Answer:
3m +2≤ 53
Step-by-step explanation:
thats just how it goes logically
9514 1404 393
Answer:
- see below for a sketch
- 12 km
Step-by-step explanation:
The distance can be calculated using the Law of Cosines. The angle internal to the triangle at Q is (180°-(146° -65°)) = 99°. Then the distance PR can be found from ...
PR² = PQ² +QR² -2·PQ·QR·cos(∠PQR)
PR² = 6² +10² -2·6·10·cos(99°) ≈ 154.77
PR ≈ √154.77 ≈ 12.44 . . . . km
The distance PR is about 12 km.
Nearly 81 moons will be required to equate the mass of moon to the mass of earth.
Step-by-step explanation:
Mass of earth is 5.972*10^24 kg.
Mass of the moon is 7.36*10^25 g = 7.36*10^22 kg
As mass of the Earth is given as 5.972 * 10^24 kg and mass of the moon is given as 7.36 * 10^22 kg, then the number of moons required to make it equal to the mass of earth can be calculated by taking the ratio of mass of earth to moon.
Mass of Earth = Number of moons * Mass of Moon
Number of Moons = Mass of Earth/Mass of moon
Number of moons = 5.972 * 10^24/7.36*10^22= 81 moons.
So nearly 81 moons will be required to equate the mass of moon to the mass of earth.
From the graph: there are 200 students overall. For the next 100, we can divide each number by 2 to find an estimate for how many with participate in MATH tutoring. 40 divided by 2 =20 students participating in MATH tutoring.
Part B- There are currently 55 students out of 200 in SCIENCE tutoring. We add 55 more for 400 = 110, add 55 more for 600=165, and then add half of 55 (27.5 or 28 because you CANNOT have half a student) for 700 students total. 55+55+55+28=193 students.
The union of the sets are the values that appear in both sets when plotted in a Venn diagram and if I'm not mistaking, I think the right answer is (B)
