The x-axis serves as a horizontal asymptote for all exponential functions. Exponential functions are of the form f(x) = ax . The domain consists of all real numbers. However, the range only consists of all numbers greater than zero. This is because no matter how large x gets, the graph will shoot upwards towards infinity. If x becomes a negative, we know that we will get f(x) = 1/a2 . The larger the negative number, the closer the function approaches zero. So, for exponential functions we will always have that restriction that the range will only include positive numbers. I hope this answers your question. I believe the statement is true.
The dimensions would be 29 by 29.
To maximize area and minimize perimeter, we make the dimensions as close to equilateral as possible.
Dividing the perimeter by the number of sides, we have
116/4 = 29
This means that both length and width can be 29.
SOLUTION
From the question, the center of the hyperbola is

a is the distance between the center to vertex, which is -1 or 1, and
c is the distance between the center to foci, which is -2 or 2.
b is given as
![\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20b%5E2%3Dc%5E2-a%5E2%20%5C%5C%20b%5E2%3D2%5E2-1%5E2%20%5C%5C%20b%3D%5Csqrt%5B%5D%7B3%7D%20%5Cend%7Bgathered%7D)
But equation of a hyperbola is given as

Substituting the values of a, b, h and k, we have
![\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B%28x-0%29%5E2%7D%7B1%5E2%7D-%5Cfrac%7B%28y-0%29%5E2%7D%7B%5Csqrt%5B%5D%7B3%7D%5E2%7D%3D1%20%5C%5C%20%5Cfrac%7Bx%5E2%7D%7B1%7D-%5Cfrac%7By%5E2%7D%7B3%7D%3D1%20%5Cend%7Bgathered%7D)
Hence the answer is
Answer:
x = 2±3sqrt(3)
Step-by-step explanation:
(x-2)^2/3=9
Multiply each side by 3
(x-2)^2/3 *3=9*3
(x-2)^2=27
Take the square root of each side
sqrt( (x-2)^2)=±sqrt(27)
x-2 = ±3sqrt(3)
Add 2 to each side
x = 2±3sqrt(3)
Can you show the multiple choice one