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3 years ago

7

WILL GIVE BRAINLIEST AND 100 PNTS!! Which of the tables represents a function? Table A Input Output 5 3 5 2 4 1 Table B Input Ou

tput 1 2 3 2 5 3 Table C Input Output 0 0 1 2 1 3 Table D Input Output 4 2 4 3 4 4 Table A Table B Table C Table D

1 answer:

expeople1 [14]3 years ago

4 0

Answer:

Table B

Step-by-step explanation:

Table A Input Output

5 3

5 2

4 1

Has an input that goes to 2 different outputs, not a function

Table B Input Output

1 2

3 2

5 3

one to one relation which is a function

Table C Input Output

0 0

1 2

1 3

Has an input that goes to 2 different outputs, not a function

Table D Input Output

4 2

4 3

4 4

Has an input that goes to 2 different outputs, not a function

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The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

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Will have a characteristic equation of the form:

$a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0$

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For real repeated roots the solution is given by:

$y(t)=c_1e^{rt} +c_2te^{rt}$

For complex roots the solution is given by:

$y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)$

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The characteristic equation is:

$r^{2} +1=0$

So, the roots are given by:

$r_1_,_2=0\pm \sqrt{-1} =\pm i$

Therefore, the solution is:

$x(t)=c_1cos(t)+c_2sin(t)$

As you can see, is the same solution provided by the problem.

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$x'(t)=-c_1sin(t)+c_2cos(t)$

Evaluating the initial conditions:

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And

$x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2$

Now we have found the value of the constants, the solution of the second-order IVP is:

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