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3 years ago

Which of the following expressions are equivalent 6^-3

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

Answer: A & D

Step-by-step explanation:

Lilit [14]3 years ago

5 0

Answer:

$6^{-3} = \frac{1}{6^3} =\frac{1}{6*6*6}=\frac{1}{216}$

Step-by-step explanation:

The expression $6^{-3}$ can be simplified using rules for negative exponents - move the base to the denominator. After following this rule, simplify.

$6^{-3} = \frac{1}{6^3} =\frac{1}{6*6*6}=\frac{1}{216}$

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krok68 [10]

Answer:

The correct answer is:

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Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

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