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Aleonysh [2.5K]
2 years ago
6

Whats the answer to #9 i need help fast!!

Mathematics
1 answer:
vekshin12 years ago
5 0
I got 784 inches cubed, I cut the figure into two, the first one was 8×4×14=448, the second was 8×3×14=336,
therefore 448+336=784
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What is the length of line segment VT?<br><br> 4 units<br> 8 units<br> 13 units<br> 14 units
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Is that just the question? I can help you out but it doesn’t make any sense.
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3x + 4y = 12<br> Convert to slope-intercept form
Zigmanuir [339]

Answer:

y=-3/4x+3

Step-by-step explanation:

subtract 3x from both sides:  4y=-3x+12

Divide both sides by 4:  y=-3/4y+3

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Paulina has read 58 pages of the 234-page book she is doing a book report on for Language Arts. She plans to finish her book by
goldfiish [28.3K]
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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
For which value of a does Limit of g(x) as x approaches alpha not exist?​
Hunter-Best [27]

The value of a where the Limit of g(x) as x approaches alpha not exist are -1 and 1

<h3>Limit of a function</h3>

The limit of a function is the limit of a function as x tends to a value.

From the given graph, you can see that the function g(x) goes large at the point where the arrows orange and purple point down from the x-coordinates -1 and 1.

Hence the value of a where the Limit of g(x) as x approaches alpha not exist are -1 and 1

Learn more on limit of a function here: brainly.com/question/23935467

#SPJ1

7 0
1 year ago
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