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RideAnS [48]
3 years ago
15

Find The sun of 2-3+9/2-27/4+.....-177147/1024

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0
Hello,

s=2-3+ \dfrac{9}{2} - \dfrac{27}{4} +...- \dfrac{177147}{1024} \\\\

=2(1- \dfrac{3}{2} + \dfrac{3^2}{2^2} - \dfrac{3^3}{2^3}+...- \dfrac{3^{11}}{2^{11}}  )\\\\
=2( \dfrac{1- (\dfrac{-3}{2})^{12}}{1+ \dfrac{3}{2} } )\\\\

=2( \dfrac{2^{12}-3^{12}}{ \dfrac{5}{2} } )\\\\

= \dfrac{4}{5*4096} *(-527345)=-102,9970703125
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Given the following linear function sketch the graph of the function and find the domain and range. f( x ) = 2/7 x− 2
Savatey [412]

Answer:

Domain: All Real Numbers Range: All Real Numbers

Step-by-step explanation:

The domain and range is going to be infinite. The linear function will be using the x and y- axis in order to continue being a function. The y-intercept will be -2 on the y-axis. I recommend using the rise-over-run method for your slope value. from the point (0, -2) on the y-axis. Go up two on the y-axis, and right 7 on the x-axis.

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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
Zigmanuir [339]

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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

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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

4 0
2 years ago
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Answer:

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conversion problem

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