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SIZIF [17.4K]
1 year ago
5

which of the following expresses the coordinates of the foci of the conic section shown below? (x-2)^2/4+(y+5)^2/9

Mathematics
1 answer:
elena55 [62]1 year ago
6 0

Step-by-step explanation:

\frac{(x - 2) {}^{2} }{4}  +  \frac{(y + 5) {}^{2} }{9}  = 1

This is the equation of the ellipse. Since the denominator is greater for the y values, we have a vertical ellipse. Remember a>b, so a

The formula for the foci of the vertical ellipse is

(h,k+c) and (h,k-c).

where c is

Our center (h,k) is (2, -5)

{c}^{2}  =  {a}^{2}  -  {b}^{2}

Here a^2 is 9, b^2 is 4.

{c}^{2}  = 9 - 4

{c}^{2}  = 5

c =  \sqrt{5}

So our foci is

(2, - 5 +  \sqrt{5} )

and

(2, - 5 -  \sqrt{5} )

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51.2

Step-by-step explanation:

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If y is proportional to the sqaure root of x and y equal x when x equal 5,evaluate y when x eqaul 9.​
Law Incorporation [45]

Answer:

y = 3\, \sqrt{5} when x = 9.

Step-by-step explanation:

The question states that y is proportional to \sqrt{x}. In other words, there is a constant a (a \ne 0) such that y = a\, \sqrt{x} for all x \ge 0.

The question also states that y = x when x = 5. Make use of this equality to find the value of a.

Since x = 5 and y = x, it must be true that y = 5. Substitute x = 5\! and y = 5\! into the equation y = a\, \sqrt{x} and solve for a:

5 = a\, \sqrt{5}.

\begin{aligned}a &= \frac{5}{\sqrt{5}} \\ &= \sqrt{5}\end{aligned}.

Thus, y = \sqrt{5}\, (\sqrt{x}) for all x \ge 0.

Substitute in the value x = 9 to find the corresponding value of y:

\begin{aligned} y &= \sqrt{5} \, (\sqrt{9}) \\ &= \sqrt{5} \times 3 \\ &= 3\, \sqrt{5}\end{aligned}.

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Marizza181 [45]
I think it is the first one

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Find the area of this shape below
rjkz [21]

Answer: 28                      

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