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Tpy6a [65]
3 years ago
9

M(x)=15x-x^2 solve for x

Mathematics
1 answer:
user100 [1]3 years ago
3 0

The maximum number of mosquitoes would exist if 7.5 inches of rain fell.


We can find that by looking for the maximum value in the equation. This value always falls at the vertex. The x-value of the vertex (which is what we are looking for) can always be found through the equation -b/2a, in which a is equal to the coefficient of x^2 (-1) and b is equal to the coefficient of x (15).


-b/2a

-(15)/2(-1)

-15/-2

7.5

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Step-by-step explanation:

Make a proportion:

4/25 = x/120

cross multiply

480 = 25x

Divide both sides by 25

X= 19.2

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Joey is twenty years younger than Becky in two years time Becky will be twice as old as Joey
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According to a commercial, 4 out of 5 dentist recommend a certain brand of toothpaste. Suppose There are 120 dentist in your are
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4/5 recommend, so about 4/5(120) = 96 doctors recommend the brand.

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3 years ago
A tree is leaning at a 5° angle from vertical, making an 85° angle with the ground. When the sun is at a 65° angle of elevation,
Vinvika [58]
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= 180° - 150° = 30°.
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3 0
3 years ago
The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

6 0
3 years ago
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