Problems that require solving the three-dimensional schrödinger equation can often be reduced to related one-dimensional problems. an example of this would be the particle in a cubical box. consider a cubical box with rigid walls (i. e., u(x, y,z)=∞ outside of the cube) and edges of length l. the general solution for this problem is

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt