int main() { if(var1>50) { num = 20.0 } else { if(var1>35) ---- Line 7 { num= 35.0; } else if (var1<80) { num=60.0; } else { num=40.0; } } What should be the outcome of the condition in line 7 in order for the value of num to be equal to 60.0?
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int main()
{
if(var1>50)
{
num = 20.0
}
else
{
if(var1>35) ---- Line 7
{
num= 35.0;
}
else if (var1<80)
{
num=60.0;
}
else
{
num=40.0;
}
}
What should be the outcome of the condition in line 7 in order for the value of num to be equal to 60.0?
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- When you borrow money to buy a house, a car, or for some other purpose, you repay the loan by making periodic payments over a certain period of time. Of course, the lending company will charge interest on the loan. Every periodic payment consists of the interest on the loan and the payment toward the principal amount. To be specific, suppose that you borrow $1,000 at an interest rate of 7.2% per year and the payments are monthly. Suppose that your monthly payment is $25. Now, the interest is 7.2% per year and the payments are monthly, so the interest rate per month is 7.2/12 = 0.6%. The first months interest on $1,000 is 1000 0.006 = 6. Because the payment is $25 and the interest for the first month is $6, the payment toward the principal amount is 25 6 = 19. This means after making the first payment, the loan amount is 1,000 19 = 981. For the second payment, the interest is calculated on $981. So the interest for the second month is 981 0.006 = 5.886, that is, approximately $5.89. This implies that the payment toward the principal is 25 5.89 = 19.11 and the remaining balance after the second payment is 981 19.11 = 961.89. This process is repeated until the loan is paid. Write a program that accepts as input the loan amount, the interest rate per year, and the monthly payment. (Enter the interest rate as a percentage. For example, if the interest rate is 7.2% per year, then enter 7.2.) The program then outputs the number of months it would take to repay the loan. (Note that if the monthly payment is less than the first months interest, then after each payment, the loan amount will increase. In this case, the program must warn the borrower that the monthly payment is too low, and with this monthly payment, the loan amount could not be repaid.)C++ programming It has been observed by management that some faculty member at the university demonstrates a lackadaisical attitude toward work. They seldom go to class yet at the end of each month they receive full salary. Management has decided that GHs 200, 300, 400, and 500 will be deducted from a faculty’s salary if he/she offends once, twice, third time and forth time respectively in a month. This means that if a faculty member offends once in a month GHs 200 will be deducted, if a faculty member offends twice in a month, GHs 500 (i.e. 200+300) will be deducted, if a faculty member offends three times in a month GHs 900 (i.e.200+300+400), will be deducted and if a faculty member offends four times in a month GHs 1,400 (i.e. 200+300+400+500). Assuming all faculty members are on a flat salary rate of GHs 2500. a. Write a program to request for the names of four faculty members and number times he/she has absented him/herself from class. One of them should have absented him/herself…نقطة واحدة Let A = {a; b; c; d} and R= {(a; a); (b; c); (c; b); (d; d)} then R is Transitive Equivalent not transitive
- Language C++ It has been observed by management that some faculty member at the university demonstrates a lackadaisical attitude toward teaching. They seldom go to class yet at the end of each month they receive full salary. Management has decided that GHs 200, 300, 400, and 500 will be deducted from a faculty’s salary if he/she offends once, twice, third time and forth time respectively in a month. This means that if a faculty member offends once in a month GHs 200 will be deducted, if a faculty member offends twice in a month, GHs 500 (i.e. 200+300) will be deducted, if a faculty member offends three times in a month GHs 900 (i.e. 200+300+400), will be deducted and if a faculty member offends four times in a month GHs 1,400 (i.e. 200+300+400+500). Assuming all faculty members are on a flat salary rate of GHs 2000. a. Write a program to request for the names of four faculty members and number times he/she has absented him/herself from class. One of them should have absented…For the C++ program show on the next page, answer the questions that follow. The names g, j, x, and y are declared multiple times. Distinguish uses of these names using line numbers where they were declared as subscripts. a. At point A in the program, what names are in scope? b. At point A in the program, what variables are live? c. At point B in the program, what names are in scope? d. At point B in the program, what variables are live?int x1 = 66; int y1 = 39; int d; _asm { } mov EAX, X1; mov EBX, y1; push EAX; push EBX; pop ECX mov d, ECX; What is d in decimal format?
- #include<iostream>using namespace std;void main(){double pi = 0, denominator = 1;int counter = 999999;for (int x = 0; x < counter; x++){if (x % 2 != 0){pi = pi - (1 / denominator);}else{pi = pi + (1 / denominator);}denominator = denominator + 2;}pi = pi * 4;cout << " So the computed value of a PI is = " << pi << endl;cout << " ";//return 0;system("pause");} Note: This a program called ComputePI to compute the value of π Tutor just have to Modify This program to use nested-if (if ... else if ... else if ... else) instead. Explain by applying a double line comment#include Servo servo; int const trigPin = 9; int const echoPin = 8; #define red 5 #define green 6 void setup() { pinMode(trigPin, OUTPUT); pinMode(echoPin, INPUT); pinMode(red,OUTPUT); pinMode(green,OUTPUT); servo.attach(10); } void loop() { int duration, distance; digitalWrite(trigPin, HIGH); digitalWrite(trigPin, LOW);// Measure the pulse input in echo pin duration = pulseIn(echoPin, HIGH);// Distance is half the duration devided by 29.1 (from datasheet) distance = (duration/2) / 29.1;// if distance less than 0.5 meter and more than 0 (0 or less means over range) if (distance = 0) { servo.write(60); digitalWrite(green,HIGH); digitalWrite(red,LOW); delay(200); } else { servo.write(10); digitalWrite(red,HIGH); digitalWrite(green,LOW); delay(80); } } TASK: create a flowchart using this code, showing the input, output and the conditions on the loop. Lastly, the flowchart must return if ever the conditions have met or notint j; switch (i) { case 0: j = 2*i; break; case 1: j = -i; break; default: j = -2*i; } What is j if i = 1? What is j if i = -1?
- Problem Statement: The mathematician Conway imagined a game, called game of life, which considered cells that are susceptible to reproduce, disappear, or survive when they obey certain rules. These cells are represented by elements on a grid of squares, where a grid has an arbitrary size. Thus, each cell (except those on the boundaries of the grid) is surrounded by eight squares that contain other cells. The rules are stated as follows: 1. Survival: Each cell that has two or three adjacent cells survives until the next generation.2. Death: Each cell that has at least four adjacent cells disappears (or dies) by overpopulation. Also, each cell that has at most one adjacent cell dies by isolation.3. Birth: Each empty square (i.e., dead cell) that is adjacent to exactly three cells gives birthto a new cell for the next generation.It is worth noting that all births and deaths occur at the same time during a generation. Write a program that simulates this game and displays all successive…Please explain the code below #include <cstdio> #include <math.h> using namespace std; int main ( ) { // radius const int RAD = 10; // tolerance const int TOL = 5; for(int x = -RAD; x <= RAD; x++){ for(int y = -RAD; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " "); printf(abs(eq) < TOL ? "*" : " "); } for(int y = -RAD+4; y <= RAD; y++){ int eq = x*x + y*y - RAD * RAD; printf(abs(eq) < TOL ? "*" : " ");…Void Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program is