s 0Miles (short), 150 Miles(medium), and 300 Miles (long).
Explanation:
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
![y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x](https://tex.z-dn.net/?f=y%27%20%3D%20-%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D%5Ccdot%20%5Csin%202x%20%2B%20%5Cfrac%7B2%7D%7Bx%7D%5Ccdot%20%5Ccos%202x)
![y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}](https://tex.z-dn.net/?f=y%27%20%3D%20%5Cfrac%7B2%5Ccdot%20x%20%5Ccdot%20%5Ccos%202x%20-%20%5Csin%202x%7D%7Bx%5E%7B2%7D%7D)
![dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx](https://tex.z-dn.net/?f=dy%20%3D%20%5Cleft%28%5Cfrac%7B2%5Ccdot%20x%20%5Ccdot%20%5Ccos%202x%20-%20%5Csin%202x%7D%7Bx%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20dx)
![dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)](https://tex.z-dn.net/?f=dy%20%3D%20%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ccdot%20%5Ccos%202%5Cpi%20-%5Csin%202%5Cpi%7D%7B%5Cpi%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%280.25%29)
![dy = \frac{1}{2\pi}](https://tex.z-dn.net/?f=dy%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D)
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
To learn more on differentials: brainly.com/question/24062595
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Answer:
Explanation gives the answer
Explanation:
% Using MATLAB,
% Matlab file : fieldtovar.m
function varargout = fieldtovar(S)
% function that accepts single structure as input, assigning each
% of the field values to user-defined variables
fields = fieldnames(S); % get the field names of the input structure
% check if number of user-defined variables and number of fields in
% structure are equal
if nargout == length(fields)
% if equal assign each value of structure to user-defined varable
for i=1:nargout
varargout{i} = getfield(S,fields{i});
end
else
% if not equal display an error message
error('The number of output variables does not equal the number of fields');
end
end
%This brings an end to the program
A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
A broken white line on the pavement show that the adjacent lane is traveling in the same direction and passing is permitted.
<h3>What does pavement markings show?</h3>
Pavement markings are known to be tools that are used to pass infor or messages to roadway users.
Note that they tell the part of the road that one need to use, give information about conditions ahead, and others
Note that A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
Learn more about pavement markings from
brainly.com/question/10179521
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