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)
X=dimes y=quarters
0.1x + 0.25y = 2.95
x+y = 16
from the second equation, x = 16 - y and thus 0.1 (16-y) +0.25y = 2.95 ==> y=9. Finally, x + 9 = 16 ==> x=7
7 dimes, 9 quarters
Answer:
u=22
Step-by-step explanation:
Triangel JKI and Triangel HKI are congruent by rule SAS (Side Angel Side)
JI and IH are congruent by rule CPCTC (Coungruent Parts of Cogruent Triangels are Cogruent).
2u=u+22
u=22
One of them is 3,8,3,8 and the other is 4,6,4,6
To find the answer you need to identify what x is.
To find the x first you need to make the equation 6x+11=29.
You then subtract 11 from each side which cancels out the 11 and makes the 29 18.
You then divide the entire equation by 6 to get the answer x=3.
You then input x for the CN line equation.
4(3)+1
Thus, your answer should be 13 (D)
