Want this question answered?
X2 (5)2 = 25 (- 5)2 = 25 An identity, almost, as any number put into this expression will yield the same value.
(5×3)÷2
152
When you have an algebraic expression with variables in it, you need to know the the value of each variable. The value of the expression is found by replacing the variables with their values and simplifying. For example, evaluate 2xy-3x if x=5 and y=-1. 2xy-3x = 2*5*(-1) -3*5 = -10-15 = -25.
5
It is 2x + 5. The value of this expression will depend on the value of x: each different value of x will give a different value for the expression.
-3
X2 (5)2 = 25 (- 5)2 = 25 An identity, almost, as any number put into this expression will yield the same value.
8
10
(5×3)÷2
A constant is a term without a variable so don't ask this user.
152
-3
-3
When you have an algebraic expression with variables in it, you need to know the the value of each variable. The value of the expression is found by replacing the variables with their values and simplifying. For example, evaluate 2xy-3x if x=5 and y=-1. 2xy-3x = 2*5*(-1) -3*5 = -10-15 = -25.
3(5-6) + 2 (54) -5 =105