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Art [367]
3 years ago
9

Can somebody please help.

Mathematics
1 answer:
ella [17]3 years ago
6 0

Answer:

Step 1: Distributed - to 8 and a

Step 2: Combined like terms or 5a and a

Step 3: Added -1.5 to each side

Step 4: Subtracted 6a from both sides

Step 5: Divided -2 from both sides

Step-by-step explanation:

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A one-parameter family of solutions of the DE P' = P( 1 - P) is given below. P = c1et/1 + c1et Does any solution curve pass thro
topjm [15]

Answer:

a. The curve P(t) = -\frac{6e^t}{5-6e^t} passes through the point (0, 6)

b. No solution of the curve P(t) passes through the point (0, 1)

Step-by-step explanation:

Consider the family of the solution of DE P' = P(1 - P) is P = \frac{c_1e^t}{1 + c_1e^t}

a. If any solution passes through the point (0, 6), then there is c_1 such that the point (0, 6) satisfies the solution P = \frac{c_1e^t}{1 + c_1e^t}

Substitute t = 0, P = 6 in P = \frac{c_1e^t}{1 + c_1e^t} and then solve the equation to obtain c_1

P(t) = \frac{c_1e^t}{1 + c_1e^t}\\P(0) = \frac{c_1e^0}{1+c_1e^0}\\ 6 = \frac{c_1}{1 + c_1}\\ c_1 =  -\frac{6}{5}

Therefore, the curve P(t) = -\frac{6e^t}{5 - 6e^t} passes through the point (0, 6)

b.  If any solution passes through the point(0, 1), then there is c_1 such that the point (0, 1) satisfies the solution P = \frac{c_1e^t}{1+c_1e^t}

P(t) = \frac{c_1e^t}{1 + c_1e^t}\\ P(0) = \frac{c_1e^0}{1 + c_1e^0}\\ 1 = \frac{c_1}{1+c_1} \\1 + c_1 = c_1

this is not possible

Hence, there is no curve P(t) that exists which passes through the point (0, 1)

4 0
2 years ago
2/5 ≤ x−4 answer, as an inequality
Greeley [361]

\dfrac{2}{5}\leq x-4\qquad|\text{add 4 to both sides}\\\\4\dfrac{2}{5}\leq x\to x\geq4\dfrac{2}{5}

7 0
2 years ago
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