Can ny1 help me with these maths questions asap please??
966822
Draw a right angled triangle.  the vertical is 35m and the sloped line, the hypotenuse, is 50m.  We have to find the angle between the string and the ground.\n\n35 is the opposite, 50 is the hyp, so it is a sine problem.  Sin x = opp/hyp = 35/50.\n\nusing the inverse sine function on the calculator, we find that the angle is 44.4 degrees\n\nhope this helps
817794
Need more info. Did the basics of this at college, for the NVQ and the NNEB. Title, details etc. If you require the help still, let me know - jesenianw@yahoo.co.uk and I'll do the best I can.
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
97110
Try\nhttp://www.math.buffalo.edu/mad/PEEPS/graham-carl.html
1128820
You won't find anything on the Internet that it explains things more clearly than your math book.  The only way to get help beyond that would be to ask questions directly to a teacher or tutor.\n\nYou probably should have been asking those questions sooner than the night before the final exam.
937250
PROBLEM 1:\nBobby is 3 years old.  His sister Nancy is four more than six times as old as Bobby.  How old is Nancy?\n\nPROBLEM 2:\nThe pizza parlor cuts large pizzas into 6 slices and small pizzas into 4 slices.  If Ryan buys 3 large pizzas and a small pizza, how many total slices will he have?\n\nPROBLEM 3:\nThere are 3 hexagons and a square drawn on a piece of paper.  How many total sides are there?\n\nPROBLEM 4:\nWhat is the base-10 equivalent of 34 base-6?\n\nPROBLEM 5:\nAdult tickets are $6 each and child tickets are $4 each.  How much will it cost for a family of 3 adults and 1 child?
663777
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
691620
(sigh)... Req = R1 + R2 + ...\n\nReq = 5 + 3 = 8.\n\nYay, I win.
858552
Have you tried these?  Surely you've been told that division by 0 isn't possible, so there's your answer for 7.  I'm not doing any others on Yahoo answers (too many at once), but if you really need help and are willing to do as much as possible for yourself I'll help you if you email h_chalker@yahoo.com.au, but nasty Maths teachers like me won't just hand you 7 answers on a plate!
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
1156124
whatever u need to ask in maths, do write to me..I'll try to help you out
1020498
http://www.inew.com/\n\nhttp://tiger.towson.edu/~mlackn1/mathwebsitestudent.htm\n\n\nthese should help
824758
Question 1 : 4^x - 2 * 5^(2x) < 10^x\n\nFactorise :\n2^(2x) - 2 * 5^(2x) < 2^x * 5^x\n\nDivide through by 2^(2x) :\n1 - 2 * (5/2)^(2x) < (5/2)^x\n\nLet y = (5/2)^x\n\nTherefore, 1 - 2y^2 < y\nor, 2y^2 + y - 1 > 0\nor, (y + 1)(2y - 1) > 0\nThus, y < -1 or y > 1/2\n\nThat is, (5/2)^x < - 1 or (5/2)^x > 1/2\n\nBut (5/2)^x is always > 0, so the only\nsolution occurs when (5/2)^x > 1/2.\n\nTake the reciprocal of both sides :\n(2/5)^x < 2\n\nTake logarithm of both sides :\nx(log2 - log5) < log2\n\nTherefore, x < log2 / (log2 - log5)\nor x < -0.7565 approximately.\n----------------------------------------------------------------------\nQuestion 2 : 7 * 3^(x + 1) - 5^(x + 2) = 3^(x + 4) - 5^(x + 3)\n\nDivide through by 3^(x + 1) :\n7 - 5 * (5/3)^(x + 1) = 27 - 25 * (5/3)^(x + 1)\n\nRearrange :\n20 * (5/3)^(x + 1) = 20\n\nTherefore, (5/3)^(x + 1) = 1 = (5/3)^0\nThus, x + 1 = 0, so, x = -1\n----------------------------------------------------------------------\nQuestion 3 : 6 * 9^(1/x) - 13 * 6^(1/x) + 6 * 4^(1/x) = 0\n\nFactorise for ease of viewing :\n6 * 3^(2/x) -13 * 2^(1/x) * 3^(1/x) + 6 * 2^(2/x) = 0\n\nDivide through by 6^(1/x), which equals 2^(1/x) * 3^(1/x) :\n6 * (3/2)^(1/x) - 13 + 6 * (2/3)^(1/x) = 0\n\nThe latter term, (2/3)^(1/x) = 1 / [ (3/2)^(1/x) ]\n\nSo now we have : 6 * (3/2)^(1/x) - 13 + 6 / [ (3/2)^(1/x) ] = 0\n\nLet y = (3/2)^(1/x)\n\nTherefore, 6y - 13 + 6 / y = 0\n\nMultiply through by y :\n6y^2 - 13y + 6 = 0\nor, (3y - 2)(2y - 3) = 0\nor, y = 2/3 or 3/2\n\nThat is, (3/2)^(1/x) = 2/3 or (3/2)^(1/x) = 3/2\n\nFirst, trying (3/2)^(1/x) = 2/3 = 1 / (3/2) = (3/2)^(-1)\nTherefore, 1 / x = -1, so x = -1.\n\nSecond, trying (3/2)^(1/x) = 3/2 = (3/2)^1\nTherefore, 1 / x = 1, so x = 1.\n\nSo there are 2 solutions, x = -1 or 1.
443281
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
971669
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
776173
1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law