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leonid [27]
3 years ago
8

What is the domain of the function shown in the mapping?

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0

Answer:

The correct is A)  {x | x = -5, -3, 1, 2, 6}

Step-by-step explanation:

Consider the provided function.

The domain of the function is the set of x values which a function can take.

Or we can say that the set of input values.

Now consider the mapping shown in figure:

Consider the "input" oval. The input oval represents the domain of the function.

The values in input oval are -3, 2, -5, 1, 6

This can be written as: -5, -3, 1, 2, 6

Therefore the domain of the function is: {x | x = -5, -3, 1, 2, 6}

Hence, the correct is A)  {x | x = -5, -3, 1, 2, 6}

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Perform the indicated operation.<br><br> (-50) ÷ (-2)<br><br> 48<br> -48<br> -25<br> 25
mezya [45]

Answer:

25

Step-by-step explanation:

Remember that multiplication and division will result in a positive number if no or two negative signs are present. So here, you ignore the negative signs.

All you really have to do is divide 50 by 2, which results in 25.

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3 years ago
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Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainde
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Answer:

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  actual roots: -1, (2 ±4i√2)/3

  no turning points; no local extrema

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Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

  ±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

  = ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

  f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

  x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

  x = (2 ± 4i√2)/3 . . . . complex roots

__

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

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NNADVOKAT [17]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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