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____ [38]
3 years ago
3

Consider the polynomial: StartFraction x Over 4 EndFraction – 2x5 + StartFraction x cubed Over 2 EndFraction + 1 Which polynomia

l represents the standard form of the original polynomial? StartFraction x cubed Over 2 EndFraction – 2x5 + StartFraction x Over 4 EndFraction + 1 –2x5 + StartFraction x cubed Over 2 EndFraction + StartFraction x Over 4 EndFraction + 1 –2x5 + StartFraction x Over 4 EndFraction + StartFraction x cubed Over 2 EndFraction + 1 1 – 2x5 + StartFraction x cubed Over 2 EndFraction + StartFraction x Over 4 EndFraction
Mathematics
1 answer:
NARA [144]3 years ago
7 0

Answer:

(B)-2x^5+\dfrac{x^3}{2}+\dfrac{x}{4}+1

–2x5 + StartFraction x cubed Over 2 EndFraction + StartFraction x Over 4 EndFraction + 1

Step-by-step explanation:

Given the polynomial: \dfrac{x}{4}-2x^5+\dfrac{x^3}{2}+1

We are required to pick an equivalent polynomial which is in standard form.

A polynomial is in standard form when it is written in descending powers of x.

Therefore, rearranging the polynomial above, we have:

-2x^5+\dfrac{x^3}{2}+\dfrac{x}{4}+1

The correct option is B.

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I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.

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The value of a second-hand car is £8000. Each year it loses 20% of its value at the start of this year. work out its value in 5
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