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3 years ago

PLEASE HELP

Mathematics

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Can you take a picture of the line or at least give the slope because without that information I won't be able to find the equation.

Step-by-step explanation:

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Write the polynomial so that the polynomial is in standard form -3x+8x²-11x +9x³

arsen [322]

Okay, here we have this:

Considering the provided polynomial, we are going to rewrite it in standard form, so we obtain the following:

So let's remember that a polynomial is in standard form when its terms are written from highest to lowest degree, so in this case when rearranging we have:

-3x+8x²-11x +9x³

=+9x³+8x²-11x-3x

The previous polynomial corresponds to the given polynomial rewritten in standard form.

4 0

1 year ago

Differential Equation

ANEK [815]

1. The given equation is probably supposed to read

y'' - 2y' - 3y = 64x exp(-x)

First consider the homogeneous equation,

y'' - 2y' - 3y = 0

which has characteristic equation

r² - 2r - 3 = (r - 3) (r + 1) = 0

with roots r = 3 and r = -1. Then the characteristic solution is

$y = C_1 e^{3x} + C_2 e^{-x}$

and we let y₁ = exp(3x) and y₂ = exp(-x), our fundamental solutions.

Now we use variation of parameters, which gives a particular solution of the form

$y_p = u_1y_1 + u_2y_2$

where

$\displaystyle u_1 = -\int \frac{64xe^{-x}y_2}{W(y_1,y_2)} \, dx$

$\displaystyle u_2 = \int \frac{64xe^{-x}y_1}{W(y_1,y_2)} \, dx$

and W(y₁, y₂) is the Wronskian determinant of the two fundamental solutions. This is

$W(y_1,y_2) = \begin{vmatrix}e^{3x} & e^{-x} \\ (e^{3x})' & (e^{-x})'\end{vmatrix} = \begin{vmatrix}e^{3x} & e^{-x} \\ 3e^{3x} & -e^{-x}\end{vmatrix} = -e^{2x} - 3e^{2x} = -4e^{2x}$

Then we find

$\displaystyle u_1 = -\int \frac{64xe^{-x} \cdot e^{-x}}{-4e^{2x}} \, dx = 16 \int xe^{-4x} \, dx = -(4x + 1) e^{-4x}$

$\displaystyle u_2 = \int \frac{64xe^{-x} \cdot e^{3x}}{-4e^{2x}} \, dx = -16 \int x \, dx = -8x^2$

so it follows that the particular solution is

$y_p = -(4x+1)e^{-4x} \cdot e^{3x} - 8x^2\cdot e^{-x} = -(8x^2+4x+1)e^{-x}$

and so the general solution is

$\boxed{y(x) = C_1 e^{3x} + C_2e^{-x} - (8x^2+4x+1) e^{-x}}$

2. I'll again assume there's typo in the equation, and that it should read

y''' - 6y'' + 11y' - 6y = 2x exp(-x)

Again, we consider the homogeneous equation,

y''' - 6y'' + 11y' - 6y = 0

and observe that the characteristic polynomial,

r³ - 6r² + 11r - 6

has coefficients that sum to 1 - 6 + 11 - 6 = 0, which immediately tells us that r = 1 is a root. Polynomial division and subsequent factoring yields

r³ - 6r² + 11r - 6 = (r - 1) (r² - 5r + 6) = (r - 1) (r - 2) (r - 3)

and from this we see the characteristic solution is

$y_c = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}$

For the particular solution, I'll use undetermined coefficients. We look for a solution of the form

$y_p = (ax+b)e^{-x}$

whose first three derivatives are

${y_p}' = ae^{-x} - (ax+b)e^{-x} = (-ax+a-b)e^{-x}$

${y_p}'' = -ae^{-x} - (-ax+a-b)e^{-x} = (ax-2a+b)e^{-x}$

${y_p}''' = ae^{-x} - (ax-2a+b)e^{-x} = (-ax+3a-b)e^{-x}$

Substituting these into the equation gives

$(-ax+3a-b)e^{-x} - 6(ax-2a+b)e^{-x} + 11(-ax+a-b)e^{-x} - 6(ax+b)e^{-x} = 2xe^{-x}$

$(-ax+3a-b) - 6(ax-2a+b) + 11(-ax+a-b) - 6(ax+b) = 2x$

$-24ax+26a-24b = 2x$

It follows that -24a = 2 and 26a - 24b = 0, so that a = -1/12 = -12/144 and b = -13/144, so the particular solution is

$y_p = -\dfrac{12x+13}{144}e^{-x}$

and the general solution is

$\boxed{y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x} - \dfrac{12x+13}{144} e^{-x}}$

5 0

3 years ago

The following foreign adoptions (in the United States) occurred during these particular years. 2006 2010China 63403465Ethiopia 7

worty [1.4K]

Answer: P(Ethiopia 2010) = 0.3580

Step-by-step explanation:

Please find the attached file for the solution.

8 0

3 years ago

Many realistic application involve sampling without replacement. For example, in manufacturing, quality control inspectors sampl

stich3 [128]

Answer:

a. possible sample size is 10

b. mean is 3

c standard deviation is 0.9

Step-by-step explanation:

Mean of any given set of numbers is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers.

Standard deviation is a measure of the amount of variation or dispersion of a set of values.

Go to attachment for detailed analysis.

4 0

4 years ago

Pls help me asap !!!!!

loris [4]

7x-1+5x-23= 180
12x-24=180
12x=204
X=17

3 0

3 years ago