Ind the jacobian of the transformation xequals=nothingu, yequals=44uv and sketch the region g: 1 less than or equals u less than or equals 3 comma 4 less than or equals 4 uv less than or equals 12g: 1≤u≤3, 4≤4uv≤12, in the uv-plane. b. then use modifyingbelow integral from nothing to nothing integral from nothing to nothing with upper r f (x comma y )dx dy equals modifyingbelow integral from nothing to nothing integral from nothing to nothing with upper g f (g (u comma v )comma h (u comma v ))startabsolutevalue upper j (u comma v )endabsolutevalue du dv∫∫rf(x, y) dx dy=∫∫gf(g(u, v),h(u, v))j(u, v) du dv to transform the integral integral from 1 to 3 integral from 4 to 12 startfraction y over x endfraction dy dx∫13 ∫412 y x dy dx into an integral over g, and evaluate both integrals.

1. minimum

2. 6400

3. eight

hope this !

c

step-by-step explanation:

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