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3 years ago

10

Which equation represents a line which is parallel to the line y=1/2x + 2

1 answer:

levacccp [35]3 years ago

8 0

Answer:

1/2

Step-by-step explanation:

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Can someone please help me on question 1 and 2 . Thank you.

crimeas [40]

Answer:

sorry I can't view

Step-by-step explanation:

i have no this app sorry

4 0

3 years ago

Given triangle, ABC is congruent to triangle EDC find A

arlik [135]

Answer:

m<A = 30 degrees.

Step-by-step explanation:

m < C in triangle ABC = m < C in EDC so

x + 13 + 42 + x + 13 = 180

2x = 180 - 13 - 13 - 42

2x = 112

x = 56.

So m < C = 56+13 = 69.

m < B = m < D = 81

M < A = 180 - 69 - 81

= 180 - 150 = 30.

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

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

7 0

3 years ago

The ordered pair (4, -8) is in the 4th quadrant if we took the opposite of the x- value and the opposite of the y-value, which q

harina [27]

9514 1404 393

Answer:

2nd quadrant

Step-by-step explanation:

Reversing the signs of both coordinates reflects the point across the origin. The quadrant diagonally opposite quadrant 4 is quadrant 2.

_____

You recall that quadrants are numbered 1 to 4 counterclockwise, starting from upper right.

8 0

3 years ago

Fifty six is what decimal part of 70

vodka [1.7K]

56 is a decimal part of 70. 56 = 4/5 = .8

5 0

3 years ago

Read 2 more answers