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Liono4ka [1.6K]

3 years ago

14

What is the value of [4-9472 0 -36 0-33 -32 0-30

1 answer:

beks73 [17]3 years ago

8 0

Answer:

Su búsqueda - ¿Cuál es el valor de [4-9472 0 -36 0-33-32 0-30 - no produjo ningún documento.

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Step-by-step explanation:

You might be interested in

Please i need help on my math

Kamila [148]

Answer:

$V = \pi (2.5)^2h$

Step-by-step explanation:

The formula for the volume of a cylinder, is V = π r^2 h. It shows that the diameter is 5. But we need the radius. The radius is half of the diameter. So, 5/2. This gets you 2.5.

The equation is now:

V = π(2.5)^2h

This is the answer.

$V = \pi (2.5)^2h$

8 0

2 years ago

kate has the following scores on her math test this quarter 89,79,83, and 94.if she want an average of 85 ,what musther score be

Georgia [21]

She must score an 80.

85*5= 425
425-89-79-83-94= 80

8 0

2 years ago

apply the function transformation of g(x)=f(x-2)+4 to f(x)=x^3-8. describe the transformation and find the final form of the fun

Fittoniya [83]

Answer:

g(x) = (x - 2)^3 - 4

Step-by-step explanation:

first, the function f is shifted to the right by 2 units and then shifted up by 4 units

4 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

3 years ago

Richard mows 1/3 of his yard in 1/2 hour. How much of his yard would Richard mow in 1 hour?

Vlada [557]

I hope this helps you

if he mows 1/3 yard in 1/2 hour

? yard in 1 hour

?.1/2=1.1/3

?=2/3 yard in 1 hour

3 0

3 years ago

Read 2 more answers