Answer: 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.
Answer:
3
Step-by-step explanation:
-4-(-7)=
-4+(7)=
3
I’m not sure if this is right but im guessing c.
That would simply be
38.14 ÷ 4 = 9.535
thus he scored 9.535 on each event
Unfortunately, multiplying and dividing decimals this effect does not occur. In some cases, the decimal number even complicates the operatsii.Dlya beginning, we introduce a new definition. We will meet him very often, and not only in this part of uroke.Znachaschaya - is all that is between the first and the last non-zero digit, including the ends. It's only about numbers, the decimal point is not uchityvaetsya.Tsifry included in the meaningful part of the number, are called significant figures. They may be repeated or even be zero.
