H E L P ! I need help on this Algebra problem!?
924033
Well, in order to solve it, you have to set it equal to something, because as of right now, it's just a random algebraic expression that means nothing.\n\nIf you set it equal to zero however...\n\n14c/15 + 7/17 = 0\n14c/15 = -7/17\n14c = -105/17\nc= -15/34
945604
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
963823
LHS = (2n)!\nRHS = 2^n( n!)^2\n\nfor n=1,\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is not satisfied.\n\nfor n = 2,\n(2n)! = (4)! = 1.2.3.4 = 24\n2^n( n!)^2 = 2^2( 2!)^2 = 16\nSo LHS > RHS\n\nYour problem has errors!!!\n\nI'll try to prove:\n (2n)! >= 2^n( n!)^2 .. .... (ii)\n\nfor n = 1, \n\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is satisfied.\n\nlet, for any arbitrary integer m, (ii) is satisfied.\n\nSo, \n (2m)! >= 2^m(m!)^2 \n\nnow, if n = m+1, (ii) becomes,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\nnow,\n(2(m+1))! = (2m+2))! = (2m+2)(2m+1) (2m)! \n= 2(m+1)(2m+1) (2m)! \n\n2^(m+1) ((m+1)!)^2  = 2.2^m  (m+1)^2 (m!)^2 \n\n\nwe have assumed, (2m)! >= 2^m(m!)^2 \nnow, multiplying by 2(m+1)(2m+1) we get,\n2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(2m+1) (m!)^2 \n\nnow, 2^m 2(m+1)(2m+1) (m!)^2  = 2.2^m (m+1)(2m+1) (m!)^2 \n\nagain 2m+1 > m+1\n\nso, 2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(m+1) (m!)^2 \n\nor,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\n(ii) is true for n = m+1\n(Proved)
820036
2p^2+2p-p-1\n=2p(p+1)-1(p+1)\n=(p+1)(2p-1)\nthat is why!!!!!
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
263286
the problem you presented above can be represented as,\n\n(8x^8/2x^2) + (16x^6/2x^2) - (12x^3/2x^2)\n\nthen you can simply divide each of the term\nyou'll get\n\n4x^6 + 8x^4 - 6x
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
864870
http://purplemath.com/modules/index.htm\n\nis an awesome resource for all kinds of algebra topics, even ones you may have forgotten and need to brush up on.\n\nTake a look at that one.\nUMRmathmajor
896553
4x^2+8x-x-2\n(4x-1)(x+2)
833237
so far, so good\n\nn!/(n-4)! = 56 [n!/(n-2)!] \n\nmultiply both sides by (n-2)!/n!\n\n[n!(n-2)!] / [n!(n-4)!] = 56\n\n(n-2)! / (n-4)! = 56\n\n(n-2)(n-3) = 56\n\nYou could multiply this out and solve the quadratic, in which case you will get one positive and one negative root, or if you remember that 56 = 8*7, then that quickly leads to n = 10
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
990847
isolate the variable\n\n-7x+1+8x=x+1\n-7x+8x=x\nx=x\nx=1\n\nthe answer is c) 1\nnow give me 10 points please!!!
70409
Go back to math class and figure it out!
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers