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3 years ago

A bacteria triples its original population in 25 hours (A=3A0). How big will its population be in 100 hours?

Mathematics

2 answers:

jeyben [28]3 years ago

5 0

Answer:

81 times the original size.

Step-by-step explanation:

AA0ktA=3A0=?=?=25hours=A0ekt

Substitute the values in the formula.

3A0=A0ek⋅25

Solve for k. Divide each side by A0.

3A0A0=e25k

Take the natural log of each side.

ln3=lne25k

Use the power property.

ln3=25klne

Simplify.

ln3=25k

Divide each side by 25.

ln325=k

Approximate the answer.

k≈0.044

We use this rate of growth to predict the number of bacteria there will be in 100 hours.

AA0ktA=3A0=?=ln325=100hours=A0ekt

Substitute in the values.

A=A0eln325⋅100

Evaluate.

A=81A0

At this rate of growth, we can expect the population to be 81 times as large as the original population.

Alexus [3.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

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3 years ago

Determine whether the given system has a unique solution, no solution, or infinitely many solutions.

victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

$\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right$

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

$\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 2: add -1 times the 1st row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 3: add 1 times the 1st row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]$

Row Operation 4: multiply the 2nd row by 3/16

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 6: multiply the 3rd row by -1/15

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 8: add 8/3 times the 3rd row to the 1st row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 9: add 1/3 times the 2nd row to the 1st row

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

which corresponds to the system

$\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right$

Equation 3 cannot be solved, therefore, the system has no solution.

7 0

4 years ago

F(x)= square root x-3

mariarad [96]

I would use a graphing tool like desmos or geogebra, or a texas instruments graphing calculator, there a number of tools out there. The first two options are free online tools. Using technology to graph is much preferred over doing it by hand.

If you must do it by hand, then the idea is to plug in various values of x to get corresponding values of y. For instance, if x = 7, then

y = sqrt(x-3)

y = sqrt(7-3)

y = sqrt(4)

y = 2

So the point (x,y) = (7,2) is on this curve. Repeat for other x values to get more points. Then plot the points all on the same xy grid. Lastly, draw a curve through those points. The more points you have, the more accurate the graph. Use graphing technology to help confirm you have the right graph.

As for the question of one-to-one, you are looking to see if the horizontal line test fails or not. After fully graphing y = sqrt(x-3), try to construct a single horizontal line that will pass through more than one point on the curve. You'll find that such a task is not possible in this case; therefore, this graph passes the horizontal line test and consequently the graph is one-to-one.

An example where a graph is not one-to-one would be a parabola. It is possible to pass a single horizontal line through more than one point on a parabola.

6 0

4 years ago