A lake near the Arctic Circle is covered by a thick sheet of ice during the cold winter months. When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a rate of 0.2 meters per week. After 7 weeks, the sheet is only 2.4 meters thick. Let y represent the ice sheet's thickness (in meters) after ___ weeks.

Complete the equation for the relationship between the thickness and number of weeks.

y=

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

step-by-step explanation:

a cube has 6 sides

the probability of getting 5 on the first cube is 1/6 since only one side of the cube has 5 on it.

there are 3 sides of the cube with even number on it [2,4,6]

therefore the probability of getting even number is 3/6

for both events to happen together, the probability would be

1/6 x 3/6 = 12/36

which simplifies to 1/12

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.