The DW Math Thread

Esnemyl - Dreamweaver

GCSE Chemistry is hard Can anybody give me a dumbed down version on how to balance chemical equations? Since its kinda maths..
I never got taught it properly because I got switched from BTEC to Edexel when I shifted schools.
Thank god it's my last year there, college + Graphics next year!

Decus - Dreamweaver

Asterelle - Sanctuary wrote: »

I woke up this morning and realize that x / sin(x) is a monotonically increasing function.... at least in the range (0 ... 180)

That implies that if a/sin(a) = b/sin(b) then a = b
By law of sines a/sin(a) = b/sin(b) = c/sin(c) so then a = b = c.

Since a + b + c = 180 we have a = b = c = 60.

Yea that was the trick to it, though you didn't have to bother with the proof. I'll post a new set of problems for this week later.



@ Nemypoo, I always found that it was best to list the amounts of atoms and ions on each side while trying to balance the equation. That way you can see the quantities involved as you try to make adjustments. It's similar to factoring in that, as you practice it more, it becomes easier and you begin to view the quantities quicker.

Here's a simple example of what I mean. Take this small reaction:

Fe + Cl2 --> FeCl3

We list the Iron and Chlorine atoms for both sides (on paper, list them across from each other):

(LH side, RH side)

Fe - 1 , 1
Cl - 2 , 3

Now, we are going to assign coefficients to the atoms and molecules to balance the quantities of Chlorine atoms first, since they are not equal. Look at the quantity you listed for each side and find the lowest common multiple (LCM). The LCM of 2 and 3 is 6, so let's pick coefficients that give us 6 of each on each side. As you add coefficients, you change the quantities written under it.:

Fe + 3Cl2 --> 2FeCl3

Fe - 1 , 2
Cl - 6 , 6

So, now that the Chlorine atoms are balanced, let's balance the Iron atoms. Since there is just 1 on the LH side and 2 on the right, we can double the quantity on the L and get the final equation:


2Fe + 3Cl2 --> 2FeCl3

Fe - 2 , 2
Cl - 6 , 6

And there it is!

---

Occasionally, you can just treat polyatomic ions as one unit, if they do not separate in the reaction. It's a matter of convenience:

CaCl2 + AgNO3 --> Ca(NO3)2 + AgCl

In this case, the Nitrate ion remains intact, so you can consider it its own quantity:

(LH side, RH side)

Ca - 1 , 1
Cl - 2 , 1
Ag - 1 , 1
NO3 - 1 , 2

Now let's look at the two quantities that are not equal: Chlorine and Nitrate. To balance Chlorine atoms, just double the number on the R side.


CaCl2 + AgNO3 --> Ca(NO3)2 + 2AgCl

Ca - 1 , 1
Cl - 2 , 2
Ag - 1 , 2
NO3 - 1 , 2

Now we move on to Silver. Double the number on the L side:

CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl

Ca - 1 , 1
Cl - 2 , 2
Ag - 2 , 2
NO3 - 2 , 2

That did the trick for us!

Remember you can never remove quantities of atoms, just increase them.

I hope that helps. My Chemistry knowledge is pretty lame though, so I make no promises b:surrender

Esnemyl - Dreamweaver

Thats ok Decus, science is my hardest thing to understand for me. Being an Aspie makes it 2x worse XD. Thanks for the insight

Decus - Dreamweaver

My pleasure. I am glad it helped. b:thanks

Xiar - Raging Tide

is there any chance Descus, you could help me with the Riemann Hypothesis?

Decus - Dreamweaver

I know as much about the Riemann zeta function as I do about the French language, which is to say I do not know much outside of some basics. I took complex analysis a long time ago and I never took a hardcore number theory class, so my knowledge is pretty poor.

Still, I can give it a shot. At the very least, if you post any questions, I know Yorkin could take a better stab at them than I can.

Shoot!

Yorkin - Dreamweaver

Decus - Dreamweaver wrote: »

Still, I can give it a shot. At the very least, if you post any questions, I know Yorkin could take a better stab at them than I can.

Shoot!

I gotta say, Dec, you give me too much credit. I just had a few lucky shots with your usual signature pictures. I am just a experimental phyisicist. Besides, which part of math did you study anyway?

For basic info on Riemann Hypothesis, beside Riemann Zeta Function, I didn't hear or learn much. It has to do something with position of non trivial zero points of function As far as I know from popular lectures, this hypothesis is unsolvable, in sense that it hasn't been mathematically proved yet, thou there are some criteria which marginally support the possibility (I think same problem is with Goedel's Theorems, Decus will correct me if I am wrong).

Check the Wikipedia for general info, I'll consult my personal library, when I get back home from college in two or three days. Well, another adventurous Christmas will be... XD

Decus - Dreamweaver

Problems for the week! I kept them fairly easy this time around to encourage more people to participate. As always, please do show work and don't cheat. That's no fun

1-

At the bank Hank cashed a check but the teller gave him cents instead of dollars, and dollars instead of cents. After buying a five cent piece of gum, Hank had exactly twice as much as his original check. How much was the check?


2-

Let P1, P2,...,Pn be points on the unit sphere. Prove that the sum of the squares of the distances between all pairs of points is at most n^2, and state conditions for this sum to be equal to n^2.


3-

A 'number' of runners entered a recent road race. These are two true statements regarding the 'number':
a) It has 3 distinct digits
b) If you add 99 the number reverses.

Here are four other statements regarding the 'number':
1) It is divisible by the sum of its digits
2) It is not prime
3) It has only one common digit with the product of its digits
4) The sum of the first and last digit is one more than the middle digit

If I told you which of these 'other' statement(s) were false, you'd be able to determine the number. How many runners entered the race?


Enjoy! b:pleased

Arctix - Dreamweaver

Decus - Dreamweaver wrote: »

3-

A 'number' of runners entered a recent road race. These are two true statements regarding the 'number':
a) It has 3 distinct digits
b) If you add 99 the number reverses.

Here are four other statements regarding the 'number':
1) It is divisible by the sum of its digits
2) It is not prime
3) It has only one common digit with the product of its digits
4) The sum of the first and last digit is one more than the middle digit

If I told you which of these 'other' statement(s) were false, you'd be able to determine the number. How many runners entered the race?


Enjoy! b:pleased

132 ? b:victory

Decus - Dreamweaver

Arctix - Dreamweaver wrote: »

132 ? b:victory

No.

See:

Decus - Dreamweaver wrote: »

As always, please do show work...

truekossy

What is 2+2? b:question

GGianAAle - Archosaur

truekossy wrote: »

What is 2+2? b:question

5! b:victory Although, I think this is too much for this thread

Decus - Dreamweaver

truekossy wrote: »

What is 2+2? b:question

Take a stab at the challenge questions

GGianAAle - Archosaur wrote: »

5! b:victory Although, I think this is too much for this thread

Kossy is allowed to troll, but you are not! b:angry

Q____Q - Dreamweaver

Decus - Dreamweaver wrote: »

Take a stab at the challenge questions

I looked at the questions.. then I saw how much effort it required to do them, and just kept scrolling

Asterelle - Sanctuary_1381265973

Bah bored.

Decus - Dreamweaver wrote: »

Problems for the week! I kept them fairly easy this time around to encourage more people to participate. As always, please do show work and don't cheat. That's no fun

1-

At the bank Hank cashed a check but the teller gave him cents instead of dollars, and dollars instead of cents. After buying a five cent piece of gum, Hank had exactly twice as much as his original check. How much was the check?

1)
Let d = original dollars and c = original cents

2*(100d + c) = d + 100c - 5
98c - 199d = 5
101*(-d) + 98 *(c-d) = 5
98*(c-2d) + 3*(-d) = 5
3*(32*(c-2d)-d) + 2 * (c-2d) = 5
3*(32c-65d) + 2*(c-2d) =5
So c-2d = 1 + 3k, 32c-65d=1 - 2k
c = 1+2d + 3k
32*(1+2d + 3k)-65d = 1 - 2k
32-d + 96k = 1 - 2k
d = 31 -98k.. if 0<d<100 then k = 0
d = 31
c = 1+2*31 + 3* 0
c = 63

$31.63

2-

Let P1, P2,...,Pn be points on the unit sphere. Prove that the sum of the squares of the distances between all pairs of points is at most n^2, and state conditions for this sum to be equal to n^2.

Geometry is boring.

3-

A 'number' of runners entered a recent road race. These are two true statements regarding the 'number':
a) It has 3 distinct digits
b) If you add 99 the number reverses.

Here are four other statements regarding the 'number':
1) It is divisible by the sum of its digits
2) It is not prime
3) It has only one common digit with the product of its digits
4) The sum of the first and last digit is one more than the middle digit

If I told you which of these 'other' statement(s) were false, you'd be able to determine the number. How many runners entered the race?


Enjoy! b:pleased

3)
if the digits are ABC

from b)
ABC + 100 = CBA + 1
A = C-1

only 4 numbers satisfy b) and 4)
122, 243, 364, and 485

122 violates a) and 485 is the only one not divisible by sum of its digits (4+8+5=17)

Answer is 485

Runemine - Dreamweaver

*Brain explodes* Now what?????????????? if i actually took time to slowly read that i might understand lol i just speed read it =P

Mitachi - Dreamweaver

Is there an easier way to work out salaries with interest over the years? For example earning $10000 the next year there will be a 6% increase on the pay, the year after that there will be another 6% increase to the previous years amount. So over 20 years how much will it be? I was thinking that you take 6 and times it by 20, but the amount seems a bit too much. Or do i have to work it out year for year?

breeeezy87

Mitachi - Dreamweaver wrote: »

Is there an easier way to work out salaries with interest over the years? For example earning $10000 the next year there will be a 6% increase on the pay, the year after that there will be another 6% increase to the previous years amount. So over 20 years how much will it be? I was thinking that you take 6 and times it by 20, but the amount seems a bit too much. Or do i have to work it out year for year?

You can't just take 6% * 20 = 120% and multiply it by the base salary. The reason is that the pay increases are compunded, so over 20 years the total interest will be greater than 120% on the base. You can use a compound interest formula T=B(i)^Y

T = total salary
B = base salary (10,000)
i = interest rate
Y = years

T=10000(1.06)^20
T=32071.35

6% may be a little high to be compounded annually for the average person but not too far off. For the most part any salaries compounded annually this way are just negated by the annual inflation rate.

Mitachi - Dreamweaver

breeeezy87 wrote: »

You can't just take 6% * 20 = 120% and multiply it by the base salary. The reason is that the pay increases are compunded, so over 20 years the total interest will be greater than 120% on the base. You can use a compound interest formula T=B(i)^Y

T = total salary
B = base salary (10,000)
i = interest rate
Y = years

T=10000(1.06)^20
T=32071.35

6% may be a little high to be compounded annually for the average person but not too far off. For the most part any salaries compounded annually this way are just negated by the annual inflation rate.

okay thanks. One more question, compound interest is the interest added every year to the new amount of money right?

Decus - Dreamweaver

Mitachi - Dreamweaver wrote: »

okay thanks. One more question, compound interest is the interest added every year to the new amount of money right?

Correct, it's the interest that is added to the principal so that you gain interest on the previous year's (in this case) interest. Sounds like that stupid meme...

The general formula, as given above, is F = P(1 + r)^n

Where F is the future amount, P is the principal, r is the interest rate (as a decimal), and n is the number of times it is compounded (accural periods).

I've seen a common version in textbooks too:

F = P[1 + (r/n)]^(nt)

Where F, P, and r are the same as above, n is the number of times per year that the interest is compounded, and t is the number of years.



EDIT: Oh snap! Time for a new challenge! I'll get right on that... ;-;

Mitachi - Dreamweaver

Decus - Dreamweaver wrote: »

Correct, it's the interest that is added to the principal so that you gain interest on the previous year's (in this case) interest. Sounds like that stupid meme...

The general formula, as given above, is F = P(1 + r)^n

Where F is the future amount, P is the principal, r is the interest rate (as a decimal), and n is the number of times it is compounded (accural periods).

I've seen a common version in textbooks too:

F = P[1 + (r/n)]^(nt)

Where F, P, and r are the same as above, n is the number of times per year that the interest is compounded, and t is the number of years.



EDIT: Oh snap! Time for a new challenge! I'll get right on that... ;-;

Thank you Decus b:kiss

Decus - Dreamweaver

Mitachi - Dreamweaver wrote: »

Thank you Decus b:kiss

Anytime, my dear.




---

New week (sort of...), new problems! I tossed in an extra super duper awesome problem I found to be really fun.

Enjoy. Usual rules apply: show work and don't cheat...

I.

From a smooth 30" diameter circular wood disk, two smaller circular disks of diameters 20" and 10" are cut. What is the largest circular disk that can be cut from one (either) piece of the remaining plywood? Ignore width of saw-blade when cutting.
Hint: think osculating circles!


II.

Fifteen college teachers (A,B,...,O) must serve on a total of twenty committees (1-20), such that:

a) Each teacher is on exactly four committees
b) Each committee has three teachers on it
c) No two committees have more than one teacher in common

List (or accurately describe) such a committee structure, or else prove it can't exist.


III.

A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after n selections (n>1), the product of the n numbers selected will be divisible by 10.



Bonus!
(You get cookies for solving it. I stumbled on this problem and could not RESIST posting it!)

A calculator is broken so that the only keys that still work are the sin, cos, tan, arcsin, arccos, and arctan buttons. The display initially shows 0. Given any positive rational number q, show that pressing some finite sequence of buttons will yield q. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

Esnemyl - Dreamweaver

Decus, I need your help again o_o
I'm stuck on science, and I have this question:

The molecular formula of Dinitrogen Tetroxide is N204, <--- numbers are nower case

a) what's the empirical formula?
b) explain why Dinitrogen textroxide is not the same as nitrogen dioxide (NO2 <-- also lower case)

I need a dumbed down way of finding out about empirical formula- something a blonde teenage girl cna understand b:avoid Since this is chemistry, you should know about this with numbers.. halp? I feel stupid not knowing this.

breeeezy87

a) If you're given the mass of each element simply multiply the mass by 1 over the atomic mass of a single atom to find the moles.

__g H x (1mol H/1.01g H) = __mol H -repeat this for each atom

After getting a final answer for each element in mols, divide that answer by the lowest answer answer in mols. (1 of the atoms will will workout to 1 each time) If the answer in red comes out as _.9999 round to nearest whole number. If you get something like _.5 or _.2 multiply each answer in red by the same number until u get all whole numbers. (10 will always work if the decimals are to the hundreth and can be the number of atoms can be reduced after).

If your given the number in % rather than in grams, just assume the total mass is 100g since the max % allowable is 100%. Therefore, 55% = 55g and so on.

If not given any mass or %, use ratios of atomic mass to solve.
2N = 28g , 4O= 64g , total mass = 92g

28gN/92g =30.43% 64gO/92g = 69.57%

Now that you have your percantages use the steps from above.

Assume the sample size is 100g.

N= 30.43% = 30.43g
O= 69.57% = 69.57g

30.43gN*(1molN/14.0gN) =2.17molN
69.57gO*(1molO/16.0gO)=4.348molO

2.17molN/2.17 = 1
4.348molO/2.17=~2

Therefore the emperical formula is NO2.

b)The makeup of NO2 consists of only 1 nitrogen and 2 oxygen, while N204 consists of 2N and 4O this changes it's molecular structure and cannot be reduced, simplified nor made the same. While 2 different molecular compunds could have the same emperical formula, they can not have the same molecular.

Kimi - Dreamweaver

Wow, I had no idea the people of Dreamweaver were so smart(Due to the **** I see in WC). I'm a disgrace to this server.b:surrender

Esnemyl - Dreamweaver

breeeezy87 wrote: »

a) If you're given the mass of each element simply multiply the mass by 1 over the atomic mass of a single atom to find the moles.

__g H x (1mol H/1.01g H) = __mol H -repeat this for each atom

After getting a final answer for each element in mols, divide that answer by the lowest answer answer in mols. (1 of the atoms will will workout to 1 each time) If the answer in red comes out as _.9999 round to nearest whole number. If you get something like _.5 or _.2 multiply each answer in red by the same number until u get all whole numbers. (10 will always work if the decimals are to the hundreth and can be the number of atoms can be reduced after).

If your given the number in % rather than in grams, just assume the total mass is 100g since the max % allowable is 100%. Therefore, 55% = 55g and so on.

If not given any mass or %, use ratios of atomic mass to solve.
2N = 28g , 4O= 64g , total mass = 92g

28gN/92g =30.43% 64gO/92g = 69.57%

Now that you have your percantages use the steps from above.

Assume the sample size is 100g.

N= 30.43% = 30.43g
O= 69.57% = 69.57g

30.43gN*(1molN/14.0gN) =2.17molN
69.57gO*(1molO/16.0gO)=4.348molO

2.17molN/2.17 = 1
4.348molO/2.17=~2

Therefore the emperical formula is NO2.

Still doesnt make sense, sorry
I have the dumb b:surrender

Escorian - Dreamweaver

Mathematics:
This comes from 2 math teachers with a combined total of 70 yrs. Experience.
It has an indisputable mathematical logic. It also made me Laugh Out Loud.
This is a strictly mathematical viewpoint..it goes like this:
What Makes 100%?
What does it mean to give MORE than 100%?
Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%.
How about achieving 103%?
But what makes up 100% in life?
Here's a little mathematical formula that might help you answer these questions:
If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Are represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
Then:
H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%
And
K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%
But ,
A-T-T-I-T-U-D-E
1+20+20+9+20+21+4+5 = 100%
And,
B-U-L-L-S-H-I-T
2+21+12+12+19+8+9+20 = 103%
AND, look how far *** kissing will take you.
A-S-S-K-I-S-S-I-N-G
1+19+19+11+9+19+19+9+14+7 = 118%
So, one can conclude with mathematical certainty, that while Hard work and Knowledge will get you close, and Attitude will get you there, its the Bull**** and *** Kissing that will put you over the top.

Now you know why some people are where they are!

Runemine - Dreamweaver

-5+3x=7x+11

How do you solve. Maybe i shouldn't have taken the accelerated course 1 test a week and can't find any help.......GAH

Never mind solved that one now lol now solve this =P

When three times a number is subtracted by 41, the result is 5 plus the number. what is the number?

Hate word problems like this

Decus - Dreamweaver

Runemine - Dreamweaver wrote: »

When three times a number is subtracted by 41, the result is 5 plus the number. what is the number?

The statement becomes:
3n - 41 = 5 + n

To solve:

2n - 41 = 5
2n = 46
n = 23

Runemine - Dreamweaver

Yep seen how i screwed that up. Now im doing fractions yay......starting Monday or Tuesday i do decimals.......AHHHHH and 3ish weeks from now i do polinumerals i got no clue how to even spell it......