find the fifth term in the expansion of (d+2e)^7.\n\nI just wnat to know, how are you able fo find J?
121662
The first term is 7C0 d^7 (2e)^0\nThe second is 7C1 d^6 (2e)^1\nThe third is 7C2 d^5 (2e)^2\nThe fourth is 7C3 d^4 (2e)^3\nSee how the pattern works -\n\nThe fifth term is 7C4 d^3 (2e)^4\n\n7C4 = 35; (2e)^4 = 16e^4; 35 times 16 = 560\n\nSo it simplifies to 560 d^3 e^4.
669034
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
384956
h=w+7\nl=3w-5\n\nh+w+l = w+7+w+3w-5 = 5w+2\n\nSo the correct answer is C.
482020
You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
278040
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
634074
7y = x + 23\n(x - 2)^2 + y^2 = 25\n\n7y = x + 23\nx = 7y - 23\n\n(x - 2)^2 + y^2 = 25\n((7y - 23) - 2)^2 + y^2 = 25\n(7y - 23 - 2)^2 + y^2 = 25\n(7y - 25)^2 + y^2 = 25\n((7y - 25)(7y - 25)) + y^2 = 25\n(49y^2 - 175y - 175y + 625) + y^2 = 25\n49y^2 - 350y + 625 + y^2 = 25\n50y^2 - 350y + 600 = 0\n50(y^2 - 7y + 12) = 0\ny^2 - 7y + 12 = 0\n(y - 4)(y - 3) = 0\ny = 4 or 3\n\nx = 7y - 23\nx = 7(4) - 23\nx = 28 - 23\nx = 5\n\nor\n\nx = 7(3) - 23\nx = 21 - 23\nx = -2\n\nANS : (3,-2) and (5,4)
423601
I got 4x² + 2.  It was actually pretty straightforward once you lay it out like a long division problem.\n\nTake 4x² and multiply it by 7x+3i (since the first term will give us 26x³).\n\n4x² * (7x+3i) = 28x³ + 12ix², leaving us with a remainder of 14x + 6i.\n\nNow take (7x+3i) times 2 to get 14x + 6i.\n\nTherefore your answer is 4x² + 2.\n\n~ ♥ ~
363968
I worked this problem. I did get both sides to zero. I don't even know how to begin explaining it.\n\nFor the sake of typing this with some readability, let (k+1) = b and (k+2) = c.\n\nFirst you subtract (2^c)/(k+3) from both sides to get 0 on one side. Then to get a common denominator, you multiply top and bottom of the first term by (k+3) and top and bottom of the third term by (k+2). You get:\n\n[ (k+3)(2^b) + (k+1)(2^b) - (k+2)(2^c) ] / [ (k+2)(k+3) ] = 0\n\nThen you distribute out the terms:\n\n[ k(2^b) + 3(2^b) + k(2^b) + (2^b) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you factor out 2^b from the first four terms:\n\n[ (2^b)(k + 3 + k + 1) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you combine like terms on the left four. On the right two, you reduce the exponent by one and multiply the term by two. Since c is one more than b, this turns (2^c) into (2^b). You can do this since it's an exponent on two:\n\n[ (2k + 4)(2^b) - 2k(2^b) - 4(2^b) ] / [ (k+2)(k+3) ]\n\nYou can factor 2^b out of the right two terms to get:\n\n[ (2k + 4)(2^b) - (2k + 4)(2^b) ] / [ (k+2)(k+3) ] = 0\n\nThe two terms in the numerator cancel out:\n\n0 / [ (k+2)(k+3) ] = 0\n\nAnd:\n\n0 = 0\n\nI must admit that I liked this problem. I don't do math for fun, but every once in a while I like a challenge.\n\nI hope this helped.
602674
8j(j-1) = 5(1-j)\n\n8j^2 - 8j = 5 - 5j     \nMultiply the terms to expand it\n\n8j^2 - 3j = 5     \nAdd 5j to both sides\n\nj^2 - (3/8)j = 5/8     \nDivide by 8, the coefficient of the leading term\n\nj^2 - (3/8)j + 9/256 = 5/8 + 9/256  \nComplete the square (to find what you want the third part of the quadratic to be, divide the middle term [-3/8 in this case] by two [to get -3/16] and square it [to get 9/256] and add it to both sides)\n\n(j - 3/16)^2 = 169/256    \nSimplify\n\nj - 3/16 = ±(13/16)    \nTake the square root of both sides, but don't forget the plus/minus (±) because think of it this way: 2 squared is 4, but (-2) squared is also 4, which is why you need that plus/minus when you take the square root, because there are two possible answers\n\nj = 3/16 ± 13/16     \nMove the 3/16 to the right side to solve for j\n\nj = 1 or j = -5/8     \nSimplify and there's your answer!\n\nNote that j has two answers because the polynomial was to the second degree (highest term was j^2), so there are two answers for j\n\nI hope this helps  :)
196080
[Xe] 4f^14  5d^4 6s^2
246959
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
493096
One thing you can use here is that 1/j = -j\nsince if j * j = -1 you can divide both sides by j\nthen j = -1/j and -j = 1/j\nYour analysis so far is correct, and if you use the substitution above, you will get 2-1/j = 2+j.\n\nHowever, if you substituted for 1/j from the beginning, \n5/z = 2 + 1/j = 2 - j\nz = 5/(2-j)\n= [5/(2-j)] * [(2+j)/(2+j)]\n= 5(2+j)/(4 - j^2) \n= 5(2+j)/5\n= 2+j
514818
So you have the formula, and you have the number to plug into it. Are you confused about how to deal with the exponents? \n\nIntensity = 5.4 x 10-10, so\n\nd= 10 * log( 5.4 x 10-10/ 10-12)\nd= 10 * log(5.4 / 10-2 )  \nd= 10 * log(540)\n\nd= 10 * 2.732                        note it is a base 10 log!!\nd= 27.32
625554
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
90848
You finally have xy=0 so x=0 or y=0 (unless it is a problem of modular algebra...)\n\n----\n\nNotice that xy doesn't "cancel": you have \n45xy=148xy\n\nBring all terms on the left:\n(45-148)xy = 0 ----> -103xy = 0\n\ndivide by -103:\nxy=0\n\nby the law of cancellation, the last equation is true for\n(x=0 and y whatever) OR (x whatever and y=0)\n\nSo there are infinite solutions for the couple (x,y)\n\n\nGraphically, the solutions are given by all the points of the x and y axes
617631
It's as simple as plugging in numbers.  The number 15 becomes the letter "a", the number 7 is "b", and the number 55 is "c".  Just substitute the number for the letter in the formula and solve.
488818
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\nThis is as far as I got:\n(2B+1)^(D-E) - F^(3G+G2) = (10J)/A\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