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Serga [27]
3 years ago
12

Solve for J: 15(j-3)+3j<45

Mathematics
1 answer:
nata0808 [166]3 years ago
3 0
15(j - 3) + 3j < 45
15j - 45 + 3j < 45
18j < 45 + 45
18j < 90
j < 90/18
j < 5
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Helpp please!!!
zavuch27 [327]

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 ...

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  • 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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