MATLAB: AN INTRODUCTION WITH APPLICATIO
MATLAB: AN INTRODUCTION WITH APPLICATIO
6th Edition
ISBN: 9781119626596
Author: GILAT
Publisher: WILEY
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Chapter 1, Problem 1P
To determine

(a)

To solve:

The equation (5 197+ 2.53)2.

Expert Solution
Check Mark

Answer to Problem 1P

Solution:

The value of equation (5 197+ 2.53)2 is 320.7937.

Explanation of Solution

Solve the given equation.

(5 19 7+ 2.5 3)2(52.7143+15.625)2320.7937

Write the MATLAB script that calculates the equation (5 197+ 2.53)2.

MATLAB Code:

(5-(19/7)+(2.5^3))^2

% Calculate the value of the equation

Save the MATLAB script with name, chapter1_56830_1_Pa.m in the current folder. Execute the script by typing the script name at the command window to get the answer for (5 197+ 2.53)2.

Result:

MATLAB: AN INTRODUCTION WITH APPLICATIO, Chapter 1, Problem 1P , additional homework tip  1

Conclusion:

Therefore, the value of equation (5 197+ 2.53)2 is 320.7937.

To determine

(b)

To solve:

The equation 7×3.1+1205155/3.

Expert Solution
Check Mark

Answer to Problem 1P

Solution:

The value of equation 7×3.1+1205155/3 is 67.3421.

Explanation of Solution

Solve the given equation.

7×3.1+ 1205155/321.7+2.19191.23367.3421

Write the MATLAB script that calculates the equation 7×3.1+1205155/3.

MATLAB Code:

7*3.1+((120^(1/2))/5)-(15^(5/3))

% Calculate the value of the equation

Save the MATLAB script with name, chapter1_56830_1_Pb.m in the current folder. Execute the script by typing the script name at the command window to get the answer for 7×3.1+1205155/3.

Result:

MATLAB: AN INTRODUCTION WITH APPLICATIO, Chapter 1, Problem 1P , additional homework tip  2

Conclusion:

Therefore, the value of equation 7×3.1+1205155/3 is 67.3421.

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Chapter 1 Solutions

MATLAB: AN INTRODUCTION WITH APPLICATIO

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