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Andreas93 [3]
2 years ago
11

I need this answered plz!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Fynjy0 [20]2 years ago
6 0

Answer:

Red: Left side equation

Blue: Right side equation

2x+7=2(x-1)

2x+7=2x-2

2x-2x=-2-7

0x=-9= 0 No solution

Step-by-step explanation:

Use Desmos graphing calculator if u need help with graphs

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) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
ANSWER RIGHT AWAY!!!!​
Burka [1]

Answer:

1st and 3rd are functions, 2nd and 4th are not functions

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Please Help, I need to know how to graph these equations
devlian [24]
There's a site called m.a.t.h.w.a.y, and you can type in those functions in order to get the graph ;)
5 0
2 years ago
PLEASE HELP I DONT WANNA FAIL MY QUIZ!!!
Gwar [14]

Answer: 2.5 ft.

Step-by-step explanation: We know that the wall is 10ft and each poster is 1 1/4 ft, so subtract both posters from the wall, 10-2.5 = 7.5, then divide 7.5 by 3 since we need 3 lengths, and we get 2.5ft. So for x, it would be 2.5ft.

x=2.5

Hope this helps!

3 0
3 years ago
Read 2 more answers
In triangle DEF, side DE is 3 inches, side EF is 4 inches, and sides DF is 5 inches. The largest angle is opposite which side.?
zimovet [89]
The longest angle would be across from the one with the longest side. In this case it would be across from DF
4 0
2 years ago
Read 2 more answers
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