Answer:
Step-by-step explanation:
First get the answer to -3n - 4 = 2
-3n - 4 + 4 = 2 + 4
-3n = 6
n = 6/-3
n = -2
That answer is the only one that is permitted. It is the only one that completely satisfies the equation.
Now when you do the inequality, look what happens.
-3n - 4 < 2 Add 4 to both sides.
-3n-4+ 4 < 2+4
-3n < 6 Now there are a bunch of ways (2) to solve this.
No matter which way you do it, the arrow will change.
-3n/-3 > 6/-3
n > - 2
That means that any number that is greater than - 2 will satisfy the inequality.
So n = 0 will work. Even n = - 1 will work. Anything bigger than -2 will work. The equation does not provide that kind of latitude.
72 minutes
Take 24 divided by 15 = 1.6 minutes per page
Then take 1.6x45=72
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:
it can be anything.
Step-by-step explanation:
Answer:
2x^2-18
Step-by-step explanation:
2x to the power of 2 minus 18
