Ask question

Ask question

All categories

3 years ago

11

At 1 p.M., the number of bacteria in a petri dish is 54. It is estimated that the population of bacteria will increase by 6% eac

h hour. Write a formula for the population of bacteria.

1 answer:

Zinaida [17]3 years ago

6 0

Answer: y = 54(1.06)^t

Step-by-step explanation:

It is estimated that the population of bacteria will increase by 6% each hour. This is an exponential growth. The formula for determining exponential growth is expressed as

y = b(1 + r)^ t

Where

y represents the population after t years.

t represents the number of years.

b represents the initial population.

r represents rate of growth.

From the information given,

b = 54

r = 6% = 6/100 = 0.06

Therefore, the formula for the population of bacteria is

y = 54(1 + 0.06)^t

y = 54(1.06)^t

You might be interested in

Write a number sentence that does not involve regrouping using this sum and two 2- digit addends

melisa1 [442]

Using the sum (addition) and two 2-digit addends, 12 + 34 can be answered without regrouping.

5 0

3 years ago

Identify az of this sequence: 0.25, 0.5, 0.75, 1, 1.25. 1.5

4vir4ik [10]

<h3><u>Question:</u></h3>

Identify a3 of this sequence: 0.25, 0.5, 0.75, 1, 1.25, 1.5,...a3=

<h3><u>Answer:</u></h3>

The third term of sequence is 0.75

$a_3 = 0.75$

<h3><u>Solution:</u></h3>

Given that, sequence is:

0.25, 0.5, 0.75, 1, 1.25. 1.5

Let us find the difference between terms

$0.5 - 0.25 = 0.25\\\\0.75-0.5 = 0.25\\\\1.25-1=0.25\\\\1.5-1.25=0.25$

This is a arithmetic sequence

Because the difference between any term and its immediately preceding term is always 0.25

In a arithmetic sequence,

$a_1 = \text{ first term of sequnece }\\\\a_2 = \text{second term of sequnce }\\\\a_3 = \text{third term of sequnece }\\\\a_n = \text{ nth term of sequnce }$

Thus, in the sequence 0.25, 0.5, 0.75, 1, 1.25. 1.5

$a_3 = \text{ third term } = 0.75$

Thus the third term of sequence is 0.75

5 0

4 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago

The pair of points lies on the same line with the given slope. Find x.

KIM [24]

If the gradient is -2 that just means we work back the way, the method i use to find the gradient is y^2-y^1/x^2-x^1 so if we sub the variables in that gives us -18-(-8)/3-x so if we add that up we get -10/3-x, so what has to be on the bottom to get -2? in this case it would be -2 so the overall answer would be -10/5 which would give us a slope of -2 when simplified :) hope this helps

8 0

3 years ago

Solve this complex fractions: (3/4)/2

strojnjashka [21]

Answer:

your answer is 0.375

Step-by-step explanation:

7 0

3 years ago