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Grace [21]
3 years ago
7

I have no idea how to do these.

Mathematics
2 answers:
Alex3 years ago
8 0

Answer:

1.  {-1 , 0, 1 }

Step-by-step explanation:

The range is the set of the second values in the ordered pairs.

Jobisdone [24]3 years ago
5 0
The range of a function is all the possible y values, and the domain of a function is all the possible x values.
For 1, the range is the list of y values: :so, it is B.
2) A
3) x can be anything. So the answer is b.
4) range refers to y values. The answer is D
5) the range goes from 6 all the way down. The answer will be "all real numbers less than or equal to 6.
7) all real numbers equal to or less than 5.
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What is the absolute value of -62/3
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Answer:

Step-by-step explanation:

The absolute value is 62/3

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Which of the following is a rational number? square root of 22, square root of 23, square root of 24, square root of 25
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<span>square root of 25 is the correct answer</span>
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X+1 if x is less than 1
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Step-by-step explanation:yes

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Wittaler [7]

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5 0
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
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