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Finger [1]
2 years ago
10

Organisms A and B start out with the same population size. Organism A's population doubles every day. After 5 days, the populati

on stops growing and a virus cuts it in half every day for 3 days. Organism B's population grows at the same rate but is not infected with the virus. After 8 days, how much larger is organism B's population than organism A's population? Answer the questions to find out. The expression showing organism A's decrease in population over the next 3 days is ( 1 2 ) ( 2 1 ​ ) 3 . This can be written as (2–1)3. Write (2–1)3 with the same base but one exponent.
Mathematics
1 answer:
Zigmanuir [339]2 years ago
3 0

Answer:

The number of times organism B's population is larger than organism A's population after 8 days is 32 times

Step-by-step explanation:

The population of organism A doubles every day, geometrically as follows

a, a·r, a·r²

Where;

r = 2

The population after 5 days, is therefore;

Pₐ₅ = = 32·a

The virus cuts the population in half for three days as follows;

The first of ta·2⁵ he three days = 32/2 = 16·a

The second of the three days = 16/2 = 8·a

After the third day, Pₐ = 8/2 = 8·a

The population growth of organism B is the same as the initial growth of organism A, therefore, the population, P₈ of organism B after 8 days is given as follows;

P₈ =  a·2⁸ = 256·a

Therefore, the number of times organism B's population is larger than organism A's population after 8 days is P₈/Pₐ = 256·a/8·a = 32 times

Which gives, the number of times organism B's population is larger than organism A's population after 8 days is 32 times.

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A small island is 4 miles from the nearest point P on the straight shoreline of a large lake. If a woman on the island can row a
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Answer:

a) t(x)  = [ √ (4)² + (x)²]/ 4    +   7/5  - x/5

b) x = 0,97 miles

c) t (min)  = 1,24 hours

Step-by-step explanation: See Annex

Figure in annex shows a clear description of the situation

Let  x be distance in miles between point P and where boat lands

A woman has to row  a distance L

L = √ (4)² + (x)²

and the part to get to the town (which she has to walk)

d = ( 7 -  x)

But we are required to give time as a function of x

distance   =  speed * time     ⇒   t  = distance / speed

Therefore

t(x)  = [ √ (4)² + (x)²]/ 4   +   ( 7  -  x  ) / 5

t(x)  = [ √ (4)² + (x)²]/ 4    +   7/5  - x/5

Taking derivatives both sides of the equation

t¨(x)   = ( 2x)*4 / 16√ (4)² + (x)²]   -  5/25  

t¨(x)   = x/  2√ (4)² + (x)²]  - 1/5

t¨(x)   = 0      x/  2√ (4)² + (x)²]  - 1/5   =  0

[ 5 x  -  2√ (4)² + (x)²]  =  0       ⇒   5x    =  2√ (4)² + (x)²]   or

   25 x² =  4  (  4 +   x²)

   25 x² = 16  +   8x²

  17x²   =   16

  x²  =   16/17      

   x²   =  0,941

  x = 0,97 miles  

And the distance walking to get to town

d =  7  -  x       d =   7  -  0,97

d  = 6,03 miles

The least travel time  is   t(x)  = [ √ (4)² + (x)²]/ 4    +   7/5  - x/5  

t (min)  =  √ 16 +  0,94) /4   +  1,4  - 0,97/5

t (min)  = 1,03  +  1,4  - 0,194

t (min)  = 1,24 hours

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2 years ago
Prompt:
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Answer:

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Step-by-step explanation:

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Answer:

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Z = 8 + 2r2

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