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3 years ago

11

96 free points! Hurry!

2 answers:

xeze [42]3 years ago

8 0

Answer:For what?

Step-by-step explanation:

lisov135 [29]3 years ago

3 0

Answer:

hstwjiwjwgswjhwjwuwbwnjwwjgegwhbw

You might be interested in

What is 3x² + 3x - 36 and x² - 3x - 28 in factored form?

satela [25.4K]

Answer:

3x² + 3x - 36=

(3x+12)(x-3)

x² - 3x - 28=

(x+4)(x-7)

Step-by-step explanation:

For both equations use the method called the "Cross Method". It is useful for these type of factorisation questions called "Trinomials".

For more info on the cross method.

Here it is:

https://www.mathsteacher.com.au/year10/ch10_factorisation/05_cross_mult_method/cross.htm

Hope you enjoyed :D

3 0

3 years ago

Read 2 more answers

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

Pascal wants to ride his bike to and from school 3 days a week. His house is 2.58 kilometers from his school. How many meters wi

klemol [59]

Answer:

I thi k the answers is 108.36

5 0

3 years ago

Please help me asap in this question

taurus [48]

Answer:

∠DKH and ∠ENH

Step-by-step explanation:

∠DKH and ∠ENH

7 0

3 years ago

PLEASE HELP FIGURE OUT HOW TO DO THIS! THANK YOU!

S_A_V [24]

Y=16 x=20
put these values in equation and see rifht or wrong

A) 16 ≠20*20 wrong
B) 16 ≠ (5/4)*20 - wrong
C) 16=(4/5)*20 - right
D) 16≠0.6*20
Answer: C) y=4/5*x

5 0

4 years ago

Read 2 more answers