Use the commutative and the associative laws to find three equivalent expressions
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n+m+29\nm+n+29\n29+n+m\nn+29+m\nm+29+n\n29+m+n\n\nHere are 6 equivelant expressions.
140974
what does a^2ba^3ab^3b =a^6b^5 or a*a*a*a*a*a*b*b*b*b*b
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1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
31994
A*B=(3i+5j)*(2i+4j)=2k
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3\nThe valence * subscript number should be equal\n\na: 3 x 2 = 6\nb: 2 x 3 = 6\n\na2b3
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multiplication\n\n\nDoug
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simple\nfirst\nwhen we apply ∑ we can write the series as\n(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3).... (a_n - a_n-1) (a_n+1 - a_n)\n\nwhen u add \nthe terms starting frm a_2(except a_1 & a_n+1) get cancelled with the correseponding terms and u get\na_n+1 + a_1
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slider is correct.  \n\nRemember the distributive property?\na(b+c) = ab + ac\n\nThis is using it in reverse.  \n\nIn part A, notice that there is an a in each term, so you can factor out the a's.  \n3a² + 2a = (3a)*a + 2 * a = a(3a + 2).\n\nIn part B, there aren't any letters in common but you can factor out numbers too.  \n12d²+6 = 6*2d² + 6*1 = 6(2d²+1)\n\nThe other parts use similar reasoning.
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1= B\n2= A\n3= D\n4= D\n5= D\n6= C
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(3x-2)(12x^2-2) + (4x^3-x^2+1)(3)\n\nThat is the un-simplified answer
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It would help if you used some parentheses to indicate which terms are being divided into which terms!  I suspect that just using the order of precedence for operations will not result in the problem you are actually trying to solve.
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it's (xy) / 2 - 1 * 3 + 1 - 2 which is (3xy) / 2 - 2 because remember the order of operations
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First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
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T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
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[It's the commutative property, named after commuters that go back and forth.]\n\n2(8)(b-3-8)(b+1)\n\n2(8)(b-11)(b+1)\n\n2 [8(b-11)] (b+1)  (associative)\n\n2 (8b-88)(b+1)  (distributive)\n\n(8b-88)(2)(b+1)  (commutative)\n\n(8b-88)[2(b+1)]  (associative)\n\n(8b-88)(2b+2) (distributive)\n\n16b^2 + 16b - 176 b - 176 (distributive)\n\n16b^2 - 160b - 176
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f(x)=a(x-x1)(x-x2) and f(x)=a(x-x0)^2 +ho \n\nthe first one: ax^2 +ax(-x2-x1) +ax1x2\nthe second one: ax^2 - 2ax x0 + ax0^2 +h0\n\nso:\n2x0 = x1+x2\nand\nax0^2 +h0 = ax1x2\n\nsubstitute the first into the second:\na( (x1+x2)/2 )^2 +h0 = ax1`x2\na/4( x1^2 + 2x1x2+ x2^2 ) + h0 = a x1x2\na/4( x1^2 + 2x1x2+ x2^2 ) -ax1x2 = h0 \n~
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1--d\n2--a\n3--b\n4--a