Since f(x, y) = 1 + y2 and ∂f/∂y = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y(0) = 0. y = Even though x0 = 0 is in the interval (−2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = ± , the solution is not defined on (−2, 2).

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation:

shorter then the tree