Arectangular fish tank measures 26 cm by 20 cm by 18 cm. the tank is filled with water to a depth of 15 cm.
a. what is the volume of the water in ml?
b. how many liters is that?
c. how many more ml of water will be needed to fill the tank to the top? explain how you know.

volume of water = 7800 mL

volume of water in liters = 7.8 L

more water need to fill the tank to top =  1560 mL

Step-by-step explanation:

given data

tank length = 26 cm

tank width = 20 cm

tank height = 18 cm

water depth = 15 cm

to find out

volume of the water , How many liters is that and How many more mL of water will be needed to fill the tank to the top

solution

volume of water in tank is

volume of water = length ×weight×height

here height of water will be take

volume of water =26 ×20×15

volume of water = 7800 mL

and

we know 1 L = 1000 mL

volume of water = 7800 mL =

volume of water in liters = 7.8 L

and

so more water to fill tank is = volume of tank - volume of water

here volume of tank is = length ×weight×height

volume of tank is = 26 × 20×18 = 9360 mL

more water need to fill = 9360 - 7800 mL

more water need to fill =  1560 mL

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.