Xy + 3ey = 3e, find the value of y'' at the point where x = 0.

y = 1

Step-by-step explanation:

If the equation is:

x*y + 3*e^y = 3*e

where x and y are variables, and e is a real number, we have that when x = 0, the value of y must be:

0*y + 3*e^y = 3*e

3*e^y = 3*e

(3*e^y)/(3*e) = 1

e^(y - 1) = 1

we know that any number A, when we have A^0 = 1

so here we have that y must be equal to 1

so y = 1

2.83941151 would be the answer you're looking for! : )

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