If I did this right, it should be:
1. 130 degrees
2. 50 degrees
3. 130 degrees
4. 50 degrees
Here's hoping I'm right. lol
Answer:
The minimum value for
is
.
Step-by-step explanation:
Given function is ![f(x) = 0.15(x + 1)^2 - 3](https://tex.z-dn.net/?f=f%28x%29%20%3D%200.15%28x%20%2B%201%29%5E2%20-%203)
We need to find the maximum value or the minimum value for the function.
Now, differentiate
w.r.t
.
![f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%5Cfrac%7Bd%7D%7Bdx%7D%280.15%28x%20%2B%201%29%5E2%20-%203%29%5C%5Cf%27%28x%29%3D%5Cfrac%7Bd%7D%7Bdx%7D%280.15%28x%2B1%29%5E2-%5Cfrac%7Bd%7D%7Bdx%7D%283%29%5C%5C)
![f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)](https://tex.z-dn.net/?f=f%27%28x%29%3D2%5Ctimes%200.15%28x%2B1%29%5Cfrac%7Bd%7D%7Bdx%7D%28x%2B1%29-0%5C%5Cf%27%28x%29%3D0.3%28x%2B1%29%281%29%5C%5Cf%27%28x%29%3D0.3%28x%2B1%29)
Now, we will equate
to find critical point.
![0.3(x+1)=0\\x=-1](https://tex.z-dn.net/?f=0.3%28x%2B1%29%3D0%5C%5Cx%3D-1)
Plug this critical point in to the function
we get,
![f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3](https://tex.z-dn.net/?f=f%28-1%29%20%3D%200.15%28-1%20%2B%201%29%5E2%20-%203%5C%5Cf%28-1%29%3D-3)
Also,
which is positive, We have minimum value.
So, the minimum value for
is
.
Answer:
the shape is semetrical
Step-by-step explanation:
360 IN.2?
Step-by-step explanation:
i don't know if correct
nakalimutan kuna yan kung pano gawin sorry