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)
-1 is your x intercept so the first point is going to be at (0,-1) and from there go up 1 and over to the right 2 and mark another point and so one
<span>For the answer to the question above,
Let d be Ted's portion; y be Tony's ; r be Terry's ; m be Tom''s
d+y+r+m=500
d+y+0+0=280
d+0+0+m=260
d+0+r+0=220
-----------
Using the Matrix function of a TI calculator to solve this system
of equations I get:
Ted = $130 ; so Ted did 130/500 = 26%
Tony= $150 ; so Tony did 150/500= 30%
Terry=$90 ; so Terry did 90/500= 18%
Tom = $130 ; so Tom did 130/500 = 26%</span>
A. 8^9/2= around 11585.2375; (sqrt8)^9= around 11585.2375. So choice A shows a pair of equivalent expressions.
B. (3sqrt125)^9=1,953,125; 125^9/3=1,953,125. So choice B also shows a pair of equivalent expressions.
C. 12^2/7= around 2.03394; (sqrt12)^7= around 5,985.96759. So choice C does not show a pair of equivalent expressions.
D. 4^1/5=around 1.31951; (sqrt4)^5=32. So choice D also doesn't show a pair of equivalent expressions.
Your answers are A and B.
I hope this helps ;)
Answer:
here you go, the answer to the whole test: https://school.ckseattle.org/documents/2019/2/8_2_The_Pythagorean_Theorem_and_Its_Converse.pdf
Step-by-step explanation:
