IF f={[1,-1],[2,3],[-3,0] and g={[0,3],[1,1],[2,-4],[3,3] f+g is?
100803
These sets are not surjective.So f+g is meaningless.
1164142
F=O \n\nbut if E=MC2 then F=ND3\n\nthen F=ND3, \nsince ND is just [(M+1)(C+1)] \nthen \nF=({[(M+1)(C+1)]}3)\n\n\nOr you could say that\nF=E+1
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
295299
F = first = 3x times 3x = 9x^2\nO = outer = 3x times 4 = 12x\nI = inner = -4 times 3x = -12x\nL = last = -4 times 4 = -16\n\nAdd like terms (they become 0) and write answer: 9x^2 - 16
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
974938
I believe this is how you do it:\n\n(f o g)(x) = f[g(x)] = 3(-x^2) - 4 = -3x^2 -4\n\nNow that this formula is found, sub in 2;\n\n(f o g)(2) = -3(2)^2 - 4 = -3(4) - 4 = -12 - 4 = -16
853859
g(x) = x^5+5x^4+10x^3+10x^2+5x+1 = (x+1)^5\n`
508385
3-4(2)+2(2)+1=0\n2(-3)+1-[3-4(-3)]=-20\n3-4[2(1)+1]=-9\n\ng(x)=(x-4)/-5\n3-4[(x-4)/-5]=x\n{[3-4(x)]-4}/-5=x
406468
D = {x | x < -3 U x > 4 }
2302
(1) = x\n(2) = / \n(3) = x
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
321226
It's false.\n\nYou have to use order of operations. You do the stuff in the parenthesis first:\n\n(3-3) - 3 = 3- (3-3)\n(0)-3=3-(0)\n\nSo you get:\n-3=3\n\nSince -3 isn't equal to 3, it's not true.
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
951524
g(x)=f(2(x+4)\n=in f(x) whereverx is there sustitute 2(x+4)\nso the answer is 2x+8\n\n2.g(x)=f(-(x-2)\n=in f(x)=x^2 wherever there is x substitute\n-(x-2)\nso [-(x-2)]^2 will be your answer\nwhich is equal to\nx^2-4x+4\n\n3.g(x)=2f(-x)-5\nin f(x)=1/x,substituting\nf(-x)=-1/x\n2f(-x)=-2/x\nso -2/x-5 is your answer
1112189
1. g(x)=2x-5\nso, g(f(x))=2f(x)-5\nor 2 (x^2+2x+3)-5\nor 2x^2+4x+1\n\n2. \na) put f(x)=y\nso y=2^x+1\n\nnow take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph\n\nIf x=0, then y=2 (2^0=1)\nif x=1, then y=3\nif x=2, then y=5\n\nb) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..