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3 years ago

I need help with a math problem.

Mathematics

2 answers:

andrey2020 [161]3 years ago

7 0

All you have to do is just add the sides, in order to find the perimeter.

4m + 3.4m + 7.3m + 11.2m = 25.9m

Rufina [12.5K]3 years ago

3 0

11.2+4+3.4+7.3+(11.2-3.4)=
11.2+4+3.4+7.3+7.8=
33.7

your perimeter is 33.7 m

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3 years ago

A hyperbola centered at the origin has a vertex at (9, 0) and a focus at (−15, 0). Which are the equations of the asymptotes? y

chubhunter [2.5K]

Well, we know is centered at the origing, thus h,k are just 0,0.

using the provided vertex and focus point, gives us a distance for "a" of 9 and a distance for "c" of 15, check the picture below.

$\bf \textit{hyperbolas, horizontal traverse axis }\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2+ b ^2}\\
\textit{asymptotes}\quad 
y= k\pm \cfrac{b}{a}(x- h)
\end{cases}\\\\
-------------------------------$

$\bf \begin{cases}
a=9\\
c=15
\end{cases}\implies c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{15^2-9^2}=b\implies \sqrt{144}=b\implies \boxed{12=b}\\\\
-------------------------------\\\\
asymptotes\qquad y=0\pm\cfrac{12}{9}(x-0)\implies y=\pm\cfrac{4}{3}x$

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4 years ago

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