When the factors of a trinomial are (x + p) and (x + q) then the constant term
1 answer:
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This question can be solved primarily by L'Hospital Rule and the Product Rule.
I) Product Rule and L'Hospital Rule:
![y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}](https://tex.z-dn.net/?f=y%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5B2xcos%28x%29-x%5E2sin%28x%29%5D-2sin%28x%29cos%28x%29%7D%7B4x%5E3%7D%20)
II) Product Rule and L'Hospital Rule:
![y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}](https://tex.z-dn.net/?f=y%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5B-2xsin%28x%29%2B2cos%28x%29%5D-%5B2xsin%28x%29%2Bx%5E2cos%28x%29%5D-%5B2cos%5E2%28x%29-2sin%5E2%28x%29%5D%7D%7B12x%5E2%7D%20%5C%5C%20y%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2cos%28x%29-4xsin%28x%29-x%5E2cos%28x%29-2cos%5E2%28x%29%2B2sin%5E2%28x%29%7D%7B12x%5E2%7D)
III) Product Rule and L'Hospital Rule:
![]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\ \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}](https://tex.z-dn.net/?f=%5Dy%3D%20%5Calpha%20%2B%20%5Cbeta%20%5C%5C%20%5C%5C%20%5Calpha%20%3D%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B-2sin%28x%29-%5B4sin%28x%29%2B4xcos%28x%29%5D-%5B2xcos%28x%29-x%5E2sin%28x%29%5D%7D%7B24x%7D%20%5C%5C%20%5Cbeta%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B4sin%28x%29cos%28x%29%2B4sin%28x%29cos%28x%29%7D%7B24x%7D%20%5C%5C%20%20%5C%5C%20y%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B-6sin%28x%29-4xcos%28x%29-2xcos%28x%29%2Bx%5E2sin%28x%29%2B8sin%28x%29cos%28x%29%7D%7B24x%7D)
IV) Product Rule and L'Hospital Rule:
![y = \phi + \varphi \\ \\ \phi = \lim_{x \to 0} \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x} \\ \varphi = \lim_{x \to 0} \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}](https://tex.z-dn.net/?f=y%20%3D%20%5Cphi%20%2B%20%5Cvarphi%20%5C%5C%20%20%5C%5C%20%5Cphi%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%20%5Cfrac%7B-6cos%28x%29-%5B-4xsin%28x%29%2B4cos%28x%29%5D-%5B2cos%28x%29-2xsin%28x%29%5D%7D%7B24x%7D%20%20%5C%5C%20%5Cvarphi%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%20%5Cfrac%7B%5B2xsin%28x%29%2Bx%5E2cos%28x%29%5D%2B%5B8cos%5E2%28x%29-8sin%28x%29%5D%7D%7B24x%7D%20)
V) Using the Definition of Limit:
I) Product Rule and L'Hospital Rule:
II) Product Rule and L'Hospital Rule:
III) Product Rule and L'Hospital Rule:
IV) Product Rule and L'Hospital Rule:
V) Using the Definition of Limit:
The perimeter of the triangle is: 13x+6
Step-by-step explanation:
The perimeter is the length of the outer boundary of any closed geometrical shape.
So,
Given
The perimeter will be:
The perimeter of the triangle is: 13x+6
Keywords: Perimeter, geometry
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