Suppose y=0. That means
<span>x+z = 18 </span>
<span>x-z = 26/5 </span>
<span>x = 58/5 </span>
<span>z = 32/5 </span>
<span>So, the largest x can be is 58/5 </span>
<span>Now, suppose we toss in y. If we subtract y/2 from both x and z, then the difference between x and z is still 26/5, and the sum is </span>
<span>(x-y/2) + y + (z-y/2) = x+z = 18 </span>
<span>For example, if y = 4, </span>
<span>x = 48/5 </span>
<span>y = 20/5 </span>
<span>z = 22/5 </span>
<span>x-z is still 26/5, and the sum is still 18. </span>
Step-by-step explanation:
let E be weight of expensive nuts, C be cheaper nuts
E*8+C*5=10*6 value equation
E+C=10 mass equation
Solve it by any method, I will do determinants
Det: 8-5=3
E= (60-50)/3=10/3 =3.33 lb
C= 80-60 /3=20/3 =6.66 lb
The Laplace transform of the given initial-value problem
is mathematically given as
<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>
Generally, the equation for the problem is mathematically given as
![&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}](https://tex.z-dn.net/?f=%26%5Ctext%20%7B%20Sol%3A-%20%7D%20%5Cquad%20y%5E%7B%5Cprime%7D%2Bs%20y%3De%5E%7B4%20t%7D%2C%20y%280%29%3D2%20%5C%5C%5C%5C%26%5Ctext%20%7B%20Taking%20Laplace%20transform%20of%20%281%29%20%7D%20%5C%5C%5C%5C%26%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%2B5%20y%5Cright%5D%3D%5Cleft%5B%5Cleft%5Be%5E%7B4%20t%7D%5Cright%5D%5Cright.%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%5Cright%5D%2B5%20L%5By%5D%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20s%20y%28s%29-y%280%29%2B5%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%28s%2B5%29%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%2B2%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%2B2%5Cright%5D%3D%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%5Cend%7Baligned%7D)
In conclusion, Taking inverse Laplace tranoform
![L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5By%28s%29%5D%3D%5Cfrac%7B1%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%5Cright%5D%2B%5Cfrac%7B17%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cright%5D%24%20%5C%5C%5C%5C)
Read more about Laplace tranoform
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Answer:
The Answer is A. 83%
Step-by-step explanation:
The Probability a certain event= 1- Probability of that certain event not to happen.
According to this, it is true that:
Probability (not rolling 5)=1- P(rolling 5) = 1=60/350= 1-6/35 =35-6/35=29/35=0.83
Answer:
If you need the explanation, just ask.
