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Liula [17]
1 year ago
14

A bacteria culture starts with 1000 bacteria and the number doubles every 40 minutes. (a) find a formula for the number of bacte

ria at time (b) find the number of bacteria after one hour (c) after how many minutes will there be 50000 bacteria?​
Mathematics
1 answer:
vfiekz [6]1 year ago
5 0

The formula for the number of bacteria at time t is 1000 x (2^t).

The number of bacteria after one hour is 2828

The number of minutes for there to be 50,000 bacteria is 324 minutes.

<h3>What is the number of bacteria after 1 hour?
</h3>

The exponential function that can be used to determine the number of bacteria with the passage of time is:

initial population x (rate of increase)^t

1000 x (2^t).

Population after 1 hour : 1000 x 2^(60/40) = 2828

Time when there would be 50,000 bacteria : In(FV / PV) / r

Where:

  • FV = future bacteria population = 50,000
  • PV = present bacteria population = 1000
  • r = rate of increase = 100%

In (50,000 / 1000)

In 50 / 1  = 3.91 hours x 60 = 324 minutes

To learn more about exponential functions, please check: brainly.com/question/26331578

#SPJ1

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Find numerical value if x=1 and y=5<br>4(x+y)​
Len [333]

Answer:

24

Step-by-step explanation:

4(x+y)

4(1+5)

4*6

24

6 0
3 years ago
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Please help I can’t figure this out!
Monica [59]

Explanation:

All values in the x-column get filled with -2.

The graph is the vertical line, x = -2.

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You are told x=-2. There is nothing to figure out. The value of y is irrelevant.

That equation describes a vertical line. The points on the line have x-coordinate -2, and any (every) y-coordinate.

4 0
2 years ago
Six times a number is 45 greater than the product of the number and -3. Find the number.
LiRa [457]

Answer:

5

Step-by-step explanation:

6a = (a*-3) + 45

then:

6a = -3a + 45

6a + 3a = 45

9a = 45

a = 45/9

a = 5

Check:

6*5 = (5*-3) + 45

30 = -15 + 45

8 0
3 years ago
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

6 0
3 years ago
HELP!!!!
egoroff_w [7]

Answer:

x axis

Step-by-step explanation:

5 0
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