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3 years ago

12

Determine how much the bill will be if there are 50 text messages

1 answer:

svlad2 [7]3 years ago

5 0

The bill will be $100 bc the rate of it is .5 per text so basically all your doing is multiplying by 2

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Compute 1 + 2 + 3 +....+ 1,997 + 1,998 + 1,999

sp2606 [1]

Answer:

1 999 000

Step-by-step explanation:

Formula:

$1+2+3+.\ .\ .+n=\frac{n\times \left( n+1\right) }{2}$

………………………………………

Then

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 1999+1\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 2000\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{3998000 }{2}$

$1+2+3+....+1997+1998+1999=1999000$

7 0

2 years ago

Which statement is true?

Vlada [557]

I think a but not sure

7 0

3 years ago

Read 2 more answers

Find the measure of LM

Karo-lina-s [1.5K]

It’s LM becuase look at the photo Lm means

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3 years ago

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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

4 years ago

Easy only shapes there might be multiple answers but it could only be 1 answer

Pavel [41]

Answer:

I think this one would be none of the above

5 0

3 years ago