$15~~,~~\stackrel{15+8}{23}~~,~~\stackrel{23+8}{31}~~,~~\stackrel{31+8}{39}~~,~~...~\hspace{10em}\stackrel{common~difference}{d=8} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\stackrel{\textit{term position}}{59}\\ a_1=\stackrel{\textit{first term}}{15}\\ d=\stackrel{\textit{common difference}}{8} \end{cases} \\\\\\ a_{59}=15+(59-1)8\implies a_{59}=15+472-8\implies a_{59}=479$

4 0

3 years ago

The half-life of a certain radioactive material is 78 hours. An initial amount of the material has a mass of 790 kg.. Write an e

adoni [48]

<span> half life i= 78 hours
amount after 78 hours is 395 kg:
395 = 790e^(k*78)

Dividing by 790 and taking natural log
ln (395/790) = (k*78)
-0.6931 = 78k
-0.00888 = k

lets calculate how much is left after 18 hours:
Amount(18)
= 790e^(-0.00888*18)
Amount = 673.301 kg
hope this helps</span>

4 0

3 years ago

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My kid asked me what was 5y-3+6 i couldnt answer it because this is all new to me we didnt do this when i was younger and i need

S_A_V [24]

Answer:

Add−3 and 6 .5y+3

4 0

3 years ago

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