Mathematics, 18.09.2019 00:30 munchyswish

The e. coli bacteria has a volume of 6 µm3 . in optimal conditions, an e. coli bacteria will double about every 30 minutes. under these conditions, how long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ? how long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

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Answer from: Netnee

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

It will take 57.6 hours to for the volume to fill the entire earth

Step-by-step explanation:

We can say that the volume of the bacteria is a geometric sequence, and each time moment is an arithmetic sequence.

Each geometric sequence has the following format:

${a, ar, ar^{2}, ar^{3},...}$

In which:

a is the first term

r is the common ratio.

We can find any term of the sequence by the following equation:

$x_{n} = ar^{(n-1)}$

Each arithmetic sequence has the following format:

{a, a+d, a+2d,...}

In which:

a is the first term

d is the difference between the terms.

We can find any term of the sequence by the following equation:

$x_{n} = a + d(n)$

How i am going to solve this problem.

We have the sequence that is the volume of the bacteria:

Obs: $1(um)^{3} = 10^{-18}m^{3}$

$V_{n} = {6*10^{-18}, 12*10^{-18},...}$

$a = 6*10^{-18}$

r = 2

And the following arithmetic sequence that are the time(in hours).

$T_{n} = {0,0.5,1,...}$

a = 0

d = 0.5

$T_{n} = 0.5n$

How long will it take for a single bacterium to grow to fill a thimble with volume 1 cm3 ?

I am going to find the value of n for which $V_{n} = 1cm^{3}$ , then i find the value at this position in the arithmetic sequence. So

$1cm^{3} = 10^{-6}m^{3}$

$V_{n} = 6*10^{-18}*(2^{(n-1)})$

$10^{-6} = 6*10^{-18}*(2^{(n-1)})$

$\frac{10^{-6}}{6*10^{-18}} = 2^{(n-1)})$

Obs: $a^{b-c} = \frac{a^{b}}{a^{c}}$

$\frac{10^{12}}{6} = \frac{2^{n}}{2}$

$2^{n} = \frac{10^{12}}{3}$

Now, we have to apply these following logarithim proprierties to find the value of n:

$log a^{n} = n log a$

$log(\frac{a}{b}) = log a - log b$

$log 10^{n} = n$

log 2 = 0.30

log 3 = 0.48

$log 2^{n} = log(\frac{10^{12}}{3})$

$n log 2 = log(10^{12}) - log 3$

0.30n = 12 - 0.48

0.30n = 11.52

n = 38.4

Lets find $T_{38}$ and $T_{39}$ in the arithmetic sequence.

$T_{n} = 0.5n$

$T_{38} = 0.5*38 = 19$

$T_{39} = 0.5*39 = 19.5$

It is going to take around 19 hours and 15 minutes for a single bacteria to grow to fill a thimble with volume 1 cm3.

How long will it take for the volume to fill the entire earth 1.08 × 108 km3 ?

$1 (km)^{3} = 10^{9}m^{3}$

$1.08*10^{8} km^{3} = 1.08*10^{17} m^{3} = 108*10^{15}m^{3}$

Lets solve the same way as the first question.

$V_{n} = 6*10^{-18}*(2^{(n-1)})$

$108*10^{15} = 6*10^{-18}*(2^{(n-1)})$

$\frac{108*10^{15}}{6*10^{-18}} = 2^{(n-1)})$

$18*10^{33} = \frac{2^{n}}{2}$

$2^{n} = 36 * 10^{33}$

Aside from the proprierties seen in the first exercise, we also have that

log a*b = log a + log b

$log (2^{n}) = log(36 * 10^{33})$

$n log 2 = log 36 + log 10^{33}$

0.30n = 1.56 + 33

0.30n = 34.56

$n = \frac{34.56}{0.30}$

n = 115.2

$T_{115} = 0.5*115.2 = 57.6$

It will take 57.6 hours to for the volume to fill the entire earth

Answer from: Quest

the answer is d. (1.2, -4.7) on edgn

hope this !

Answer from: Quest

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)

Need 45 ! compare the rates at which each of the three students read: which student read at a fa

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