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Neporo4naja [7]
3 years ago
12

How do find the volume of a cube as a monomial

Mathematics
1 answer:
Drupady [299]3 years ago
7 0
The volume of a cube as a monomial is v=a3.
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The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000​
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Answer:

here i finished!

hope it helps yw!

Step-by-step explanation:

The doubling period of a bacterial population is 15 minutes.

At time t = 90 minutes, the bacterial population was 50000.

Round your answers to at least 1 decimal place.

:

We can use the formula:

A = Ao*2^(t/d); where:

A = amt after t time

Ao = initial amt (t=0)

t = time period in question

d = doubling time of substance

In our problem

d = 15 min

t = 90 min

A = 50000

What was the initial population at time t = 0

Ao * 2^(90/15) = 50000

Ao * 2^6 = 50000

We know 2^6 = 64

64(Ao) = 50000

Ao = 50000/64

Ao = 781.25 is the initial population

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Find the size of the bacterial population after 4 hours

Change 4 hr to 240 min

A = 781.25 * 2^(240/15

A = 781.25 * 2^16

A= 781.25 * 65536

A = 51,199,218.75 after 4 hrs

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If ex + e⁻¹= 2, then x =..
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Answer:

\rm x = \dfrac{2e - 1}{ {e}^{2} }

Step-by-step explanation:

\rm Solve \:  for \:  x: \\  \rm \longrightarrow ex +  {e}^{ - 1}  = 2 \\  \\   \rm \longrightarrow e x +  \dfrac{1}{e}  = 2 \\  \\  \rm Subtract \:   \dfrac{1}{e} \:  from \:  both  \: sides: \\   \rm \longrightarrow e x  +  \dfrac{1}{e} -  \dfrac{1}{e}  = 2  -  \dfrac{1}{e}  \\  \\   \rm \longrightarrow ex =  \dfrac{2e}{e} -  \dfrac{1}{e}   \\  \\   \rm \longrightarrow ex =  \dfrac{2e - 1}{e}  \\  \\  \rm Divide \:  both  \: sides  \: by  \: e: \\   \rm \longrightarrow  \dfrac{ex}{e}  =  \dfrac{2e - 1}{e \times e}  \\  \\   \rm \longrightarrow x =  \dfrac{2e - 1}{ {e}^{2} }

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