Chapter 1, Problem 1.3P

Question

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Answer 1

Question

Chapter 1, Problem 1.3P

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

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Answer 2

Question

Chapter 1, Problem 1.3P

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

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Answer 3

Question

Chapter 1, Problem 1.3P

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

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Answer 4

Question

Chapter 1, Problem 1.3P

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

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Answer 5

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

Answer 6

Expert Solution & Answer

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.

Answer 7

Expert Solution & Answer

Answer 8

Expert Solution & Answer

Answer 9

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer 10

Answer to Problem 1.3P

Solution:

The required result is verified.

Answer 11

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value 1. If the expression is false, then it will have the logical value 0. The given expression “3==5+2” is false. Hence, the output will be 0.

Consider the value of a is 0 and the value of b is 1. The output will be,

b<a+11<0+1

The obtained output is false. Hence, it will have the logical value 0.

The given expression “10>5+2” is false. Hence, the output will be 1.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Add “2” to this logical value. The final output is 3.

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1&&2<4”. The expression becomes,

c==d−1&&2<40==1−1&&2<4

In this expression, the order of preference is “−, <, ==, &&”. Hence,

0==(1−1)&&2<40==0&&(2<4) (0==0)&&1 1&&1 1

Consider the value of c is 0 and the value of d is 1 in the expression “'c'=='d'-1||2>4”. The expression becomes,

c==d−1||2>40==1−1||2>4

In this expression, the order of preference is “−, <, ==, ||”. Hence,

0==(1−1)||2>4 0==0||(2>4)(0==0)||0 1||0 1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

xor(c==d−1,2>4)xor(0==1−1,2>4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2>4)xor(0==0,(2>4))xor((0==0),0)xor(1,0)1

Consider the value of c is 0 and the value of d is 1 in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

xor(c==d−1,2<4)xor(0==1−1,2<4)

In this expression, the order of preference is “−, <, ==, xor”. Hence,

xor(0==(1−1),2<4)xor(0==0,(2<4))xor((0==0),1)xor(1,1)0

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value 1. Now “(1>2)” is false. Hence, it will have the logical value 0.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 1

Matlab, Fourth Edition: A Practical Introduction to Programming and Problem Solving, Chapter 1, Problem 1.3P , additional homework tip 2

Therefore, the required variable is created.