how do i slove this:h(x)=(x++2)^3-5(x+2^2+3(x+2)-1?
918705
Don't you see the pattern?\nIt's right in front of your eyes.  Let g(x) be what is in the parentheses.  So g(x) = x+2\nNow replace x+2 with just x, and that will be f(x).  \nSo f(x) = x^3 - 5x^2 + 3x - 1\nAgain, this is assuming that h(x) = f(g(x)) (based on your other problems)
25946
=3fg*(f^2g)+3fg*5f-3fg*4\n=3(f^(1+2g))*g+15(f^2)g-12fg
565612
Take all the terms that include H to one side.\n\nS - 2LW = 2LH + 2WH.\n\nNow, factor out the H from the right side to get:\n\nS - 2LW = (2L + 2W)*H.\n\nNow, divide both sides by what it takes to leave H by itself.  In this case, get rid of the (2L + 2W)\n\n(S - 2LW)/(2L + 2W) = H
775656
2(x - 5)^2 = 3\nSo (x - 5)^2 = 3/2\nThus x - 5 = ± √(3/2)\n= ± √6 / 4\n\nThus x = 5 ± √6 / 4
1122155
not sure what box method is...the answer is\n(2x+5)*(x-3)
554172
Um...\n\n2(-X) + 3(-X) + X + 8\n\n-2X - 3X + X + 8\n\n-5X + X + 8\n\n-4X + 8\n\n8 - 4X
559365
f(x)=x+3\nif x=0, f(0)=0+3=3\nif x=2, f(2)=2+3=5\nif x=-2, f(-2)=-2+3=1\n\nfor g(x)=2x+1\nif x=0, g(0)=2*0+1=1\nif x=2, g(2)=2*2+1=5\nif x=5, g(5)=2*5+1=11\n\nfor h(x.y)=2x+y\nif (x,y)=(0,0) h(0,0)=2*0+0=0\nif (x,y)=(2,1) h(2,1)=2*2+1=5\nif (x,y)=(1,3) h(1,3)=2*1+3=5
240687
4m+H-(4m-h) =\n4m+H-4m+h=\n4m-4m+H+h=\n0+H+h=\nH+h
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
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
302578
commutator: [x, H] = (xH - Hx)\n\n\napply to Hilbert vector ψ\n\n(xH - Hx)ψ\n\n\n= x(p²/2m + V)ψ - (p²/2m + V)xψ\n\n= x*p²ψ/2m + x*V*ψ - p²(x*ψ)/2m - V*x*ψ\n\n= x*p²ψ/2m - p²(x*ψ)/2m \n\n= (x*p²ψ - p²(x*ψ))/2m\n\n\n1-dim case: p = -ih∂\n\n\n= -h²(x*∂²ψ - ∂²(x*ψ))/2m\n\n= -h²(x*∂²ψ - ∂(∂x*ψ + x*∂ψ))/2m\n\n= -h²(x*∂²ψ - (∂∂x*ψ + ∂x*∂ψ + ∂x*∂ψ + x*∂∂ψ))/2m\n\n= -h²(x*∂²ψ - 2*∂ψ - x*∂²ψ)/2m\n\n= h²(∂ψ)/m\n\n= ih(-ih∂ψ)/m\n\n= ih*pψ/m\n\n\n=> [x, H] = -ihp/m
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)
846663
(x+3)(x-1)\n\n x= -3\nx=1
195265
lim h->0 (f(x+h)-f(x))/h\n\n[(x+h)^2-3(x+h)+1-(x^2-3x+1)]/h\n=(x^2+2xh+h^2-3x-3h+1\n    -x^2+3x-1)/h\n=(2xh+h^2-3h)/h\n=(2x+h-3)\n\nThere should be a limit as h->0 in front of each line above, so let h=0 in the final line to give you 2x-3.\n\nThis is called a derivative by the way, and if you haven't learned them yet there are easy rules to follow for computing the derivative of polynomials like this one.
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
672142
it's\n[x(x+2)]/[x*x] -2/3x=\n\n[x+2]/x - 2/3x=\n\n[3[x+2]-2]/3x=\n\n[3x+4]/3x=\n\nx + [4/3]x