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Aly's ~ Help me with math thread ; ;
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By Odin.Registry 2013-02-01 16:12:22
but according to wolfram alpha the correct answer is:
 You need to watch out where the x belongs in the equation.
It should be
not
Well ;-; when you see 2/3x it's normally 2/(3x) not (2/3)x
but yea, I noticed that right after I hit submit <_<
edit: oops, made a new post...
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By Siren.Kalilla 2013-02-01 16:18:35
You're correct, but we have to assume that aly would specify that.
So far I haven't seen her post an equation that is like that yet.
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By Odin.Registry 2013-02-01 16:21:09
Good point. ;x
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By Bahamut.Raenryong 2013-02-01 16:58:56
Don't let fractions intimidate you. Fractions are actually really easy and convenient when you know how to manipulate them.
Fraction rules below!
Addition/subtraction: Bring to a common denominator and then add/subtract numerators.
Eg: 3/4 + 1/2 => multiply the two denominators for an easy common denominator (as the multiple of two numbers must by definition have each of those numbers as a whole number factor);
6/8 + 4/8 = 10/8 (= 5/4)
But wait, how do you know to multiply the numerator by something too?
I can multiply any number by 1 and still have the same result. No matter what. This is likely a very easy concept to understand. Well, I simply want to re-express 3/4 for instance by multiplying it by 1. 1 as a fraction is 1/1... 2/2 is also a fraction that is equivalent to 1. Therefore I can multiply by 2/2 (aka 1) with no problem. Refer to the multiplication section if confused!
Multiplication
Very easy. Multiply numerators, multiply denominators.
Eg: 3/4 * 1/2 = 3/8. This may initially look "weird" because 3/8 is less than 3/4, but 1/2 is also less than 1 and so the result will be less than the original amount.
Division
Reciprocation - multiplication and division are inherently linked. For instance, if I want to divide something by 2, 8 for instance, I can also get the same result by multiplying it by 1/2.
8 / 2 = 4
8/1 * 1/2 = 8/2 = 4
Likewise dividing something by 1/2 is the same as multiplying it by 2.
Therefore, when dividing fractions, simply do 1/(the other fraction), basically inverting it.
3/4 / 1/2 = 3/4 * 2/1 = 6/4 (= 3/2)
Learn these simple rules and why they work and you will be confident with working fractions. With the amount of power and simplicity fractions have, you definitely want to master them.
Also, when typing fractions, as hinted above please make a clear distinction between say 3/2x and 3/(2x). These are very different quantities! Quote: Well it IS a rush. I think a lot of girls appreciate maths (I know mine does *giggles) but I think folks in general get scared by it. I don't think they teach estimating or "sanity checking" anymore in school. I always have some basic idea of what the answer should be before I get started in most cases.
Luckily math is rather simple in that it has rules and you can build almost anything just like legos! Don't let math get complicated, just because it might get complex!
There is a stigma against math which kind of irks me to be honest. It's one of these things which is often taught badly but is just as important as any other language in my opinion - however many people seem to almost take pride in not being good at it! Kinda anti-intellectualism in a way, but that's a discussion for another day (and is not meant to reflect on Aly etc here).
And yeah, you should ALWAYS look at your answer just to see if it's reasonable. It's often somewhat obvious if you've made a mistake (and in bigger calculations, have a rough idea of the order of magnitude you might expect. In physics, if you get a velocity above the speed of light in a vacuum for instance, you are normally [but not always] wrong).
And yeah, I think you've actually snatched up the other girl on the planet who appreciates it :(
 Carbuncle.Nezea
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By Carbuncle.Nezea 2013-02-01 20:23:49
The reason there is a stigma against math is because people don't even have the slightest clue what it actually is. (Point in case: I tell people I have a degree in mathematics and get asked every single time if I am going to be an accountant). The way it's taught in grade school is just awful and misleading. The focus is far too much on memorizing things, making it about "numbers" and useless skills that you will never have to use in real life, when it should be more about changing the way you think about things.
Learning how make logically sound arguments, as well as how to communicate them clearly and concisely, would be a far more valuable skill in general than knowing how to compute the vertex of a parabola, for instance. In fact if the emphasis were more on learning the way of thinking, then I feel that people would be able to figure out such trivial things for themselves anyway. I would love to see some kind of abstract algebra section (e.g. group or ring theory) introduced into the high school curriculum, so that people could get a better idea of what the true flavor of mathematics really is. I strongly feel that it would be a much more popular subject if people were given this opportunity. But currently the curriculum is just getting even more stupid each year, so I don't suppose I can expect to see that happen in my lifetime.
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By Shiva.Nikolce 2013-02-04 09:13:26
This is 'murica... we don't even teach kids how to balance a checking account or understand credit card/loan interest....
Ring theory!? I would be happy if we could graduate one kid that didn't run out and buy a mountain bike they couldn't afford with the first 29% interest credit card offer they got in the mail...
/screams in pain
NO!!!! WAIT! NO! I'm sorry! Don't put that THING on me!!! I won't derail anymore I promise! AAAAAAAAAAAAAAAAAAA! /dies
 Leviathan.Veltan
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By Leviathan.Veltan 2013-02-05 15:08:41
I hope I don't bugger up Aly's thread in case she needed it or was using it as a reference, but I didn't want to make a new thread for a, i assume, simple question.
Assuming an unlimited bankroll, it is theoretically possible to make a specified amount of money on every hand of blackjack. So, for example, if I wanted to make $50 dollars per hand, the first hand I would bet $50. The second hand if I won the first I would bet 50 again, but if I lost hand 1 I would bet have to bet 150 to make 50$ per hand, bet*2 =sum of previous bets + required winning/hand(number of hand).
Is my logic flawed in this? Also is there a way to simplify the formula for something like x=req. winning/hand and y = hand number?
Phoenix.Kirana
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By Phoenix.Kirana 2013-02-05 15:36:22
I hope I don't bugger up Aly's thread in case she needed it or was using it as a reference, but I didn't want to make a new thread for a, i assume, simple question.
Assuming an unlimited bankroll, it is theoretically possible to make a specified amount of money on every hand of blackjack. So, for example, if I wanted to make $50 dollars per hand, the first hand I would bet $50. The second hand if I won the first I would bet 50 again, but if I lost hand 1 I would bet have to bet 150 to make 50$ per hand, bet*2 =sum of previous bets + required winning/hand(number of hand).
Is my logic flawed in this? Also is there a way to simplify the formula for something like x=req. winning/hand and y = hand number? Assuming you keep playing until you win, and have enough money to keep doubling your bet until you win, logically you could essentially guarantee yourself to "win" x amount of dollars for each hand that you play. The problem lies in having enough money to keep doubling your bet.
If we say that 'x' is the amount of money you want to win per hand and 'y' is the hand number:
Bet amount = x(2^(y-1))
Let's say you want to win $50 per hand. If you go on a losing streak of 10 hands (making this the 11th hand), you need to have brought 50(2^10)= $51200 with you to keep trying.
Leviathan.Veltan
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By Leviathan.Veltan 2013-02-05 15:52:53
That makes sense, thanks!
EDIT: makes sense but doesn't seem to calculate right. Assuming x is 50, i.e. i want to win 50$ a hand, I know for certain that the 2nd bet shoud be 150 and the 3rd should be 350.
Phoenix.Kirana
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By Phoenix.Kirana 2013-02-05 15:56:05
Another flaw in that strategy is that Casinos generally have maximum bets, and your doubling will quickly reach that maximum.
I also assume that any good casino knows about this strategy and sets the min/max bets so that you only have to lose a few times before doubling over the max.
 Cerberus.Eugene
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By Cerberus.Eugene 2013-02-05 16:06:25
Bahamut.Raenryong said: »Fractions are actually really easy and convenient when you know how to manipulate them. I'd much rather work with fractions than decimals.
Leviathan.Veltan
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By Leviathan.Veltan 2013-02-05 16:07:21
oh I figured it out, it is:
Bet = A *(2^(x)-1)
Thanks for pointing me in the right direction kirana.
Phoenix.Kirana
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By Phoenix.Kirana 2013-02-05 16:31:45
oh I figured it out, it is:
Bet = A *(2^(x)-1)
Thanks for pointing me in the right direction kirana. the -1 needs to be inside the parens for the exponent. A *(2^(x-1))
 Garuda.Chanti
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By Garuda.Chanti 2013-02-05 17:12:23
Your logic is OK, I don't know about your knowledge of blackjack payouts.
This IS a roulette strategy using only the even money bets. It is flawed only by the lack of that unlimited bankroll and the green slots on the wheel.
Necro Bump Detected!
[91 days between previous and next post]
Gilgamesh.Alyria
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By Gilgamesh.Alyria 2013-05-07 14:47:09
Ok back again for help.
Solve the system of equations using elimination.
6X + 8y = 12
3x - 2y = -6
Of course I have the answer (-2/3,2), just need to know how to get it. And if there is a short way to do this.
I tried using sites but...they are a bit confusing...
 Ragnarok.Pythean
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By Ragnarok.Pythean 2013-05-07 14:52:47
6X + 8y = 12 (1)
3x - 2y = -6 (2)
Eqn (2) x 4 => 12x - 8y = -24 (3)
Eqn (1)+(3) 18x = -12
Divide by 18 x = -2/3
Substitute into (1) -4 + 8y = 12
so 8y = 16
y = 2
Can then check in eqn (2) if you want.
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Gilgamesh.Alyria
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By Gilgamesh.Alyria 2013-05-07 14:55:50
Hmmm
A Lil more details please.
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By Fenrir.Leoheart 2013-05-07 14:57:56
Ok back again for help.
Solve the system of equations using elimination.
Of course I have the answer (-2/3,2), just need to know how to get it. And if there is a short way to do this.
I tried using sites but...they are a bit confusing...
Eliminated that for you :)
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Quetzalcoatl.Xueye
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By Quetzalcoatl.Xueye 2013-05-07 14:58:15
Figure out how many x's are in one y.
After that, you can figure it out like a normal question.
Gilgamesh.Alyria
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By Gilgamesh.Alyria 2013-05-07 15:02:02
Yes...I don't know what you are saying sect :(
Ragnarok.Pythean
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By Ragnarok.Pythean 2013-05-07 15:03:11
6X + 8y = 12 (1) < number these equations 1 and 2
3x - 2y = -6 (2)
Eqn (2) x 4 => 12x - 8y = -24 (3) < Here we are multiplying equation 2 by 4 to make the number of y's the same as equation 1
Eqn (1)+(3) 18x = -12 < Here he are adding equation 1 and 3, the y is eliminated since -8y + 8y = 0
x = -2/3 < we are dividing the above line by 18 to find x
Substitute into (1) -4 + 8y = 12 < now we know what x is we can replace 12 x with - 4
so 8y = 16 < rearranging and solving
y = 2
Can then check in eqn (2) if you want.
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Cerberus.Thongypoo
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By Cerberus.Thongypoo 2013-05-07 15:03:26
Ok back again for help.
Solve the system of equations using elimination.
6X + 8y = 12
3x - 2y = -6
Of course I have the answer (-2/3,2), just need to know how to get it. And if there is a short way to do this.
I tried using sites but...they are a bit confusing...
They key to elimination/addition type problems are you want to manipulate one of the equations to essentially make the opposite of the x term or y term. The easier one to do here is to simply multiply all of the bottom equation by 4 giving you 12x-8y=-24. Then you add the two equations together. This gives you 18x +0y=-12 or simply 18x=-12 then divide boy sides by 18 and you get x = -12/18 simplified to -2/3.
Then you plug that -2/3 value into one of the equations. 6(-2/3)+8Y=12 = -4 + 8Y=12 which is then simplified to 8Y=16 by adding four to both sides. Then you divide both sides by 8 giving you Y = 2 so your final values are (-2/3,2).
Note: I randomly chose the first equation it would work in the original bottom equation too.
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Asura.Tiran
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By Asura.Tiran 2013-05-07 15:04:22
Been awhile since I've had to do these equations, but I'll try to explain it as best i can.
6x + 8y = 12
3x - 2y = -6
step 1: Identify the variable you wish to start with or eliminate first. I like to start with the first one in the equation, in this case 6x.
Step 2: In order to eliminate this variable (6x) you must use its opposite (-6x). To get this, you look at the second equation and identify what will make 3x change to -6x. In this case multiply by -2. You must apply the -2 to the whole equation. -2(3x- 2y = -6). Which results in -6x + 4y = 12
6x + 8y = 12
-6x + 4y = 12
Step 3: Eliminate the X variable by adding the equations together.
12y = 24 is the resulting equation.
Step 4: solve for Y
Y=2
Step 5: plug Y=2 back into the original equation. 6x + 8(2) = 12
Step 6: solve for X.
6x + 16 = 12 -> 6x = -4 -> x= -2/3
Hope this helps
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Bahamut.Bojack
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By Bahamut.Bojack 2013-05-07 15:05:18
6X + 8y = 12 (1)
3x - 2y = -6 (2)
Eqn (2) x 4 => 12x - 8y = -24 (3)
Eqn (1)+(3) 18x = -12
Divide by 18 x = -2/3
Substitute into (1) -4 + 8y = 12
so 8y = 16
y = 2
Can then check in eqn (2) if you want.
He multiplied the second equation all by 4 to change the -2y into -8y so that when he added the 2 equations together the 8y and -8y cancel each other out and eliminates y. It's just a basic 'solve for x' scenario after that. After x was solved he just put it in one of the equations and used it to solve for y.
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By Cerberus.Commandervimes 2013-05-07 15:05:32
rearrange one equation for X equals, then replace the X in the second equation with the new term
6X + 8y = 12 EQ.1
3x - 2y = -6 EQ.2
EQ.1 can become X=(12-8Y)/6
EQ.2 becomes 3*((12-8Y)/6)-2Y = -6
multiply out brackets
6 -4Y -2Y = -6
do the basic operations
-6 Y = -12
(divide by the 6)
which becomes: Y = 2
then substitute this value back into one of the starting equations to find X
6X +8*2=12
6X=12-16
6x = -4
X = -4/6 = -2/3
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Ragnarok.Pythean
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By Ragnarok.Pythean 2013-05-07 15:08:16
Cerberus.Commandervimes said: »rearrange one equation
Technically this is solving by substitution not elimination, but it still works.
Gilgamesh.Alyria
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By Gilgamesh.Alyria 2013-05-07 15:10:26
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By Odin.Godofgods 2013-05-07 15:19:07
6X + 8y = 12 (1) 3x - 2y = -6 (2) Eqn (2) x 4 => 12x - 8y = -24 (3) Eqn (1)+(3) 18x = -12 Divide by 18 x = -2/3 Substitute into (1) -4 + 8y = 12 so 8y = 16 y = 2 Can then check in eqn (2) if you want.
i used to be able to do that in mere seconds with no problems... now i cant even remember how lol :(
Sylph.Knala
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By Sylph.Knala 2013-05-07 15:27:09
Another way to go about this is solving for one variable then substituting in the equation.
That is to say in the first equation you solve for y or you can solve for x in this case we will go with x.
6x + 8y - 8y = 12 - 8y
(6x)/6 = (12 - 8y)/6
x= 2 - 4/3y
Now substitute it into the second equation
3x - 2y = -6 becomes
3(2 - 4/3y) - 2y = -6
6 - 4y - 2y - 6 = -6 - 6
(-6y)/-6 = -12/-6
y=-2
Now substitute the value for y where you solved for x earlier
x = 2 - 4/3(-2)
x = 2 - (2 and 2/3)
x = -2/3
Ok I am taking math this term and I know people like to get bent over math but I sometimes need the help....
So I hope this thread doesn't go into some flame/douche fest...I will topicban you from my thread if you are going to destroy or start fighting. And don't nitpick of me putting parenthesis or not, I type it exactly what it looks like from the program and book I use....
I am more a visual learner and the book doesnt give an example of this specific problem.
So my first problem I have is a linear equation with fractions...
y/5 + 3/4 = y/2 + 3/4
I have the answer already...just I need to know how to get the answer the book gives.
Answer: y = 0
Thank you to those who help.
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