Answer:
True
Explanation:
Kepler's second law states that for a planet orbiting a star, if we join the planet and the star by an imaginary line, such line sweeps equal areas in equal times.
Let's imagine the planetary orbit is elliptical. When the planet is closest to the star it travels a large distance in its orbit but takes little time to travel. When the planet is furthest from the star, it travels a small distance in the orbit but in a long time so it ends up sweeping the same are as in the first case.
Therefore, this means when the planet is closest to the star it must have a greater speed than when it's furthest. We know this thanks to the speed formula v = d/t (velocity equals distance divided by time).
For a great distance and small time, speed is a large value.
For a small distance and a great time, speed is a small value.
The answer would be the last option
ST=21, SP=3b-11, PT=B+4
ST=SP+PT
21=(3b-11)+(b+4)
21=3b-11+b+4
21=4b-7
21+7=4b
4b=28
b=7
It could either be of these Halite, Calcite,Granite igneous rock,Basalt-igneous
rock,Metamorphic,Gabbro-igneous,Igneous rock,Metamorphic
rocks,Sedimentary rock, or Obsidian rock.
The slope of 3x+6y=12 is y=2
