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3 years ago

5

The collection of whole numbers is an example of a set. OA. True O B. False

1 answer:

Pani-rosa [81]3 years ago

6 0

Answer:

It is false

Step-by-step explanation:

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Why did my DAD leave me is it because he needed to get some milk or to get some more pop from the store? Why would I know Im jus

mart [117]

Answer:

I have 1 question, and 1 question only, WHO t.f says "pOp" like nooooo its "SODA"

and your dad sucks for leaving but im sure he has a good reason

Step-by-step explanation:

8 0

2 years ago

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A certain bay with very high tides displays the following behavior. In one 12-h period the water starts at mean sea level, rises

Firdavs [7]

Answer:

the required equation is; y = 21 sin(πt/6)

Step-by-step explanation:

Given the data in the question;

Water rises above sea level = 21 ft

Water drops below sea level = 21 ft

so

maximum = 21 ft and also

minimum = 21 ft

Amplitude = ( maximum + minimum ) / 2

Amplitude = ( 21 + 21 ) / 2

Amplitude = ( 21 + 21 ) / 2 = 42/2 = 21 ft

Period = 12 hours

and we know that; period = 2π/b

so

12 = 2π/b

12b = 2π

b = 2π / 12

b = π/6

Standard equation for simple harmonic is; y = asin(bt)

we substitute

y = 21 sin(πt/6)

Hence the required equation is; y = 21 sin(πt/6)

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

2 years ago

Convert decimal to fractions 092 =

eduard

92/1000
hope this helps

4 0

2 years ago

Read 2 more answers

you increase your walking speed from 1 m/s to 3 m/s in a period of 1 s. What is your acceleration? A) 2 m/s ^2 B) 3 m/s ^2 C) 5

arsen [322]

Acceleration is to change in speed divided by the time for the change. Change in speed is end speed minus started speed.

3m/s-1m/s and equals to 2m/s.

The answer should be A. (2m/s^2.)

Hope it helped you!

Have a great day!

8 0

2 years ago

Read 2 more answers