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kkurt [141]
2 years ago
14

Bacteria colonies increase by 37% every 2 days. If you start with 55 microorganisms, how large would the colony be after 10 days

Mathematics
1 answer:
ahrayia [7]2 years ago
3 0

The number of microorganisms that I will have after 10 days would be: 245.41 microoganisms.

<h3>How to calculate the number of microorganisms that I will have after 10 days?</h3>

To calculate the number of microorganisms that I will have after 10 days, I must perform the following operations:

We need to find 37% of 55 as shown below:

  • 55 ÷ 100 = 0.55
  • 0.55 x 37% = 20.35
  • 55 + 20.35 = 75.35

So after 2 days I will have 75.35 microorganisms.

  • 75.35 ÷ 100 = 0.75
  • 0.7535 x 37% = 27.87
  • 75.35 + 27.87 = 103.22

On the fourth day I will have: 103.22 microorganisms.

  • 103.22 ÷ 100 = 1.0322
  • 1.0322 x 37% = 38.19
  • 103.22 + 38.19 = 141.41

On the sixth day I will have 141.41 microorganisms.

  • 141.41 ÷ 100 = 1.4141
  • 1.4141 x 37% = 52.32
  • 141.41 + 52.32 = 193,73

On the eighth day I will have 193.73 microorganisms.

  • 193.73 ÷ 100 = 1.9373
  • 1.9373 x 37% = 71.68
  • 193.73 + 71.68 = 265.41

On the tenth day I will have 245.41 microoganisms.

Learn more about percentages in: brainly.com/question/13450942

#SPJ1

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A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}. The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}.

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

c^{2} = a^{2} + b^{2} - 2ab Cos C

So that;

7^{2} = 6^{2} + (6.5)^{2} - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = \frac{29.25}{78}

         = 0.375

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   = 67.9757

C = 68^{o}

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{6.5}{Sin B} = \frac{7}{Sin 68}

SIn B = \frac{6.5 *Sin 68}{7}

         = 0.861

B = Sin^{-1} 0.861

   = 59.43

B = 59.4^{o}

Thus to determine the value of A, we have;

A + B + C = 180^{o}

A + 59.4^{o} + 68^{o} = 180^{o}

A = 180^{o} - 127.4

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A = 52.6^{o}

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

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