One would rewrite this as: 73/(12-7)
B = boys g = girls
.................................
b + g = 195 b = 195 - g
<u>b</u> = <u>8</u>
g 7
Substitute b for 195 - g in the 2nd equation:
<u>b</u> = <u>8</u>
g 7
<u>195 - g</u> = <u>8</u>
g 7
Cross Multiply
8g = 7 . ( 195 - g )
8g = 1365 - 7g
8g + 7g = 1365
15g = 1365
g = <u>1365</u>
15
g = 91
Substitute g for 91 in one of the equations:
b + g = 195
b + 91 = 195
b = 195 - 91
b = 104
S = 104 boys.
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)
