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Sergeeva-Olga [200]

2 years ago

12

Use the Associative Property of Addition to write an expression equivalent to (u + 2) + 8.

1 answer:

Alla [95]2 years ago

7 0

Answer:

D

Step-by-step explanation:

First of, associative property of addition is rearranging the parenthesis and not moving the numbers

A is incorrect because you can not take out the parenthesis in the equation

B is inccorrect because they change the position of the number

C is incorrect because they change the position of number

D is correct. It only change the position of the parenthesis

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Which of the following is equivalent to 0 equals 3 x squared minus 12 x minus 15 when completing the square?

skad [1K]

Answer:

x = 5, -1

Step-by-step explanation:

0 = 3x^2 - 12x - 15 or

3x^2 - 12x - 15 = 0

By solving for x, use the Quadratic Formula.

… x = 5 and/or -1

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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