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Kisachek [45]
2 years ago
7

I do need help with this problem that includes pi. ^_^

Mathematics
1 answer:
Troyanec [42]2 years ago
5 0

Solution:

<u>Note that:</u>

  • Circumference = 50.24 ft = 2πr = πd
  • Dia. = 16 ft.

<u>Divide the diameter both sides.</u>

  • 50.24 ft = πd
  • => 50.24/d ft = πd/d
  • => 50.24/d ft = π

<u>Substitute the diameter in the equation.</u>

  • => 50.24/16 = π (Option A)
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q(x)= x 2 −6x+9 x 2 −8x+15 ​ q, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, minus, 8, x, plus, 1
AURORKA [14]

According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.

<h3>What is the behavior of a functions close to one its vertical asymptotes?</h3>

Herein we know that the <em>rational</em> function is q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15), there are <em>vertical</em> asymptotes for values of x such that the denominator becomes zero. First, we factor both numerator and denominator of the equation to see <em>evitable</em> and <em>non-evitable</em> discontinuities:

q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15)

q(x) = [(x - 3)²] / [(x - 3) · (x - 5)]

q(x) = (x - 3) / (x - 5)

There are one <em>evitable</em> discontinuity and one <em>non-evitable</em> discontinuity. According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.

To learn more on rational functions: brainly.com/question/27914791

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Step-by-step explanation:

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