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3 years ago

12

Which is table shows a proportional relationship?????????????????? ??????????? ??????????? ??????????? ??????????? ??????????? ?

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2 answers:

lubasha [3.4K]3 years ago

8 0

Divide each y-coordinate by its corresponding x-coordinate except for the point (0, 0). If every division gives you the same answer, it is proportional.

Table A:
2/(-2) = -1
3/(-1) = -3
Since -1 and -3 are different, this is not proportional.

Table B:
-3/(-1) = 3
3/1 = 3
6/2 = 3
All divisions give the same answer, 3. This is proportional.

Kisachek [45]3 years ago

7 0

Table b is correct. I cant give a full answer due to circumstances but there you go.

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3 years ago

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2 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

3 years ago

What’s the correct answer

tamaranim1 [39]

The correct answer might be A

5 0

3 years ago

Finnegan needs average of 70 on four tests in science to make the honour roll. What is the lowest score he can receive on the fo

Lapatulllka [165]

<span><span>1. </span>Finnegan needs to get the average of 70 in 4 tests in science to be able to make it to honour roll.
Find the lowest score he received on the fourth test if his first tests are 65, 60 and 92.
let x be the missing number
=> 65 + 60 + 92 + x / 4 = 70
=> 217 + x / 4 = 70
=> 217 + x = 4 x 70
=> 217 + x = 280
=> x = 280 – 217
=> x = 63
Thus the lowest score he get is 63.</span>

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3 years ago