Answer:
x = 13
Step-by-step explanation:
First start with the blank equation
![a^{2} + b^{2} = c^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D%20%2B%20b%5E%7B2%7D%20%3D%20c%5E%7B2%7D)
Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.
![5^{2} + 12^{2} = c^{2}](https://tex.z-dn.net/?f=5%5E%7B2%7D%20%2B%2012%5E%7B2%7D%20%3D%20c%5E%7B2%7D)
Then solve
25 + 144 = ![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
169 =![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
![\sqrt{169} = \sqrt{c^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B169%7D%20%3D%20%5Csqrt%7Bc%5E%7B2%7D%20%7D)
c = 13
We will use substitution for the partial integration:
![u=arc tan ( \frac{1}{x}), dv = dx \\ du= \frac{-dx}{(1+ x^{2}) x^{2} } , v=x](https://tex.z-dn.net/?f=u%3Darc%20tan%20%28%20%5Cfrac%7B1%7D%7Bx%7D%29%2C%20%20%20dv%20%3D%20dx%20%5C%5C%20du%3D%20%5Cfrac%7B-dx%7D%7B%281%2B%20x%5E%7B2%7D%29%20x%5E%7B2%7D%20%20%7D%20%2C%20v%3Dx%20)
The integral becomes:=
![x arc tan x- \int {x \frac{-dx}{(1+ x^{2}) x^{2} } } \, dx = \\ =x arc tan x+ \int { \frac{dx}{(1+ x^{2} )x} } \, dx](https://tex.z-dn.net/?f=x%20arc%20tan%20x-%20%5Cint%20%7Bx%20%5Cfrac%7B-dx%7D%7B%281%2B%20x%5E%7B2%7D%29%20x%5E%7B2%7D%20%20%7D%20%7D%20%5C%2C%20dx%20%3D%20%5C%5C%20%3Dx%20arc%20tan%20x%2B%20%5Cint%20%7B%20%5Cfrac%7Bdx%7D%7B%281%2B%20x%5E%7B2%7D%20%29x%7D%20%7D%20%5C%2C%20dx%20)
2nd integration:
we will add and subtract x^2 to the numerator:
![\int { \frac{1+ x^{2} - x^{2} }{(1+ x^{2} )x} } \, dx= \\ \int { \frac{1}{x} } \, dx- \int { \frac{x}{1+ x^{2} } } \, dx = \\ ln(x) - \int { \frac{x}{1+ x^{2} } } \, dx](https://tex.z-dn.net/?f=%20%5Cint%20%7B%20%5Cfrac%7B1%2B%20x%5E%7B2%7D%20-%20x%5E%7B2%7D%20%7D%7B%281%2B%20x%5E%7B2%7D%20%29x%7D%20%7D%20%5C%2C%20dx%3D%20%5C%5C%20%20%5Cint%20%7B%20%5Cfrac%7B1%7D%7Bx%7D%20%7D%20%5C%2C%20dx-%20%5Cint%20%7B%20%5Cfrac%7Bx%7D%7B1%2B%20x%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%20%3D%20%5C%5C%20ln%28x%29%20-%20%5Cint%20%7B%20%5Cfrac%7Bx%7D%7B1%2B%20x%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%20%20%20)
u-substitution:
![u=1+ x^{2} , du=2xdx, xdx= \frac{du}{2}](https://tex.z-dn.net/?f=u%3D1%2B%20x%5E%7B2%7D%20%2C%20du%3D2xdx%2C%20xdx%3D%20%5Cfrac%7Bdu%7D%7B2%7D%20)
...=
![ln(x)- \frac{1}{2} \int { \frac{1}{u} } \, du=ln(x)- \frac{1}{2} ln(u)= \\ ln(x)-ln( \sqrt{1+ x^{2} )} =ln \frac{x}{ \sqrt{1+ x^{2} } }](https://tex.z-dn.net/?f=ln%28x%29-%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cint%20%7B%20%5Cfrac%7B1%7D%7Bu%7D%20%7D%20%5C%2C%20du%3Dln%28x%29-%20%5Cfrac%7B1%7D%7B2%7D%20ln%28u%29%3D%20%5C%5C%20ln%28x%29-ln%28%20%5Csqrt%7B1%2B%20x%5E%7B2%7D%20%29%7D%20%3Dln%20%5Cfrac%7Bx%7D%7B%20%5Csqrt%7B1%2B%20x%5E%7B2%7D%20%7D%20%7D%20%20%20)
Finally:...=
Answer:
However much he had on his card in the first place.
Step-by-step explanation:
Say he had $500 on his card. He took 0 rides ( no rides ) so he doesn't lose any money. Leaving him with his starting amount, $500.
Compose the result function for y=2x - 7 by replacing function designators with the actual function y= 2x - 7
The answer you are looking for is c). 1