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Troyanec [42]
3 years ago
11

In a lab experiment, 60 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to doub

le every 21 hours. How long would it be, to the nearest tenth of an hour, until there are 107 bacteria present?
Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

17.5 hours

Step-by-step explanation:

107 = 60(2^n)

n(lg2) = lg(107/60)

n = 0.8345763908 × 21

n = 17.52610421 hours

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2x + 9 = 5x

5x = 2x + 9

Subtract sides 2x

- 2x + 5x = - 2x  + 2x + 9

3x = 9

Divide sides by 3

\frac{3x}{3}  =  \frac{9}{3}  \\

x = 3

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CHECK :

2x + 9 = 5x

2(3) + 9 = 5(3)

6 + 9 = 15

15 = 15

So the value is correct.

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