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Assoli18 [71]
3 years ago
13

In an experiment, the population of bacteria is increasing

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
6 0

25 million

number of bacteria doubles every minute

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What is the correct answer?​
vladimir1956 [14]

Answer:

17.82

Step-by-step explanation:

since this is a right triangle you can use one of the trigonometric ratios, in this case it would be cosine because you have values for the adjacent and hypotenuse sides

cos(27°) = x/20

x = 20 · cos(27°)

x = 17.82

5 0
3 years ago
3,10,17,24,31,38 are composite
rewona [7]
Prime: 3,17,31

Composite: 10,24,38

Prime numbers can only be divided by one and itself, while composite numbers can be divided by a number other than one and itself. 
3 0
4 years ago
Read 2 more answers
I need to know how to find the perimeter please.
shutvik [7]
First, take all the sides you already have, then add them up. (You should get 12) To find the other two sides, do some more simple addition. The 2+3+2 of the the top side add up to six (the bottom side) and sides equal each other so add another three. Total is 12+6+3=21
5 0
3 years ago
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you had eight hours until you have to be on set if you will spend 3/4 of an hour driving to the TV station how much time does th
Inessa05 [86]

Answer:

7 hours and 15 minutes

Step-by-step explanation:

8 - 3/4 = 7 1/4 hours

7 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
4 years ago
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