Let f(x)=x^2+9,g(x)=frac{x}{x-9},a=f(g(10)),a | vedclass

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(B) Given $f(x)=x^2+9$ and $g(x)=\frac{x}{x-9}$.First,calculate $a$:$g(10) = \frac{10}{10-9} = 10$$a = f(10) = 10^2+9 = 109$.Next,calculate $b$:$f(3)

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$8$

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(B) Given $f(x)=x^2+9$ and $g(x)=\frac{x}{x-9}$.First,calculate $a$:$g(10) = \frac{10}{10-9} = 10$$a = f(10) = 10^2+9 = 109$.Next,calculate $b$:$f(3) = 3^2+9 = 18$$b = g(18) = \frac{18}{18-9} = \frac{18}{9} = 2$.The ellipse equation is $\frac{x^2}{109}+\frac{y^2}{2}=1$. Here $A^2=109$ and $B^2=2$.Since $A^2 > B^2$,the eccentricity $e$ is given by $e^2 = 1 - \frac{B^2}{A^2} = 1 - \frac{2}{109} = \frac{107}{109}$.The length of the latus rectum $l$ is given by $l = \frac{2B^2}{A} = \frac{2(2)}{\sqrt{109}} = \frac{4}{\sqrt{109}}$.Thus,$l^2 = \frac{16}{109}$.Finally,calculate $8e^2+l^2$:$8e^2+l^2 = 8\left(\frac{107}{109}\right) + \frac{16}{109} = \frac{856+16}{109} = \frac{872}{109} = 8$.

Let $f(x)=x^2+9$,$g(x)=\frac{x}{x-9}$,$a=f(g(10))$,and $b=g(f(3))$. If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$,then $8e^2+l^2$ is equal to.

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