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3 years ago

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A population of largemouth bass is 950 this year and the rate of decrease is 35% per year. How many bass will remain after 15 ye

ars of decay at this rate?
425
617
1
525
332

1 answer:

GalinKa [24]3 years ago

6 0

Hi! I think it’s 425

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If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams

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Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

$N=N_0e^{-\lambda t}$

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

$\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34$ , t= 10

$N=N_0e^{-\lambda t}$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow \frac34 =e^{-\lambda .10}$

Taking ln both sides

$\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|$

$\Rightarrow ln|\frac34|=-10\lambda$

$\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}$

Now , N₀= 512 grams, N= 121.5 grams, t=?

$N=N_0e^{-\lambda t}$

$\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}$

$\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}$

$\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}$

Taking ln both sides

$\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|$

$\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}$

$\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}$

$\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}$

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

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Answer:

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Step-by-step explanation:

We re-write the equation of the line in the format:

$\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}$

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

$\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>$

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

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Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

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Then to find the distance we just use the distance formula:

$\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

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