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Daniel [21]
3 years ago
12

In an insect colony there are 270 insects after 9 days. If there were initially 80 insects how long will it take the population

to grow to 800 insects?
Mathematics
1 answer:
Semenov [28]3 years ago
7 0

Answer:

The time it will take the population to grow to 800 insects is 17 days.

Step-by-step explanation:

The growth function of the insects is exponential.

The exponential growth function is:

y=a(1+r)^{t}

Here,

<em>y</em> = final value

<em>a</em> = initial value

<em>r</em> = growth rate

<em>t</em> = time taken

It is provided that there were 270 insects after 9 days and initially there were 80 insects.

Compute the value of <em>r</em> as follows:

y=a(1+r)^{t}\\\\270=80(1+r)^{9}\\\\3.375=(1+r)^{9}\\\\\ln(3.375)=9\cdot \ln(1+r)\\\\0.135155=\ln(1+r)\\\\1+r=1.14471\\\\r=0.145

Now, compute the time it will take the population to grow to 800 insects as follows:

800=80(1+0.145)^{t}\\\\10=(1.145)^{t}\\\\\ln(10)=t\cdot \ln(1.145)\\\\t=17.00522\\\\t\approx 17

Thus, the time it will take the population to grow to 800 insects is 17 days.

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A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position
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Answer:

233.48s  

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Step-by-step explanation:

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y=A cos(\omega t+\phi)+C

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A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

\omega= angular speed.

in this case:

\omega=\frac{2\pi}{T}

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\omega=\frac{2\pi}{360}

\omega = \frac{\pi}{160}

and

\phi= phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

y=-20 cos(\frac{\pi}{180}t)+22

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

-20 cos(\frac{\pi}{180}t)+22>13

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

-20 cos(\frac{\pi}{180}t)+22=13

and solve it:

-20 cos(\frac{\pi}{180}t)=13-22

-20 cos(\frac{\pi}{180}t)=-9

cos(\frac{\pi}{180}t)=\frac{9}{20}

and we can take the inverse of cos to get:

\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})

which yields two possible answers: (see attached picture)

so

\frac{\pi}{180}t=1.104 or \frac{\pi}{180}t=5.179

so we can solve the two equations. Let's start with the first one:

\frac{\pi}{180}t=1.104

t =1.104(\frac{180}{\pi})

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and the second one:

\frac{\pi}{180}t=5.179

t=5.179(\frac{180}{\pi})

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[0, 63.25]  for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25]  for a test value of 1

-20 cos(\frac{\pi}{180}(1))+22>13

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

-20 cos(\frac{\pi}{180}(70))+22>13

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

-20 cos(\frac{\pi}{180}(300))+22>13

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

233.48(\frac{1min}{60s})=3.84 min

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

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