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:
The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.
Step-by-step explanation:
The equation of the parabola is:

Compute the first order derivative of <em>y</em> as follows:

![\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7Bd%7Dy%7D%7B%5Ctext%7Bdx%7D%7D%3D%5Cfrac%7B%5Ctext%7Bd%7D%7D%7B%5Ctext%7Bdx%7D%7D%5B0.00035x%5E%7B2%7D%5D)

Now, it is provided that |<em>x </em>| ≤ 605.
⇒ -605 ≤ <em>x</em> ≤ 605
Compute the arc length as follows:


Now, let



Plug in the solved integrals in Arc Length and solve as follows:


Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.
Domain:
{-1, 3, 6}
Range:
{4, 5, 6}
Answer:
T = 60 degrees
Step-by-step explanation:
The dotted line is the height so it is a right angle
We are able to use trig functions since this is a right triangle
cos T = adj side / hyp
cos T = a/b
cos T = 8 sqrt(2) / 16 sqrt(2)
cos T = 1/2
Taking the inverse of each side
cos^-1 ( cosT) = cos^-1 ( 1/2)
T = 60 degrees
Right 2 up 3 because when the change happens within x the opposite result happens but for vertical it remains the same . So the answer is <span>right 2, up 3</span>