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Nina [5.8K]
4 years ago
8

The equation that models Saturn's elliptical orbit around the sun is.... ,

Mathematics
2 answers:
Artyom0805 [142]4 years ago
8 0

Answer: 1.5131 billion mi.


Mandarinka [93]4 years ago
4 0

Answer:

1.5131 billion mi.

Step-by-step explanation:

The general equation of the ellipse with center (h,k) is

\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} =1

Here the equation of ellipse is given as

\frac{(x+0.0831)^2}{2.0449}+ \frac{y^{2}}{2.0380} =1 \\

Therefore, the center of the given ellipse is (-0.0831,0)

The focus is (0,0)

a is the radius along major axis

Here a= √2.0449 =1.43

The distance between focus and center of the given ellipse = 0.0831

Since the farthest distance between Saturn and the Sun is required, we have to add up the radius of the ellipse along the major axis and the distance between the focus and center.

Distance between sun and saturn = 1.43 + 0.0831 = 1.5131

∴ The farthest distance between the Sun and saturn is 1.5131 billion miles.





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2. 5²- 4 +3 • (9 • 2) /2=
stepladder [879]

2) 48

3) 72

4) 48

5) 105

6) 47

7) 37

8) 400

9) 17

10) 13

I hope this helps! Let me know if I get anything wrong so I can fix it.

5 0
3 years ago
Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

7 0
3 years ago
5. [02.06]
Ne4ueva [31]

Answer:

2.14

Step-by-step explanation:

7 0
4 years ago
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what are your initial's thoughts about saving.what do you want to learn about saving 4 sentences at least
adelina 88 [10]

Answer: i think that is ur own opinion

Step-by-step explanation:

6 0
3 years ago
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PLEASE HELP!
Murljashka [212]

Answer: i think its b

Step-by-step explanation:

6 0
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