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tangare [24]
3 years ago
7

The population of bald eagles is decreasing 4% each year. In 2015, there were 2800 bald eagles.

Mathematics
2 answers:
Vladimir [108]3 years ago
7 0

Answer:

1120 Eagles

Step-by-step explanation:

If the population of Eagles decreases by 4% percent each year, then in 2030 (which is 15 years later), the population would decrease by (4 × 15)%

that is 60%

This indicates that the population of bald eagles in the year 2030 would be 60% less of what it was in the year 2015.

so we have, \frac{60}{100} × 2800

which gives us 1680

this means that the number of bald eagles in the year 2030 would be 1680 less than the number of bald eagles in the year 2015

hence 2800 - 1680

then the answer is 1120 Bald Eagles

nikdorinn [45]3 years ago
6 0

Answer: the approximate population of bald eagles be in the year 2030 is 15178

Step-by-step explanation:

The population of bald eagles is decreasing 4% each year. This means that the rate is exponential.

We would apply the formula for exponential decay which is expressed as

A = P(1 - r)^t

Where

A represents the population after t years.

t represents the number of years.

P represents the initial population.

r represents rate of growth.

From the information given,

P = 2800

r = 4% = 4/100 = 0.04

t = 2030 - 2015 = 15 years

Therefore

A = 2800(1 - 0.04)^ 15

A = 2800(0.96)^ 15

A = 15178

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Answer:

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To find the inverse of a function, swap the places of the x and y variables and solve for y again.

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4 years ago
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

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Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

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Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

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y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

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y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

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Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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A dime is a ten cent coin while a quarter is a 25 cent coin. The total of the money should be $8.40. Let x be the number of dime and y be the number of quarter. The equation that would best represent the given problem is,
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Answer: im on this one too

Step-by-step explanation:

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