??? HELPP????????????

Mathematics

1 answer:

vladimir2022 [97]3 years ago

5 0

Answer:

ok so first put a arrow on the seven then put an arrow on the 6 but 4 pionts foward

Step-by-step explanation:

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Simplify.<br><br>Rewrite the expression in the form a^n<br><br><br>a^-13/a^-6 = ??

masya89 [10]

Answer:

a^-7

Step-by-step explanation:

we need to know that :

so the ans is

$a^{(-13)-(-6)} = a^{-7}$

by the way

$\muti{a^{b} }{a^{c} } =a^{b+c} \\$

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Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity

Papessa [141]

If you notice, the positive fraction, is the one with the "x" variable in it, that simply means, the hyperbola is opening over the x-axis, so is a horizontal one, therefore

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

$\bf \cfrac{(x-5)^2}{81}-\cfrac{(y-1)^2}{144}=1\implies \cfrac{(x-5)^2}{9^2}-\cfrac{(y-1)^2}{12^2}=1~~
\begin{cases}
a=9\\
b=12
\end{cases}
\\\\\\
c=\sqrt{9^2+12^2}\implies c=15\\\\
-------------------------------\\\\
vertices~~(5\pm 9,1)\implies 
\begin{cases}
(14,1)\\
(-4,1)
\end{cases}~~foci~~(5\pm 15,1)\implies 
\begin{cases}
(20,1)\\
(-10,1)
\end{cases}$

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Read 2 more answers

If anyone knows pls answer<br><br><br>

svetlana [45]

Answer:

Step-by-step explanation: