Always
Because the slant length is the longest length when looking at the Pythagoras theorm
In this case the slant is y, with x and y as the height and base
y²=
![x^{2}+y^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%2By%5E%7B2%7D%20)
Hence y the slant is the sum of the height and and base and will always be larger
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
:
![I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cfrac1%5Cphi%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7B%5Ctan%5E%7B-1%7D%28x%5E%7B%5Cphi-1%7D%29%7D%7B%281%2Bx%29%5E2%7D%20%5C%2C%20dx)
Split the integral at x = 1. For the integral over [1, ∞), substitute
:
![\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_1%5E%5Cinfty%20%5Cfrac%7B%5Ctan%5E%7B-1%7D%28x%5E%7B%5Cphi-1%7D%29%7D%7B%281%2Bx%29%5E2%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B%5Ctan%5E%7B-1%7D%28x%5E%7B1-%5Cphi%7D%29%7D%7B%5Cleft%281%2B%5Cfrac1x%5Cright%29%5E2%7D%20%5Cfrac%7Bdx%7D%7Bx%5E2%7D%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B%5Cpi2%20-%20%5Ctan%5E%7B-1%7D%28x%5E%7B%5Cphi-1%7D%29%7D%7B%281%2Bx%29%5E2%7D%20%5C%2C%20dx)
The integrals involving tan⁻¹ disappear, and we're left with
![I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cfrac%5Cpi%7B2%5Cphi%7D%20%5Cint_0%5E1%20%5Cfrac%7Bdx%7D%7B%281%2Bx%29%5E2%7D%20%3D%20%5Cboxed%7B%5Cfrac%5Cpi%7B4%5Cphi%7D%7D)
Answer: As the number of cookies eaten increases, the calories consumed increases.
This is because the number of cookies and calories consumed have a directly proportional relationship, meaning that if one value increases, so does the other.
Answer:
Step-by-step explanation:
I dont have cashapp sorry
Initial is 12
Percent 1.05(I think but you might have to convert it)
Y=12(1.05)(5)
Y=63