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If R = {(x,y)| y = 3x + 2}, then R -1 = (1) x = 3y + 2 (2) y = (x – 2)/3 (3) {(x,y)| y = 3x + 2} (4) {(x,y)| y = (x – 2)/3} (5) {(x,y)| y – 2 = 3x} (6) {(x,y)| y = x/3 – 2}

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What is the first line of this proof? (1) Let x R. (2) Let x R -1. (3) Let (x,y) R. (4) Let (x,y) R -1.

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What is the first line of this proof? (1) Let x R.(2) Let x R -1. (3) Let (x,y) R.(4) Let (x,y) R -1. (5) Let (x,y) Dom(R) (6) Let (x,y) Dom(R -1 ) (7) Let x Dom(R) (8) Let x Dom(R -1 )

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Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then S R = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8)

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Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then R S = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8)

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Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then R R = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,1), (2,3), (1,3)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8)

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Let A = {1,2,3} with relations defined on A given by R = {(1,1), (2,3)} and S = {(1,3), (2,1), (3,2)}. Then S S = (1) { 1, 2, 2, 3} (2) { 3, 2, 3, 1} (3) {(2,1), (3,3)} (4) {(1,1), (2,1)} (5) {(1,2), (2,3), (3,1)} (6) {(1,3), (2,2)} (7) {(1,1), (2,3), (1,3), (2,1), (3,2)} (8)

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Assume f: A B and g: B C are functions. Which are defined? (1) f g (2) g f(3) both(4) neither.

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