math is simplify, simplify, simplify.

Take for example:
2x + 6y = 4
It would look so much nicer if you could divide it by 2, but wait you can!
x + 3y = 2
Both of those equations are equivalent.

For a fraction example
(x+y)/2 + y = 5
take the denominator and multiply all of the equation by it, so, multiply the equation in this case by 2 and you get
x + y + 2y = 10 -> x + 3y = 10

Basically simplify as you would in one variable algebra. Simplifying helps a lot. Take it from someone who's taken Calculus 3.

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Sat Jul 23, 2011 9:44 pm
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Post Re: Homework Thread
:lol: Fractions are your friends!
Get used to them: they never leave ;)
zeekgenateer wrote:
In regards to fractions, the best way to deal with anything in math is simplify, simplify, simplify.
^This, this and this! (Calc 2)

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Sat Jul 23, 2011 10:23 pm
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Post Re: Homework Thread
Turn any decimals into fractions. .6 = (3/5) and use this note if this attaches right. I'm about to work on the other note.

Attachment:
math help.JPG
math help.JPG [ 97.1 KiB | Viewed 2590 times ]


Diss, Sleet, Aaron, THF, Brent, if y'all know how to combine double posts with attachments without screwing things up, go ahead and edit my double post for me. I'm sorry! D:

NOTE: On that second note, the line should read, "This is any number you can divide 2, 2, and 80 with, and the resulting numbers will still be WHOLE numbers.

A whole number is, for example, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 anything without decimals or fractions!

Attachment:
more math help.JPG
more math help.JPG [ 140.11 KiB | Viewed 2581 times ]

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Last edited by Dissension on Sun Jul 24, 2011 12:37 am, edited 1 time in total.

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Sat Jul 23, 2011 11:21 pm
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Post Re: Homework Thread
I just missed math. I feel awful and lame and useless.

Anyone need anything still explained? :lol:

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Sun Jul 24, 2011 2:19 am
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Wow, you guys are good! ...

... But how am I supposed to categorize this neatly into the index at the second post? :lol: Oh well, will try to figure this out later!

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Sun Jul 24, 2011 5:33 am
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Post Re: Homework Thread
so, erm, college started...I have no idea if anybody knows how to work this out, but it's worth a shot, thanks in advance.

Quote:
A straight line has gradient 'm' and passes through the point (x,y).
Find, in terms of ax+by+c=0 an equation of the line when:

m=2,(x+y)=(5,1)


Honestly, graphs and other things similar confuse me the most. I'm really unlucky that this is the first topic, and what we'll be tested on to see if we can handle AS maths. (also, I live in England, in case I confuse anybody)

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Sun Sep 11, 2011 9:48 am
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Post Re: Homework Thread
I would start with the form y = mx + b. You already have m, so that just leaves b to be found, which is the y-intercept, or the value of y when x = 0.

It passes through (5,1) with a slope of 2, so when x increases/decreases, y increases/decreases by twice as much. So to make x = 0, you decrease x by 5, which means you decrease y by 10, giving you (0, -9). So the y-intercept is -9, giving you y = 2x - 9. Then you just move it all over to one side (because they ask you to) to give you 2x - y - 9 = 0.

Does that help?

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Sun Sep 11, 2011 11:10 am
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Post Re: Homework Thread
that helps sooooo much. Thanks a ton Sleet. :mrgreen:

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Sun Sep 11, 2011 11:27 am
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Post Re: Homework Thread
This isn't really homework but in a recent astronomy test the question "what is the most magnificent this in the universe?" came up. I guessed it and got it right but I just wanted to see if this question can be done by others.(I was the only one in my year to get it right)

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Sat Sep 24, 2011 9:47 pm
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Post Re: Homework Thread
That's subjective and philosophy, not astronomy. There is no right or wrong answer!

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Sat Sep 24, 2011 11:41 pm
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Post Re: Homework Thread
That's what I thought at first but it was a one word answer question in an astronomy test.



For those who dont want to guess the answer was Quasar apparently.

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Sun Sep 25, 2011 8:08 am
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Post Re: Homework Thread
What exactly makes them more magnificent than anything else?

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I, too, am curious. Please explicate.

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Sun Sep 25, 2011 5:00 pm
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Post Re: Homework Thread
According to one of our textbooks it says they are the brightest things known to exist. So maybe magnificent means bright in this case. It is still a poorly worded question.

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Sun Sep 25, 2011 7:34 pm
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Post Re: Homework Thread
That's probably it. But yes, that is poorly worded.

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Sun Sep 25, 2011 8:01 pm
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Post Re: Homework Thread
Interesting... A homework thread?

3^x+6x=39 If the root can be approximated to 2.82,
What is the root of this function (approximated to hundredth place: 2 digits)?
3^x+2x=13
No calculators!

I have no idea how to approach this problem.


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Post Re: Homework Thread
Well, you can rearrange them:

3^x = 39 - 6x
3^x = 13 - 2x

You can notice that the right side can have a 1/3 factored out:

3^x = 1/3 (39 - 6x)

Then it can be multiplied over:

3*3^x = 39 - 6x

Which is, as you should know...:

3^(x+1) = 39 - 6x

Does that help? That's a good way to rearrange it.

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Wed Nov 30, 2011 11:20 pm
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Post Re: Homework Thread
That didn't really help me but thank you anyway.
I found out the answer is 2, it was a trick question.
I don't think there is an algebraic solution...


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Post Re: Homework Thread
I just started a paper on another one of T.S. Eliot's poems | "The Hallow Men" |

It's incredible hard to understand

Can anyone tell what this poem is about!!!!?!?!?!????!?!!

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