Saturn orbits the sun at a distance of 1.43 × 1012 m. the mass of the sun is 1.99 × 1030 kg.
use to determine saturn’s orbital period in earth years. is it:
a)29.6 earth years
b)296 earth years
c)874 earth years

Saturn’s orbital period is 29.6 years

Step-by-step explanation:

It is given that,

The mass of the sun, m = 1.99 × 10³⁰ kg

Saturn orbits the sun at a distance of, a = 1.43 × 10¹² m

Using third law of Kepler's :

Where,

T is the orbital time period

G is the universal gravitational constant

M is the mass of the sun

a is the distance

So,

To convert seconds into year divide the time by 3.154 × 10⁷

So, T = 29.55 Years

or T = 29.6 Years

Hence, the Saturn's orbital period is 29.6 Earth years       

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.