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Ray Of Light [21]

2 years ago

11

I’m having trouble with this answer

1 answer:

Pachacha [2.7K]2 years ago

5 0

Answer:-414

Step-by-step explanation:

x/-18=23

+18= +18

x= -18 times 23

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Help plzz division math prob

leva [86]

g(x) = x - 3 = 0

g(x) = x = 3

f(x) = 2x^3 + x - 4

f(3) = 2(3)^3 + 3 - 4

f(3) = 2(27) - 1

f(3) = 54 - 1

f(3) = 53

The remainder when f(x) is divided by x - 3 is <u>53</u>.

3 0

2 years ago

Solve for m.t=mr^2/3

spin [16.1K]

t = mr^2/3

Divide both sides by r^2/3

m = t / r^2/3

6 0

2 years ago

Read 2 more answers

in a hotel ,100 bulbs each of 60W are lightened for 6 hours a day and 40 fans each of 50Wvate run for 10 hours a day . calculate

alekssr [168]

Answer:

1680 kwhr

Step-by-step explanation:

See attached for explanation. I converted watt*hr to kW*hr by dividing by 1000.

Fans Bulbs

40 100

50 60 W

10 6 hours/day

20000 36000 W*hr/day

30 30 days/month

600000 1080000 W*hr/month

600 1080 kWhr/month

Total = 1680 kWhr/month

5 0

2 years ago

PLS HELP!!!! A rectangle shown has a length of 11cm and a width of 4.2cm. A circle is drawn inside that touches the rectangle at

oee [108]

Answer:

C

Step-by-step explanation:

3 0

3 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