Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
We conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.
Therefore, the solution to f(x) = g(x) is:
Step-by-step explanation:
Given the table
x f(x) = 2ˣ - 1 g(x) = 1/2x
-2 -3/4 -1
-1 -1/2 -1/2
0 0 0
1 1 1/2
2 3 1
If we carefully observe, we can determine that
at x = 0, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.
In other words,
at x = 0
Thus,
at x = 0
f(x) = g(x)
Also at x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.
In other words,
at x = -1
Thus,
at x = -1
f(x) = g(x)
Summary:
Thus, we conclude that at x = 0 and x = -1, the value of f(x) = 2ˣ - 1 and g(x) = 1/2x is the same.
Therefore, the solution to f(x) = g(x) is:
2/5 is the answer unless your multiplying
True
The two shorter lengths do not add up to more than the longest length. 3+3 is less than 9. Therefore, even if the two shorter lengths lay on top of the longer side, the two ends cannot meet to form a closed 3 sided figure
Multiply top by top
2 * -3 = -6
Multiply bottom
7 * 8 = 56
The answer is -6/56
Simplify = -3/28
