Answer: -a^2+c\left(a-c\right)-n^2a=3\quad :\quad c=-\frac{-a+\sqrt{-3a^2-4an^2-12}}{2},\:c=-\frac{-a-\sqrt{-3a^2-4an^2-12}}{2}
Step-by-step explanation: YW! plese mark branlest!
Answer:
believe it's G, I hope that's correct for you
Answer: The numbers are showing in descending order 3.6-1.8= 1.8 = 1.8 is half of 3.6 - therefore if 1200 is half again = 0.8 then what lies between 1.8 and 0.8? Answer to 1000 is 0.8 +5 = 1.3
Step-by-step explanation:
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)
Equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
In this case:
<span>center (2,7) and radius 4 so h = 2, k = 7 and r = 4
</span>Equation:
(x - 2)^2 + (y - 7)^2 = 4^2
(x - 2)^2 + (y - 7)^2 = 16
Hope it helps.
