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ludmilkaskok [199]
2 years ago
8

In the figure below, BD and EC are diameters of circle P. What is the arc measure of ADE in degrees? IMAGE BELOW

Mathematics
1 answer:
Pani-rosa [81]2 years ago
8 0

Answer:

The arc measure of ∠ADE is equal to 333 degrees.

Step-by-step explanation:

Please refer to the attached diagram,

Let the unknown angle ∠APE = x

Since BD is the diameter of the circle then

∠APE + ∠DPE + ∠APB = 180°

∠APE + 63° + 90° = 180°

x + 63° + 90° = 180°

Solving for x

x = 180° - 90° - 63°

x = 90° - 63°

x = 27°

Since a circle has 360° then

∠ADE + ∠APE = 360°

∠ADE + x = 360°

∠ADE + 27° = 360°

∠ADE = 360° - 27°

∠ADE = 333°

Therefore, the arc measure of ∠ADE is equal to 333 degrees.

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Answer:

J 672 cubic inches

Step-by-step explanation:

The volume of a rectangular prism is

V = l*w*h

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2 years ago
Suppose a certain population satisfies the logistic equation given by dP
Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

\frac{dP}{dt}=10P-P^2

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

\frac{dP}{10P-P^2}=dt

Integrate both sides.

\int \frac{dP}{10P-P^2}=\int dt             .... (1)

Using partial fraction

\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}

A=\frac{1}{10},B=\frac{1}{10}

Using these values the equation (1) can be written as

\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt

\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt

On simplification we get

\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C

\frac{1}{10}(\ln \frac{P}{10-P})=t+C

We have P(0)=1

Substitute t=0 and P=1 in above equation.

\frac{1}{10}(\ln \frac{1}{10-1})=0+C

\frac{1}{10}(\ln \frac{1}{9})=C

The required equation is

\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})

Multiply both sides by 10.

\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}

e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}

\frac{P}{10-P}=\frac{1}{9}e^{10t}

Reciprocal it

\dfrac{10-P}{P}=9e^{-10t}

P(t)=\dfrac{10}{1+9e^{-10t}}

The population when t = 3 is

P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}

Using calculator,

P=9.999\approx 10

Therefore, the population when t = 3 is 10.

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What postulate or theorem can be used to prove that these two triangles are congruent?
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Answer:

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Step-by-step explanation:

Let us revise the cases of congruence  

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In the given figure

∵ There is a pair of vertically opposite angles

∵ The vertically opposite angles are congruent ⇒ (1)

∵ There are two angles have the same mark

∴ These marked angles are congruent ⇒ (2)

∵ There are two sides have the same mark

∴ These two marked sides are congruent ⇒ (3)

→ From (1), (2), and (3)

∴ The two triangles have 2 angles and 1 side congruent

→ By using case 4 above

∴ The two triangles are congruent by the AAS postulate of congruency.

AAS postulate can be used to prove that these two triangles are congruent

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