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3 years ago

right triangle ABC is shown below. A similar right triangle would be created by a run of 4 and rise of______

Mathematics

2 answers:

Svetllana [295]3 years ago

4 0

The rise would have to be 12.

This is because in order to be "Similar" triangles the sides need to be proportionate.

(1/3 = 4/12)

julia-pushkina [17]3 years ago

3 0

Answer:

Examine the right triangle ABC. Which rise and run would create a similar right triangle on the same line?

a rise of 6 and a run of 8

a rise of 8 and a run of 6

a rise of 6 and a run of 5

a rise of 5 and a run of 6 B is correct

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F (x) = x^2 + 2x - 5<br> g(x) = 2x + 4<br> What is (f• g)(x)?

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Answer:

Step-by-step explanation:

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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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4 years ago

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