Answer:
x = 3
Step-by-step explanation:
5(x+1) = 20
x+1 = 20/5
or, x+1 = 4
or, x = 4-1
so, x = 3
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)
Answer:
x=5
Step-by-step explanation:
For triangle A to be a scale model of triangle B, x must equal 5. This is because all side lengths must be reduced by the same ratio. In this case that ratio is 2.5. 12.5/2.5=5
Answer:
Area of Shaded Region 
Area of Semicircle 

Step-by-step explanation:
Area of Shaded Region = Area of Sector - Area of Semicircle
<u>Area of Sector</u>
Radius of the sector =20cm

<u>Area of Semicircle</u>
Since AB is the diameter of the semicircle
Radius of the Semicircle=20/2=10cm
Area of semicircle

Therefore, area of Shaded Region

Area of Square =20 X 20 
