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Debora [2.8K]
3 years ago
10

Hey does anyone know if the answer is nonexistent?

Mathematics
2 answers:
Trava [24]3 years ago
5 0

Answer:

Non-existent is the answer.

Marrrta [24]3 years ago
4 0

Answer: Choice D is correct. The limit does not exist.

========================================================

Explanation:

Let's see what happens when we approach 0 from the left side.

So we could start at x = -1 and get closer to 0, but never actually get to that exact value.

If x = -1, then

\frac{x}{|x|}=\frac{-1}{|-1|}=\frac{-1}{1} = -1

Now let's try x = -0.5

\frac{x}{|x|}=\frac{-0.5}{|-0.5|}=\frac{-0.5}{0.5} = -1

You should find that any negative x value will lead to an output of -1.

You can make a table of values to help see this.

-----------------------

In short, as x approaches 0 from the left, the y value will approach -1.

We would then write

\displaystyle \lim_{x\to 0^{-}}f(x) = -1

This is the left hand limit which I'll abbreviate as LHL.

-------------------------

For similar reasoning, the RHL (right hand limit) is

\displaystyle \lim_{x\to 0^{+}}f(x) = 1

Because the LHL and RHL aren't the same value, this means that the limit at x = 0 does not exist. The two sides do not approach the same meeting point. It's like two cars, on two separate roads, that don't meet up. The roads need to connect.

Check out the graph below.

Side note: we avoid division by zero errors when we specifically state that x is nonzero for the first piece of that piecewise function.

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Ejercicio Nº 4 Desde la parte superior de un acantilado de 80 metros de altura se dispara horizontalmente una piedra a razón de
vaieri [72.5K]

Answer:

1) La piedra permanece en el aire en 4 segundos.

2) La piedra alcanza una distancia horizontal de 32 metros.

3) La velocidad de la piedra con la que alcanza el suelo es aproximadamente 40.792 metros por segundo.

Step-by-step explanation:

El problema nos indica un caso de tipo parabólico, el cual consiste en la suma de un movimiento horizontal a velocidad y un movimiento uniforme acelerado por la gravedad desde el reposo.

1) El tiempo total que la piedra permanecería en el aire es tiempo requerido entre la parte superior del acantilado y el fondo. La ecuación cinemática que vamos a utilizar es la siguiente:

y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2} (Ec. 1)

Donde:

y_{o} - Altura inicial, medida en metros.

y - Altura final, medida en metros.

v_{o,y} - Velocidad vertical inicial de la piedra, meadida en metros por segundo.

g - Aceleración gravitacional, medido en metros por segundo al cuadrado.

t - Tiempo, medido en segundos.

Si sabemos que y_{o} = 80\,m, v_{o,y} = 0\,\frac{m}{s} y g = -10\,\frac{m}{s^{2}}, entonces encontramos la siguiente función cuadrática:

-5\cdot t^{2}+80 = 0 (Ec. 2)

El tiempo en el que la piedra permanece en el aire es:

t = 4\,s

La piedra permanece en el aire en 4 segundos.

2) La distancia horizontal es descrita por la siguiente fórmula cinemática:

x = x_{o}+v_{o,x}\cdot t (Ec. 3)

Donde:

x_{o} - Posición horizontal inicial, medido en metros.

x - Posición horizontal final, medido en metros.

v_{o,x} - Velocidad horizontal inicial de la piedra, medida en metros por segundo.

Si sabemos que x_{o} = 0\,m, v_{o,x} = 8\,\frac{m}{s} and t = 4\,s, entonces la distancia horizontal alcanzada por la piedra es:

x = 0\,m + \left(8\,\frac{m}{s}\right)\cdot (4\,s)

x = 32\,m

La piedra alcanza una distancia horizontal de 32 metros.

3) En primer lugar, determinamos los componentes vertical y horizontal de la velocidad final de la piedra por medio de las siguientes fórmulas cinemáticas:

Velocidad final horizontal (v_{x}), medida en metros por segundo.

v_{x} = \frac{x-x_{o}}{t} (Ec. 4)

Velocidad final vertical (v_{y}), medida en metros por segundo.

v_{y} = v_{o,y}+g\cdot t (Ec. 5)

Si x = 32\,m, x_{o} = 0\,m. t = 4\,s, v_{o,y} = 0\,\frac{m}{s} y  g = -10\,\frac{m}{s^{2}}, los componentes de la velocidad final de la piedra son:

v_{x} = \frac{32\,m-0\,m}{4\,s}

v_{x} = 8\,\frac{m}{s}

v_{y} = 0\,\frac{m}{s}+\left(-10\,\frac{m}{s^{2}} \right) \cdot (4\,s)

v_{y} = -40\,\frac{m}{s}

Por último, determinamos la velocidad final de la piedra por Teorema de Pitágoras:

v = \sqrt{v_{x}^{2}+v_{y}^{2}} (Ec. 6)

v = \sqrt{\left(8\,\frac{m}{s} \right)^{2}+\left(-40\,\frac{m}{s} \right)^{2}}

v \approx 40.792\,\frac{m}{s}

La velocidad de la piedra con la que alcanza el suelo es aproximadamente 40.792 metros por segundo.

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What would be the total length of the 2 1/4-inch barrettes if Mikayla places them end to end?
Virty [35]
2 barrettes that measure 1/4-inch each. So we need to add them together

1/4+1/4=2/4-inch
this can be simplified to 1/2-inch

Answer=1/2-inch


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Answer:

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Step-by-step explanation:

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Answer:

x^{2} + 2

Step-by-step explanation:

Group:

(2x^{3} + 4x) (-5x^{2} - 10)

2x (x^{2} + 2)     -5 (x^{2} + 2)

Common binomial factor = (x^{2} + 2)

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