Somebody please help me.

A biologist recorded a count of 337 bacteria present in a culture after 5 minutes and 699 bacteria present after 15 minutes.

hoa [83]

The initial population was 234.

Explanation

Formula for the exponential growth is: $A= P*e^r^t$ , where P is the initial amount, A is the final amount, r is the rate of growth and t is the time duration.

There was 337 bacteria after 5 minutes and 699 bacteria after 15 minutes. So, the equations will be......

$337=P*e^5^r .................................. (1)\\ \\ 699=P*e^1^5^r ................................. (2)$

Now dividing equation (2) by equation (1) , we will get .......

$\frac{699}{337}=\frac{e^1^5^r}{e^5^r} \\ \\ e^1^0^r = \frac{699}{337}$

Taking 'natural log' on both sides.........

$ln (e^1^0^r) = ln (\frac{699}{337})\\ \\ 10r= 0.7295....\\ \\ r= 0.07295.... \approx 0.073$

Now, plugging this $r=0.073$ into equation (1), we will get......

$337= P*e^5^(^0^.^0^7^3^)\\ \\ 337= P*1.4405 \\ \\ P= 233.946 \approx 234$

So, the initial population was 234.

7 0

3 years ago

(5a-2)(5a+2)-3(1-a)-3a=

attashe74 [19]

So first we need to know the order of operations
P - Parenthesis
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction

Then the special cases:
(I have attached a picture below)

Lets get into action!
(5a - 2)(5a + 2) - 3( 1 - a ) - 3a
25a² - 4 - 3 + 3a - 3a
25a² - 7
25a² - 7 is the answer

8 0

3 years ago

Read 2 more answers

3a) What is the side length of the following square?*

skelet666 [1.2K]

YhgfdrAnswer:

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

30 tens times 5 hundreds= 50 tens and 3 hundreds true or false?

Pavlova-9 [17]

Answer:

true

Step-by-step explanation:

Somebody once told me the world is gonna roll me

I ain't the sharpest tool in the shed

She was looking kind of dum with her finger and her thumb

In the shape of an "L" on her forehead

Well the years start coming and they don't stop coming

Fed to the rules and I hit the ground running

Didn't make sense not to live for fun

Your brain gets smart but your head gets dum

So much to do, so much to see

So what's wrong with taking the back streets?

You'll never know if you don't go

You'll never shine if you don't glow

Hey now, you're an all-star, get your game on, go play

Hey now, you're a rock star, get the show on, get paid

And all that glitters is gold

Only shooting stars break the mold

It's a cool place and they say it gets colder

You're bundled up now, wait 'til you get older

But the meteor men beg to differ

Judging by the hole in the satellite picture

The ice we skate is getting pretty thin

The water's getting warm so you might as well swim

My world's on fire, how about yours?

That's the way I like it and I'll never get bored

Hey now, you're an all-star, get your game on, go play

Hey now, you're a rock star, get the show on, get paid

All that glitters is gold

Only shooting stars break the mold

Hey now, you're an all-star, get your game on, go play

Hey now, you're a rock star, get the show, on get paid

And all that glitters is gold

Only shooting stars

Somebody once asked could I spare some change for gas?

I need to get myself away from this place

I said, "Yup" what a concept

I could use a little fuel myself

And we could all use a little change

Well, the years start coming and they don't stop coming

Fed to the rules and I hit the ground running

Didn't make sense not to live for fun

Your brain gets smart but your head gets dum

So much to do, so much to see

So what's wrong with taking the back streets?

You'll never know if you don't go (go!)

You'll never shine if you don't glow

Hey now, you're an all-star, get your game on, go play

Hey now, you're a rock star, get the show on, get paid

And all that glitters is gold

Only shooting stars break the mold

And all that glitters is gold

Only shooting stars break the mold

6 0

3 years ago

Read 2 more answers

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago