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Dennis_Churaev [7]
2 years ago
7

Describe Freshwater wetlands. A good description.

Advanced Placement (AP)
2 answers:
juin [17]2 years ago
7 0

Answer: Freshwater wetlands are not connected to the ocean. They can be found along the boundaries of streams, lakes, ponds or even in large shallow holes that fill up with rainwater. Freshwater wetlands may stay wet all year long, or the water may evaporate during the dry season.

uysha [10]2 years ago
4 0

Answer:

hrghehhehr

Explanation:

hdhehejhehehehehehehehehehheheheheghhhdhdbrghrhrhrhrhrrhrhrhrhrhrhrhrhrgrgrgrgrhrg

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Circles A and B are externally tangent to each other. From center B a tangent is drawn to circle A intersecting the circle at C.
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The are of the circles is 39.14
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3 years ago
State one difference between the eclipse of the sun and Moon​
beks73 [17]

Answer:

One difference is that the moon leaves a shadow on the earth's surface that eclipse is called a solar eclipse. and when the the earth passes between the sun and the moon it casts a shadow on the moon making a lunar eclipse

Explanation:

i hope this helps

4 0
2 years ago
2. Find the solution of each differential equation. (a) y/2y-8y = 0 (b) 25y/- 20y + 4y = 0 (c) y + 2y + 2y = 0 2. Find the solut
jek_recluse [69]

Each of these ODEs is linear and homogeneous with constant coefficients, so we only need to find the roots to their respective characteristic equations.

(a) The characteristic equation for

y'' - 2y' - 8y = 0

is

r^2 - 2r - 8 = (r - 4) (r + 2) = 0

which arises from the ansatz y = e^{rx}.

The characteristic roots are r=4 and r=-2. Then the general solution is

\boxed{y = C_1 e^{4x} + C_2 e^{-2x}}

where C_1,C_2 are arbitrary constants.

(b) The characteristic equation here is

25r^2 - 20r + 4 = (5r - 2)^2 = 0

with a root at r=\frac25 of multiplicity 2. Then the general solution is

\boxed{y = C_1 e^{2/5\,x} + C_2 x e^{2/5\,x}}

(c) The characteristic equation is

r^2 + 2r + 2 = (r + 1)^2 + 1 = 0

with roots at r = -1 \pm i, where i=\sqrt{-1}. Then the general solution is

y = C_1 e^{(-1+i)x} + C_2 e^{(-1-i)x}

Recall Euler's identity,

e^{ix} = \cos(x) + i \sin(x)

Then we can rewrite the solution as

y = C_1 e^{-x} (\cos(x) + i \sin(x)) + C_2 e^{-x} (\cos(x) - i \sin(x))

or even more simply as

\boxed{y = C_1 e^{-x} \cos(x) + C_2 e^{-x} \sin(x)}

3 0
1 year ago
Which of the following was not one of Napoleon bonaparte's activities?
strojnjashka [21]
An establishment of the directory (i think)
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skad [1K]

Answer:

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Explanation:

4 0
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