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3 years ago

14

Help ASAP! 25 points! I will also give brainliest!

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

756 m²

Step-by-step explanation:

I just know the answer XD

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(5x10^3)x(9x10^7) can someone please answer this for me.

Anna11 [10]

4.5 times 10^10

1. Multiply 5times ten to the third

2. Multiply 9 times ten raised to the seventh

Then multiply them together
5000 times 90000000

5 0

3 years ago

Fine fare sells apple juice boxes in packages of 4 juice boxes for $2.00. What is the unit price of each juice box

Zinaida [17]

Answer:

its 8 because if you do 4 times 2 equal 8

4 0

3 years ago

Hi could someone tell me how to solve this, you don’t even have to tell me the answer I’m just confused on this math and need he

BabaBlast [244]

Answer:

132 ft

Step-by-step explanation:

The radius is 21 ft

We want to find the circumference

C = 2 * pi *r

Letting pi = 22/7

C = 2 * 22/7 * 21

C = 132 ft

8 0

3 years ago

Read 2 more answers

Please help me! what does x=

marta [7]

58 degrees bra cause the arc is 58 degrees and due to congruent angles x is also 58 degrees

8 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