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nydimaria [60]
4 years ago
14

I need the answers:

Advanced Placement (AP)
1 answer:
IRINA_888 [86]4 years ago
3 0
It should be A since fish are the omly animals there without lungs...
You might be interested in
If the pond is resampled a year later, which of the following would best indicate that the pond had been adversely affected by a
Usimov [2.4K]

This question is not complete; here is the complete question:

A one-hectare pond is sampled in early September. The sample yields 1 small catfish as well as 17 benthic invertebrates that represent 10 species.

If the pond is resampled a year later, which of the following would best indicate that the pond had been adversely affected by adjacent development?

A. An uncommon species has become more numerous

B. An increase in low-tolerance species has occurred

C. A decrease in high-tolerance species has occurred

D. Phylogenetic diversity has occurred

E. The biodiversity of the pond has decreased

The answer to this question is E. The biodiversity of the pond has decreased

Explanation:

In ecology, an ecosystem can be affected by adjacent developments including human developments or even other ecosystems and this can lead to positive or negative consequences.

In this context, one example that shows the was an adverse effect is "The biodiversity of the pond has decreased" because the reduction in diversity represents a major thread for an ecosystem as this makes species more vulnerable to disappear. Also, this can be caused by factors such as pollution caused by human development or major predator of the adjacent ecosystem feeding on organisms in the main ecosystem. Besides this, in the first sample, there were multiple species and a reduction in the second sample shows the ecosystem was weakened.

3 0
3 years ago
Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0​
skad [1K]

Answer:

Option A.  y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}

Explanation:

This is a second order DE, so we'll need to integrate twice, applying initial conditions as we go.  At a couple points, we'll need to apply u-substitution.

<u>Round 1:</u>

To solve the differential equation, write it as differentials, move the differential, and integrate both sides:

y''=e^{-2t}+10e^{4t}

\frac{dy'}{dt}=e^{-2t}+10e^{4t}

dy'=[e^{-2t}+10e^{4t}]dt

\int dy'=\int [e^{-2t}+10e^{4t}]dt

Applying various properties of integration:

\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt

Prepare for integration by u-substitution

\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt, letting u_1=-2t and u_2=4t

Find dt in terms of u_1 \text{ and } u_2

u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt     u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt

\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt\\\int dy'=\int e^{u_1} (-\frac{1}{2} du_1) + 10\int e^{u_2}  (\frac{1}{4} du_2)\\\int dy'=-\frac{1}{2} \int e^{u_1} (du_1) + 10 *\frac{1}{4} \int e^{u_2}  (du_2)

Using the Exponential rule (don't forget your constant of integration):

y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1

Back substituting for u_1 \text{ and } u_2:

y'=-\frac{1}{2} e^{(-2t)} + 10 *\frac{1}{4}e^{(4t)} +C_1\\y'=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\

<u>Finding the constant of integration</u>

Given initial condition  y'(0)=0

y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\0=y'(0)=-\frac{1}{2} e^{-2(0)} + \frac{5}{2}e^{4(0)} +C_1\\0=-\frac{1}{2} (1) + \frac{5}{2}(1) +C_1\\-2=C_1\\

The first derivative with the initial condition applied: y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\

<u>Round 2:</u>

Integrate again:

y' =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\\frac{dy}{dt} =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\dy =[-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int [-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int -\frac{1}{2} e^{-2t} dt + \int \frac{5}{2}e^{4t} dt - \int 2 dt\\\int dy = -\frac{1}{2} \int e^{-2t} dt + \frac{5}{2} \int e^{4t} dt - 2 \int dt\\

y = -\frac{1}{2} * -\frac{1}{2} e^{-2t} + \frac{5}{2} * \frac{1}{4} e^{4t} - 2 t + C_2\\y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + C_2

<u />

<u>Finding the constant of integration :</u>

Given initial condition  y(0)=1

1=y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + C_2\\1 = \frac{1}{4} (1) + \frac{5}{8} (1) - (0) + C_2\\1 = \frac{7}{8} + C_2\\\frac{1}{8}=C_2

So, y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}

<u>Checking the solution</u>

y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}

This matches our initial conditions here y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1

Going back to the function, differentiate:

y' = [\frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}]'\\y' = [\frac{1}{4} e^{-2t}]' + [\frac{5}{8} e^{4t}]' - [2 t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} [e^{-2t}]' + \frac{5}{8} [e^{4t}]' - 2 [t]' + [\frac{1}{8}]'

Apply Exponential rule and chain rule, then power rule

y' = \frac{1}{4} e^{-2t}[-2t]' + \frac{5}{8} e^{4t}[4t]' - 2 [t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} e^{-2t}(-2) + \frac{5}{8} e^{4t}(4) - 2 (1) + (0)\\y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2

This matches our first order step and the initial conditions there.

y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0

Going back to the function y', differentiate:

y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2\\y'' = [-\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2]'\\y'' = [-\frac{1}{2} e^{-2t}]' + [\frac{5}{2} e^{4t}]' - [2]'\\y'' = -\frac{1}{2} [e^{-2t}]' + \frac{5}{2} [e^{4t}]' - [2]'

Applying the Exponential rule and chain rule, then power rule

y'' = -\frac{1}{2} e^{-2t}[-2t]' + \frac{5}{2} e^{4t}[4t]' - [2]'\\y'' = -\frac{1}{2} e^{-2t}(-2) + \frac{5}{2} e^{4t}(4) - (0)\\y'' = e^{-2t} + 10 e^{4t}

So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.

7 0
2 years ago
Allie is shopping when she finds a pair of running shoes priced at $90. When Allie uses her debit card to pay, it is declined be
Nadusha1986 [10]

Answer:

The economic concept of scarcity.

Explanation:

In economics, <em>scarcity</em><em> </em>represents the phenomenon of <em>limitless</em> <em>wants</em> suppressed by <em>limited</em><em> </em><em>resources</em>.

In this case, Allie feels she needs $90 shoes while she has not got the resources required to buy them.

This typical economic problem can be solved by moderating one's wants and clearly identifying what is priority from what is not, then intelligently making decisions on what available resources should be spent.

7 0
4 years ago
12+9-8=?<br><br>Who is part of the LGBTQ+ <br><br>do not answer if no.<br><br>I AM!!!
Delicious77 [7]
12+9-8=13

I AM!!!!!!!!️‍❤️







4 0
3 years ago
Read 2 more answers
(01.02 MC)
sergiy2304 [10]

Answer:

Meddlesome and eager.

Explanation:

The given text is taken from Jane Austen's novel titled 'Sense and Sensibility.' The novel is centered around the Dashwood family.

Mrs Jennings is a widow of a man who left ample 'juncture' for her. She spends most of her time visiting her children's household, where they are married off. She is a lady who kvells to make pair of young people, specifically of Elinor and Marianne.

From the first paragraph of the text one can infer her to be meddlesome and eager.

Meddlesome is a quality of interfering into other's business. Mrs. Jennings is a character who loves to interfere in the business of young people and is eager to attach them.

Thus the correct answer is the fourth option.

4 0
3 years ago
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