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3 years ago

I need help with this I don't get it at all

Mathematics

2 answers:

iren [92.7K]3 years ago

8 0

Answer:

the answer is C

Step-by-step explanation:

because the negative needs to be on the inside not on the outside

enyata [817]3 years ago

8 0

Answer:

Step-by-step explanation:

you cant get the answer if the number is negative

since it's a square it's either positive or 0(zero)

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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

PLEASE ANSWER FAST HAVE TO FINISH TEST BY THE END OF THE DAY:

Dennis_Churaev [7]

Step-by-step explanation:

0.05*6=0.3*5000=1500

3 0

2 years ago

An example of dependent events is drawing a green marble out of one jar and then drawing a....?

gayaneshka [121]

Dependent events involve a first event that affects the outcome of a second event. This means that the probability for the second event has changed because of the occurrence of the first event. This is usually present in probabilities without replacement because each event has a different condition. The answer is C)red marble out of the same jar,without replacing the first marble.

4 0

2 years ago

Please answer this question!

Reil [10]

The answer is D, 2,500

5 0

2 years ago

In the given figure PQ is a transversal intersecting lines AB and CD. Find the measures of angles x and y. Hence prove AB//CD.

tekilochka [14]

Explanation:

From the given figure ,

AB || CD

PQ is a transversal

Given angles are 50° and 130°

50° and x are linear pair

⇛ 50°+x = 180°

⇛ x = 180°-50°

⇛ x = 130°

Therefore, "x" = 130°

and

130° and y are vertically opposite angles which are equal.

⇛ "y" = 130°

Therefore, y = 130°

Additional comment:-

If two parallel lines Intersected by a transversal then
Corresponding angles are equal.
Vertically Opposite angles are equal.
The sum of two adjacent angles is equal to 180° then they are called Linear Pair.

3 0

2 years ago