When working with polygons the main properties which are important are: The number of sides of the shape. Theangles<span> between the sides of the </span>shape<span>. The </span>length<span> of the sides of the </span>shape<span>.</span>
I would start by using completing the square to rewrite the function 'y' in vertex form:
![y+6+\frac{1}{4} = (x- \frac{1}{2})^2 \\ \\ y+\frac{25}{4} = (x- \frac{1}{2})^2](https://tex.z-dn.net/?f=y%2B6%2B%5Cfrac%7B1%7D%7B4%7D%20%3D%20%28x-%20%5Cfrac%7B1%7D%7B2%7D%29%5E2%20%5C%5C%20%20%5C%5C%20y%2B%5Cfrac%7B25%7D%7B4%7D%20%3D%20%28x-%20%5Cfrac%7B1%7D%7B2%7D%29%5E2)
Now since you want the inverse, swap the x,y variables
![x+\frac{25}{4} = (y- \frac{1}{2})^2](https://tex.z-dn.net/?f=x%2B%5Cfrac%7B25%7D%7B4%7D%20%3D%20%28y-%20%5Cfrac%7B1%7D%7B2%7D%29%5E2)
Make a substitution to introduce parameter 't'.
![t = y - \frac{1}{2}](https://tex.z-dn.net/?f=t%20%3D%20y%20-%20%5Cfrac%7B1%7D%7B2%7D%20)
Therefore
You have to use Trigonometric ratios here. I'll help you with a question, and you try to do the other two.
a. You are given the hypotenuse, and told to figure out the opposite. The trigonometric function that deals with that is sin(x), which is opposite over hypotenuse. So:
![sin(51)= \frac{y}{12}](https://tex.z-dn.net/?f=sin%2851%29%3D%20%5Cfrac%7By%7D%7B12%7D%20)
Solve for y:
![y = 12sin(51)](https://tex.z-dn.net/?f=y%20%3D%2012sin%2851%29)
Simplify:
![y = 8.043](https://tex.z-dn.net/?f=y%20%3D%208.043)
For these problems, you have to remember the ratios Sine, Cosine, and Tangent. An easy way is to make a mnemonic device. A good one that a lot of people use is SohCahToa. Which is Sine (Opposite, Hypotenuse), Cosine (Adjacent, Hypotenuse) and Tangent (Opposite, Adjacent). Remember trigonometry is just a glorified field of ratios of sides to angles. There are many more trigonometric ratios including inverse trigonometric ratios, reciprocal trigonometric ratios, and hyperbolic trigonometric ratios (which show up during differential calculus). But for now, focus on this. Haha.
Distributive property: <em>a(b + c) = ab + ac</em>.
We have 18 + 24
18 = 6 · 3
24 = 6 · 4
Therefore
18 + 24 = 6 · 3 + 6 · 4 = 6 · (3 + 4)
<h3>Answer: 18 + 24 = 6 · (3 + 4)</h3>
The Laplace transform of the given initial-value problem
is mathematically given as
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
<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)
![\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Let%20%7D%20%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%3D%5Cfrac%7Ba_%7B0%7D%7D%7Bs-4%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Bs%2B5%7D%20%5C%5C%26%5CRightarrow%202%20s-7%3Da_%7B0%7D%28s%2Bs%29%2Ba_%7B1%7D%28s-4%29%5Cend%7Baligned%7D)
![put $s=-s \Rightarrow a_{1}=\frac{17}{9}$](https://tex.z-dn.net/?f=put%20%24s%3D-s%20%5CRightarrow%20a_%7B1%7D%3D%5Cfrac%7B17%7D%7B9%7D%24)
![\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctext%20%7B%20put%20%7D%20s%20%26%3D4%20%5CRightarrow%20a_%7B0%7D%3D%5Cfrac%7B1%7D%7B9%7D%20%5C%5C%5CRightarrow%20%5Cquad%20y%28s%29%20%26%3D%5Cfrac%7B1%7D%7B9%28s-4%29%7D%2B%5Cfrac%7B17%7D%7B9%28s%2Bs%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)
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
Read more about Laplace tranoform
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