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2 years ago

Anything that applies. sec0 is undefined for 0 =

Mathematics

1 answer:

emmainna [20.7K]2 years ago

5 0

If the sec θ is not defined, then the value of angle θ will be (2n + 1)π/2.

<h3>What is trigonometry?</h3>

The connection between the lengths and angles of a triangular shape is the subject of trigonometry.

The trigonometric expression is given below.

⇒ sec θ

If the sec θ is not defined, then the value of angle θ will be (2n + 1)π/2.

Where n is the any integer.

More about the trigonometry link is given below.

brainly.com/question/22698523

#SPJ1

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Find the ratio of 1 1/2 hours to 25 minutes.

murzikaleks [220]

Answer:

18:5???

Step-by-step explanation:

1 1/2hrs = 90 mins

90 / 5 = 18

25 / 5 = 5

I'd say: 18 : 5??

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4 years ago

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14 2/3 + 1 7/9 - 11 14/29

NARA [144]

Answer:exact form 1295/261

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Step-by-step explanation:

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3 years ago

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What's the least common multiple of 8, 11, and 12?

Lena [83]

8 = 2 x 2 x 2
11 = 11
12 = 2 x 2 x 3

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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

SOMEONE PLEASE HELP IM GONNA POKE MY BRAIN OUT IF I CAN'T FIGURE OUT HOW TO SOLVE THIS !!!!!!!!! PLEAAAAAAAAAAAAASE HELP !!!

Novosadov [1.4K]

Answer:

Step-by-step explanation:

∠F=\frac{78+54}{2}=\frac{132}{2}=66°

∠Q=180-66=114°

8 0

3 years ago