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1 year ago

If someone could please answer all 3 points (20 points)

Mathematics

1 answer:

Aleksandr [31]1 year ago

7 0

The missing parts of the table for the function are 49, 1 and 49 respectively

<h3>How to complete the missing parts of the table?</h3>

An exponential function is a type of function which involves exponents. A simple exponential function is of the form y = bˣ

Given: the exponential function y = (1/7)ˣ

In order to find the missing parts, we have to substitute the relevant values into the function. Thus:

For x = -2:

y = (1/7)ˣ

Substitute x = -2 into the function:

y = (1/7)⁻² = 49

For x = 0:

y = (1/7)ˣ

Substitute x = 0 into the function:

y = (1/7)⁰ = 1

For x = 2:

y = (1/7)ˣ

Substitute x = 2 into the function:

y = (1/7)² = 1/49

The missing part is 49

Therefore, the missing parts of the table are 49, 1 and 49 respectively

Learn more about exponential function on:

brainly.com/question/27161222

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2 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

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Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

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2 years ago