I need help please, thank you

What can be designed to create annoying glitches or destroy data

Yanka [14]

Viruses can cause this.

3 0

3 years ago

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

5 0

3 years ago

Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type

sattari [20]

Solution :

#include $$$$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal $\text{To}$ Binary(char* hexdec)

{

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$ (sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

break;

case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

break;

case 'F':

case 'f':

strcat(string, "1111");

break;

default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

//Input 2 numbers p and n.

int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

4 0

3 years ago

Whereas enterprise network convergence focuses on the consolidation of traditionally distinct voice, video, and data communicati

Vinil7 [7]

Answer: Unified communication

Explanation: Unified communication is the communication technique in which merges various communication routines into a individual business.It works as the phone system which helps in increment of productivity in an organizational business.

The unifying of voice,data/information,video etc is done for optimizing and improving the business that also results in faster communication rates, secure communication etc.

4 0

3 years ago

A domestic air carrier airplane lands at an intermediate airport at 1815Z. The latest time it may depart without a specific auth

serious [3.7K]

Answer:

A domestic air carrier airplane lands at an intermediate airport at 1815Z. The latest time it may depart without a specific authorization from an aircraft dispatcher is <u>1915Z (1 hour)</u>.

Explanation:

Under the domestic operations, an airplane landed on intermediate airport can remain their for not more than one hour so the time would me 1 hour.

Here the time represented by 1815Z and 1915Z is in Zulu Time Zone as depicted from letter "Z". The first two digits represent the hour (0-24) and the next two represent the minutes (0-59).

Here the landing time is 6:15 pm while departing time is after one hour that is 7:15 pm (1915Z).

i hope it will help you.

6 0

3 years ago