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Firdavs [7]
3 years ago
11

Which statement is NOT true regarding regular expression quantifiers? Group of answer choices They act on the preceding pattern

element They can operate on groups and character classes In order to use a plus sign "+" quantifier as a literal character, you must first 'escape' it using a backslash '\' character The question mark quantifier "?" will match the preceding element exactly one time
Computers and Technology
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

"The question mark quantifier "?" will match the preceding element exactly one time" is the correct answer to the given question .

Explanation:

The Quantifiers are defined as it determine how the several occurrences of a set of symbols, categories, or the characters should be found throughout the input to searching the matches.

  • The * quantifier in the regular expression corresponds the zero or more then zero to the previous item. it is represented by  {0,} quantifier
  • The + quantifier in the regular expression appears to fit in one or more times with the previous item. it equates with {1} quantifier.
  • The? Quantifier in regular expression compares zero or once of a previous item. It equates with {0,1}.
  • All the option are true regarding regular expression quantifiers so these option are incorrect according to the question
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Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?
frutty [35]

Answer:

The root is:

c=1.5708

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function  [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

%       - a y b are the initial and the final value respectively

%       - delta is the tolerance or error.

%           - c is the root

%       - yc = f(c)

%        - err is the stimated error for  c

ya = feval(f, a);

yb = feval(f, b);

if  ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for  k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if  yc == 0

 a = c;

 b = c;

elseif  yb*yc > 0

 b = c;

 yb = yc;

else

 a = c;

 ya = yc;

end

if  b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

 function_handle with value:

   @(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

   1.5708

err =

  5.8208e-11

yc =

 -3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

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3 years ago
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