bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.
we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.
we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.
![\bf \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}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%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%5C%5C%20%5Ctextit%7Basymptotes%7D%5Cquad%20y%3D%20k%5Cpm%20%5Ccfrac%7Bb%7D%7Ba%7D%28x-%20h%29%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

3
(
y
−
3
)
y
3
(
−
3
)
That is your answer
Answer:
D
Step-by-step explanation:
Multiply each term in the first expression by each term in the second expression and combine like terms.
12a3+2a2+23a+18
-hope it helps
12 - (-2) (-4)
-2 * -4 = ?
A negative times a negative equals a positive value.
-2*-4 = 8
12 - 8 = ?
12-8 = 4
Because 4+8 = 12
Final answer: 4