Answer:
Analyzed and Sketched.
Step-by-step explanation:
We are given ![y=\frac{\ln\left(5x)}{x^2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%5Cln%5Cleft%285x%29%7D%7Bx%5E2%7D)
To sketch the graph we need to find 2 components.
1) First derivative of y with respect to x to determine the interval where function increases and decreases.
2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.
![y'=\frac{1-2\ln\left(5x)}{x^3}=0](https://tex.z-dn.net/?f=y%27%3D%5Cfrac%7B1-2%5Cln%5Cleft%285x%29%7D%7Bx%5E3%7D%3D0)
is absolute maximum
![y''=\frac{6\ln\left(5x)-5}{x^4}=0](https://tex.z-dn.net/?f=y%27%27%3D%5Cfrac%7B6%5Cln%5Cleft%285x%29-5%7D%7Bx%5E4%7D%3D0)
is the point concavity changes from down to up.
Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.
The graph is given in the attachment.
17 bc I did it trust me pls
Answer:
c = 6
Step-by-step explanation:
(I'm going to call the unidentified inner angle Henry, 'kay?)
First, we need to find Henry.
64 + Henry = 180
(64 - 64) + Henry = (180 - 64)
Henry = 116
Now to c.
Henry + 32 + 5c + 2 = 180
(116 + 32 + 2) + 5c = 180
(150 - 150) + 5c = (180 - 150)
5c/5 = 30/5
c = 6