<u>Swap files </u>contain(s) remnants of word processing documents, e-mails, Internet browsing activity, database entries, and almost any other work that has occurred during past Windows sessions.
Explanation:
I order to simulate extra space an operating system like windows use hard disk space in order to stimulate extra memory.When a computer system is running low with the memory space the swap file swaps a section of RAM onto the hard disk in order to obtain a free memory space.
This Process at times results in slowing down of the computer computer considerably.
<u>The combination of RAM and swap files is known as virtual memory.</u>
It is due to the use of virtual memory that our computer is able to run more programs than it could run in RAM alone.
Answer:
percentage
Explanation:
percentage, character, interger
Answer:
Check the explanation
Explanation:
<em>Cube.m:</em>
mass = input("Enter the mass of cube [kilograms]: ");
if(mass<=0)
disp("Error: Mass must be greater than zero grams")
else
fprintf("The length of one side of cube is %.2f inches",2.7*mass);
end
<em>Output1</em>
octave:2> source ( Cube.m Enter the mass of cube [kilograms]: octave:2>-3 Error: Mass must be greater than zero grams
<em />
Answer: a)To have the computer close the current form when the user clicks the Exit button, the Me.Close() statement should be entered in a button’s Click event procedure
Explanation: Me.close() statement is used for the assuring that execution of the program has stopped along with stopped computer event.This statement is puts the current form into the form.closing.
Other statements are incorrect because title bar can be vanished from application,size of the form can be change while execution of application and both minimize and maximize button are in the form of set and can be removed togather only .Thus, the correct option is option(a)
Answer:
1 PROCESSOR :
(1 × 2.56 × 10^9) + (12 × 1.28 × 10^9) + (5 × 2.56 × 10^8) / 2 GHz = 9.6 s
2 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×2 + (5 × 2.56 × 10^8) / 2 GHz = 7.04 s
Speed -up is 1.36
4 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×4 + (5 × 2.56 × 10^8) / 2 GHz = 3.84 s
Speed -up is 2.5
5 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×8 + (5 × 2.56 × 10^8) / 2 GHz = 2.24 s
Speed -up is 4.29
Explanation:
The following formula is used in this answer:
EXECUTION TIME = CLOCK CYCLES / CLOCK RATE
Execution Time is equal to the clock cycle per clock rate
