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mario62 [17]
3 years ago
13

I need help pls i don't know what to do at all pls help

Mathematics
2 answers:
MariettaO [177]3 years ago
8 0

Answer:

Add all the outer lines and they may add up to be the perimeter, if you need the area multiply 4 times 8 and add that to 5.6*4/2 +4.5*4/2 plus four times eight

Step-by-step explanation:

ozzi3 years ago
7 0

Answer:

I remember these form when I was like 10 but ok.

Step-by-step explanation:

pretty sure you multiply the base times the height. But first you have to take out all triangwkes which leaves you with two of them.

so 8x4 for the top one and then 14x4.5 for the bottom one.

then for the triangles the one on the left is 5.6x4

the right one is 4.5x14

then divide both triangles answers into two. That should be the triangle areas. Then add the two triangles pieces together.

once don’t add the squares areas with the triangles and that’s THE AREAAAA

im sry I’m a lil rusty so if it’s wrong pwease forgive me :/

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A number line with a closed circle on 25, with shading to the right.

Explanation:

A closed circle means "greater than or equal to" or "less than or equal to".

A open circle means "less than" or "greater than"

Therefore, since the question says: "who are at least 25 years old", a closed circle is the correct answer.

And says it says "at least 25 years old" it means that it shaded to the right, and that the circle is on the number 25.

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3 years ago
Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2
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The Laplace transform of the given initial-value problem

y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<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}

\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}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\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}

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]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

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Angle 7 and Angle 2 are also congruent, so if Angle two forms a straight line with Angle 1, Angle 7 should be able to do that too.

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