kindly pls help me find the answer to this math problem. my boyz in asia or us...help : >?
414919
goyalnarinder2 is correct.\n\nIf you also want to understand it, not only have the answer, then you should first know how to get rid of those brackets, for instance:\n\n(3x-1)(2x+3)= 3x(2x+3)-1(2x+3) \n\nGot the picture? It's called distributivity, the fact that you can distribute the factor 2x+3 to each of the terms of the first bracket - these would be 3x and -1.\n\nThen u continue distributing:\n3x(2x+3)=3x.2x+3x.3=6x^2+9x \n-1(2x+3)=-2x-3\n\nThat is how u get to:  \n6x^2+9x-2x-3=6x^2+7x-3 and this is just the result of the product of the first 2 brackets.\n\nAre u still w me? :-)\n\nYou now try to do the second product - which is (x-4)(x-1) - by yourself and see if you get this:\n(x-4)(x+1)=x^2-3x-4\n\nThen u add the 2 things (or rather substract...):\n(3x-1)(2x+3)-(x-4)(x+1)=(6x^2+7x-3)-(x^2-3x-4)\nand here u take care of the signs in the second bracket, which u should reverse (because say if I take -3 apples from u what I do is really give u +3 apples, got it?)\nSo u continue w:\nthe above = 6x^2+7x-3-x^2+3x+4=5x^2+10x+1\n\nHope this helped. Good luck!
511709
Here is a link you may find helpful.\n\nhttp://www.math.buffalo.edu/mad/00.INDEXmad.html
902088
Well done! Not one is wrong. The answer to the third one is y = 11. If y = 11/12, you'd end up with 11=132!
570829
u should probably ask that to a math tutor or on askusnow.info
903533
ignore the decimals when u originally do the problem, then however many deciams there are in both of the decimals you're multiplying is how many there will b in the answer. i.e. if you multiply 6.78x4.234 together the answer will have 5 numbers behidnthe dceimal point
779807
Yiu would first decide how many decimal places you are going to consider.  The usual is 3.  Then work it from there...;
1181326
Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
862589
maybe you could try practising. For maths, the phrase practice makes perfect is very true. hmm...short cuts in multiplication and division...maybe you could try learning the abaccus (a mathematical instrument used by the chinese => functions like a calculator, but manually powered, no batteries)...I only know how to add and subtract using it...
1090524
it would be 22 choose 4 =22!/(4!18!) minus the ways in which Mr and Mrs Math can be together = 20 choose 2 = 20!/(2!18!)\n=(22.21.20.19)/(4.3.2)-...\n...(20.19)/(2)\n=((20.19)/2).((22.21/4.3)-1)\n=190.75/2\n=7125\nSorry had to two line one bit due to Yahoos dodgy cutting off of things
981364
Gopal has the correct answer! \n\nSo... becareful with some other results... and... never use:\n0.833 instead of 5/6... it's a mistake!\n\nFinished Algebra with 19!
700886
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
132964
You are solving for k so it you need to rewrite the problem to look like this k=z/(xy) where x=3, y=8, z=6 substitute the letters for numbers so k=6/(3*8) k=6/24 or k=1/4 to check just solve for any other letter such as z by using what you know of k and you should come out with z=6
725814
This is a very difficult counting problem.  I will show one possible grouping and count the number of ways it can be done.\n\nTo put just 1 boy in group 1, there are 10 ways to pick the boy and "combination of 15 things taken 4 at a time" ways to pick the girls.  This is 10 * (15*14*13*12 / 1*2*3*4) = 10*1365 = 13650 ways to pick 4 girls and 1 boy.\n-----------------\nThere are many grouping scenarios; let us start with 1 boy, 1 boy, 1 boy, 2 boys, 5 boys.\n\nWe know there are 13650 ways to pick a single boy for group 1.\n\nTo then pick a single boy for group 2 from the remaining children, there are 9*(11*10*9*8 / 1*2*3*4) = 9 * 330 = 2970 ways.\n\nIf a single boy is then selected for group 3, there are 8 * (7*6*5*4 / 1*2*3*4) = 8*35 = 280 ways.\n\nGroup 4 must have at least 2 boys.  There are 7 boys left, and we must choose 2 of them.  So there are 7*6/1*2 = 21 ways to choose those boys with the remaining girls.\n\nThe last group would have 5 boys.\n\nSo just for this one possible way of grouping - 1 boy, 1 boy, 1 boy, 2 boys, 5 boys, there are \n13650 * 2970 * 280 * 21 * 1 = 238,378,140,000 ways to arrange them.\n\nTHIS MAY NOT BE THE CORRECT FIGURE FOR THIS GROUPING SCENARIO - THERE MAY BE DUPLICATES, IN WHICH CASE THIS FIGURE IS TOO HIGH.  I did not know how to remove them from the figure.  The 13650 is correct for the first, but then we need to look at how many ways are there to grab 4&1, 4&1, 3&2, 0&5 from the remaining children.\n\n\nWe must also account for other grouping scenarios, such as 2, 2, 2, 2, 2 boys; 1,1,2,2,4 boys; 1,1,1,3,4 boys; 1,2,2,2,3 boys, and some others.  For each of these grouping possibilities, the above such calculations would have to be done.  The sum of all the grouping possibilities is the total.\n\nSo this is a very difficult counting problem.  I would think you should search for more tools from your professor/teacher, or book.  There may be some more advanced way of saving time rather than go through each grouping scenario one at a time; though that will give the correct answer.
276680
Ok, If A is a point on the circle and C is the middle of the circle then AC is the RADIUS of the circle.  Which means any point on the circle to C WILL ALWAYS be 12.  So if the excess of the circle to B is 8, then Add the RADIUS to the distance to B, and we arrive at the result of 20. Which is AC + (Distance of circle to B)\n\n20
345909
Rearranged letters you have China which is a name of country.\n2 does not belong to the line because when you subtract two contiguous numbers you must get a number which can be divided by 3. (2-4=-2;-2 cannot be divided by 3 equally)
946717
I think the simplest way to solve this is to move the decimal place over one to the left.   If for example they wanted 1% you would move it over two to the left.  If they wanted .01% three to the left.  If they wanted .001% four to the left.  You get the point. \n\nSo 10%= $2500
667073
Let x = number of boys and \ny = number of dogs, then\nx + y = 40 and\n2x + 4y = 100\nDivide the second equation by two and then subtract it from the first\nx + y = 40\nx + 2y = 50\n-y = -10\ny = 10\nx + y = 40\nx + 10 = 40\nx = 30\n30 boys, 10 dogs