Double trouble question?
411019
Take a good look at models and actresses and pageant queens. All skinny. Women have been bamboozeled into thinking that the models and actresses are normal and they are not. Yet go into any grocery store and look at the women. Do they look like models and actresses or do they look like normal women? Advertisement tries to get normal women to think they are not normal.
614837
I already posted this solution (or how to get it).  Do a search of yahoo answers....
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
934652
So let's say the problem looks like this\n\n   11\nx 12\n--------\n        <-----the answer from 11*2\n       <------answer from 11*1\n\nfirst hide the 1 from the 12 and work out the problem as 11*2\nthen work on multiplying the 1 with 11 but make sure you start putting your answer down below your first answer and the answer starts in the column that is below the 1 from the 12.\n\nfinally when the two answers are found, you can add them and get your final answer. SOOO here's what it looks like\n\n      11\n   x  12\n--------------\n           22    (this is from 11x2)\n +  11       (this is from 11 x1)\n------------\n     132 = final answer!
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
957682
half it then times it by 2! that will give u the answer
763076
1st Problem:\n\nLet's see, if we sell 198 thousand candy bars, we are giving them away, which is terrific for volume, but bad for profit.    \n\nRevenue = x P(x), = 9x-(x^2/22)  differentiate to get 9-(2x/22) so x/11=9 x=99\n\nPrice would be  4.50
795059
You came up with x > -1 OR x < 7.  The "or" is key.  It can be less than 7, or  greater than -1.  There's no real number that doesn't fit the solution.\n\nFor the second one, you came up with x >= 0.8 AND x >= 2.  Similarly, the "and" is the key here.  Anything greater than or equal to 2 is going to be greater than 0.8, fulfilling both conditions.  The same cannot be said if the numbers are reversed.\n\nTo sum it up, if the problem has an "or" condition, every solution must solve at least one inequality.  If the condition is "and", every solution must solve ALL inequalities.
206381
why the same question twice??? :o)
500368
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
1063248
To solve correctly, distribute two to x and -2, add four to both sides, and divide both sides by x:\n\n2(x - 2) = x + 3\n2x - 4 = x + 3\n2x = x + 7\nx = 7\n\nThe error that was made was trying to simply remove the parentheses instead of actually distributing 2 to both of its contents.\n\n-----\n\nIf this is not what you meant and I have read it backwards, this is the other interpretation:\n\n-----\n\nTo solve correctly, add two to both sides, then divide both sides by x:\n\n2x - 2 = x + 3\n2x = x + 5\nx = 5\n\nThe error that was made was trying to change 2x [2 times x] into 2(x - 2) [2 times the difference of x and 2]. This is not allowed.
1179373
You ought to be able to "logic" it out.  I think you might be making it more complicated, in your mind, than it really is.\n\n"y" is an unknown number.  2y is double that number.  \n\nThat means that there are two of "y" is 2y, or that "y" can go into 2y twice.
244823
The amount (A) will be doubling, so substitute "2P" in for A. Then you can drop the "P" from each side by division and solve for t.\n\nA = Pe^(rt)\n2P = Pe^(rt)\n2 = e^(rt)\nln 2 = (rt) ln e\nln 2 = .054t (ln e = 1)\n(ln 2)/.054 = t\n(approx.) 12.836 = t\n\nDoes that sound correct? (It's been awhile since I've done these kinds of problems.)
754767
Assuming that the two numbers are positive, the best you can do is take double and triple 850.  Both numbers will be divisible by 850, and since neither can be divisible by the other they need each to be multiplied by numbers which are not divisible by the other.  The smallest numbers that work are 2 and 3, so 850 * 2 and 850 * 3 are the smallest solutions to your problem.
43253
Depending on what you need to solve, you need two angles and a side, or two sides and an angle. Hit me with an E-Mail if you are stuck with something.\n\nkeptcompany@yahoo.com
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.