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atroni [7]
3 years ago
7

Using the similar triangles, which equation could be used to find the slope of line AB?

Mathematics
1 answer:
NARA [144]3 years ago
6 0

Answer:

A) slope = p/q

Step-by-step explanation:

Slope of any given can be found using the formula, \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1}.

In this case, using similar triangles, the rise over the run of any of the given similar triangles will give us the slope of line AB.

This means, slope of line AB would be:

\frac{rise}{run} = \frac{m}{n} = \frac{p}{q}.

✅Therefore, slope of line AB = p/q

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a. Distance = speed * time

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The total time that the bee is inflight = 19 - 16 = 3 minutes

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So for the return flight the time taken is (180-t) seconds.

So d = 6(180 - t).

and  we can find the distance  from the hive to flowerbed by solving the the 2 equations highlighted.

b,      d = 9t

         d = 6(180 - t).

Therefore:

9t  = 6(180 - t)

9t = 1080 - 6t

9t + 6t = 1080

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c.  d =  9t.

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1. Determinar la ecuación canónica de la parábola con vértice en (-2,4) y foco en (1,4) 2. Determinar el foco y el vértice de la
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Answer:

1. La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Vértice = (-1, 3), enfoque = (-5/2, 3)

3. y = 12x no es una parábola

4. y = -8x, no es una parábola

Step-by-step explanation:

1. La ecuación estándar de una parábola es y = a · x² + b · x + c

El vértice V es (h, k)

El foco (h + p, k)

Por lo tanto, tenemos en comparación k = 4, h = -2

h + p = 1

p = 1 - h = 1 - (-2) = 3

Lo que da la ecuación como (y - k) ² = 4 · p · (x - h)

Al ingresar los valores de k, h y p, tenemos

(y - 4) ² = 4 × 3 × (x - (-2)) = 12 × (x + 2)

12 · x + 24 = (y - 4) ²

x = 1/12 × (y - 4) ² - 2

La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Determinar el foco y el vértice de la parábola (y - 3) ² = -6 · (x + 1)

Reescribimos la ecuación en forma de vértice de la siguiente manera;

-6 · x -6 = (y - 3) ²

x = -1 / 6 × (y - 3) ² - 1

La ecuación de una parábola en forma de vértice es x = a · (y - k) ² + h

Con el vértice = (h, k)

Comparando, tenemos, h = -1 yk = 3, el vértice = (-1, 3)

También la ecuación de la parábola en forma cónica es (y - k) ² = 4 · p · (x - h)

Comparando con (y - 3) ² = -6 · (x + 1), tenemos 4p = -6, p = -3/2

El foco está en (h + p, k) que es (-1 + -3/2, 3) = (-5/2, 3)

Vértice = (-1, 3), Enfoque = (-5/2, 3)

3. Para la parábola, y = 12 · x, tenemos;

En comparación con la forma de la ecuación, y = a · x² + b · x + c

b = 12, a = 0, c = 0

Dado que el vértice = (h, k), tenemos;

h = -b / (2 × 0), h = ∞

k = a · h² + b · h + c = ∞

No hay vértice

Foco x valor = Vértice x valor = ∞

No hay foco

Directrix = (k - 1) / (4 · a) = (k - 1) / (4 × 0) = ∞, sin directriz

y = 12x no es una parábola

4. Para y = -8x, tampoco es una parábola como se muestra arriba.

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