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kicyunya [14]
3 years ago
9

Helpp please!!!

Mathematics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

  h, j2, f, g, j1, i, k, l (ell)

Step-by-step explanation:

The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).

We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).

The horizontal asymptotes are ...

  • h(x): 16x/(-4x) = -4
  • j1(x): 2x^2/x^2 = 2
  • i(x): 3x/x = 3
  • l(x): 15x/(2x) = 7.5
  • g(x): x^2/x^2 = 1
  • j2(x): 3x^2/-x^2 = -3
  • f(x): 0x^2/(12x^2) = 0
  • k(x): 5x^2/x^2 = 5

So, the ordering least-to-greatest is ...

  h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)

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Step-by-step explanation:

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Step-by-step explanation:

Given

B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\

Required [Missing from the question]

Every __ seconds, the bacteria loses \frac{1}{2}

First, we model the function from t/3 to t.

B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\

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B(t) = 8500 * (\frac{8^\frac{1}{3}}{27^\frac{1}{3}})^t

Evaluate each exponent

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B(0) = 8500 * (\frac{2}{3})^0

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When 1/2 disappears, we have:

B(r) = \frac{B(0)}{2}

B(r) = \frac{8500}{2}

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Substitute r for t

B(r) = 8500 * (\frac{2}{3})^r

Substitute B(r) = 4250

4250 = 8500 * (\frac{2}{3})^r

Divide both sides by 8500

\frac{4250}{8500} =  (\frac{2}{3})^r

\frac{1}{2} =  (\frac{2}{3})^r

Take log of both sides

log(\frac{1}{2}) = log (\frac{2}{3})^r

Apply law of logarithm

log(\frac{1}{2}) = r\ log (\frac{2}{3})

Make r the subject

r = log(\frac{1}{2}) / log (\frac{2}{3})

r = \frac{-0.3010}{-0.1761}

r = 1.71

<em>Hence, it reduces by 1/2 after every 1.71 seconds</em>

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