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Licemer1 [7]
3 years ago
9

CAN SOMEONE HELP ASAP (Economics)

Mathematics
2 answers:
bonufazy [111]3 years ago
6 0
Substitution effect
income effect
positive externality
negative externality
lutik1710 [3]3 years ago
4 0
The andswers are 1-b,2-a,3-d,4-c hope it helped
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What percentage of one hour is fifteen minutes?
Scilla [17]

 1 hour = 60 minutes

15 minutes = (60/15) = 1/4th of an hour

1/4 = 0,25

<span>15 minutes is 25% of an hour</span>

8 0
4 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
I need help and ill give brainliest
elena-s [515]

Answer:

(-2,7) reflects to (-2,-7)

(3,9) reflects to (3, -9)

(7,2) reflects to (7,-2)

(-3,9) reflects to (-3,-9)

Step-by-step explanation:

in  reflection across the x-axis the x value of the coordinate stays the same and the y value is negated to show it been reflected

7 0
3 years ago
Differentiate from first principles <br> y=3x
wel

\frac{d}{dx} 3x = 3

\frac{d}{dx} ax = a

8 0
3 years ago
What value should be added to the expression to create a perfect square?
raketka [301]
25 would be your final answer.
5 0
4 years ago
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