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What is an expression for subtract 16?
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
2y squared plus 13y plus 15?
It is a quadratic expression which when factored is (2y+3)(y+5)
What is y squared minus y equal?
y² - y all depends on the value of y. If you want to factor the expression, then we have y(y - 1)
What is value of the expression when x 5 and y -3?
8
What is the value of the expression x equals 4 and y equals 3?
The value is 1 if (x, y) = (4, 3) and 0 otherwise.
What is the value of the expression when y equals 8?
Depends what the expression is. Just plug in 8 for y in the expression and that'll do it for you.
What is x-y-86(2x 4)?
It is an algebraic expression that can be simplified depending on the plus or minus value of 4
What is an expression for subtract 16?
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
What is the answer to -2a plus y?
They are two terms of an algebraic expression which can be arranged as : y-2a
What is the minimum value of 2x plus 2y?
That completely depends on the values of 'x' and 'y'. The value of that expression at any instant is exactly double the sum of 'x' and 'y', but as soon as either of them changes, the value of the expression likewise instantly changes. The only possible statement is that the value of the expression is minimum when the sum of 'x' and 'y' is minimum.
2y squared plus 13y plus 15?
It is a quadratic expression which when factored is (2y+3)(y+5)
What is y squared minus y equal?
y² - y all depends on the value of y. If you want to factor the expression, then we have y(y - 1)
What is the sum of x plus y?
That depends on the value of x and y. As an expression, "x + y" can't be simplified.
When a pre increment is followed by a post increment?
You cannot follow a prefix increment with a postfix increment. int x = 40; int y = ++x++; // error This has to be an error because the postfix operator has higher precedence than the prefix operator. Thus the above code is equivalent to: int x = 40; int y = ++(x++); // error The expression (x++) is evaluated first. This increments x to 41 but the expression evaluates to 40 (the original value of x). Thus the prefix operator is essentially trying to evaluate the expression ++40 rather than ++x. This cannot work because the value 40 is not a modifiable lvalue. It has to be modifiable because we want to increment the value and return a reference to the modified value. But there's nothing to refer to here. The value is temporary and will fall from scope immediately after we use it. That is, the prefix operator may well be able to increment the temporary value to 41, but that value immediately falls from scope. With nothing to refer to, the prefix expression cannot be evaluated. The only way we can use both operators together is if we reverse the precedence using parenthesis: int x = 40; int y = (++x)++; Now the prefix operator is evaluated first, returning a reference to x which (now) holds the value 41. The postfix operator then increments x to 42 but returns the original value of 41 which is then assigned to y. Thus when all statements have been executed, y holds the value 41 while x holds the value 42.
What is the answer to 13 y-8?
21
What is value of the expression when x 5 and y -3?
8
What is a indendent variable?
You may be referring to an independent variable, which is a quantity in a mathematical expression whose value does not depend upon variations in other quantities in that mathematical expression. For example, in the expression y=2x the variable y depends on the value of x and so is not independent, while in the expression z=x+y the variable z is dependent, but the variables x and y are both independent.
