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LiRa [457]
3 years ago
9

Lin counts 5 bacteria under a microscope. She counts them again each day for four days, and finds that the number of bacteria do

ubled each day-from 5 to 10, then from 10 to 20, and so on. Is the population of bacteria a function of the number of days? If so, is it linear? Explain your reasoning.
Mathematics
1 answer:
Liula [17]3 years ago
6 0

Answer:

The population of bacteria can be expressed as a function of number of days.

Population = \[5*2^{n-1}\] where n is the number of days since the beginning.

Step-by-step explanation:

Number of bacteria on the first day=\[5 * 2^{0} = 5\]

Number of bacteria on the second day = \[5 * 2^{1} = 10\]

Number of bacteria on the third day = \[5*2^{2} = 20\]

Number of bacteria on the fourth day = \[5*2^{3} = 40\]

As we can see , the number of bacteria on any given day is a function of the number of days n.

This expression can be expressed generally as \[5*2^{n-1}\] where n is the number of days since the beginning.

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alisha [4.7K]

Answer:

7 : 1

Step-by-step explanation:

Paul's car goes 40 kilometers on 7 tanks.

Distance    :     Tanks

     49        :         7

Fuel efficiency means the distance traveled by the car in one tank.

Therefore, we will reduce the ratio by dividing both sides by 7.

Distance    :     Tanks

    \frac{49}{7}           :         \frac{7}{7}

    7            :         1

Therefore, car's fuel efficiency is 7 kilometers in one tank or the ratio between Distance traveled and fuel tanks is 7 : 1.  

8 0
3 years ago
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marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}

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Answer:

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Step-by-step explanation:

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8 0
2 years ago
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