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3 years ago

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Can i have some help

1 answer:

olga2289 [7]3 years ago

3 0

Answer:

you can ask questions that are on tests or live exams. I have reported this to brainly. Please don't do that again.

Step-by-step explanation:

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PLEASE HELP I WILL GIVE BRAINLIEST MUST USE WORK :)

Nikolay [14]

slope = (y2-y1)/(x2-x1)

m =(-2-7)/(8-0)

m =(-9/8)

slope is -9/8

the y intercept is 7 (0,7)

y = mx +b

y = -9/8 x+7

Answer: y = -9/8x +7

6 0

3 years ago

What is the slope of 3,5 and -3,-5

olya-2409 [2.1K]

The slope is 5/3 five over 3. You youse the formula y2 - y1 over x2-x1. Plug into calculator,and then reduce.

5 0

3 years ago

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

In the circle below, suppose m WXU<br> 216° and m XWV= 137º. Find the following.

sladkih [1.3K]

Answer:

incompleted question

Step-by-step explanation:

......?

8 0

3 years ago

An object is launched at 9.8 meters per second (m/s) from a 308.7-meter tall platform. The function that represents the object's

Sedaia [141]

this is the full answer for you're Q

please give me good

8 0

3 years ago