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3 years ago

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A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

ivides into two cells every 20 minutes. The initial population of a culture is 60 cells. Find the relative growth rate.

1 answer:

qwelly [4]3 years ago

7 0

Answer:

Step-by-step explanation:

Given that,

Cells are divided into two cell every 20mins, this implies that the half-life is 20mins

Initial population growth is 60cells

Growth rate r?

Population growth is modeled as

P(t) = Po•exp(rt)

Where,

Po is the initial growth at t= 0

Po= 60

P(t) is population at any time t.

Since the population doubles every 20mins(⅓hr)

Then, P(⅓) = 2×60 =120

Then, P(t) = 120cells

So, applying the formula

P(t) = Po•exp(rt)

120 = 60•exp(⅓r)

120/60 = exp(r/3)

2 = exp(r/3)

Take In of both sides

In(2) = r/3

Cross multiply

r = 3In(2)

r = 2.079 /hour

The growth rate is 2.079/hour.

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Solve<br><img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

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Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

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