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What else can I help you with?
How do you get the constant in algebraic expression?
By itself you cannot. You need to have a set of values for the variables and the expression which you need to solve.
How do you convert the square root of x into an expression?
Write it as sqrt(x) or x1/2 or x0.5
What is equivalent radical expression for the exponential expression below?
To convert an exponential expression to an equivalent radical expression, you can use the relationship ( a^{m/n} = \sqrt[n]{a^m} ). For example, the expression ( x^{3/2} ) can be rewritten as ( \sqrt{x^3} ) or ( \sqrt{x^3} = x^{3/2} ). If you provide a specific exponential expression, I can give you its corresponding radical form.
Why is the difference between expression and equation?
Simply put, an expression does not need an equal sign, while an equation does. Equation: x+y=z Expression: x+y
How do you translate an expression?
If you are talking about language translation, then translating expressions requires some creativity. You would need to understand the core intention of the expression and then choose a similiar expression in the target language.
How to convert prefix to postfix?
convert to perfixed to postfixed
Prefix to postfix conversion using C programming?
#include<stdio.h> #include<conio.h> #include<string.h> char symbol,s[10]; int F(symbol) { switch(symbol) { case '+': case '-':return 2; case '*': case '/':return 4; case '^': case '$':return 5; case '(':return 0; case '#':return -1; default :return 8; } } int G(symbol) { switch(symbol) { case '+': case '-':return 1; case '*': case '/':return 3; case '^': case '$':return 6; case '(':return 9; case ')':return 0; default: return 7; } } void infix_to_postfix(char infix[],char postfix[]) { int top=-1,j=0,i,symbol; s[++top]='#'; for(i=0;i<strlen(infix);i++) { symbol=infix[i]; while(F(s[top])>G(symbol)) { postfix[j]=s[top--]; j++; } if(F(s[top])!=G(symbol)) s[++top]=symbol; else top--; } while(s[top]!='#') { postfix[j++]=s[top--]; } postfix[j]='\0'; } void main() { char infix[30],postfix[30]; clrscr(); printf("Enter the valid infix expression\n"); scanf("%s",infix); infix_to_postfix(infix, postfix); printf("postfix expression is \n %s", postfix); getch(); }
Why do compilers convert infix expressions to postfix?
people almost exclusively use infix notation to write mathematical expressions, computer languages almost exclusively allow programmers to use infix notation. However, if a compiler allowed infix expressions into the binary code used in the compiled version of a program, the resulting code would be larger than needed and very inefficient. Because of this, compilers convert infix expressions into postfix notation expressions, which have a much simpler set of rules for expression evaluation. Postfix notation gets its name from the fact that operators in a postfix expression follow the operands that they specify an operation on. Here are some examples of equivalent infix and postfix expressions Infix Notation Postfix Notation 2 + 3 2 3 + 2 + 3 * 6 3 6 * 2 + (2 + 3) * 6 2 3 + 6 * A / (B * C) + D * E - A - C A B C * / D E * + A C * - Where as infix notation expressions need a long list or rules for evaluation, postfix expressions need very few.
Write an algorithm in c to convert an infix expression to a postfix expressionexecute your algorithm with the following infix expression as your input. m nk pgh a bc?
An algorithm can not be written with the following infix expression without knowing what the expression is. Once this information is included a person will be able to know how to write the algorithm.
Sample program of postfix to infix?
You can convert from postfix to infix through the use of stacks. Consider the following expression conversion:54+67*+ -> ((5+4)+(6*7))The way this can be achieved is that whenever you encounter an operator, pop the last two expressions and join them using the operator. Remember to include the open braces before the first expression and a close braces after the second expression. Check the given link below for the program:
Convert infix to prefix to postfix?
(a + b) * c / ((x - y) * z)
What is the difference between prefix and postfix increment operator in c plus plus?
Both the prefix and the postfix increment operators increment the operand. The difference is what is the value of the expression during the evaluation of the expression. In the prefix form, the value is already incremented. In the postfix form, it is not. int a = 1; int b = ++a; // both a and b are now equal to 2 int a = 1; int b = a++; // a is equal to 2 and b is equal to 1
What is a 'post fix expression' in java programming?
Postfix expressions are expressions where the operator is at the end of the expression. These include the "++" (increment) and "--" (decrement) operators. Most Java expressions use in-fix notation (e.g. "a + b") but the increment and decrement operators can be postfix ("e.g. "a++" to increment variable a) or even prefix (e.g. "++a").
Evaluate a post fix expression?
Okay, here is a postfix expression: 3 4 * 5 6 * + the evaluation: 3*4 + 5*6 12 + 30 42
Minimum size of stack to evaluate postfix expression?
Scan the postfix expression from left to right and count the number of values and the number of operators. The maximum value of their difference is the required stack size. Eg: 1 2 3 + 4 + * 1 2 3 2 3 2 1 The maximum is 3.
Which data structure is needed to convert infix notations to post fix notations?
stack is the basic data structure needed to convert infix notation to postfix
How do you convert infix to postfix without using data structures?
Without data-structures you cannot even store expressions, let alone convert or evaluate them.
