This data gives the length of the radius, in feet, of several circles Length ft: 1/2 3/16 1/2 1/2 3/8 3/16 1/2 3/16 7/16 create a line plot, hover over each number on the number line then drag up to plot the data​

b

step-by-step explanation:

a difference is the result of subtraction. while each of these expressions has subtraction in it, a difference will be the result when subtraction is the last operation applied in the order of operations for the expression. this means subtraction must be outside of the parenthesis. only b has this.

3

step-by-step explanation:

well kind of. you get friend requests. and you can accept them or not so yes.

step-by-step explanation:

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.