If you do in fact mean 
(as opposed to one of these being the derivative of 
at some point), then integrating twice gives
From the initial conditions, we find
Eliminating 
, we get
![C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)](https://tex.z-dn.net/?f=C_1%20%3D%20-%5Cdfrac%7B%5Cln%286%29%7D5%20%3D%20-%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%20%5Cimplies%20C_2%20%3D%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29)
Then
![\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7Bf%28x%29%20%3D%20%5Cln%7Cx%7C%20-%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%5C%2Cx%20%2B%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%7D)
The answer is C. 4/5 I already gave you a explanation on why in your last question
Answer:
(f-g)(x)=-x^(2)+2x+8
the solutions are:
<em><u>x=4 or x=-2</u></em>
Step-by-step explanation:
(f-g)(x)=2x+1-(x^(2)-7)
(f-g)(x)=-x^(2)+2x+1+7
(f-g)(x)=-x^(2)+2x+8
does this help or should I solve for the zeros/solutions of this quadratic equation?
then:
-x^(2)+2x+8=0
-(x^(2)-2x-8)=0
x^(2)-2x-8=0
(x-4)(x+2)=0
<em><u>x=4 or x=-2</u></em>
Answer:
30 shelves is the answer so it's B.
Step-by-step explanation:
1. PQR
2. PS
3.Idek too broad of a question
