Graph-2x+y=8 I need to know the answer

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

Over the interval [0, 2π), what are the solutions to cos(2x) = cos(x)? Check all that apply.

valentinak56 [21]

Answer:

Step-by-step explanation:

given the expression;

cos(2x) = cos(x)

According to trig identity;

cos(2x) = cos(x+x)

cos(2x) = cos x cos x - sinx sinx

cos(2x) = cos²(x)-sin²(x)

cos(2x) = cos²(x)-(1-cos²x)

cos(2x) = cos²(x)+cos²x-1

cos(2x) = 2cos²(x)-1

2cos²(x)-1 = cos(x)

let P = cosx

2P²-1 = P

2P²-P-1 = 0

Factorize;

2P²-2P+P-1 = 0

2P(P-1)+1(P-1) = 0

2P - 1 = 0 and P-1 = 0

P = 1/2 and 1

cosx = 1/2 and cos x = 1

x = arccos 1/2

x = π/3

Also;

x = arccos1

x = 0

Hence the value of x are 0 and π/3

Also the angle = π+ π/3 = 4π/3

The angles are 0, π/3 and 4π/3

4 0

3 years ago

Read 2 more answers

Please help i need to complete this by tomorrow

nekit [7.7K]

Answer:

ED: 6.5 cm BE = 14.4 cm

Step-by-step explanation:

Add 20 and 5 to get side AC.

25/20 gives us the scale factor to go from 25 triangle ABE to triangle ACD.

Multiply 26 by the SF to get AD.

Subtract length of AD by AE to get ED. You should get 6.5 cm (don't forget units!!!!)

To go from triangle ACD to triangle ABE, we can do 20/25 to get the scale factor.

Using that number, multiply it with 18 to get BE.

The answer should be 14.4 cm.

This is possible because the two triangles ACD and ABE are similar via the angle angle angle similarity theorem depicted in my screenshot.

3 0

3 years ago

Read 2 more answers

Angle Sum Theorem Acellus<br> 20 120 y = ?

Dmitry [639]

We know

Sum of two interior angles=exterior angle

$\\ \sf\longmapsto 20°+120°=$

$\\ \sf\longmapsto$

Hope it helps

6 0

3 years ago

I miiiiighhhhtttt need more helpppp

Julli [10]

Answer: 5
Explanation: 16 divided by 2 equals 8 and 10 divided by 2 is 5, just trust me

8 0

3 years ago