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dezoksy [38]
4 years ago
10

A scientist begins an experiment with a petri dish containing a culture of bacteria cells and records the cell growth every hour

. At 1 hour, the bacteria had grown to 880 cells, and at 2 hours, had grown to 3,520 cells. Which function can be used to represent the relationship between the number of bacteria cells, y, and the time in hours, x?

Mathematics
2 answers:
KengaRu [80]4 years ago
7 0

Answer: Hello!

we know that in the first hour, we had 880 cells, and in the second hour we have 3520 cells. The first step is see how much the population changed in this one hour, this can be done by looking into the quotient:

3520/880 = 4

then the population of cells cuatriplied in one hour, and is safe to asume that every hour that pases, the population will catriplied  again.

so if the populationat hour 1 is: 880

at hour 2 is: 4*880

at hour 3 is = 4*4*880 and so on.

This can be described with an exponential function as follows

P(t) = 880*4^{x-1}

where 880 is the initial population, and x is represents the hours.

egoroff_w [7]4 years ago
5 0
These are the choices y = 4 x y = 220(4) x y = 880(4) x y = 880(0.25) x Right..?

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I need help please help me thanks please
Vitek1552 [10]

so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})

\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{4}(x-\stackrel{x_1}{1})\\\\\\y-3=4x-4\implies y = 4x-1\implies \blacktriangleright  y = 4x+(-1)\blacktriangleleft

4 0
3 years ago
Which of the following is a valid use of the associative property of addition?
Natasha_Volkova [10]

Answer:

A

Step-by-step explanation:

(25+15)-5=40-5=35

25+(15-5)=25+10=35

3 0
3 years ago
Read 2 more answers
Obtain the general solution:<br><br> (x-y)(4x+y)dx + x(5x-y)dy = 0
storchak [24]

Answer:

y=2x

y=-x

d=0

Step-by-step explanation:

sorry iam not sure i just tray to help

7 0
3 years ago
(50 POINTS) The figure is transformed as shown in the diagram. Describe the transformation. A) dilation, then reflection B) refl
Helga [31]

Answer:

Hi there I was just working on this question on UsaTestPrep and idk if the answer is really correct but here: C: reflection then rotation.

Step-by-step explanation:

Extra Info:

Dilation is when the shape changes in size, which doesn't happen.

Translation is when it just moves across. It may look like that is happening here, but I'll explain.

The triangle is first reflected along the line . Imaging putting a mirror on the x=0 line (the y axis, basically), and you'd see 1 to 2.

Then, it is rotated around the point . Get a bit of tracing paper, draw over 2, and then, holding the tracing paper down at  (works well with a pencil), rotate it round and you will find it fits perfectly over 3.

Also is this the diagram?

6 0
3 years ago
At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr
Dennis_Churaev [7]
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
</span>
Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250

At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
</span>
A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5

So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50 </span>
4 0
3 years ago
Read 2 more answers
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