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3 years ago

12

What would be a model that could be used to describe this situation?

1 answer:

Taya2010 [7]3 years ago

4 0

I confused ! I don't get what your trying to say please explain better !

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Please help im so confused??

Ann [662]

V = PI x r^2 x h/3

3.14 x 7^2 x 24/3

49 x 8 = 392

answer is 392PI units^3

answer is B

5 0

3 years ago

1/3(22) - 1 -1\2(12)

irinina [24]

The answer is (1/3)...

6 0

3 years ago

Brainist to who answers this!

yanalaym [24]

Answer:

13t² + 2t + 9

Step-by-step explanation:

(9t² - 3t + 5) - (-4t² + 5t + 4)

Solving like terms

-(-4t²) becomes + 4t²

(9t² + 4t²) + (-3t + 5t) + (5+4) =

13t² + 2t + 9

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

4 0

3 years ago

Read 2 more answers

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago

if the velocity of a body changes from 13m/s to 30m/s while undergoing constant acceleration whats the average velocity of the b

34kurt

Average speed = (1/2) (beginning speed + ending speed)
= (1/2) ( 13 m/s + 30 m/s )
= (1/2) ( 43 m/s )
= 21.5 m/s

3 0

3 years ago

Read 2 more answers