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2 years ago

10

I need help with number 8 and 9! :) Thanks!

1 answer:

Arturiano [62]2 years ago

5 0

8 = D
9 = A
i think that 9 might be C as well but i just went for my gut feeling

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Which expression is equivalent to (-4y-x)-(y-9x) ?

777dan777 [17]

Answer:

-5y + 8x

Step-by-step explanation:

$(-4y-x)-(y-9x)\\\\-4y-x-(y-9x)\\\\\rightarrow-(y-9x)*-1=-y+9x\\\\-4y-x-y+9x\\\\-4y-y+9x-x\\\\\boxed{-5y+8x}$

7 0

2 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

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2 years ago

Where is the point of discontinuity of y =<br> -4x+81/x-2<br> ?

Genrish500 [490]

Answer:

x = 2

Step-by-step explanation:

y = $\frac{-4x+81}{x-2}$

If the denominator is equal to zero , then y is undefined.

this discontinuity can be found by equating the denominator to zero and solving for x.

x - 2 = 0 ⇒ x = 2 is the point of discontinuity

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1 year ago

4581 divided 32 in Long division

lubasha [3.4K]

Answer: 143.15625

Step-by-step explanation:

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2 years ago

Does anyone know how to solve 54 = 36 - 9b

vodomira [7]

Answer:

-2

Step-by-step explanation:

54 - 36 = 18

18 % -9 =-2

6 0

2 years ago

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