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A: Please find the answer to the above question below:
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Q: What’s the correct answer
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A: Lets see the solution.
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A: The performance of a computer system can be defined as the efficiency with which a computer system m...
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Q: What’s the correct answer
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A:
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- (b) This code has a problem. What is it? How is it called? double x = 1.5, y = 3.4; double* p = &x; double* p2 = new double[12]; *p += y; p = &y; *p *= 2; for (int i{ 0 }; i < 12; ++i) { p2[i] = 1.6; } p2 = p; delete p2;In a try block, what kinds of statements may be found?1.Run show_bytes for this two values: 90 and -90 print the result (for int float and pointer). 2. F or x = 0x66 and y = 0x59 x & y = ? x && y = ? x | y = ? x || y = ? ~x | ~ y = ? !x || !y = ? x & !y = ? x && ~y = ? Written answers only (NO CODE) 3. A. Ox503c + 0x8 4. D. Ox50ea – Ox503C = = 1. Run show_bytes for two different values - Upload the output*
- Given f(x) = x+ x2/2+ x3/3+.. x"/n. Using while do, write a program to compute f(x) for ...... given x and n. The user should enter x and a positive integer n. Save your file as Q3.Java and upload it.which block completes the code to generate 'Remainder: 0.0' as output? (replacing the XXX) using: def Remainder(a, b): if b == 0: return(b % a) else: return(a % b)XXXfinally: print('Remainder:', z)JAVA, PLEASE WRITE IN 50 MIN Write a program that takes an integer from the user using Scanner and prints if the number is prime or not. If the user enters a non numeric input, the program must ask for another input, until a numeric input is entered. Note that you must use try-catch block.
- This Code print the first occurrences of x, Modify the code to print the last occurrences of x and print how many times it exists? ( with explanation for each line if possible) many thanks.What will be output to the screen when the following code is executed:def PrintMe(x): try: return s * 5 except: print("Error has occured") finally: print("Doing some cleanup")Trace the following code and give the output int a = 6, b = 8; int c = (a < b) ? a : b; int v = c; while (v%a != 0 || v%b != 0) v += c; cout << v;
