Jayne stopped to get gas before going on a road trip. the tank already had 4 gallons of gas in it. which best describes why the graph relating the total amount of gasoline in the tank, y, to the number of gallons that she added to it, x, will be continuous or discrete? the graph will be continuous because the amount of gas that she added to the tank does not need to be an integer amount. the graph will be continuous because we are not told a maximum value for the amount of gas. the graph will be discrete because there are already exactly 4 gallons of gas in the tank, so to fill it up will take a whole number of gallons of gas. the graph will be discrete because there is an end to the amount of gas she can use, as the tank will be completely full at some point.

Jayne stopped to get gas before going on a road trip. The tank already had 4 gallons of gas in it. Which best describes why the graph relating the total amount of gasoline in the tank, y, to the number of gallons that she added to it, x, will be continuous or discrete?

The graph will be continuous because the amount of gas that she added to the tank does not need to be an integer amount.

The graph will be continuous because we are not told a maximum value for the amount of gas.

The graph will be discrete because there are already exactly 4 gallons of gas in the tank, so to fill it up will take a whole number of gallons of gas.

The graph will be discrete because there is an end to the amount of gas she can use, as the tank will be completely full at some point.

b, i believe, but not sure.

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.