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3 years ago

10

The water in a swimming pool evaporates at a rate of 1/6 inch per day. The water level in the pool has dropped 3 inches since th

e last time the pool was filled. How many days have passed since the pool was filled.

2 answers:

garik1379 [7]3 years ago

8 0

The answer is 18 days some one help me to

Ne4ueva [31]3 years ago

3 0

18 days

1/6 x 6 = 6/6 = 1
1/6 x 18 = 18/6 = 3

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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.

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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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Arte-miy333 [17]

Step $1$

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<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

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$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

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Step $2$

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

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$AC=(13*13)/5=33.8\ units$

Step $4$

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we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

<u>the answer is</u>

$BC=31.2\ units$

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