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photoshop1234 [79]
4 years ago
13

I'm studying a new bacteria that doubles in numbers every 25 minutes. If i start with 50 bacteria, how long until i have 5 milli

on of them
Mathematics
1 answer:
dimaraw [331]4 years ago
6 0

Answer:

415.63 minutes

Step-by-step explanation:

Growth can be represented by the equation A=A_0e^{rt}. We can find the rate at which it grows by using t=25 minutes and \frac{A_{0}}{A} =2 or double the amount at that time. The first step we always take is to divide A_0 by A.

[tex]\frac{A_{0}}{A}=e^{rt}\\2=e^{r(25)}

2=e^{(25)r}

To solve for r, we will take the natural log of both sides and use log rules to isolate r.

ln 2=ln e^{(25)r}\\ln 2=25r (ln e)\\\frac{ln2}{25} =r

We know lne=1 so we were able to cancel it out and divide both sides by 25.

We solve with a calculator \frac{ln2}{25} =r\\0.0277=r

We change 0.0277 into a percent by multiplying by 100 to get 2.77% as the rate.

The equation is A=A_0e^{0.0277t} .

We repeat the step above substituting A=5,000,000, A_0=50, and r=0.02777. Then solve for t.

5000000=50e^{0.0277t}\\\frac{5000000}{50} =e^{0.0277t}\\100000=e^{0.0277t}\\ln100000=lne^{0.0277t}\\ln100000=0.02777t(lne)\\\frac{ln100000}{0.0277} =t

t=415.63 minutes

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Can someone help me with this question?
meriva
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3 years ago
Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x
tester [92]

Answer:

The solutions on the given interval are :

0

\pi

\cos^{-1}(\frac{1}{4})

-\cos^{-1}(\frac{1}{4})+2\pi

Step-by-step explanation:

We will need the double angle identity \sin(2x)=2\sin(x)\cos(x).

Let's begin:

2\sin(2x)=\sin(x)

Use double angle identity mentioned on left hand side:

2\cdot 2\sin(x)\cos(x)=\sin(x)

Simplify a little bit on left side:

4\sin(x)\cos(x)=\sin(x)

Subtract \sin(x) on both sides:

4\sin(x)\cos(x)-\sin(x)=0

Factor left hand side:

\sin(x)[4\cos(x)-1]=0

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

\sin(x)=0 \text{ or } 4\cos(x)-1=0

The first is easy what angles \theta are y-coordinates on the unit circle 0. That happens at 0 and \pi on the given range of x (this x is not be confused with the x-coordinate).

Now let's look at the second equation:

4\cos(x)-1=0

Isolate \cos(x).

Add 1 on both sides:

4 \cos(x)=1

Divide both sides by 4:

\cos(x)=\frac{1}{4}

This is not as easy as finding on the unit circle.

We know \arccos( ) will render us a value between 0 and 2\pi.

So one solution on the given interval for x is x=\cos^{-1}(\frac{1}{4}).

We know cosine function is even.

So an equivalent equation is:

\cos(-x)=\frac{1}{4}

Apply \cos^{-1} to both sides:

-x=\cos^{-1}(\frac{1}{4})

Multiply both sides by -1:

x=-\cos^{-1}(\frac{1}{4})

This going to be negative in the 4th quadrant but if we wrap around the unit circle, 2\pi , we will get an answer between 0 and 2\pi.

So the solutions on the given interval are :

0

\pi

\cos^{-1}(\frac{1}{4})

-\cos^{-1}(\frac{1}{4})+2\pi

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3 years ago
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