Answer:
y = ±√(x-16)
Step-by-step explanation:
To find the inverse of an equation, swap the x- and y-variables:
y = x² + 16
x = y² + 16
Then, solve for the y-variable:
x = y² + 16
y² = x - 16
y = ±√(x-16)
<h2>
Hello!</h2>
The answer is: ![g(x)=-\sqrt[3]{x-1}](https://tex.z-dn.net/?f=g%28x%29%3D-%5Csqrt%5B3%5D%7Bx-1%7D)
<h2>
Why?</h2>
Let's check the roots and the shown point in the graphic (2,-1)
First,
![0=-\sqrt[3]{x-1}\\\\0^{3}=(-\sqrt[3]{x-1})^{3}\\\\0=-(x-1)\\\\x=1](https://tex.z-dn.net/?f=0%3D-%5Csqrt%5B3%5D%7Bx-1%7D%5C%5C%5C%5C0%5E%7B3%7D%3D%28-%5Csqrt%5B3%5D%7Bx-1%7D%29%5E%7B3%7D%5C%5C%5C%5C0%3D-%28x-1%29%5C%5C%5C%5Cx%3D1)
then,
![g(0)=-\sqrt[3]{0-1}\\g(0)=-(-1)\\g(0)=1\\y=1](https://tex.z-dn.net/?f=g%280%29%3D-%5Csqrt%5B3%5D%7B0-1%7D%5C%5Cg%280%29%3D-%28-1%29%5C%5Cg%280%29%3D1%5C%5Cy%3D1)
So, we know that the function intercepts the axis at (1,0) and (0,1), meaning that the function match with the last given option
(
)
Second,
Evaluating the function at (2,-1)
![y=-\sqrt[3]{x-1}\\-1=-\sqrt[3]{2-1}\\-1=-\sqrt[3]{1}\\-1=-(1)\\-1=-1](https://tex.z-dn.net/?f=y%3D-%5Csqrt%5B3%5D%7Bx-1%7D%5C%5C-1%3D-%5Csqrt%5B3%5D%7B2-1%7D%5C%5C-1%3D-%5Csqrt%5B3%5D%7B1%7D%5C%5C-1%3D-%281%29%5C%5C-1%3D-1)
-1=-1
It means that the function passes through the given point.
Hence,
The equation which represents g(x) is ![g(x)=-\sqrt[3]{x-1}](https://tex.z-dn.net/?f=g%28x%29%3D-%5Csqrt%5B3%5D%7Bx-1%7D)
Have a nice day!
If a^2 + b^2 = c^2 then it is a right triangle.
22^2 + 26^2 ?=? 34^2
484 + 676 ?=? 1156
1160 ?=? 1156
This is NOT TRUE so it is not a right triangle.
The frist is 2.7
the 2nd is 2.34
the 4th is 8
the 5th is 4
I think the 3rd is 11over8
Answer:
(0,-1)
(1,-5)
(2,-9)
Step-by-step explanation:
y = −4x −1
Let x=0
y = 0-1 = -1 (0,-1)
Let x = 1
y = -4(1) -1 = -4-1 =-5 (1,-5)
Let x = 2
y = -4(2) -1 = -8-1 =-9 (2,-9)