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antoniya [11.8K]
3 years ago
7

A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 128 million kilometers.

Its orbital eccentricity is 0.3. Find the planet's orbital period in months
Mathematics
1 answer:
Jet001 [13]3 years ago
3 0
The answer for this problem would be:

Orbital period if distance = 128,000,00 km and mass = Sun? 

<span>This is approximately be .85 astronomical units if converted km to au.
</span>
We will be using Kepler's third law which is:

<span>T^2 = r^3 </span>

<span>T^2 = .85^3 </span>

<span>T^2 = .61
</span>
get the square root of .61
so the answer will be:

<span>T = .78 Years which is in months is 9.36 months.</span>
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Answer:

<h2>x = -8 and y= 0 → (-8, 0)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}y=\dfrac{1}{4}x+2&(1)\\3y=-\dfrac{3}{4}x-6&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\3\left(\dfrac{1}{4}x+2\right)=-\dfrac{3}{4}x-6\qquad\text{use the distributive property}\\\\(3)\left(\dfrac{1}{4}x\right)+(3)(2)=-\dfrac{3}{4}x-6\\\\\dfrac{3}{4}x+6=-\dfrac{3}{4}x-6\qquad\text{subtract 6 from both sides}\\\\\dfrac{3}{4}x=-\dfrac{3}{4}x-12\qquad\text{add}\ \dfrac{3}{4}x\ \text{to both sides}\\\\\dfrac{6}{4}x=-12\qquad\text{multiply both sides by 4}\\\\6x=-48\qquad\text{divide both sides by 6}\\\\x=-8

\text{Put the value of x to (1):}\\\\y=\dfrac{1}{4}(-8)+2\\\\y=-2+2\\\\y=0

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