Erin had a container of water that was partially full. She used a garden hose to fill the rest of the cintaiber at a constant rate. Erin recorded the height of the water at the end of diffrent intervals of
time in this table.

Part A

what was the height in centimeters, of the water before erin started to fulk the rest of the container?Round your answer to the nearest whole number.

Part B

After how many minutes was the height of the water double what it was before Erin started to fill the rest of the cintaiber?Round your answer to the nearest whole minute.

Part C

Does this equation model this relationship:h=2.7m+35?Explain why or why not.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58

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.

3.73 * 10^7 is the anser for ttm ur welcome < 3

step-by-step explanation: