contradition equation 5(w+3)+3(w+1)=14?
925167
Hi Karen \n\n5(w+3)+3(w+1)=14\n\n● Step 1\n5(w+3) = \n(5 * w ) + ( 5 * 3 ) =\n5w + 15\n\n● Step 2\n3(w+1) =\n( 3 * w ) + (3 * 1) =\n3w +3 \n\n● Step 3\n5w + 15 + 3w +3  = 14\n( 5w + 3w ) = 14 - 15 - 3\n8w = -1 -3 \n8w = -4\nw = - 4 / 8\nw = -1/2  Or  - 0.5 \n\nGood Luck dear ♣
1097866
(x + 5)^2 + 1 = 37\n(x + 5)^2 = 37 - 1 = 36\nx + 5 = +- 6\nHere you have two solutions:\nx = -11\nand\nx = 1
1025267
Here is how:\n\n6+6+3\n6+5+4\n5+5+5\n\nnow change the first two around until you have all the possiblities.
757630
h=w+7\nl=3w-5\n\nh+w+l = w+7+w+3w-5 = 5w+2\n\nSo the correct answer is C.
898439
18 +  4(w - 1) = 5w + 6\n\n18 + 4w - 4 = 5w + 6\n\n14 + 4w - 4 = 5w + 6\n\n14 + 4w - 4w = 5w + 6 - 4w\n\n14 = x + 6\n\n14 - 6 = x + 6 - 6\n\n8 = x\n\n- - - - - - - - - - - - - - - - \n\n5(y + 4) = 10\n5y + 20 = 10\n\n5y + 20 - 20 = 10 - 20\n\n5y = - 10\n\n5y/5 = -10/5\n\ny = - 2\n\n- - - - - - - - - - - - - - - - - -\n\nn + 5 = 9n - 11\n\nn + 5 - n = 9x - 11 - n\n\n5 = 8n - 11\n\n5 + 11 = 8n - 11 + 11\n\n16 = 8n\n\n16/8 = 8n/8\n\n2 = n\n\n- - - - - - - - - - - - - - - - - - - - - -\n\n- 50 - 6 = - 8x + x\n\n- 56 = 7x\n\n- 56/7 = 7x/7\n\n- 8 = x\n\n- - - - - - - - - - - - - - - - - - - - - - - -\n\n- 11x = - 5 + 38\n\n- 11x = 33\n\n- 11x/ -11 = 33/- 11\n\nx = - 3\n\n- - - - - - - - -s-
905618
(A) 3+2(x-1)\n\n3+2x+2\n2x+5\n\n(B) 5-2(x+3)\n\n5-2x-6\n-2x-1\n\n(C) 1+2(x-1)\n\n1+2x-2\n2x-1\n\n(D) 6-3(w-2)\n\n6-3w+6\n-3w+12
913663
Let a = 3, b - 5, and n = 3.  Then a^n + b^n = 3^3 + 5^3 = 27 + 125 = 152 = 441 - 289 = 21^2 - 17^2, which are coprime squares.  So maybe you left out something?  Check your question.
735319
easy 24. \n5 (x3) =15\n6 (x3) = 18\n7 (x3) = 21\n8 (x3) = 24
913862
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
771554
(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
19885
(5) x (2x^2-3x+5) I think
235893
(15y^3 + 12y^2)+ (5y+4)\nGCF 3y^2(5y+4)+1(5y+4)\n\n(5y+4)(3y^2+1)
57681
1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1\n\nYou can prove it by induction:\nFor n=1, 1(1!) = 2!-1 = 1\nNow assume 1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1 for n. Then show that it is true for n+1 as well: \n\n1(1!) + 2(2!) + ... n(n!) = (n+1)!-1\nAdd (n+1)(n+1)! to both sides:\n1(1!) + 2(2!) + ... n(n!) + (n+1)(n+1)! \n= (n+1)!-1 + (n+1)(n+1)!\n= (n+1)!(1+(n+1)) - 1\n= (n+1)!(n+2) - 1\n= (n+2)! - 1\n\nSince the equation is true for n=1, and when it is true for n it is also true for n+1, it must be true for all positive integers.
776941
-25 < 5(w + 3)\n-25< 5w +15\n-40< 5w\n-8<w\n\n-9q + 18 -3 > 6q\n15 > 15q\n1> q\n\n\n19 < 7 + 1/3k + 2 -2/3k\nmultiply by 3 to get rid of denominators\n57 < 21 +k +6 -2k\n30< -1k\ndivide by -1\n-30<k
985090
l = 5+w\n\nl*w = 36\n\np = 2l + 2w
952696
n + (n - 5) + (n + 6) \n3n + 1
1014029
1,2,3,5,16,63, ...\n3=(1+2)*1\n5=(2+3)*1\n16=(3+5)*2\n63=(5+16)*3