Sides of a triangle in terms of $x$ are $2x^2-x+3, x^2-x$ and $x^2 +3x-9$. Find its perimeter.
Given:
Sides of a triangle in terms of $x$ are $2x^2-x+3, x^2-x$ and $x^2 +3x-9$.
To do:
We have to find its perimeter.
Solution:
The perimeter of a triangle is the sum of its sides.
Therefore,
Perimeter of the triangle $=(2x^2-x+3)+(x^2-x)+(x^2 +3x-9)$
$=(2x^2+x^2+x^2)+(-x-x+3x)+(3-9)$
$=4x^2+(3x-2x)+(-6)$
$=4x^2+x-6$
The perimeter of the triangle is $4x^2+x-6$.
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