Raíz cuadrada de 864
Para resolver este problema debemos tomar en cuenta los datos que nos dan y la ecuación de una hipérbola. Comencemos con los datos:
The general equation of ellipse is given as, x2a2+y2b2=1 x 2 a 2 + y 2 b 2 = 1 , where, a is length of semi-major axis and b is length of semi-minor axis.
centro: (0,0)
focus : (0, -√28) , (0, √28)
eje conjugado = 2√3
por los focos podemos ver que la hipérbola se dirige hacia el eje y, por lo que debemos tomar la siguiente forma de la ecuación de la parábola:
de los focos podemos obtener que: √28
y del eje conjugado podemos saber que al dividir la longitud del eje conjugado dentro de 2 obtenemos b, así que:
b = √3
podemos utilizar la siguiente fórmula para obtener a:
si despejamos a en la ecuación obtenemos lo siguiente:
ahora podemos sustituir los valores:
a=5
así que media vez conozcamos a, podemos sustituir los datos en la ecuación de la hipérbola así que obtenemos lo siguiente:
si graficamos la hipérbola, queda como en el documento adjunto.
To learn more hipérbola
Visit : brainly.com/question/24223341
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Answer:
Step-by-step explanation:
First, find the <em>rate of</em><em> </em><em>change</em><em> </em>[<em>slope</em>]:


Now input the slope into the Point-Slope Formula. In this formula, all negative symbols give the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL inserting the coordinates into the formula with their CORRECT signs:
![y + 3 = -1\frac{3}{5}[x - 2]\:or\:y - 5 = -1\frac{3}{5}[x + 3]](https://tex.z-dn.net/?f=y%20%2B%203%20%3D%20-1%5Cfrac%7B3%7D%7B5%7D%5Bx%20-%202%5D%5C%3Aor%5C%3Ay%20-%205%20%3D%20-1%5Cfrac%7B3%7D%7B5%7D%5Bx%20%2B%203%5D)
I am joyous to assist you anytime.
Check the picture below, so the circle looks more or less like that one.
well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%203%20-5%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%205%20%2B%209%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-2%7D%7B2%7D~~%2C~~%5Ccfrac%7B14%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%28-1~~%2C~~7%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B3%20-%20%28-5%29%5D%5E2%20%2B%20%5B5%20-%209%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B5%29%5E2%2B%28-4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B8%5E2%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B80%7D%5Cimplies%20d%3D4%5Csqrt%7B5%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bhalf%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%5Csqrt%7B5%7D%7D%7B2%7D%5Cimplies%20%5Cunderset%7Bradius%7D%7B2%5Csqrt%7B5%7D%7D%7D)
The area is 28.26
to find the radius you have to divide the diameter by 2 which you’ll get 3
then plug it into the equation
A=3.14•3^2
Answer:
3g^2 (2g^6-g^2+3) the ^ symbole is means to the power
Step-by-step explanation: