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V125BC [204]
3 years ago
6

12x^+2x^2=2y^2+28y-6

Mathematics
1 answer:
Ket [755]3 years ago
5 0

Hello,

There is no X intercept but there is a Y

Y intercept =  (0,-7 + 2 square root of 13),(0,-7,- 2 square root 13)

Center (-7,0)

Vertices: (−7+√x2+52,0),(−7−√x2+52,0)

(-7+x2+52,0),(-7-x2+52,0)

Foci: (−

42−√42(x2+52) 6 ,0),(−

42+√42(x2+52)

6

,0)

(-42-42(x2+52)6,0),(-42+42(x2+52)6,0)

I honestly dont know what your trying to solve so im guessing thats what you wanted...

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Answer:

5.2

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
There are currently 17 frogs in a (large) pond. The frog population grows exponentially, tripling every 6 days. How long will it
Tema [17]

Answer:

t=11,11 days

Step-by-step explanation:

F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating \int\limits^ {} \, dF/F =K\int\limits^ {} \, dt⇒ LnF=Kt+c,F=ce^{Kt}; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), F=ce^{K0} = F=c=17; F=17e^{6K} =51⇒e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183, so F=17e^{0,183t}, now if F=130, t=? we have:

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Answer:

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Katyanochek1 [597]
X=3 is a vertical line . The question is "Is this line a vertical asymptotefor the function f(x)?". Vertical asymptote is vertical line near which the function grows without bound.
The function f(x)=5x-15/x-3 is not asymptotic to the line x=3, because for x=3 f(x)=5*3-15/3-3=0/0=0 . So, there is a value for x=3 and the function f(x) does not grow near x=3 without bound.

3 0
3 years ago
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