Given the equation p2 = a3, what is the orbital period, in years, for the planet saturn? (saturn is located 9.5 au from the sun.)

29.28 years

Explanation:

From Kepler's third law the square of orbital period of revolving celestial body is proportional to the cube of semi -major axis from the body  it is revolving about.

P² =A³

Where, P is the orbital period in years and A is the semi-major axis in AU (Astronomical units)

It is given that, For Saturn, A = 9.5 AU. We need to find P

⇒P² = (9.5 AU)³

⇒P² = 857.38

⇒P = 29.28 years

Thus, the orbital period of Jupiter is 29.28 years around the Sun.

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1