Answer:C
Step-by-step explanation: When a table represents a nonlinear function, the rate of change is not constant. A wouldn't be the answer because the rate of change is always +10 (you would add 10 to get from -9 to 1; you would add 10 to get from 1 to 10) . It wouldn't be B because the rate of change is always -2, and the rate of change for D is always +3. For C, however, the rate of change is not constant all the way through (to get from 0 to 6, you would add 6, but to get from 6 to 16 you add 10).
Answer:
(3x-1) + (3x-1) + (2x+1) = 55
Step-by-step explanation:
1. 3x + 3x + 2x = 8x
2. -1 + -1 + 1 = -1
3. 8x -1 = 55
4. 8x = 55 + 1
5. 8x = 56/8
6. x = 7
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)
Answer:
See below.
Step-by-step explanation:
A year has 12 months and 365 days.
3 months: 3/12 year = 1/4 year
55 days: 55/365 year = 11/73 year
1 month: 1/12 year
7 months = 7/12 year
120 days = 120/365 year = 24/73 year
