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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You find the variable x or y in one of the lines of the problem and then substitute the value of the variable you first found into the other line of the problem to find the other variable. Then, you plug all of the values you found into either line/equation of the problem to ensure that one side of the equation you pick is equal to what is on the other side of the "=" symbol.
The most cookies Heidy can make is 36 cookies.
<h3>How to calculate how many cookies can Heidy make?</h3>
To know how many cookies Heidy can make, you have to take into account the following information:
12 cookies need the following ingredients:
- 125g butter
- 200g flour
- 50g sugar
In the case in which Heidy has more ingredients, we must carry out the following operations:
Divide the quantities, in the reference quantity we have:
- 500g of butter ÷ 125g of butter = 4
- 700g flour ÷ 200g flour = 3.5
- 250g of sugar ÷ 50g of sugar = 5
According to the above, we must take into account the lowest value of all because if that ingredient is enough, we can infer that the rest of the ingredients also.
So the number of cookies Heidy can make are:
12 × 3.5 = 42
Learn more about ingredients in: brainly.com/question/26532763
Answer:
( 2,1) is the center of dilation and -2 is the scale factor
Step-by-step explanation:
We can use the formula
A' = k( x-a) +a, k( y-b)+b where ( a,b) is the center of dilation and k is the scale factor
(0,0) becomes (6,3)
( 6,3) = k( 0-a) +a, k( 0-b)+b
6 = -ka+a
3 = -kb+b
We also have
(4,0) becomes (-2,3)
( -2,3) = k( 4-a) +a, k( 0-b)+b
-2 =4k -ka+a
3 = -kb+b
Using these two equations
6 = -ka+a
-2 =4k -ka+a
Subtracting the top from the bottom
-2 =4k -ka+a
-6 = ka -a
-------------------
-8 = 4k
Divide by 4
-8/4 = 4k/4
-2 = k
Now solving for a
6 = -ka +a
6 = - (-2)a +a
6 = 2a+a
6 = 3a
Divide by 3
6/3 =3a/3
2=a
Now finding b
3 = -kb+b
3 = -(-2)b+b
3 = 2b+b
3 = 3b
b=1
<span>B. sam forgot to multiply.
The correct answer is
</span> x^2*y=(7^<span>2)*(9)=49*9=441 </span>
