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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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The content of this lesson will be on number (not shape). Writing proofs can be very difficult, especially for those people who rush maths (lazy people)

The content of this lesson will be on number (not shape). Writing proofs can be very difficult, especially for those people who rush maths (lazy people)

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