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3 years ago

A geologist in South America discovers a

Mathematics

1 answer:

diamong [38]3 years ago

7 0

Answer:

This is a radioactive decay / half-life problem.

Initial amount of C 14 = 100% Present Amount = 57%

k = .0001 (that value should be negative)

Nt = No * e^ (k*t)

You need to solve that equation for "time" (equation attached)

time = natural log (Ending amount / Starting Amount) / k

time = natural log (57% / 100%) / -.0001

time = ln (.57) / -.0001

time = -.56211891815 / -.0001

time = 5,621.2 years (age of the bird skeleton)

These problems are quite complicated but I think I know this pretty well.

Need to know more? Visit my website (it's on the graphic).

Step-by-step explanation:

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3 years ago

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3 years ago

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Olin [163]

Answer:

m^2/n

Step-by-step explanation:

The following laws of exponent are useful to this problem:

$\displaystyle \large{ {(mn)}^{b} = {m}^{b} {n}^{b} } \\ \displaystyle \large{ {n}^{ - b} = \frac{1}{ {n}^{b} } }$

We are given the expression:

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } }$

Use the first law of exponent above.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{(8)( \frac{1}{4}) } {n}^{ ( - 4)( \frac{1}{4}) } } \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } {n}^{ - 1} }$

Make sure to recall the important necesscary fundamental math such as operation with negative numbers/integers, basic division, fraction, etc.

From the expression, apply the second law of exponent to n^-1.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n}^{1} } )} \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n} } )} \\$

Multiply m in.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = \frac{ {m}^{2} }{ {n} } }$

Thus the answer is m^2/n.

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2 years ago