C; this is because the line of beat fit isn’t linear but curved in that one graph.
Your answer here should be D because you multiply base times height times length and you get 30 m^3
I think that it is A. because it is less than the otehrs
<u>Answer:</u>
Planet A is inner
Planet A is Mars
Planet B is Outer
Planet B is Uranus
<u>Solution:</u>
We know that the inner planets are the planets which are close to the sun. They are relatively small, mostly rocky composition, and have few or no moons.
On other hand, the outer planets are the planets which are far away from the sun. They are mostly huge, ringed, gaseous and have several moons.
In the given problem,
Planet A has rocky mantle and iron core, less no of Moons and no rings, also due to 96% of carbon dioxide, 3% nitrogen and 1% other gases this is denser, and Hence Planet A is inner planet. As the distance from the sun is 1.5 AU and no of moons are 2, hence Planet A is Mars.
On the other hand, Planet B is gaseous with hydrogen and helium gas, hence it is also denser and it has large no of moons and faint rings. So Planet B is Outer planet. As the distance of the planet is 19.22 AU and has 27 moons, hence Planet B is Uranus.
Answer:
10m.
Step-by-step explanation:
Since there is no remainder we simply need to find the least common multiple of each of the lengths of Janice and Jasmin's strings until we find a multiple that matches for each. Like so...
2*1 = 2 5*1 = 5
2*2 = 4 5*2 = 10
2*3 = 6
2*4 = 8
2*5 = 10
Finally, we have found the first common multiple which is 10m. That means that the shortest equal length for both Janice's and Jasmin's ribbon is 10m.
