With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that
![I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7B%5Csqrt%5B%5Cphi%5D%7Bx%7D%20%5Ctan%5E%7B-1%7D%28x%29%7D%7B%281%2Bx%5E%5Cphi%29%5E2%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bx%5E%7B%5Cphi-1%7D%20%5Ctan%5E%7B-1%7D%28x%29%7D%7Bx%20%281%2Bx%5E%5Cphi%29%5E2%7D%20%5C%2C%20dx)
Replace
:

Split the integral at x = 1. For the integral over [1, ∞), substitute
:

The integrals involving tan⁻¹ disappear, and we're left with

Answer:
C = 4.94 M = 18.38
Step-by-step explanation:
the easiest way is to graph both equaitons and see where they intersect,
see the image
or solve by substitution
6C + 47M=893.50 C = (893.50 - 47M)/6
25C + 22M=527.86
25((893.50 - 47M)/6) + 22M = 527.86
25/6 [(893.50) - 47M] + 22M = 527.86
25/6 [ - 47M] + 22M = 527.86 - 25/6 [(893.50)
22M - 25(47M)/6 = -3195.06
22M - 195.83M = -3195.06
-173.83M = -3195.06
M = 18.38
C = (893.50 - 47(18.38)/6
C = 4.94
Answer:
1,000
Step-by-step explanation:
If your asking for the area you multiply 40 times 25