#1) Graph I; #2) Graph O; #3) Graph C; #4) Graph E; #5) Graph N; #6) Graph G; #7) Graph M; #8) Graph H; #9) Graph T; #10) Graph S; #11) Graph P; #12) Graph A

He acquired a pane in his stomach.

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Explanation:

#1) We desire a graph wright here -4 is circled and not filled in, and -1 is circled and filled in; from -4, the graph goes to the right and also from -1 it goes to the left, meeting in between the 2 numbers and not extending additionally. This is graph I.

#2) Solving the first inequality, we subtract 4 from each side; this provides us x > -1. Solving the second inehigh quality, we add 2 to each side; this offers us x

#3) We want a graph where -2 is circled and filled in and also 3 is circled and not filled in; from -2, the graph is shaded left and from 3 the graph is shaded best. This is graph C.

#4) Solving the initially inehigh quality, divide both sides by -3; this offers us t

#5) Solving the first inehigh quality, subtract 5 from each side; this offers us 2n > -4. Divide both sides by 2; this gives us n>-2. Solving the second inehigh quality, subtract 4 from each side; this provides us 3n>3. Divide both sides by 3, and we have actually n>1. This is and; this implies we want an inehigh quality reflecting n>-2 and n>1. It has to be true for both numbers; this means we desire a graph of n>1. We desire the number 1 circled and also not filled in, and also the graph shaded to the right; this is graph N.

#6) Solving the first inequality, subtract 9 from each side; this gives us -4u>-8. Divide both sides by -4, offering us u

#7) Solving the first inehigh quality, subtract 20 from each side; this provides us 12≤3x, or 3x≥12. Divide both sides by 3 and we have actually x≥4. Solving the second inetop quality, subtract 1 from each side; this offers us 16>-8x, or -8x-2. We want a graph wbelow 4 is circled and also filled in and -2 is circled and also not filled in; since this is "or" we include both answers. This indicates everything from -2 to the best is shaded; this is graph M.

#8) Solving the initially inehigh quality, subtract 8 from each side; this provides us -2k-3. Solving the second inehigh quality, subtract 1 from each side; this offers us 3k

#9) Solving the first inetop quality, divide both sides by 5; this gives us w+4≥1. Subtract 4 from each side and also we have w≥-3. Solving the second inetop quality, divide both sides by 2; this provides us w+4

#10) Solving the initially inehigh quality, divide both sides by 3; this offers us 6-y≤2. Subtract 6 from both sides; this gives us -y≤-4. Divide both sides by -1 and also we have actually y≥4. Solving the second inequality, subtract 6 from both sides; this provides us -y≥2. Divide both sides by -1 and also we have y≤-2. Tright here is no means to graph whatever that is larger than a positive and less than an adverse at the very same time, so this is graph S, the empty collection.

#11) Solving the first inehigh quality, subtract 2x from each side; this gives us x-9. Divide both sides by 3 and we have x>-3. We want a graph wbelow -3 is circled and also not filled in, and everything on both sides is shaded; this is graph P.

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#12) To fix initially equation, multiply both sides by 2; this provides us x≤-4. To deal with the second equation, multiply both sides by -2; this gives us x≤0. We desire whatever less than or equal to -4 or much less than or equal to 0; this implies we want everything smaller sized than 0, with 0 filled in. This is graph A.