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Rudik [331]
3 years ago
7

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).
Mathematics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

(3x, 6)

(3, 8)

Step-by-step explanation:

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Answer:

P(A) = 0.45

P(B) = 0.32

Step-by-step explanation:

Given

P(A) > P(B)

P(A\ u\ B) = 0.626

P(A\ n\ B) = 0.144

Required

Find P(A) and P(B)

We have that:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B) --- (1)

and

P(A\ n\ B) = P(A) * P(B) --- (2)

The equations become:

P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B) --- (1)

0.626 = P(A) + P(B) - 0.144

Collect like terms

P(A) + P(B) = 0.626 + 0.144

P(A) + P(B) = 0.770

Make P(A) the subject

P(A) = 0.770 - P(B)

P(A\ n\ B) = P(A) * P(B) --- (2)

0.144 = P(A) * P(B)

P(A) * P(B) = 0.144

Substitute: P(A) = 0.770 - P(B)

[0.770 - P(B)] * P(B) = 0.144

Open bracket

0.770P(B) - P(B)^2 = 0.144

Represent P(B) with x

0.770x - x^2 = 0.144

Rewrite as:

x^2 - 0.770x + 0.144 = 0

Expand

x^2 - 0.45x - 0.32x + 0.144 = 0

Factorize:

x[x - 0.45] - 0.32[x - 0.45]= 0

Factor out x - 0.45

[x - 0.32][x - 0.45]= 0

Split

x - 0.32= 0 \ or\ x - 0.45= 0

Solve for x

x = 0.32\ or\ x = 0.45

Recall that:

P(B) = x

So, we have:

P(B) = 0.32 \ or \ P(B) = 0.45

Recall that:

P(A) = 0.770 - P(B)

So, we have:

P(A) = 0.770 - 0.32 \ or\ P(A) =0.770 - 0.45

P(A) = 0.45 \ or\ P(A) =0.32

Since:

P(A) > P(B)

Then:

P(A) = 0.45

P(B) = 0.32

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