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3 years ago

-8,-8,-6,-2,4,12,22 quadratic sequence

Mathematics

1 answer:

cestrela7 [59]3 years ago

7 0

9514 1404 393

Answer:

a[n] = n^2 -3n -6

Step-by-step explanation:

First differences are ...

-8 -(-8) = 0

-6 -(-8) = 2

-2 -(-6) = 4

Second differences are ...

2 -0 = 2

4 -2 = 2

Constant second differences indicate a degree 2 (quadratic) sequence.

The general formulation can be written as ...

an = a1 +(n -1)(d1 +(n -2)/2(d2)) . . . . where a1 is the first term; d1 is the first first difference; d2 is the second difference

= -8 +(n -1)(0 +(n -2)/2(2)) = -8 +(n -1)(n -2)

an = n^2 -3n -6 . . . . . formula for the n-th term

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2 years ago

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3 years ago

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Answer:

Three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{8}$ .

Step-by-step explanation:

We are given a fraction $\frac{14}{2}$ .

We need to find the three equivalent ratios/fractions of $\frac{14}{2}$ .

The common factor of 14 and 2 is 2.

So, dividing top and bottom by 2, we get

14÷2 = 7 and 2÷2 that is $\frac{7}{1}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

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So, we get another equivalent fraction $\frac{21}{6}$ .

Multiplying $\frac{7}{1}$ fraction by a common number 3 in top and bottom, we get

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So, we get another equivalent fraction $\frac{28}{4}$ .

Therefore, three equivalent ratios of $\frac{14}{2}$ are $\frac{7}{1}$ , $\frac{21}{3}$ and $\frac{28}{4}$ .

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