An albatross is 5 meters above the surface of the ocean. How does its distance from the surface compare to the vertical distance from the eyes of: The jumping dolphin? The flying seagull? The octopus?

The jumping dolphin --  changes

The flying seagull -- same

The octopus --  same

Step-by-step explanation:

The distance of the Albatross from the sea surface when compared to the vertical distance from the eyes of the :

The jumping dolphin :

As the dolphin jumps out from the sea, the vertical distance of the albatross from the surface seems to be decreasing and then increasing as seen by the jumping dolphin.

The flying seagull :

The distance of the albatross is at the same height as seen by the seagull since it is flying at the same level, i.e. at the level of the sea surface

The octopus :

The octopus remains at the same depth below the water surface. So the distance of the albatross for the octopus remains the same at 5 m above the ocean surface.

part d

step-by-step explanation:

theresa talked her parents into getting sd cards for her phone and her brother’s phone. inserting an sd card into a phone gives it more storage. they both get 8-gigabyte sd cards. how many bits of storage come with each sd card? type your answer in both scientific and standard notation.

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.