The sides of a rectangle are x and 3-2x.Express the rectangle's area as a function of x.Express the rectangle's perimeter as a function of x. Explain why x cannot equal 2.
Area= L x W
A= x (3-2x)
A= 2x^2 + 3x
Perimeter= 2(L + W)
P= 2(x+3-2x)
P= 2(3-x)
P= 6-2x
Would saying that there can't be a negative side (-2), be correct?
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You are mostly correct with your calculations of the area and perimeter of the rectangle. The area of a rectangle is indeed the product of its length and width, which in this case are x and (3-2x), respectively. Therefore, the area can be expressed as A = x(3-2x). Simplifying this expression, we get A = 2x^2 + 3x.
Similarly, the perimeter of a rectangle is computed by summing the lengths of all its sides. In this case, the lengths of the sides are x and (3-2x), so the perimeter can be expressed as P = 2(x + (3-2x)). Simplifying this expression, we get P = 2(3-x) = 6 - 2x.
Now, concerning the statement that x cannot equal 2, let's analyze why. Looking at the expressions for the area and perimeter, we can see that the value of x is involved in some denominator or cannot be evaluated at x=2, which would make a term in the formula undefined.
For the area, the expression 3-2x in the parentheses would simplify to 3-2(2) = 3-4 = -1, which would result in a negative value for the width. Since the sides of the rectangle cannot be negative, this means that x=2 is not a valid solution for the rectangle's dimensions.
Therefore, to ensure that the rectangle is well-defined and has positive dimensions, we should exclude x=2 from the allowed values of x.
A = x(3-2x) = 3x-2x^2
P is ok
your conclusion is correct, except that 3-2x = -1 when x is 2.