0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
What is the statement that stores the remainder of dividing the variable you by j in a variable named k?
k = you - int (you / j) * j; You can also use, if your language supports the modulus (%) operator... k = you % j;
Mathematical formulation of assignment problem?
Let there be n jobs which are to be assigned to n operators so that one job is assigned to only one operator. i = Index for job, i = 1, 2, … n j = Index for operators, j = 1, 2, … n Cij = Unit cost for assigning job 'i' to operator 'j' 1 if job i is assigned to operator j Xij = 0 Otherwise The objective is to minimize the total cost of assignment. If job I is assigned to operator 1, the cost is (C11X11). Similarly, for job 1, operator 2 the cost is (C12X12). The objective function is: Minimize = ∑ni=1 ∑nj=1 Cij Xij …(1) Since one job (i) can be assigned to any one of the operators, we have following constraint set: ∑ni=1 Xij = 1; for all j;j = 1, 2, ... n …(2) Similarly for each operator, there may be only one assignment of job. For this, the constraint set is: ∑ni=1 Xij = 1; for all i;i = 1, 2, ... n …(3) The non-negativity constraint is: Xij > 0 …(4) Minimize Z = ∑ni=1 ∑nj=1 Cij Xij Subject to ∑ni=1 Xij = 1; for all j;j = 1, 2, ... n ∑nj=1 Xij = 1; for all i;i = 1, 2, ... n Xij > for all i and all j. JANAK RAJ UPRETI TIMT COLLAGE RAJOURI GARDEN NEW DELHI 17/41 WEST PUNJABI BAGH S-6.. 9311864681 WITH ... DINESH GARG...9899558427
Write a program to overload preincrement and postincrement operator?
#include<iostream.h> #include<conio.h> class num { private: int a,b,c,d; publ;ic: num(int j,int k,int m,int l) { a=j; b=k; c=m; d=l; } void show(void); void operator++(); }; void num::show() { cout<<................................................................... } void num::operator++() {++a;++b;++c;++d; } main() { clrscr(); num X(3,2,5,7); cout<<"b4 inc of x"; X.show(); ++X; cout<<"after inc of x"; X.show(); return 0; }
Write the relationship between Impedance Resistance and Reactance using Operator J?
The operator, 'j', is used to indicate a phasor quantity that has been rotated, counterclockwise, through an angle of 90 degrees.So, if (for example) the operator is applied to a voltage U, then it is written as jU, which indicates means that the voltage lies along the vertical positive axis. A further operation by j, results in jjU, or j2 U, which means that the voltage lies along the horizontal negative axis -so, j2 is equivalent to -1U (or j is equivalent to the square-root of -1) or, simply, -U.A further operation by j, results in the voltage lying along the negative vertical axis: that is: jjjU = jj2U=-jU.But to answer your question, for inductive reactance (XL), we express impedance (Z) a follows: Z = R+jXL and, for capacitive reactance (XC), we express impedance as Z = R - jXC (the L and C should be subscripts).Strictly-speaking the operator j doesn't actually apply to impedance, because impedance, resistance, and reactance or not phasor (vector) quantities. So if we wanted to be strictly accurate, the above equations should be written as:(E/I) = (UR/I) + j (UL/I) and (E/I) = (UR/I)- j (UC/I)...but this is being rather pedantic.
Without using mod operator find the even or odd?
import java.util.*; class Demo { public static void main(String args[]) { System.out.println("Even Numbers Are "); for (int i=1;i<=20;i++) { System.out.print(" "+(2*i)); } System.out.println("\n \n Odd Numbers Are "); for(int j=0;j<=20;j++) { System.out.print(" "+((2*j)+1)); } } }
What has the author D J Bradley written?
D. J. Bradley has written: 'THE OPERATOR'
What has the author J Draper written?
J. Draper has written: 'The application of information theory to human operator problems'
What is the statement that stores the remainder of dividing the variable you by j in a variable named k?
k = you - int (you / j) * j; You can also use, if your language supports the modulus (%) operator... k = you % j;
What is i plus 1 operator in C plus plus?
The increment operator in C++ is defined by operator++(). All arithmetic types (char, int, float, double, long, short, long long and long double) and all pointer types except void* are supported by operator++(). User-defined types can overload operator++() to provide support where required. operator++() has two versions, prefix increment and postfix increment. Prefix increment behaves as one would expect, incrementing the operand by 1 and returning the modified value. Postfix increment also increments the operand, however, the return value is the pre-incremented value. To understand the difference between prefix and postfix, consider the following: int i = 0; int j = ++i; // i=1, j=1 int i = 0; int j = i++; // i=1, j=0
What is difference between equals and equals equals operators?
The equals operator is used for assigning a value to a variable. ex String name = "efu"; the equals equals operator is used for comparaing if the given value is equal to another value which is assigned for a variable. ex int i = 0; int j = 5; if( i == 0){ System.out.println(j+"can't be divided by"+i); }else{ System.out.println("answer is " +j/i); }
What education is needed to become a carnival worker?
None whatsoever. However, if you can, a class-J cotton candy device operator certification can help.
What has the author I J Maddox written?
I. J. Maddox has written: 'Elements of functional analysis' -- subject(s): Functional analysis 'Infinite matrices of operators' -- subject(s): Infinite matrices, Operator theory, Summability theory
What are different types of operators?
The different types of operators are as follows: *Arithmatic operator *Relational operator *Logical operator *Assignment operator *Increment/Decrement operator *Conditional operator *Bitwise operator *Special operator
Mathematical formulation of assignment problem?
Let there be n jobs which are to be assigned to n operators so that one job is assigned to only one operator. i = Index for job, i = 1, 2, … n j = Index for operators, j = 1, 2, … n Cij = Unit cost for assigning job 'i' to operator 'j' 1 if job i is assigned to operator j Xij = 0 Otherwise The objective is to minimize the total cost of assignment. If job I is assigned to operator 1, the cost is (C11X11). Similarly, for job 1, operator 2 the cost is (C12X12). The objective function is: Minimize = ∑ni=1 ∑nj=1 Cij Xij …(1) Since one job (i) can be assigned to any one of the operators, we have following constraint set: ∑ni=1 Xij = 1; for all j;j = 1, 2, ... n …(2) Similarly for each operator, there may be only one assignment of job. For this, the constraint set is: ∑ni=1 Xij = 1; for all i;i = 1, 2, ... n …(3) The non-negativity constraint is: Xij > 0 …(4) Minimize Z = ∑ni=1 ∑nj=1 Cij Xij Subject to ∑ni=1 Xij = 1; for all j;j = 1, 2, ... n ∑nj=1 Xij = 1; for all i;i = 1, 2, ... n Xij > for all i and all j. JANAK RAJ UPRETI TIMT COLLAGE RAJOURI GARDEN NEW DELHI 17/41 WEST PUNJABI BAGH S-6.. 9311864681 WITH ... DINESH GARG...9899558427
Write a program to overload preincrement and postincrement operator?
#include<iostream.h> #include<conio.h> class num { private: int a,b,c,d; publ;ic: num(int j,int k,int m,int l) { a=j; b=k; c=m; d=l; } void show(void); void operator++(); }; void num::show() { cout<<................................................................... } void num::operator++() {++a;++b;++c;++d; } main() { clrscr(); num X(3,2,5,7); cout<<"b4 inc of x"; X.show(); ++X; cout<<"after inc of x"; X.show(); return 0; }
Write the relationship between Impedance Resistance and Reactance using Operator J?
The operator, 'j', is used to indicate a phasor quantity that has been rotated, counterclockwise, through an angle of 90 degrees.So, if (for example) the operator is applied to a voltage U, then it is written as jU, which indicates means that the voltage lies along the vertical positive axis. A further operation by j, results in jjU, or j2 U, which means that the voltage lies along the horizontal negative axis -so, j2 is equivalent to -1U (or j is equivalent to the square-root of -1) or, simply, -U.A further operation by j, results in the voltage lying along the negative vertical axis: that is: jjjU = jj2U=-jU.But to answer your question, for inductive reactance (XL), we express impedance (Z) a follows: Z = R+jXL and, for capacitive reactance (XC), we express impedance as Z = R - jXC (the L and C should be subscripts).Strictly-speaking the operator j doesn't actually apply to impedance, because impedance, resistance, and reactance or not phasor (vector) quantities. So if we wanted to be strictly accurate, the above equations should be written as:(E/I) = (UR/I) + j (UL/I) and (E/I) = (UR/I)- j (UC/I)...but this is being rather pedantic.
Without using mod operator find the even or odd?
import java.util.*; class Demo { public static void main(String args[]) { System.out.println("Even Numbers Are "); for (int i=1;i<=20;i++) { System.out.print(" "+(2*i)); } System.out.println("\n \n Odd Numbers Are "); for(int j=0;j<=20;j++) { System.out.print(" "+((2*j)+1)); } } }
