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3 years ago

10

Y=(x) = (1/8)^x Find f(x) when x =(1/3) Round your answer to the nearest thousandth.

1 answer:

asambeis [7]3 years ago

5 0

Answer:

Step-by-step explanation:

y(x) = (1/8)^x

y(1/3) = (1/8)^(1/3) = 0.5

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Find the vertices and locate the foci for the hyperbola whose equation is given. 49x2 - 100y2 = 4900

Fiesta28 [93]

<h2>Answer:</h2>

A hyperbola is the set of all points in a plane such that the difference of whose distances from two distinct fixed points called foci is a positive constant. In this problem, we have the following equation:

$49x^2-100y^2=4900$

What if we divide the whole equation by $49 \times 100=4900$ ? Well the result is:

$\frac{1}{4900}(49x^2-100y^2=4900) \\ \\ \frac{49x^2}{4900}-\frac{100y^2}{4900}=\frac{4900}{4900} \\ \\ \frac{x^2}{100}-\frac{y^2}{49}=1 \\ \\ \boxed{\frac{x^2}{10^2}-\frac{y^2}{7^2}=1}$

The standard form of the equation of the hyperbola given that the vertex lies on the origin is:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

SO THE VERTICES ARE:

$\boxed{(a,0)=(10,0) \ and \ (-a,0)=(-10,0)}$

Calculating the foci:

$We \ know \ that \ foci \ are: \\ \\ (-c,0) \ and \ (c.0) \\ \\ Also: \\ \\ c=\sqrt{a^2+b^2} \\ \\ c=\sqrt{100+49} \therefore c=\sqrt{149}$

SO THE FOCI ARE:

$\boxed{(-\sqrt{149},0) \ and \ (\sqrt{149},0)}$

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