False!
if you plug in x, the equation would be 512+14= 23 which is incorrect. x would have to equal 526 for this statement to be considered true.
To answer this question first you have to count the difference between the highest number with the lowest number.
In this case, the difference is: 40-12=28
After that, you need to determine the leaps in the sequence. In this case, there are 4 leaps.
After that divide the difference with the leaps : 28:4=7
Therefore the sequence will be 12 , 19, 26, 33, 40
Answer:
Answer is <u>C</u>
Step-by-step explanation:
This is because Lithium =2.8.1
Sodium=2.8.8.1
<em>Valence</em><em> </em><em>electron</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>electron</em><em> </em><em>located</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>last</em><em> </em><em>shell</em>
Hope this is correct and helpful
HAVE A GOOD DAY!
Answer:
Step-by-step explanation:
![L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B4%7D%7B%28s-3%29%5E%7B3%7D%7D%20%5D%3D)
Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3
Translation theorem:
![L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B4%7D%7B%28s-3%29%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2%28s%2B3%29%2B4%7D%7Bs%5E%7B3%7D%7D%20%5D)
Separate the fraction in a sum:
![e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B10%7D%7Bs%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2s%7D%7Bs%5E%7B3%7D%7D%2B%5Cfrac%7B10%7D%7Bs%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20%28L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%7D%5D%2B%20L%5E%7B-1%7D%5B%5Cfrac%7B10%7D%7Bs%5E%7B3%7D%7D%5D%29)
The formula for this is:
![L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7Bn%21%7D%7Bs%5E%7Bn%2B1%7D%7D%20%5D%3Dt%5E%7Bn%7D)
Modify the expression to match the formula.
![e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%20%5Cfrac%7B10%7D%7B2%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29%3De%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%205%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29)
Solve
![e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%205%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29%3De%5E%7B3t%7D%282t%2B5t%5E%7B2%7D%20%29)
