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Ksju [112]
3 years ago
5

Find the area of the figure below

Mathematics
2 answers:
Evgen [1.6K]3 years ago
8 0

Answer:

a

Step-by-step explanation:

AnnZ [28]3 years ago
4 0
The answer is to this would be a
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Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).
andre [41]
Check the picture below.  So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

\bf \textit{hyperbolas, vertical traverse axis }\\\\&#10;\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ({{ h}},{{ k}})\\&#10;vertices\ ({{ h}}, {{ k}}\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{{{ a }}^2+{{ b }}^2}&#10;\end{cases}\\\\&#10;-------------------------------\\\\

\bf \begin{cases}&#10;h=0\\&#10;k=0\\&#10;a=4\\&#10;c=5&#10;\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1&#10;\\\\\\&#10;c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\\\\\\&#10;\sqrt{5^2-4^2}=b\implies \boxed{3=b}&#10;\\\\\\&#10;\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}

7 0
3 years ago
Find the new price for the markup given. Round to the nearest cent if necessary. $3.00 marked up 72%
Marrrta [24]

Answer:

$5.16

Explanation:

Okay, so the first step is to move the decimal point of the 72% over to the left twice. Since it is a whole number, the decimal is at the back. So, the number went from 72 to .72.

After that, multiply 3.00 by .72 Like this:

3.00 * .72 = 2.16

The last step is to add the original number to that number. Like this:

3.00 + 2.16 = 5.16

Hope this helps !

6 0
2 years ago
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