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ASHA 777 [7]
2 years ago
8

.Help Please! Is this a monomial, binomial, trinomial, or polynomial? Explain

Mathematics
1 answer:
zhenek [66]2 years ago
4 0

Answer:

Polynomial

Step-by-step explanation:

=> x³ - 4x² + 1 + 3x² + x

=> x³ - x² + x - 1

This is a polynomial expression, as, there are 4 terms in the equation, <em>v</em><em>iz</em><em>.</em><em>,</em><em> </em>x³, x², x and -1.

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em>

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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
What’s the answer to this!!!
aliina [53]

Answer:

8.54

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Geometry Question Help if you can please
docker41 [41]

Answer:

≈ 78.55 cm

Step-by-step explanation:

Using the ratio of arc (AB) to circumference (C) is equal to the ratio of angle subtended at centre by arc AB to 360°, that is

\frac{AB}{C} = \frac{110}{360} , that is

\frac{24}{C} = \frac{110}{360} ( cross- multiply )

110C = 8640 ( divide both sides by 110 )

C = \frac{8640}{110} ≈ 78.55 cm ( to 2 dec. places )

5 0
3 years ago
Simplify 63 ÷ 4 + 2 x 9(32 x 8 – 17 x 4).
Vadim26 [7]

Answer:

98.66

Step-by-step explanation:

Trust Me i got it right on a test .

7 0
3 years ago
PLEASE HELPP
Igoryamba

Answer:

There are 12 bottles of water in a full pack.

Step-by-step explanation:

12

12

12

11

11

9

-------

67

5 0
2 years ago
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