Answer: ![\dfrac{x^2+1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%2B1%7D%7B2%7D)
Step-by-step explanation:
Given
![f(x)=\sqrt{2x-1}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%7B2x-1%7D)
We can write it as
![\Rightarrow y=\sqrt{2x-1}](https://tex.z-dn.net/?f=%5CRightarrow%20y%3D%5Csqrt%7B2x-1%7D)
Express x in terms of y
![\Rightarrow y^2=2x-1\\\\\Rightarrow x=\dfrac{y^2+1}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20y%5E2%3D2x-1%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7By%5E2%2B1%7D%7B2%7D)
Replace y be x to get the inverse
![\Rightarrow f^{-1}(x)=\dfrac{x^2+1}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20f%5E%7B-1%7D%28x%29%3D%5Cdfrac%7Bx%5E2%2B1%7D%7B2%7D)
To prove, it is inverse of f(x). ![f(f^{-1}(x))=x](https://tex.z-dn.net/?f=f%28f%5E%7B-1%7D%28x%29%29%3Dx)
![\Rightarrow f(f^{-1}(x))=\sqrt{2\times \dfrac{x^2+1}{2}-1}\\\\\Rightarrow f(f^{-1}(x))=\sqrt{x^2+1-1}\\\\\Rightarrow f(f^{-1}(x))=x](https://tex.z-dn.net/?f=%5CRightarrow%20f%28f%5E%7B-1%7D%28x%29%29%3D%5Csqrt%7B2%5Ctimes%20%5Cdfrac%7Bx%5E2%2B1%7D%7B2%7D-1%7D%5C%5C%5C%5C%5CRightarrow%20f%28f%5E%7B-1%7D%28x%29%29%3D%5Csqrt%7Bx%5E2%2B1-1%7D%5C%5C%5C%5C%5CRightarrow%20f%28f%5E%7B-1%7D%28x%29%29%3Dx)
So, they are inverse of each other.
1012 divided by 120 is 8.43 so 9 boxes will be needed
Answer:
y +8 = -4(x -8)
Step-by-step explanation:
You recognize that the given equation is in slope-intercept form:
y = mx + b
with m = 1/4 and b = 5.
A perpendicular line will have a slope that is the negative reciprocal of this value of m, so the desired slope is ...
-1/m = -1/(1/4) = -4
The point-slope form of the equation for a line is ...
y -k = m(x -h) . . . . . for slope m through point (h, k)
Using m=-4 and (h, k) = (8, -8), the point-slope form of the equation for the line you want is ...
y +8 = -4(x -8)
A ) A graph of the equation is in the attachment.
B ) The vertex
( 0, -10 )C ) The graph of y = x² is translated for 10 units down.
<h2>
Answer:</h2><h2>c = -24</h2><h2 /><h2>Hope this helps!!</h2>