5. what is printed by the following code fragment?

int x = 5, y = 3, *p = & x, *q = & y;

std: : cout < < "x = " < < x < < ", y = " < < y < < '\n';

x = y;

std: : cout < < "x = " < < x < < ", y = " < < y < < '\n';

x = 7;

std: : cout < < "x = " < < x < < ", y = " < < y < < '\n';

*p = 10;

std: : cout < < "x = " < < x < < ", y = " < < y < < '\n';

p = q;

*p = 20;

std: : cout < < "x = " < < x < < ", y = " < < y < < '\n';

Output:

x = 5, y = 3

x = 3, y = 3

x = 7, y = 3

x = 10, y = 3

x = 10, y = 20

Explanation:

Here x and y is variables are assigned with value of 5 and 3, the pointers p, q are assigned to base addresses of y and x. While executing the first cout statement, the values of 5,3 is printed for x,y.  Then next statement is   x = y which means y value is assigned to x and the second cout statement is executed, so here the values of x,  y  are  equal to 3  respectively.

Since the variable p points to base address of x and its value is be changed 10, so x = 10.Now when the third cout statement is executed so  x  changed to 10 and y is 3.Now the value of q(base address of y) is assigned to p and the value of p(using dereferencing operator *) gets assigned the value of 20.So while executing the fourth cout statement, the values of x and y are 10 and 20 respectively. The output is given below.

given the options, i would say your best bet is c. the modem supports up to 56 kbps, which is too low for video streaming. dial up modems use actual phone lines, which only can go up to 56 kbps as that's more than sufficient for traditional phone calls. this is also the reason why when a dial-up modem connects, you hear all those sounds. it's literally using the phone line to establish a connection.

they really not common the os x is just a better version of the os like in iphone theirs xr and their xs