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Rashid [163]
3 years ago
5

2/3 x-4 = -2A -3 B -4C -9D 3​

Mathematics
1 answer:
Luden [163]3 years ago
5 0

Answer:

i think its d

Step-by-step explanation:

cause i solved it and got the same answer

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G(x) = -2x^3 – 15x^2 + 36x
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Consider the function G(x) = -2x^3 - 15x^2 + 36x. First, factor it:

G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).

The x-intercepts are at points \left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).

1. From the attached graph you can see that

  • function is positive for x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);
  • function is negative for x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).

2. Since

G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x) the function is neither even nor odd.

3. The domain is x\in (-\infty,\infty), the range is y\in (-\infty,\infty).

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