The half-life of carbon-14 is 5600 years. if a piece of charcoal made from the wood of a tree shows only 62% of the carbon-14 expected in living matter, when did the tree die?

Following are the calculation when the tree dies:

Given:

Carbon-14 half-life = 5600 years

To find:

The tree dies in years=?

Solution:

Using formula:

Using natural log:

Therefore, the final answer is "3862 years".

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The amount of carbon-14 present obeys an exponential decay law.

A(t) = A(0)e-^kt

Get k from the half-life:

0.5 = 1e-^k(5600)

k = ln(0.5) / (-5600 years) = 1.2378 x 10^-4 years^-1

If only 73% of the carbon-14 remains after time T, then

0.73 = 1e^-kT

T = ln(0.73) / (-1.2378 x 10^-4 years^-1) = 2542.5 years

The tree has been dead for about 2542.5 years.

step-by-step explanation:

4  

25

5258y

54.

t5

5r

54

5856

542

5458

86

.8558.65

565

565

55898

786

8/95

5896

8656265

556478

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)