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bixtya [17]
3 years ago
15

Which answer describes the type of sequence?

Mathematics
1 answer:
jeyben [28]3 years ago
6 0

Answer:

The sequence is a geometric sequence.


Step-by-step explanation:


The given sequence is

\frac{1}{6},\frac{1}{2},\frac{3}{2},\frac{9}{2},...



We examine to see whether there is a common difference or a common ratio.


We first check for a common difference by subtracting the subsequent terms.


\frac{1}{2}-\frac{1}{6}=\frac{2}{3}



\frac{3}{2}-\frac{1}{2}=1


The two differences are not equal.


Hence the sequence is not arithmetic.




We now look out for a common ratio.


\frac{1}{2}\div \frac{1}{6}=3



\frac{3}{2}\div \frac{1}{2}=3




\frac{9}{2}\div \frac{3}{2}=3



Since there is a common ratio of 3, the sequence is geometric.








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Find the quotient. Justify your answer. x^5+2x^4-7x^2-19x+15/x^2+2x+5
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Answer:

\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3

Step-by-step explanation:

To find the quotient we need to apply long division on \frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}, as follows:

  1. Divide \frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}

Divide the leading coefficients of the numerator x^5+2x^4-7x^2-19x+15 and the divisor x^2+2x+5

\frac{x^5}{x^2}=x^3

Quotient = x^3

Multiply x^2+2x+5 by x^3 = x^5+2x^4+5x^3

Subtract x^5+2x^4+5x^3 from x^5+2x^4-7x^2-19x+15 to get new remainder

Remainder = -5x^3-7x^2-19x+15

Therefore

\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3+\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}

      2. Divide \frac{-5x^3-7x^2-19x+15}{x^2+2x+5}

Divide the leading coefficients of the numerator -5x^3-7x^2-19x+15 and the divisor x^2+2x+5

\frac{-5x^3}{x^2}=-5x

Quotient = -5x

Multiply x^2+2x+5 by -5x = -5x^3-10x^2-25x

Subtract -5x^3-10x^2-25x from -5x^3-7x^2-19x+15 to get new remainder

Remainder = 3x^2+6x+15

Therefore

\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}=-5x+\frac{3x^2+6x+15}{x^2+2x+5}

\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3-5x+\frac{3x^2+6x+15}{x^2+2x+5}

     3. Divide \frac{3x^2+6x+15}{x^2+2x+5}

Divide the leading coefficients of the numerator 3x^2+6x+15 and the divisor x^2+2x+5

\frac{3x^2}{x^2}=3

Quotient = 3

Multiply x^2+2x+5 by 3 = \:3x^2+6x+15

Subtract \:3x^2+6x+15 from 3x^2+6x+15 to get new remainder

Remainder = 0

Therefore

\frac{3x^2+6x+15}{x^2+2x+5}=3

\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3

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3 years ago
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