how to use barcode in c#.net Worked-Out Solutions to Exercises: s 11 to 19 in Software

Encoder QR in Software Worked-Out Solutions to Exercises: s 11 to 19

11
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6 We have standard names for the sets of rational and irrational numbers: Q and S, respectively These sets are disjoint; they have no elements in common If x is an element of Q and y is an element of S, then x is never equal to y In logical form along with set notation, we can write [(x Q) & (y S)] x y 7 We can write the statement as mathematical verse by reading it out loud and taking careful note of each symbol Here s the logical statement again, for reference ( a, b, c) : [(a b) & (b c)] (a = c) When we break up the statement into parts and write them down on separate lines, we come up with the following: For all a, b, and c: If a is larger than or equal to b, and b is smaller than or equal to c, then a is equal to c This little poem might be cute, but it doesn t state a valid mathematical law Suppose that a = 5, b = 3, and c = 7 In that case, a is larger than or equal to b and b is smaller than or equal to c However, a is not equal to c 8 Our task is to simplify the equation to a form where x appears all by itself on the left side of the equality symbol, and a plain numeral appears all by itself on the right Here s the equation again, for reference: x + 4 = 2x We can subtract x from both sides, getting x + 4 x = 2x x which simplifies to 4=x We can reverse the order to get the solution in its proper form: x=4 The original equation holds true only when x is equal to 4
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622 Worked-Out Solutions to Exercises: s 11 to 19
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9 We must simplify the inequality so that y appears all by itself on the left side of the not equal symbol, and a plain numeral appears all by itself on the right Here s the inequality again, for reference: y /2 4y + 7 First, let s multiply through by 2 That gives us (y /2) 2 (4y + 7) 2 which multiplies out to y 8y + 14 Now, let s subtract 8y from each side That gives us y 8y 8y + 14 8y which simplifies to 7y 14 We can divide this through by 7 to get ( 7y)/( 7) 14/( 7) which simplifies to y 2 The original inequality holds true for all values of y except 2 10 We must simplify the inequality so that z appears all by itself on the left side of the smaller than or equal symbol, and a plain numeral appears all by itself on the right Here s the inequality again, for reference: z /( 3) 6z + 6 Let s multiply through by 3, remembering that we must reverse the sense of the inequality whenever we multiply through by a negative That gives us [z /( 3)] ( 3) (6z + 6) ( 3) which simplifies to z 18z 18
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If we add 18z to each side, we get z + 18z 18z 18 + 18z which simplifies to 19z 18 We finish by dividing each side by 19 That leaves us with z 18/19 The original inequality holds true for all values of z larger than or equal to 18/19
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1 See Table B-1 2 See Table B-2 3 See Table B-3
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Table B-1
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Statements 4x + 4 = 2x 2 2x + 4 = 2 2x + 6 = 0
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Solution to Prob 1 in Chap 12
Reasons This is the equation we are given Subtract 2x from each side Add 2 to each side
Table B-2
Statements x/3 = 6x + 2 x = 3(6x + 2) x = 18x + 6 17x = 6 17x 6 = 0 17x + 6 = 0
Solution to Prob 2 in Chap 12
Reasons This is the equation we are given Multiply through by 3 Distributive law applied to the right side Subtract 18x from each side Subtract 6 from each side Multiply through by 1 and apply the distributive law to the right side, obtaining a more elegant equation
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