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Aleonysh [2.5K]
3 years ago
15

Suppose that the functions g and f are defined as follows.

Mathematics
1 answer:
Sholpan [36]3 years ago
8 0

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Given:

\to g(x)= -3+2x^2 \\\\\to f(x)=9-6x

Calculating the value of (\frac{g}{f})(2):

\to (\frac{g}{f}) (2)= (\frac{g(x)}{f(x)}) \times  (2)

               =\frac{2x^2-3}{9-6x} \times 2\\\\=\frac{2\cdot 2^2-3}{3(3-2\cdot 2)} \\\\=\frac{2\cdot 4-3}{3(-1)}\\\\ =\frac{8-3}{-3}\\\\  =\frac{5}{-3}\\\\  = - \frac{5}{3}\\\\

x=- \frac{\sqrt{3}}{2} and x=\frac{\sqrt{3}}{2} are not in the domain.

Domain: x \belong (- \infty, -\frac{\sqrt{3}}{2})\cup (- \frac{\sqrt{3}}{2},  \frac{\sqrt{3}}{2} )\cup (\frac{\sqrt{3}}{2}, \infty)

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$\left( \frac{m+h}{2}, \frac{n}{2}\right  )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right  )

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Multiply by 2 on both sides of the equation, we get

n = y

y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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