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allochka39001 [22]
2 years ago
10

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

0,0), where unit distance on the coordinate plane is measured in miles A receiver is located somewhere in the first quadrant. The receiver computes that the difference in the distances from the receiver to these transmitters is 100 miles.
What is the standard form of the hyperbola that the receiver sits on if the transmitters behave as foci of the hyperbola?
Mathematics
2 answers:
irina1246 [14]2 years ago
8 0

Answer:

sénsya po need ko po talaga ng points

USPshnik [31]2 years ago
7 0

LORAN follows an hyperbolic path.

The equation of the hyperbola is: \mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

The coordinates are given as:

\mathbf{(x,y) = (-60,0)\ (60,0)}

The center of the hyperbola  is  

\mathbf{(h,k) = (0,0)}

The distance from the center to the focal points is given as:

\mathbf{c = 60}

Square both sides

\mathbf{c^2 = 3600}

The distance from the receiver to the transmitters  is given as:

\mathbf{2a = 100}

Divide both sides by 2

\mathbf{a = 50}

Square both sides

\mathbf{a^2 = 2500}

We have:

\mathbf{b^2 = c^2 - a^2}

This gives

\mathbf{b^2 = 3600 - 2500}

\mathbf{b^2 = 1100}  

The equation of an hyperbola is:

\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}

So, we have:

\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}

\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

Hence, the equation of the hyperbola is: \mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}

Read more about hyperbolas at:

brainly.com/question/15697124

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sertanlavr [38]

Given:

Consider the given expression is

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The radical form of given expression.

Solution:

We have,

(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}

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