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Hello thickhead:
therefore required sum is 1000*1001/2-1023=49977
That's how we solved that but you've got to calculate as accurate as you think. That answer (49977) just can't be right.
Last edited by ElainaVW (2016-08-16 17:17:25)
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I understand your point.The values to be subtracted are
that is a G.P. in The value to be subtracted is The sumisLast edited by thickhead (2016-08-16 17:19:56)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hello thickhead:
That's wrong too. All the terms are integers so you can't get an irrational answer.
Mistake is when you did
1000*1001/2-1023=49977
your calculation got all squished together
1000 * 100 (1/2) -1023 = 49977
the right way
(1000 * 1001) / 2 - 1023 = 499,477
Last edited by ElainaVW (2016-08-16 17:48:20)
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I was giving logic and not calculation. I had agreed on bobbym's answer and proceeded for the logic.Sometimes keyboard fails.You may find that in the later wrong version I wrote the correct value of 499477!
Last edited by thickhead (2016-08-16 18:08:00)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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@bobby: For number 4 I went back and I found that if you set the entire thing equal to some number n and put in exponent form you can solve for both x and y^5z with some manipulation. It can be done using trial and error, but there is a cleaner way.
@thickhead and ElainaVW:
I understand how to do number 5 now - I personally think it's more important to understand how to do it than to know the exact answer when you're learning, but in applied mathematics a small mistake can be fatal.
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Hi aleph_zero;
Glad you figured it out.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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7. Let
and denote those areas within the ellipse that are in the first, second, third, and fourth quadrants, respectively. Determine the value of .8. An ellipse and a hyperbola have the same foci, A and B, and intersect at four points. The ellipse has major axis 50, and minor axis 40. The hyperbola has conjugate axis of length 20. Let P be a point on both the hyperbola and ellipse. What is PA * PB?
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Hi aleph_zero,
(4) p=43;q=6 ;clue: The given expression
Last edited by thickhead (2016-08-17 03:02:37)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hi aleph_zero;
Regrettably my first answer was an error because I used 50 = semi major axis and 40 the semi minor axis. This has been corrected. Sorry for the confusion.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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The answer to 8 is 500...
9. Suppose 0< a,b,c < 1 and ab + bc + ca = 1. Find the minimum value of a + b + c + abc.
I've tried using AM-GM on this and I can get an inequality for ab+bc+ca and a+b+c in respect to abc but I'm not sure how to answer the problem...
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Regarding the problem:tough maths proof by 010595 ( now a closed thread) the only way other than using maxima seems to be 'The isosceles triangle involved therein will have maximum area when it becomes equilateral" the same principle as I used in problem (9) here.
Last edited by thickhead (2016-08-18 18:31:37)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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I wasn't able to find more than one ordered pair for number 6, x = -1 and y = 0. What did you find ?
I had thought it was meant for bobbym. Now I realize it is for me. I agree, I bungled the solution.
Last edited by thickhead (2016-08-18 21:45:31)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hello:
@bobbym, it is amazing that you didn't get the right answer for 8. Your diagram shows you were working on the wrong hyperbola!
The red one is the right hypie.
The correct solution is not yet given so I'll put it down.
This solution uses the defining properties of both ellipse and hyperbola.
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Hi;
You are correct for everything except the amazing part. I suppose you want some sort of explanation for the oversight of the wrong hyperbola...
1) As Elizabeth Stapel states, hyperbolas ( hyperboli ? ) are not often seen in problems. She is correct, this was my first time working on one of them.
2) Because of that, I had to head to the internet for formulas and definitions. As you know, I was in trouble already with that approach. They gave many conflicting formulas and I switched a and b.
3) There is a fellow on the SE who uses your solution as a refutation to mine, have you seen it?
Overriding principle
4) When working on problems without using the EM way I faced the same confusion that is common on math forums. If I would used your approach I could have found that constant
without knowing anything more about a darn hyperbola. Ole Bush Buffalo would saySometimes you get the bear and sometimes, he gets you.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Next time just email me before you attempt to solve a problem that is a bit too tough for ya. I'll be glad to assist an old guy.
Last edited by ElainaVW (2016-08-19 04:28:34)
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Hmmm, I will keep that in mind. All this talk about bears and hyperbolas has made me hungry. I will feeble my way over to the kitchen and see what I can cook up. Us old guys get hungry about now.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hello:
The correct solution is not yet given so I'll put it down.
That is not factually correct. I had given the solution and now in more detail.
Last edited by thickhead (2016-08-19 05:22:48)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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That is not factual.
I didn't say that the right answer wasn't posted. I said the correct solution hadn't been posted yet.
The solution you propose is longer and requires more work.
This solution uses the defining properties of both ellipse and hyperbola. This is shorter.
It can be shortened more because you don't even need to know
. He could've got that with EM. It is obvious that solving 2 linear equations is terse and therefore best.Last edited by ElainaVW (2016-08-19 05:52:21)
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Actually, I liked my method best of all but it only lacked the virtue of being accurate. To me the highest accolade one can give any solution is one of being correct.
Hi EVW;
You know my ways and my methods. I have long argued against calling one solution superior to another based on criterion like shorter, more elegant, prettier etc. If a method yields the right answer I am happy and if it does not...
Now, if you could say that solution B uses less memory or is faster to compute then of course I am interested. Measuring solutions in terms of clock cycles or ram makes sense.
He could've got that with EM.
That would have been an interesting exercise and I am sorry that I did not go down that road.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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That is not factual.
I didn't say that the right answer wasn't posted. I said the correct solution hadn't been posted yet.
The solution you propose is longer and requires more work.
This solution uses the defining properties of both ellipse and hyperbola. This is shorter.
It can be shortened more because you don't even need to know
. He could've got that with EM.
Last edited by thickhead (2016-08-19 16:30:38)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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