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2 years ago

12) 5.8+ 7 what is the answer am not sure

Mathematics

2 answers:

makkiz [27]2 years ago

8 0

im assuming you're adding the two, it's 12.8.

VikaD [51]2 years ago

5 0

Answer:

12,8

Step-by-step explanation:

Just line them up and you will arrive at your answer:

5,8

+ 7

_____

12,8

I am joyous to assist you anytime.

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What is the total cost of a $12.49 item with a sales tax of 5%?

Ivan

Answer: $13.11

Step-by-step explanation:

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3 years ago

The class has collected $75.50 for a pizza party. A large pizza costs $9.50. Which one-step-inequality could be used to find the

Veseljchak [2.6K]

Answer:

7 large pizzas

Step-by-step explanation:

multiply 9.50 by every number on the calculator and get 9.50×7=66.5

or just divide 75.50 by 9.50

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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

Someone help me please

Y_Kistochka [10]

Step-by-step explanation:

the answer is

About 50 inches

4 0

3 years ago

The coordinates below represent two linear equations. How many solutions does this system of equations have? Line 1 x y –1 6 0 4

iren2701 [21]

I think you need to retype this one. The formatting makes this question difficult to understand.

8 0

3 years ago

12) 5.8+ 7 what is the answer am not sure​

12) 5.8+ 7 what is the answer am not sure