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3 years ago

Insert 6 A . M's between 15 and -13

Mathematics

1 answer:

Archy [21]3 years ago

4 0

Answer:

15, 11, 7, 3, -1, -5, -9, -13

Step-by-step explanation:

Inserting Arithmetic Means (AM's)

Suppose we want to insert n equally-spaced numbers inside a given interval with endpoints [x,y].

To place n numbers, we need to open n+1 spaces, at the end of which, a new number will be placed. The given interval is then divided into n+1 rational numbers of a value (y-x)/(n+1).

We need to insert 6 AM's between 15 and -13. Each AM will be inserted at a space of (-13-15)(6+1)=-28/7=-4.

Thus the sequence is formed subtracting 4:

15-4=11

11-4=7

7-4=3

3-4=-1

-1-4=-5

-5-4=-9

-9-4=-13

The sequence is (numbers in bold are the AM's):

15, 11, 7, 3, -1, -5, -9, -13

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In this case we need to find the intercepts of each quadratic equation and transform each quadratic equation into standard form. The x-intercept correspond with each of the roots of the polynomial and the y-intercept is found by evaluating the expression at x = 0.

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Case 1

x-Intercept

x = - 4 or x = 6

Standard form

f(x) = (x + 4) · (x - 6)

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x-Intercept

x = 1 / 4 or x = 3

Standard form

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y-Intercept

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Insert 6 A . M's between 15 and -13​

Insert 6 A . M's between 15 and -13