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3 years ago

X + 2y = 33 x-y = 11

Mathematics

1 answer:

Tasya [4]3 years ago

7 0

Answer:

$x=\frac{55}{3},\:y=\frac{22}{3}$

Step-by-step explanation:

Solve by Substitution ;

$\begin{bmatrix}x+2y=33\\ x-y=11\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:x+2y=33:\quad x=33-2y\\\\\mathrm{Subsititute\:}x=33-2y\\\\\begin{bmatrix}33-2y-y=11\end{bmatrix}\\\\Simplify\\\begin{bmatrix}33-3y=11\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:33-3y=11:\quad y=\frac{22}{3}\\\mathrm{For\:}x=33-2y\\\\\mathrm{Subsititute\:}y=\frac{22}{3}\\x=33-2\times\frac{22}{3}\\\\x=\frac{55}{3}\\\\x=\frac{55}{3},\:y=\frac{22}{3}$

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Answer:

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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.

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In which:

a is the first term

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

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Each arithmetic sequence has the following format:

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In which:

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

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How i am going to solve this problem.

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

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

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

$r = 2$

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

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

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

$0.30n = 12 - 0.48$

$0.30n = 11.52$

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$T_{n} = 0.5n$

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