Water enters a lake at the rate of 2 cubic kilometers per day. Water evaporates from the lake at the rate of 1.2 cubic kilometers per day, and the dam at the end of the lake regulates the flow so that the volume in the lake stays at a constant 245 cubic kilometers. The water entering the lake contains 2.9 tons of salt per cubic kilometer, and the water leaving the lake is well mixed. The differential equation for the amount of salt in the lake, measured in tons, with time measured in days, is. a) Q' = 8.41 - 0.00 163 Q b) Q' = 8.7 - 0.00816 Qc) Q' = 4.2 - 0.0049 Qd) Q' = 5.8 -0.00327 Q

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

answer: i ! the answer is   9000 seconds, also d

hope i

-lavira : )