The count in a bacteria culture was 800 after 15 minutes and 6443 after 25 minutes. assume the growth can be modelled exponentially by a function of the form q(t) = a e^{r t}, where t is in minutes.

Hope this helps to guide you

The count in a bacteria culture was 700 after 20 minutes and 1000 after 40 minutes. What was the initial size of the culture?

490 microorganisms.

Explanation:

I will assume exponential growth for bacteria. This means that we can model the growth with an exponential function:

f

(

t

)

=

A

0

e

k

t

where

k

is the growth constant and

A

0

is the initial amount of bacteria.

Sub the two known values into the function to get two equations:

700

=

A

0

e

20

k

(1)

1000

=

A

0

e

40

k

(2)

Divide (2) by (1) to find

k

:

1000

700

=

A

0

e

40

k

A

0

e

20

k

10

7

=

e

40

k

−

20

k

=

e

20

k

Take the natural log of both sides to isolate

k

:

ln

(

10

7

)

=

ln

e

20

k

ln

(

10

7

)

=

20

k

k

=

ln

(

10

7

)

20

Now that we have the growth constant,

k

, we can substitute one of the points in to solve for the initial amount,

A

0

:

(

40

,

1000

)

1000

=

A

0

e

ln

(

10

7

)

20

⋅

40

A

0

=

1000

e

0.0178

⋅

40

=

490

Estimate the following (by rounding off to nearest hundred (a)6,941 (b)6,320 (c) 416 (d)126 (e)109 (f)202 (g) 146 (h)7,699 (I)8,311 (j)986

Step-by-step explanation:

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.