This problem was given to me on Saturday and I'm having trouble figuring it out.
137528
give parameters for x, y, and z.  Otherwise you cannot prove it, as no parameters are given for the equation as you described it.
286491
They were having problems since yesterday afternoon. They are aware of it and are working on it. You can report it on the forum, link found in the upper right corner of the page, it's the second thread.
114438
Goto this site and look at problem #24 and you will see the solution to this problem...in the end the answer sec x.\n\nhttp://noether.uoregon.edu/~montgome/math251/homework/Section3.8.pdf\n\nI hope this helps...\n\nDon
19743
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
271664
Here is a great article with sample problems.\n\nHope this helps!
57031
Basically: There is an initial $30 charge.  It should have been $25, so $5 must be returned and accounted for.  $3 is given to the 3 friends, $2 is kept by the manager - there you have the $5.  \n\nThe trick to this riddle is that the addition and subtraction are done at the wrong times to misdirect your thinking - and quite successfully for most.  Each of the 3 friends did indeed pay $9, not $10, and as far as the friends are concerned, they paid $27 for the night.  But we know that the clerk will tell us that they were charged only $25 and when you add the $3 returned with the $2 kept by the bellhop, you come up with $30.
359885
sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
284862
I'll solve it for you. Send it over.\n\nstill\nshell\ngrill\ncliff\n give us a clue
457204
Since benabean took the absolute difference in the periods, the value of K IS relevant; in fact, crucial to the solution.  My solution:\nTake the ratio of the 2 periods: T2/T1 = √(g1/g2) = 1.00095426\nMultiply this by the seconds in a day (86400)→ 86482.4\n\nSo, a difference of 82.4 sec lost per day.
321442
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
87535
when you figure it out let me know it is a serious problem for me aswell. e mail at krystaalvblue@ yahoo.com
473416
bring it to ur teacher, and hopefully if they cant fix it theyll give u another that u can use during the exam
447345
When you retry the problem remember to move only yout z^-2 up don't move the 4 that is a constant in the denominator. \n\nyou should get this . (12x^5z^3/30x^7y^7) * (15x^2z^10/4y^3)\n\nremember to simplify as best you can prior to multiplying.\n\ntry from here and let me know what you get.
559475
You wrote:\n"If Line AP of the shape is 12 units and Line PD is 21 units, then Line AD is ____ units. Round your answer to?"\n\n\nThat problem can't be solved as written.
90027
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
101971
That problem is written somewhat wrong.  You can re-write that problem as \nx - .3x = 1925 \nsubtract .3x from 1 x and you bet\n.7x=1925\ndivide both sides by .7 and you get\nx = 2750
675863
this has happened to me before on my 89. it's usually when the expression is too complicated--that is, it would take the calculator more memory than it has to solve it. so i'd recommend trying to first simplify the expression and then enter it. For instance, don't leave things in terms like sin(2) when you can just get an approximate number for that. maybe that's it?