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#1 Re: Help Me ! » How much cardboard do I have? » 2009-02-14 09:44:36

Good to have you with us, JurassicPlank3. I believe muddyshoes couldn't imagine (when he asked the question) how much trouble he would cause with his problem. I consider that an engineer could do a better job. Thank you for your input.

#2 Re: Help Me ! » Simple arithmetic » 2009-02-14 09:39:44

My daughter's maths teacher apparently showed them a solution without solving equation. My daughter said she understood his explanation (kind of short in fact) but she couldn't reproduce it for me. In my opinion, trying to solve this problem without equation is against the nature. But I cannot say I know everything. So I agree 100% with you Jane, and thanks for reply.

#3 Re: This is Cool » Mathematical terms in other languages » 2009-02-13 15:00:07

That's nice, Jane. I know that you can find them.

#4 Re: Help Me ! » How much cardboard do I have? » 2009-02-13 14:46:41

I missed in my calculation one thing and you, janetpc2008, are right. So, I would say that muddyshoes has an estimation now.

#6 Re: This is Cool » Mathematical terms in other languages » 2009-02-12 16:08:44

I can add to your list romanian version for each term. Anyone interested?

#7 Re: Help Me ! » Simple arithmetic » 2009-02-12 12:37:02

Yes, it is clear. Many thanks. What made you choose 60 as units high? I realized that 12 minutes would be a good time unit after I solved the problem using equations- but saying that 12 comes from 3 times 4 is not right.
12 minutes is actually 1/5 of one hour. What's the trick I miss?

#8 Help Me ! » Simple arithmetic » 2009-02-11 12:28:34

mathsmypassion
Replies: 7

Two candles, same height, are lit at the same time. The first takes 4 hours to burn and the second takes 3 hours. When is the first candle twice the height of the second one?

I need a solution that doesn't use equation. Thanks.

#9 Re: Help Me ! » How much cardboard do I have? » 2009-02-10 11:30:49

I am going to offer you an aproximation of the amount of cardboard you have. This is the best I can do.
If the core is 2 inches in diameter than the length of the roll is aproximately:

L = pi × (2 × n + (n-1)×n/16) inches
where n is the number of times the cardboard is rolled (count the circles inside)
For 128 circles (and this the theoretical number of circles) the length would be around 3996.1 inches = 333 feet.

The total area of the roll :
L × 30 inches²  or 3996.1 × 30 inches ² = 832.52 feet ² if it is too hard to count the circles.
I would take 5% away of this number, just in case.
I would say you have almost 790.9 feet² of cardboard.

#10 Re: Help Me ! » How to work out percentages. still stuck lol » 2009-02-10 11:04:17

Understanding percentage is  a bit of  a challenge. Once you get them you will never forget.

For the first problem:
The  initial amount is $60.
10% of this amount is ... $6. Clear enough?
So $9 = $6 + $3  in fact 10% + 5% (1/2 of 10%) = 15%
The answer: 15% of $60 is $9
This is the best way to start to understand percentages, by thinking.

#11 Re: Help Me ! » New job requires basic math ,quick thinking ! » 2009-02-09 17:58:59

Work out the multiplications by 20 (you have 20 chips in one row) and 100 first.

Use flash cards as a training tool. Repeat the multiplications as often as possible and practice using flash cards by yourself. You don't need another person to test you. Have these cards handy.

If these multiplications work well step further starting with 2 times table, 3 and so on. Use flash cards. Stop at 9 × 2, 9 ×3 ...

Next step: addition. You have to be able to do 5  × 200 + 5 × 10 +5 ×6
Do these additions on paper first. Write on paper the result only (nothing else). Having them on paper helps you visualise the calculation.
In time you will be able to do them in your head.  Good luck!

#12 Re: Help Me ! » Missing Term » 2009-02-08 12:04:12

Answer for the second one:

220 = 22 × 10

144 = 18 × 8
112 = 16 × 7
What is missing?
     = 20 × 9 = 180
My answer is 180.

#13 Re: Help Me ! » Euler's formula help » 2009-01-28 16:47:58

Initially you had 30 faces. By truncating the vertices of type 2, which are 12 in total,  actually we add another 12 faces (in fact pentagons ) to the 3-d shape.

#14 Re: Help Me ! » Euler's formula help » 2009-01-28 08:07:55

I got it.
Every face has 4 vertices. So 4F = 120.
So 3m + 5n = 120.
I hope you understand why.
we have now simultaneous equation.

m + 3n = 56
3m + 5n = 120

Solve it!

#15 Re: Help Me ! » Euler's formula help » 2009-01-28 08:00:43

1) I can give you a little idea. F = 30 so V - E = -28

But V is made up from two types of vertices. Let's say there are m of type one and n of type two.
So V = m + n
Every edge joins two vertices.
Some of them are type one, they belong to 3 edges in the same time, and the others are type two and they can be found on 5 edges.
So , actually, when you count the edges, you count 3 times every of the vertices type one, and 5 times every of the vertices type two.
I would say 2E = 3m + 5n .

I put 2E because I counted both vertices at the ends of one edge.
We have now one equation:
m + n - (3m/2 +5n/2) = -28
same as:
m + 3n = 56

We need another equation. I can't think of any other so far.

#16 Re: Help Me ! » Induction. o_O » 2009-01-28 07:20:32

2+4+....+2k+(2k+2)>K² + (2k +2) = (k+1)² +1 >(k+1)² QED

#17 Re: Help Me ! » commutitive group » 2009-01-28 07:15:28

Thanks luca-deltodensco. Much better the way you wrote it.
Ricky, your solution is very ellegant. The way gemma tried to solve it made me consider a different solution than yours.

#18 Re: Help Me ! » the concept of "a polynomial over Galois Field" » 2009-01-26 07:26:06

GF(p) represents a Galois field when p is prime. In our problem p = 4 that is not a prime number.

#19 Re: Help Me ! » commutitive group » 2009-01-26 07:09:53

(x1*x2)*(x1*x2) = e
x1*(x2*x1)*x2 =e  / *x2 (operate to the right)
x1*(x2*x1)*e =x2
Now do the same thing to the left with x1
x1* /   x1*(x2*x1)=x2
e*(x2*x1) =x1*x2 or
x2*x1 = x1*x2
Q.E.D.

#20 Re: Help Me ! » what is the lowest and Unique value from 0 to 1000? » 2009-01-25 14:49:43

Hi rickzeen,

As I said before there is no right or wrong answer. There are in total 100 001 numbers to choose from between 0.00 and 1000.00 (same number as between 0 and 100001 if multiply all our numbers by 100).  If I can be of any help for you , let me know.
Sorry for being so late with my reply.

#21 Re: Help Me ! » Unbounded function » 2009-01-22 09:32:03

For a rigouros proof it may work proof by contradiction (reductio ad absurdum). I have a feeling that will do it.

#22 Re: Help Me ! » what is the lowest and Unique value from 0 to 1000? » 2009-01-22 09:11:09

Can you have more than 2 digits after decimal point?
Because 10^-55 has 55 disgits after decimal point.

#23 Re: Help Me ! » what is the lowest and Unique value from 0 to 1000? » 2009-01-22 08:24:24

This is gambling anyway.
If 0 is allowed, I would choose it.
I give you two reasons for that. Some would avoid the number (they may consider it trivial) and others (very many) wouldn't imagine that 0 is in the range of numbers between 0 and 1000.

#24 Re: Help Me ! » Asymptotes and Slants » 2009-01-22 08:08:33

To find horizontal asymptotes you have to calculate the limits to +  ∞ and - ∞. If any of them is a finite number, that's a HA. In your case, both limits are 1 so y = 1 is HA (to + ∞ and - ∞).

For slants: if a function does not have HA, it may have slants asymptotes.
For example: f(x) = (x^2 + 1)/(x + 1)
In order to find the SA follow the steps:
1. calculate the limit to +∞ for f(x)/x. If this limit if finite (a real number ≠ 0), let's call it m, than m would represent the slope of the SA. Go to step 2. If the limit is not finite , the function does not have SA to +∞. Go to step 4.
2. Find now the limit to +∞ for f(x) - mx and call this real number n.
3. The SA is y = mx + n
4. Do the same thing for  -∞ . You might get a different asymptote.

#25 Re: Help Me ! » Prove » 2009-01-15 12:26:45

Sorrry Jane, but the intial problem is equivalent with:

if: 0<A,B,C<pi/2
  A+B+C<pi
then: tanA +tanB +tanC>tanAtanBtanC found by Kurre

Another way to prove that tanA +tanB +tanC>tanAtanBtanC is (not much difference compared with Kurre's solution):

tanA +tanB +tanC- tanAtanBtanC = (sinA/cosA + sinB/cosB) + tanC(1 - tanAtanB) = sin(A+B)/(cosAcosB) + tanCcos(A+B)/(cosAcosB) = sin(A+B+C)/(cosAcosBcosC) >0 when 0<A,B,C<pi/2 and A+B+C<pi

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