All categories

3 years ago

A sample of bacteria is growing at a continuously compounding rate. The sample triples in 10 days.

Mathematics

1 answer:

WARRIOR [948]3 years ago

5 0

Answer:

The number of bacteria after days is given by

where is the initial number of bacteria.

Step-by-step explanation:

The number of bacteria in the sample triples every 10 days, this means after the first 10th day, the number of bacteria is

where is the initial number of bacteria in the sample.

After the 2nd 10th days, the number of bacteria is

after the 3rd day,

and so on.

Thus, the formula we get for the number of bacteria after the nth 10-days is

where is is the nth 10-days.

Since, is 10 days, we have

You might be interested in

Solve y + 4 > - 2.2. Check your solution.

maks197457 [2]

Pretty sure its y > -6.2
This is because you subtract 4 from both sides and get y > -6.2
Please correct me if im wrong thx!

6 0

3 years ago

Read 2 more answers

Enter the domain and range of the function g(x) = 9x2 − 5 as an inequality, using set notation and using interval notation. Comp

tia_tia [17]

The domain and the range of a function are the set of input and output values, the function can take.

The domain and the range of $g(x) = 9x^2 - 2$ is $(-\infty, \infty)$ .
The parent function $f(x) =x^2$ is vertically compressed by 9, then shifted down by 5 units to get $g(x) = 9x^2 - 2$

Given

$g(x) = 9x^2 - 2$

Domain and range

There is no restriction as to the input and the output of function g(x).

This means that the domain and the range are $(-\infty, \infty)$

$(-\infty, \infty)$ is in interval notation

The corresponding set notation is: $- \infty < x < \infty$

The parent function

We have:

$f(x) = x^2$

First, the parent function is vertically compressed by a factor of 9.

The rule of this transformation is:

$(x,y) \to (x,9y)$

So, we have:

$f'(x) = 9x^2$

Next, the function is shifted down by 5 units.

So, we have:

$g(x) = f'(x) - 5$

$g(x) = 9x^2 - 5$

Read more about functions at:

brainly.com/question/21027387

4 0

2 years ago

Factor the expression. r^2 – 49 a. (r – 7)(r 9) b. (r 7)(r 7) c. (r – 7)(r 7) d. (r – 7)(r – 7)

gayaneshka [121]

$r^2-49=r^2-7^2\\\\/use\ a^2-b^2=(a-b)(a+b)/\\\\=\boxed{(r-7)(r+7)}$

6 0

3 years ago

This is a 3 part 1 question each part on how to locate the plane and a small explanation report working out the recovery details

sesenic [268]

Third leg.

The crew flies at a speed of 560 mi/h in direction N-20°-E.

The wind has a speed of 35 mi/h and a direction S-10°-E.

We then can draw this as:

We have to add the two vectors to find the actual speed and direction.

We will start by adding the x-coordinate (W-E axis):

$\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}$

and the y-coordinate (S-N axis) is:

$\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}$

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

$\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}$

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

$\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}$

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

$\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}$

The angle, from E to N, can be calculated as:

$\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}$

If we want to express it from N to E, we substract the angle from 90°:

$\beta=90\degree-\alpha=90-49.56=40.44\degree$

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.

7 0

1 year ago

Solve for f. 8 = 2f + 2

tiny-mole [99]

Answer:

f = 3

Step-by-step explanation:

8 = 2f + 2

-2 -2

6 = 2f

/2 /2

3 = f

5 0

3 years ago