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4 years ago

Plz help me with this

Mathematics

1 answer:

Furkat [3]4 years ago

6 0

The triangle is upon a straight angle. So subtract 139 from 180.

180-139=41

Add 103+41=144

Subtract 180-144=36

x=36

103+41+36=180

You had to find the angle that was not mentioned (41), then use that angle to complete the problem. A triangle always adds up to 180.

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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)

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3 years ago