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3 years ago

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Please help The question is on Mathamatics

1 answer:

sweet [91]3 years ago

4 0

Answer:

A
(The probability of randomly selecting a picture that shows Justin with his friends is greater than the probability of randomly selecting a picture that shows Justin with his family.)

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An organism that feeds upon primary consumers is known as a _______.

Snezhnost [94]

It's known as a heterotroph. They can't make their own food.

4 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

4 years ago

Help me please ......

oksano4ka [1.4K]

-4 = x - 7
-4 + 7 = x
3 = x

Therefore x = 3

4 0

3 years ago

In a board game, students draw a number do not replace it, and then draw a second number. Determine the probability of each even

murzikaleks [220]

The figure depicting the board game is attached below.

Answer:

Step-by-step explanation:

Kindly note that selections done without replacement.

Count of numbers on the board game = 8

Count of odd numbers = (1, 9, 1) = 3

Count of digit 6 = 3

Probability = required outcome / Total possible outcomes

P(picking an odd number) = 3 / 8

Without replacement

Numbers left on board game = 8 - 1 = 7

P(picking a 6) = 3 / 7

Hence,

P(picking an odd number then, a 6) = 3/8 * 3/7 = 9 / 56

7 0

3 years ago

Evaluate the imaginary number i^3

Pavlova-9 [17]

The answer to the problem i^3 is -i

6 0

3 years ago