(7-3)^2 plus (7-1)^2 = 16+36
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∙ 2016-10-01 14:01:34You don't "solve" an expression. You can solve an equation; an expression can be simplified or otherwise manipulated, and if you know the value of "x" (in this case), you can evaluate its value.
The answer is b+1. Therefore the algebraic expression for this is b+1
You have to substitute a value for the letter variable in the expression. This is what we call evaluating the algebraic expression. An example would be 3x+1=7, when x=2.
14
When simplified it is just: 1
You don't "solve" an expression. You can solve an equation; an expression can be simplified or otherwise manipulated, and if you know the value of "x" (in this case), you can evaluate its value.
1/2w+2
b + 1 is b plus 1 as an algebraic expression.
It evaluates to: 12+(-1)+(-1)+3 = 13
Put n = 1, 2, 3, 4 etc in the expression 5n + 2 and evaluate to get the sequence.
3(10x + y) + 2x + 7y (multiply the 3 through the parentheses then combine like terms) = 30x + 3y + 2x + 7y = 32x + 10 y Evaluate an expression for some values of x and y, means to substitute those values into the expression, such that when x = 3 and y = 1 the value of the given expression is 106: 32x + 10y = 32(3) + 10(1) = 96 + 10 = 106
2!
xy + x + y + 1 is an expression, not an equation. An expression cannot be solved.
You have to substitute a value for the letter variable in the expression. This is what we call evaluating the algebraic expression. An example would be 3x+1=7, when x=2.
You cannot. What you have in the question is an expression. An expression cannot be proven. You need an equation (or inequality).
The answer is b+1. Therefore the algebraic expression for this is b+1
You have to substitute a value for the letter variable in the expression. This is what we call evaluating the algebraic expression. An example would be 3x+1=7, when x=2.