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3 years ago

What is the eighth term of the arithmetic sequence? An= 10n – 14?

Mathematics

1 answer:

Alja [10]3 years ago

3 0

Answer:

A8 = 66 [8th term]

Step-by-step explanation:

Given the explicit arithmetic sequence An = -14 + 10n → -4 + 10(n-1)

To find the 8th term. An nth term is where n is n in the sequence. So substitute 8 for n, and simplify.

So A8 = -4 + 10(8-1) → -4 + 10(7) → -4 + 70 → 70 - 4 → 66.

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a) The polynomial in expanded form is $f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ .

b) The slant asymptote is represented by the linear function is $y = -x + 1$ .

c) There is a discontinuity at $x = 2$ with a slant asymptote.

a) In this question we are going to use the Factor Theorem, which establishes that polynomial are the result of products of binomials of the form $x-r_{i}$ , where $r_{i}$ is the i-th root of the polynomial and the grade is equal to the quantity of roots. Therefore, the polynomial $f(x)$ has the following form:

$f(x) = (x-6)\cdot (x+1)\cdot (x+3)$

And the expanded form is obtained by some algebraic handling:

$f(x) = (x-6)\cdot (x^{2}+4\cdot x +3)$

$f(x) = x\cdot (x^{2}+4\cdot x + 3)-6\cdot (x^{2}+4\cdot x +3)$

$f(x) = x^{3}+4\cdot x^{2}+3\cdot x -6\cdot x^{2}-24\cdot x -18$

$f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ (1)

The polynomial in expanded form is $f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18$ .

b) In this question we divide the polynomial found in a) (in factor form) by the polynomial $x^{2}-x -2$ (also in factor form). That is:

$g(x) = \frac{(x-6)\cdot (x+1)\cdot (x+3)}{(x-2)\cdot (x+1)}$

$g(x) = \frac{(x-6)\cdot (x+3)}{x-2}$ (2)

The slant asymptote is defined by linear function, whose slope ( $m$ ) and intercept ( $b$ ) are determined by the following expressions:

$m = \lim_{x \to \pm \infty} \frac{g(x)}{x}$ (3)

$b = \lim_{x \to \pm \infty} [g(x)-x]$ (4)

If $g(x) = \frac{(x-6)\cdot (x+3)}{x-2}$ , then the equation of the slant asymptote is:

$m = \lim_{x \to 2} \frac{(x-6)\cdot (x+3)}{x\cdot (x-2)}$

$m = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x^{2}-2\cdot x} \right)$

$m = 1$

$b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x-2}-x \right)$

$b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x - 18-x^{2}+2\cdot x}{x-2}\right)$

$b = \lim_{n \to \infty} \left(\frac{-x-18}{x-2} \right)$

$b = -1$

The slant asymptote is represented by the linear function is $y = -x + 1$ .

c) The number of discontinuities in rational functions is equal to the number of binomials in the denominator, which was determined in b). Hence, we have a discontinuity at $x = 2$ with a slant asymptote.

We kindly invite to check this question on asymptotes: brainly.com/question/4084552

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