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IRISSAK [1]
3 years ago
14

When asked to find an equivalent expression to 3(8+7) 3(8+7) , James said 24 + 21. a) What property did James use? b) Do you agr

ee with James? Explain.
Mathematics
1 answer:
ryzh [129]3 years ago
4 0

Answer:

  a) the distributive property

  b) Yes

Step-by-step explanation:

a) The distributive property of multiplication over addition tells you ...

  a(b+c) = ab +ac

James has a=3, b=8, c=7, so he can fill in the values to get ...

  3(8+7) = 3·8 +3·7

Taking this one step into the evaluation, James gets ...

  3(3+7) = 3·8 +3·7 = 24 +21 . . . . . . the result James reports

___

b) Our result agrees with James' result, so we agree with James.

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b

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1 = 1.61

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1.61 x 30 =

48.3

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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
NEED THE ANSWER ASAP PLEASE
BARSIC [14]

Answer:

y = 2x -8

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Answer:

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Step-by-step explanation:

Let width = x, then length = 3x.

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x = 32 / 8 = 4 m.

The length = 3*4 = 12m and width = 4 m.

6 0
1 year ago
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