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Kitty [74]
3 years ago
6

Find the 102nd term,given the first term and the common difference A1=52 and d=12

Mathematics
1 answer:
alina1380 [7]3 years ago
7 0
This is the concept of sequences, we are required to calculate for the 102nd term of the arithmetic sequence;
the nth term of the arithmetic sequence is given by:
nth=an+d(n-1)
where;
an=1st term
d=common difference
n=the nth term
thus, the 102nd term will be:
102nd=52+12(102-1)
=52+12*101
=1,264
The answer is 1,264

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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
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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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