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Marysya12 [62]
4 years ago
11

Number 5 pretty please

Mathematics
1 answer:
d1i1m1o1n [39]4 years ago
6 0
Try and use a calc ;)
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0.6766

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6.8%*9.95=0.6766

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−−→ LO bisects ∠ NLM , LM = 18 , NO = 4 , and LN = 10 . LO → bisects ∠ NLM , LM = 18 , NO = 4 , and LN = 10 . What is the value
Sindrei [870]

Answer:

7.2

Step-by-step explanation:

Given that in triangle LMN, LO is angle bisector of angle L.

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Or cross multiply to get

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What is the actual question?
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Read 2 more answers
A polynomial f (x) has the
Helga [31]

a) The polynomial f(x) in <em>expanded</em> form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The <em>rational</em> function g(x) in <em>factored</em> form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no <em>slant</em> asymptotes.

c) There is one <em>evitable</em> discontinuity at x = - 1, and one <em>definitive </em>discontinuity at x = 2, where there is a <em>vertical</em> asymptote.

<h3>How to analyze polynomial and rational functions</h3>

a) In the first part of this question we need to determine the equation of a polynomial in <em>expanded</em> form, derived from its <em>factor</em> form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in <em>expanded</em> form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The <em>rational</em> function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the <em>slant</em> asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the <em>slant</em> asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no <em>slant</em> asymptotes.

c) There is <em>vertical</em> asymptote at a x-point if the denominator is equal to zero. There is one <em>evitable</em> discontinuity at x = - 1, and one <em>definitive </em>discontinuity at x = 2, where there is a <em>vertical</em> asymptote.

To learn more on asymptotes: brainly.com/question/4084552

#SPJ1

4 0
2 years ago
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