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noname [10]
3 years ago
7

Write the standard equation for the hyperbola with the following conditions: vertices: (-2, -3) and (6, -3) foci: (-4, -3) and (

8, -3)

Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0
Hello,

Vertices are on a line parallele at ox (y=-3)

The hyperbola is horizontal.

Equation is (x-h)²/a²- (y-k)²/b²=1

Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and  2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)

Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6

c²=a²+b²==>b²=36-4²=20

Equation is:
\boxed{ \dfrac{(x-2)^2}{16} - \dfrac{(y+3)^2}{20} =1}





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Here’s a revenue and expenses for the month.caculate whether Mia had a profit or a loss
jeyben [28]

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Mia had a profit

Step-by-step explanation:

4 0
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Read 2 more answers
What are the coordinates of point B on AC Such that the ratio of AB to AC is 5:6
Oksi-84 [34.3K]

Answer:

\vec B = \left(x_{A}+\frac{5}{6}\cdot (x_{C}-x_{A}), y_{A}+\frac{5}{6}\cdot (y_{C}-y_{A}), z_{A}+\frac{5}{6}\cdot (z_{C}-z_{A})\right)

Step-by-step explanation:

Let suppose that A, B, and C have the following points with respect to origin in the Euclidean space:

\vec A = (x_{A}, y_{A}, z_{A})

\vec B = (x_{B}, y_{B}, z_{B})

\vec C = (x_{C}, y_{C}, z_{C})

Besides, let consider that locations of A and B are currently known. The ratio is:

\frac{AB}{AC} = \frac{5}{6}

AB = \frac{5}{6} \cdot AC

Vectorially speaking, expression can be rewritten in the following terms:

\overrightarrow{AB} = \frac{5}{6}\cdot \overrightarrow{AC}

(x_{B}-x_{A}, y_{B}-y_{A}, z_{B}-z_{A}) = \frac{5}{6}\cdot (x_{C}-x_{A}, y_{C}-y_{A}, z_{C}-z_{A})

Now, each side of the equation is summed vectorially by \vec A and coordinates of point B are finally found:

\vec B = \left(x_{A}+\frac{5}{6}\cdot (x_{C}-x_{A}), y_{A}+\frac{5}{6}\cdot (y_{C}-y_{A}), z_{A}+\frac{5}{6}\cdot (z_{C}-z_{A})\right)

3 0
3 years ago
Can someone answer this <3
lara [203]
The degree is equal to 6!
4 0
3 years ago
Read 2 more answers
Please explain in depth what the answer is and how i show work! &lt;3 ;( <br> (picture shown below!)
Andru [333]
Answer: D

To show work is plug the the x-coordinates for the points given.

Example:
x=2
y=-(1/4)x^2
y=-(1/4)2^2
y=-(1/4)4
y=-1

If you would like further explain action feel free to ask. :)
6 0
3 years ago
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