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Fantom [35]
3 years ago
13

Use Pythagorean theorem to find right triangle side le

Mathematics
2 answers:
Elis [28]3 years ago
4 0
Ok, so the Pythagorean theorem is a^2 + b^2= x^2

I think it would be A
posledela3 years ago
3 0
The answer should be 10 but it is not in the choices
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What are the domain and range of f (x)=16
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Range: {16 }

Domain:(- ∞ ,  ∞ ),{ x| x∈  R }

Step-by-step explanation: Find the domain by finding where the function is defined. The range is the set of values that correspond with the domain.

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The equation for the line of best fit for a data set is y = 2x + 1.5. If the point (1, 4) is in the data set, what is the
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The equation for the line of best fit for a data set is y = 2x + 1.5. If the point (1, 4) is in the data set, what is the

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4 0
3 years ago
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
goldenfox [79]

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.

4 0
3 years ago
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