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Illusion [34]
2 years ago
13

What is the vertex for this function? y = 3x^2 + 12x + 5

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
8 0
That’s weird, it should be (-2,-7). Try (-2,7)
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The answer is Y=-5/2x-1
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A triangle is placed in a semicircle with a radius of 7mm , as shown below. Find the area of the shaded region. Use the value 3.
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3 years ago
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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3 years ago
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a_sh-v [17]

Answer:

D

Step-by-step explanation:

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