im having problems with this math question?
924254
f(x) = 3cos(2x)+4sin(2x)\n√(3² + 4²) = 5\nDivide and mutlipy by 5\nSo f(x) = 5*(3/5 cos(2x) + 4/5 sin(2x))\nNow because (3/5)² + (4/5)² = 1 there exists an angle such that cos ε = 3/5 and sin ε = 4/5 (OR vice versa if you prefer)\n\nSo f(x) = 5*(cosεcos(2x) + sinεsin(2x)))\n= 5 cos(2x - ε) where ε ≈ 53°07'48" ≈ 0.9273 rad\n\nie f(x) = 5cos(2x - 0.9273)\n\nYou could also have got to f(x) = 5 sin (2x + 0.6435) if you used sinε = 3/5 and cosε = 4/5 It is easy to show that these are the same as (2x + 0.6435) - (2x - 0.9273) = 1.5708 = π/2\n\nand if x - y = x + (-y) = π/2\nthen x = π/2 - (-y)\nSo sin x = cos (-y) = cos y as the cosine function is an odd function (ie f(-x) = f(x))\n\nIn general:\nif(x) = AsinY + BcosY\n= √(A² + B²)(A/√(A² + B²) * sinY + B/√(A² + B²) * cosY\n= √(A² + B²)(sinY cosε + cosY sinε)\n= √(A² + B²)sin(Y + ε) where ε = arctan (B/A)\n(also = √(A² + B²)cos(Y - ε) where ε = arctan (A/B)\n\nand other variations can be more useful depending on the signs of A and B and a desire to have 0 < ε < π/2 (ie ε acute)\n\nThere are 4 possibilities f(x) = Rsin(Y + ε)\nf(x) = Rsin(Y - ε)\nf(x) = Rcos(Y + ε)\nf(x) = Rcos(Y - ε), the latter two if A and B are opposite in sign.
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
784978
Depends on what calculator you use...some are +/- and some are just -. So it depends...Sometimes you have to push 2nd function to get it.
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.
111021
If you are sloving for X then anything greater than 2 will work. basically it is 6 time what (X) will be greater than 12. Well 6x2=12 so that won't work.
1081089
[ (50000 x 2.4)+(25000 x .8) ] / (50000+250000)=1.8\n\nThe remainder of the info in the problem is not necessary.
202280
http://www.aaamath.com/ Try this. Hope it works!!
618077
go to www.curnaturecalculus.com  They have all the answers.\n\nSplooch!
966057
well well..the second time i see the same question!!\nshow some of ur work buddy.. ur a degree student and u dont simply put the questions on others..u wud never pass that way...\nso prove that u have tried ur best to solve this problem..show some coding that u have made!
849743
This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
1025264
pls specify ur question, didnt quite understand it\n\nIf there is "OR" u have to add and if it is "And" u have to multiply.
843943
4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
12336
Try the BBC's Maths series and MathsFile as well as Math.com (see links below)
157275
dont understand question sorry
950875
plsmake ur question clear.im not able 2 understand it