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3 years ago

Evaluate 5c+cd5c+cd5, c, plus, c, d when c=\dfrac15c=

Mathematics

1 answer:

zepelin [54]3 years ago

5 0

Answer:

Step-by-step explanation:

The function is given as 5c+cd, c=1/5 and d=15d.

We evaluate by substituting the given values of c and d in the function:

$5c+cd, c=1/5 \ \ \, d=15\\\\=5\times \frac{1}{5}+\frac{1}{5}\times 15$

$=1++3\\\\=4$

After substituting, we find the function's equivalent as 4.

Hence, the value of the function is 4

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Is Markiplier the best man?

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Two of the numbers were switched in a multiplication equation, so the answer would be commutative- multiplication.

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The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

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The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

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Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

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$90000 = P_0 \times e^{2.08 \; 5/3}$

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3 years ago

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The experimental probability for ones was less than and for fours it was greater than the theoretical probability.

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4 years ago