No. 4  if (a2 +  c2 ) : (ab+cd) = (ab +cd) : (b2 + d 2) prove that a,b,c,d are in proportion.?
960637
(a2+c2):(ab+cd)=(ab+cd):(b2+d2)\n(a2+c2)/(ab+cd)=(ab+cd)/(b2+d2)\n(a2+c2)(b2+d2)=(ab+cd)^2\nPlug in some values for a2,b2,c2,d2,a,b,c,d say 1 for all values\n(1+1)(1+1)=(1+1)^2\n4=4\nHence they are in proportion.
1168391
c=1\na=2\nn=3\nCan = 1X2X3 = 6\nA=1 \nn=2 \ny=3\nb=4\no=5\nd=6\ny=7\n1X2X3X4X5X6X7= 5040\n\ng=1\no o = 4\nd=3\n\n1X4X3= 12\n\nA=1\nT=2 \n1X2=2\nM=1\na=2\nt=3\nh=4\n1X2X3X4= 24
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
771168
1= B\n2= A\n3= D\n4= D\n5= D\n6= C
1030960
AB'C'D + AB'CD = AB'D(C'+C) = AB'D\n\n(AB'+D)(AB'+D+CF')\n= AB' + DAB' + D + AB'CF' + DCF'\n= AB'(1+D+CF')+D(1+CF')\n= AB'+D\n\nI'm not quite sure what you mean by the second question, but the dual of an expression is found using de Morgan's law:\n\nA' + B' = (AB)'\n\nHowever, the expression you're interested in, A(B+C'+1) is trivially equal to A, and so has no interesting dual.
534750
1/2 + 1/2b + 2b^2 + 6bc + c^2\n\n(1/4b)(1/4b) + 4bc + (bc)(bc)\n1/2 + 1/4b +1/4b + b^2 +4bc + b^2 + c^2 + 2bc\n1/2 + 1/2 + 2b^2 + 6bc + c^2\n\nI think this is right... email me if I'm wrong?
939068
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
759105
False... y has to be 2...\n(3,2) is a solution, but not (3,1).
57033
the answer is 3 because a-b|a-b & a-b|a so a-b|b
909722
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
589588
T1 = 2\n\nT2= 4\n\n>d= T2-T1\n=4-2\n2.\n\nT4= 8\n\nT3=6\n\n>d=T4-T3\n=8-6\n=2.\n\nGet it?
739570
d2 - =int(c2/10)\n\neasy
775420
you can write \nP=d product((x-a_i)^k_i)) poduct((x-b_j)^2 + c_j^2)\n Then I give you 3 hints:\n1) d must be positive\n2) all k_i are even numbers\n3) use the formula (a^2+b^2)(c^2+d^2) =\n= (ac-bd)^2 + (ad+bc)^2
631400
Factor out the y:\n\n(a-2) * y = c\n\nThen divide both sides by (a-2):\n\ny = c / (a-2)
864410
The proposition is\n\nSQRT(a) + SQRT(b) = SQRT(a+b)\n\nIf the equality is true, then we can rais both sides to the second power, yielding\n\n(SQRT(a) + SQRT(b))^2 = (a+b)\n\nIf we put c=SQRT(a) and d=SQRT(b) to simplify the equation, we have \n\n(c + d)^2 = a+b\nand expanding\nc^2 + 2cd + d^2 = a+b\n\nBut c^2 = SQRT(a) ^2, and d^2 = SQRT(b) ^2, so in the end\nc^2 = a and d^2 = b\nso back substituting we have\na +2cd +b = a+b\nand 2cd must be equal to zero, and that is possible only if either c or d is zero, and that is possible only if either a or b is zero.
812747
a + b + c = 180\n52 + 34 + c = 180\nc = 94
336800
let a = b\na² = ab (Multiply both sides by a)\na² + a² - 2ab = ab + a² - 2ab (Add {a² - 2ab} to both sides)\n2(a² - ab) = a² - ab (Factor the left, and collect like terms on the right)\n2 = 1 (Divide both sides by {a² - ab})\n\nOnce you've got that settled, you can start on 2 + 2 = 5:\n\n2 = 1 + 1, so therefore\n\nx = 2 + 2 goes to\n\nx = 2 + (1 + 1)\n\nBecause you now have proof that 1 = 2\n\nx = 2 + (1 + 2)\n\nx = 2 + 3\n\nx = 5\n\nSubstituting the original equation for x:\n\n2 + 2 = 5\n\nBut also, since 1=2 and 2=1 then 1+1=4 and 2+2=2