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3 years ago

13

Please help :( Find X. Round to the nearest tenth.

1 answer:

g100num [7]3 years ago

6 0

90^2 = 55^2 + 50^2 - 2(55)(50)cosX
(90^2 - 55^2 - 50^2)/(2(55)(50)) = cosX
X = 62.1°

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Let L be a tangent line to the hyperbola x y = 2 at x = 9 . Find the area of the triangle bounded by L and the coordinate axes.

mafiozo [28]

Answer:

$A = 4$

Step-by-step explanation:

The equation of the slope of the tangent line L is obtained by deriving the equation of the hyperbola:

$y = \frac{2}{x}$

$y'=-2\cdot x^{-2}$

The numerical value of the slope is:

$y' = -2 \cdot (9)^{-2}\\y' = -\frac{2}{81}$

The component of the y-axis is:

$y = \frac{2}{9}$

Now, the tangent line has the following mathematical model:

$y = m \cdot x + b$

The value of the intercept is found by isolating it within the equation and replacing all known variables:

$b = y - m \cdot x$

$b = \frac{2}{9}-(-\frac{2}{81} )\cdot (9)\\b = \frac{4}{9}$

Thus, the tangent line is:

$y = -\frac{2}{81}\cdot x + \frac{4}{9}$

The vertical distance between a point of the tangent line and the origin is given by the intercept.

$d_{y} = \frac{4}{9}$

In order to find horizontal distance between a point of the tangent line and the origin, let equalize y to zero and clear x:

$-\frac{2}{81}\cdot x + \frac{4}{9}=0$

$-\frac{2}{9}\cdot x + 4 = 0$

$x = 18$

$d_{x} = 18$

The area of the triangle is computed by this formula:

$A = \frac{1}{2}\cdot d_{x}\cdot d_{y}$

$A = \frac{1}{2}\cdot (18)\cdot (\frac{4}{9} )$

$A = 4$

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Brilliant_brown [7]

1) 8

2) $-\frac{3}{4}=-\frac{6}{8}$ , so it is equal to $-\frac{6}{8}+\frac{5}{8}$

3) 1/8

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ANSWER

See attachment.

EXPLANATION

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The graph of this function is shown in the attachment.

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