Inverse variation:
![y=\frac{k}{x}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bk%7D%7Bx%7D)
k is a constant
Use the given information (when x=4 y=4) to find the value of the constant:
![\begin{gathered} 4=\frac{k}{4} \\ \\ \text{Multiply both sides of the equation by 4:} \\ 4\times4=4\times\frac{k}{4} \\ \\ 16=k \\ \\ \\ k=16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%204%3D%5Cfrac%7Bk%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20%5Ctext%7BMultiply%20both%20sides%20of%20the%20equation%20by%204%3A%7D%20%5C%5C%204%5Ctimes4%3D4%5Ctimes%5Cfrac%7Bk%7D%7B4%7D%20%5C%5C%20%20%5C%5C%2016%3Dk%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20k%3D16%20%5Cend%7Bgathered%7D)
Then, the given relationship has the next equation:
![y=\frac{16}{x}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B16%7D%7Bx%7D)
Use the equation above to find y when x=2:
![\begin{gathered} y=\frac{16}{2} \\ \\ y=8 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D%5Cfrac%7B16%7D%7B2%7D%20%5C%5C%20%20%5C%5C%20y%3D8%20%5Cend%7Bgathered%7D)
Then, when x=2, y=8
So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
Answer:
the answer is D: undefined becous it's lies at Y-axis
Answer:
a = 14
Step-by-step explanation:
Area; 196
hope this helps.
Answer:
x=2
Step-by-step explanation:
![x^{2} -14=-10\\](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-14%3D-10%5C%5C)
Add 14 to both sides
![x^{2}=4](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D4)
Square root it
x=2