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2 years ago

I am trying to solve this problem then simplify

Mathematics

2 answers:

Alborosie2 years ago

5 0

Answer:

3x^2+11x+6

Step-by-step explanation:

3x +2

x 3x^2 2x

+3 9x 6

3x^2+9x+2x+6
=

3x^2+11x+6

Archy [21]2 years ago

3 0

6 is the correct answer

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Step-by-step explanation:

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Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

y''(t) + p(t) y'(t) + q(t) y(t) = f(t)

so the first thing you need to do is divide both sides of your equation by t :

y'' + (2t - 1)/t y' - 2/t y = 7t

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and W denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

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2 years ago

If no vertical line intersects the graph of a relation in more than one point, then the relation is a function.

ki77a [65]

If I'm understanding this correctly you mean:
If a vertical line does not connect to more than one point in a relation then that relation is a function.
Then your answer is A-True.
That is called the vertical line test. The vertical line cannot touch more than one point at the same time if it is a function. If it does touch then it is not a function.

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3 years ago

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BartSMP [9]

Answer: 43.25?

Step-by-step explanation:

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