Step-by-step explanation:
![{b}^{2} = {16}^{2} + {12}^{2} \\ {b}^{2} = 256 + 144 \\ b = 20](https://tex.z-dn.net/?f=%20%7Bb%7D%5E%7B2%7D%20%20%3D%20%20%7B16%7D%5E%7B2%7D%20%20%2B%20%20%7B12%7D%5E%7B2%7D%20%20%5C%5C%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20256%20%20%2B%20144%20%5C%5C%20b%20%3D%2020)
The statement that can be used to fill in the numbered
blank space is ∡ABD
≅ ∡DBC. The correct answer between all the choices given is the
third choice. I am hoping that this answer has satisfied your query and it will
be able to help you in your endeavor, and if you would like, feel free to ask
another question.
Answer: The second derivative would be
![f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)
Step-by-step explanation:
Since we have given that
![f(x)=5e^{-x}-2e^{-5x}](https://tex.z-dn.net/?f=f%28x%29%3D5e%5E%7B-x%7D-2e%5E%7B-5x%7D)
We will find the first derivative w.r.t. 'x'.
So, it becomes,
![f'(x)=-5e^{-x}+10e^{-5x}](https://tex.z-dn.net/?f=f%27%28x%29%3D-5e%5E%7B-x%7D%2B10e%5E%7B-5x%7D)
Then, we will find the second derivative w.r.t 'x'.![f''(x)=5e^{-x}+10\times -5e^{-5x}\\\\f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D%2B10%5Ctimes%20-5e%5E%7B-5x%7D%5C%5C%5C%5Cf%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)
Hence, the second derivative would be
![f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)