The slope of an absolute function at the point where it evaluates to zero is undefined.
Here, when x=1, so |4x-4|=|4-4|=|0|, the slope is undefined. This is where the vertex of the "V" on the graph of the absolute value function.
Answer:
Step-by-step explanation:
Given that:
The differential equation; ![(x^2-4)^2y'' + (x + 2)y' + 7y = 0](https://tex.z-dn.net/?f=%28x%5E2-4%29%5E2y%27%27%20%2B%20%28x%20%2B%202%29y%27%20%2B%207y%20%3D%200)
The above equation can be better expressed as:
![y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0](https://tex.z-dn.net/?f=y%27%27%20%2B%20%5Cdfrac%7B%28x%2B2%29%7D%7B%28x%5E2-4%29%5E2%7D%20%5C%20y%27%2B%20%5Cdfrac%7B7%7D%7B%28x%5E2-%204%29%5E2%7D%20%5C%20y%3D0)
The pattern of the normalized differential equation can be represented as:
y'' + p(x)y' + q(x) y = 0
This implies that:
![p(x) = \dfrac{(x+2)}{(x^2-4)^2} \](https://tex.z-dn.net/?f=p%28x%29%20%3D%20%5Cdfrac%7B%28x%2B2%29%7D%7B%28x%5E2-4%29%5E2%7D%20%5C)
![p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \](https://tex.z-dn.net/?f=p%28x%29%20%3D%20%5Cdfrac%7B%28x%2B2%29%7D%7B%28x%2B2%29%5E2%20%28x-2%29%5E2%7D%20%5C)
![p(x) = \dfrac{1}{(x+2)(x-2)^2}](https://tex.z-dn.net/?f=p%28x%29%20%3D%20%5Cdfrac%7B1%7D%7B%28x%2B2%29%28x-2%29%5E2%7D)
Also;
![q(x) = \dfrac{7}{(x^2-4)^2}](https://tex.z-dn.net/?f=q%28x%29%20%3D%20%5Cdfrac%7B7%7D%7B%28x%5E2-4%29%5E2%7D)
![q(x) = \dfrac{7}{(x+2)^2(x-2)^2}](https://tex.z-dn.net/?f=q%28x%29%20%3D%20%5Cdfrac%7B7%7D%7B%28x%2B2%29%5E2%28x-2%29%5E2%7D)
From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2
When x = - 2
![\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}](https://tex.z-dn.net/?f=%5Clim%20%5Climits_%7Bx%20%5Cto-2%7D%20%28x%2B%202%29%20p%28x%29%20%3D%20%20%5Clim%20%5Climits_%7Bx%20%5Cto2%7D%20%28x%2B%202%29%20%5Cdfrac%7B1%7D%7B%28x%2B2%29%28x-2%29%5E2%7D)
![\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}](https://tex.z-dn.net/?f=%5Cimplies%20%20%5Clim%20%5Climits_%7Bx%20%5Cto2%7D%20%20%5Cdfrac%7B1%7D%7B%28x-2%29%5E2%7D)
![\implies \dfrac{1}{16}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B1%7D%7B16%7D)
![\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}](https://tex.z-dn.net/?f=%5Clim%20%5Climits_%7Bx%20%5Cto-2%7D%20%28x%2B%202%29%5E2%20q%28x%29%20%3D%20%20%5Clim%20%5Climits_%7Bx%20%5Cto2%7D%20%28x%2B%202%29%5E2%20%5Cdfrac%7B7%7D%7B%28x%2B2%29%5E2%28x-2%29%5E2%7D)
![\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}](https://tex.z-dn.net/?f=%5Cimplies%20%20%5Clim%20%5Climits_%7Bx%20%5Cto2%7D%20%20%5Cdfrac%7B7%7D%7B%28x-2%29%5E2%7D)
![\implies \dfrac{7}{16}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B7%7D%7B16%7D)
Hence, one (1) of them is non-analytical at x = 2.
Thus, x = 2 is an irregular singular point.
The simplified answer is 27m^2
Answer:
2 triangles ?
Step-by-step explanation:
sorry if i got this wrong, when you cut it through the middle connecting the ends it give you two triangles?
Answer:
yes
Step-by-step explanation:
euhm ... did u just asked if a figure with h=9inches and b=12inches is rectangular???
Does it matter what the diagonals are etc?
if the frame has angles of 90° and it has two different lengths then it has to be rectangular. The sides of a square would have the same lengths.