Arectagular, above ground pool is 18 feet long, 9 feet wide, and 54 inches deep. one cubic foot of water weighs about 62.4 pounds. what is the total weight of the water in the pool when it is completely full?

45,489.6 pounds!

Step-by-step explanation:

To solve this problem, first we need to find the volume of the pool.

We know that the dimensions of the pool are:

Width = 9ft

Length = 18ft

Height = 54in = 4.5 ft

So the volume of the pool is:

Volume = Width x Length x Height = 9ftx18ftx4.5ft = 729ft^3

We know that one cubic foot of water weights about 62.4 pounds. To fill the  pool, we will need 729ft^3 of water.

So the weight of the water is:

Weight = 729ft^3 x 62.4 pounds/ft = 45,489.6 pounds!

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64