For this case we have the following function:
f (x) = root (x)
We have the following information:
Reflections
To graph y = -f (x), reflect the graph of y = f (x) on the x-axis. (Vertical reflection)
f (x) = - root (x)
Vertical translations
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
f (x) = - root (x) +3
Answer:
f (x) = - root (x) +3
option C
Answer:
Step-by-step explanation:
Stephan can run
3 miles in 15.75 minutes
1 miles in 15.75/3 minutes
1 miles in 5.25 minutes
Kelsha can run
5 miles in 226 minutes
1 miles in 226/5 minutes
1 miles in 45.2 minutes
Stephan can run faster, since 5.25 < 45.2
Answer:
so if x = 0, x = 2, and x = 4
3(0) + 2 = 0 + 2 = 2, so when x = 0, y = 2
3(2) + 2 = 6 + 2 = 8, so when x = 2, y = 8
3(4) + 2 = 12 + 2 = 14, so when x = 4, y = 14
Answer:
Step-by-step explanation:
We have to take the derivatives for both functions and replace in the differential equation. Hence
for y=e^{-5x}:
![y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-5x%7D%5C%5Cy%27%28x%29%3D-5e%5E%7B-5x%7D%5C%5Cy%27%27%28x%29%3D25e%5E%7B-5x%7D%5C%5C)
for y=e^{6x}:
![y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B6x%7D%5C%5Cy%27%28x%29%3D6e%5E%7B6x%7D%5C%5Cy%27%27%28x%29%3D36e%5E%7B6x%7D%5C%5C)
Now we replace in the differential equation y'' − y' − 30y = 0
for y=e^{-5x}:
![25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0](https://tex.z-dn.net/?f=25e%5E%7B-5x%7D%2B5e%5E%7B-5x%7D-30e%5E%7B-5x%7D%3D0%5C%5C25%2B5-30%3D0)
for y=e^{6x}:
![36e^{6x}-6e^{6x}-30=0\\36-6+30=0](https://tex.z-dn.net/?f=36e%5E%7B6x%7D-6e%5E%7B6x%7D-30%3D0%5C%5C36-6%2B30%3D0)
Now, to know if both function are linearly independent we calculate the Wronskian
![W(f,g)=fg'-f'g](https://tex.z-dn.net/?f=W%28f%2Cg%29%3Dfg%27-f%27g)
![W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0](https://tex.z-dn.net/?f=W%28e%5E%7B-5x%7D%2Ce%5E%7B6x%7D%29%3D%28e%5E%7B-5x%7D%29%286e%5E%7B6x%7D%29-%28-5e%5E%7B-5x%7D%29%28e%5E%7B6x%7D%29%5Cneq%200)
I hope this is useful for you
Best regard