The legs of an isosceles triangle measures 2x^4+2x-1 units. The perimeter of the triangle is 5x^4-2x^3+x-3 units. Write a polynomial that represents the measure of the base of the triangle.
the answer is 46
I don't know how you actually found a value :/
2x^4 + 2x -1 add 2x^4 + 2x-1 because these are two legs of the triangle.
4x^4 + 4x -2
Now subtract this from the perimeter to get the length of the base. Actually with signed numbers, we don't subtract, we add the opposite.
5x^4-2x^3 + x - 3
+ -4x^4 -4x +2
x^4 -2x^3 -3x -1 This is the answer.
It is not possible to come up with a 46
To find the measure of the base of the isosceles triangle, we need to first understand that an isosceles triangle has two equal sides. Let's assume that both legs of the triangle measure 2x^4+2x-1 units. We can represent the length of the base of the triangle as "b."
Given that the perimeter of the triangle is 5x^4-2x^3+x-3 units, we can write an equation for the perimeter using the lengths of the triangle's sides:
Perimeter = Length of Leg 1 + Length of Leg 2 + Length of Base
Using the equation, we have:
5x^4-2x^3+x-3 = (2x^4+2x-1) + (2x^4+2x-1) + b
Since the triangle is isosceles, the sum of the lengths of the legs will be twice the length of one leg:
5x^4-2x^3+x-3 = 2(2x^4+2x-1) + b
Now, we can simplify the equation:
5x^4-2x^3+x-3 = 4x^4+4x-2 + b
Combining like terms, we get:
x^4-2x^3+x-1 = b
Therefore, the polynomial that represents the measure of the base of the isosceles triangle is x^4-2x^3+x-1 units.