Ask question

Ask question

All categories

2 years ago

8

−15(x−4)=−2 Hurry! Brainliest if right!

2 answers:

devlian [24]2 years ago

6 0

Step-by-step explanation:

The answer is 62/15.

I can't explain it bc I'm taking a test rn but still-

Ronch [10]2 years ago

5 0

Answer:

4x-17

Step-by-step explanation:

You might be interested in

Solve the following dividend problem.alameda tent company has 75,000 shares of stock outstanding. what total dividend declaratio

butalik [34]

75,000×2=150,000
...............

5 0

3 years ago

Read 2 more answers

If a figure is translated 5 units down and 3 units right, how do you represent it algebraically

qwelly [4]

Answer

(x+3) -5

Step-by-step explanation:

plug the values into the parent transformation equation to its rightful places

6 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

2 years ago

Please help! Only answer if you know how to do it.

Mashutka [201]

Your answer would 699.3 that is it

7 0

2 years ago

THESE ARE THE QUESTIONS<br> WILL MARK BRAINLIEST 99 POINTS

Arisa [49]

1. You are given PR and RS, so simply add the two measurements to get PS.

28.2+30.7= 58.9

2. Subtract QV from QT to get TV.

78-56= 22

3. Again, Subtract QV from QT to get TV.

74-36= 38

4. Similar to question 1, add PR and RS to get PS.

40+53= 93

4 0

3 years ago

Read 2 more answers