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A. xB. −xC. x^5D. x − 1E. x^(−
- a)x
- b)−x
- c)x5
- d)x − 1
- e)x(−1)
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A. xB. −xC. x^5D. x − 1E. x^(−a)xb)−xc)x5d)x &...
Understanding the Question
The question likely revolves around identifying which expression is a polynomial function. Polynomials are mathematical expressions that consist of variables raised to whole-number powers and coefficients.
Analysis of Each Option
- A. x
- This is a polynomial of degree 1.
- B. −x
- This is also a polynomial of degree 1.
- C. x5
- This is a polynomial of degree 5.
- D. x − 1
- This is a polynomial of degree 1. It can be expressed as x1 + (-1).
- E. x(−a)
- This is not a polynomial because the exponent is negative, which does not meet the criteria for polynomial expressions.
Conclusion
The correct answer is option D (x − 1) because it is a polynomial expression. All other options either have negative exponents or are not in the standard polynomial form. In polynomial functions, each term must be a non-negative integer exponent, which is satisfied by option D.
This detailed breakdown illustrates why option D is the only valid polynomial expression in the given set.
The question likely revolves around identifying which expression is a polynomial function. Polynomials are mathematical expressions that consist of variables raised to whole-number powers and coefficients.
Analysis of Each Option
- A. x
- This is a polynomial of degree 1.
- B. −x
- This is also a polynomial of degree 1.
- C. x5
- This is a polynomial of degree 5.
- D. x − 1
- This is a polynomial of degree 1. It can be expressed as x1 + (-1).
- E. x(−a)
- This is not a polynomial because the exponent is negative, which does not meet the criteria for polynomial expressions.
Conclusion
The correct answer is option D (x − 1) because it is a polynomial expression. All other options either have negative exponents or are not in the standard polynomial form. In polynomial functions, each term must be a non-negative integer exponent, which is satisfied by option D.
This detailed breakdown illustrates why option D is the only valid polynomial expression in the given set.
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Community Answer
A. xB. −xC. x^5D. x − 1E. x^(−a)xb)−xc)x5d)x &...
The given inequality is:
x3 < x2
x3 < x2
To understand this inequality, let's consider the following cases:
Case 1: x > 0
If x is a positive number, raising it to any positive power will yield a positive result. Therefore, x3 and x2 will both be positive. However, since x3 is less than x2 in this case, we can conclude that x3 - x2 < 0.
If x is a positive number, raising it to any positive power will yield a positive result. Therefore, x3 and x2 will both be positive. However, since x3 is less than x2 in this case, we can conclude that x3 - x2 < 0.
Case 2: x = 0
If x is equal to zero, both x3 and x2 will be zero. In this case, x3 - x2 = 0 - 0 = 0.
If x is equal to zero, both x3 and x2 will be zero. In this case, x3 - x2 = 0 - 0 = 0.
Case 3: x < 0
If x is a negative number, raising it to an odd power will result in a negative number, while raising it to an even power will yield a positive number. Therefore, x3 will be negative, and x2 will be positive. Since x3 is less than x2 in this case, we can conclude that x3 - x2 < 0.
If x is a negative number, raising it to an odd power will result in a negative number, while raising it to an even power will yield a positive number. Therefore, x3 will be negative, and x2 will be positive. Since x3 is less than x2 in this case, we can conclude that x3 - x2 < 0.
Based on the analysis of the inequality, we can see that x3 - x2 < 0 for all values of x except when x equals zero. Now, let's examine the options:
A. x: This option does not necessarily have to be negative since x can be positive.
B. -x: This option does not necessarily have to be negative either since x can be positive.
C. x5: This option does not have to be negative either. It can be positive or negative depending on the value of x.
D. x - 1: This option must be negative. Since we know that x3 - x2 < 0, we can rewrite the inequality as x3 < x2.
By subtracting x2 from both sides, we get x3 - x2 < 0. Factoring out an x, we have x(x2 - 1) < 0. Solving the equation x2 - 1 = 0, we find that x = -1 or x = 1. Therefore, for x < -1 or -1 < x < 1, x(x2 - 1) < 0. This means that x - 1 is negative for x < -1 or -1 < x < 1.
B. -x: This option does not necessarily have to be negative either since x can be positive.
C. x5: This option does not have to be negative either. It can be positive or negative depending on the value of x.
D. x - 1: This option must be negative. Since we know that x3 - x2 < 0, we can rewrite the inequality as x3 < x2.
By subtracting x2 from both sides, we get x3 - x2 < 0. Factoring out an x, we have x(x2 - 1) < 0. Solving the equation x2 - 1 = 0, we find that x = -1 or x = 1. Therefore, for x < -1 or -1 < x < 1, x(x2 - 1) < 0. This means that x - 1 is negative for x < -1 or -1 < x < 1.
E. x(-1): This option does not necessarily have to be negative either. It can be positive or negative depending on the value of x.
In conclusion, the option that must be negative is D. x - 1.
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