To find the value of y in the expression 15y-4 = 41, you need to isolate y. Start by adding 4 to both sides of the equation to get 15y = 45. Then, divide both sides by 15 to solve for y, which equals 3. Therefore, the value of y in the expression 15y-4 when it equals 41 is 3.
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
It is a quadratic expression which when factored is (2y+3)(y+5)
y² - y all depends on the value of y. If you want to factor the expression, then we have y(y - 1)
8
The value is 1 if (x, y) = (4, 3) and 0 otherwise.
Depends what the expression is. Just plug in 8 for y in the expression and that'll do it for you.
Question is y+12=? where value of y is 29. y+12 = 29+12 = 41 The answer is 41.
In mathematics, "y plus 3" is an algebraic expression representing the sum of a variable 'y' and the constant 3. This expression cannot be simplified further unless the value of 'y' is known. If 'y' is a known value, then the sum can be calculated by adding 3 to that value. If 'y' is unknown, then the expression "y + 3" remains as is, representing a sum of 'y' and 3.
It is an algebraic expression that can be simplified depending on the plus or minus value of 4
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
They are two terms of an algebraic expression which can be arranged as : y-2a
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.
It is a quadratic expression which when factored is (2y+3)(y+5)
y² - y all depends on the value of y. If you want to factor the expression, then we have y(y - 1)
That depends on the value of x and y. As an expression, "x + y" can't be simplified.
21
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.