The variable x starts with the value 0. The variable y starts with the value 5. Add 1 to x. Add 1 to y. Add x and y, and store the result in y. Display the value in y on the screen.

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Answer 1

Textbook Question

Chapter 1, Problem 1PTR

The variable x starts with the value 0.

The variable y starts with the value 5.

Add 1 to x.

Add 1 to y.

Add x and y, and store the result in y.

Display the value in y on the screen.

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9*…

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Answer 2

Textbook Question

Chapter 1, Problem 1PTR

The variable x starts with the value 0.

The variable y starts with the value 5.

Add 1 to x.

Add 1 to y.

Add x and y, and store the result in y.

Display the value in y on the screen.

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

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Answer 3

Textbook Question

Chapter 1, Problem 1PTR

The variable x starts with the value 0.

The variable y starts with the value 5.

Add 1 to x.

Add 1 to y.

Add x and y, and store the result in y.

Display the value in y on the screen.

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 1, Problem 1PTR

The variable x starts with the value 0.

The variable y starts with the value 5.

Add 1 to x.

Add 1 to y.

Add x and y, and store the result in y.

Display the value in y on the screen.

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

Answer 8

Expert Solution & Answer

Program Description Answer

The value of y will be displayed on the screen as “7”.

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1

y = 6

Step5: add the value of “x” and “y” and store the result in “y”.

y = x + y

y = 1 + 6

y = 7

Step6: display the output.

The value of “y” is 7

The output for the given statement:

The value of y is 7.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Program Description Answer

The value of y will be displayed on the screen as “7”.

Answer 12

Explanation of Solution

Analyzing the given English statement:

The program is to add the two values. Initially declare and initialize with the value to the variables. Then, increment both the variables by 1. Addition operation is performed, finally display the output.

The step by step operations are as follows:

Step1: the “x” variable stores the value of 0.

x = 0

Step2: the “y” variable stores the value of 5.

y = 5

Step3: the variable “x” is incremented by 1.

x = x + 1

x = 0 + 1

x = 1

Step4: the variable “y” is incremented by 1.

y = y + 1

y = 5 + 1