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denpristay [2]
3 years ago
13

Use algebraic means to show that x^2 + y^2 = 8 is not a function. Explain your process.

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

For a single value of x function has more than one corresponding value of y which satisfies the equation.

Step-by-step explanation:

Function: A relationship between a set of inputs and a set of possible outputs, where exactly one output is associated with each input.

It means for an equation to represent a function any single value of  x there should be only one corresponding value of  y which satisfies the equation.

Now consider the given equation.

x^2 + y^2 = 8

If we put x=0 then we get two value of y i.e \sqrt8 and -\sqrt8 which satisfy the equation  and therefore the equation is not a function.

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Determine whether Point A lies on the circle whose center is Point C and whichcontains the Point P. Justify your answer mathemat
Arada [10]

The Solution:

Given:

Center = (0,0)

Point A = (-5,2) being a point on the circle.

We are required to check if point P = (2,-5) is on the circle.

Solving the given problem graphically, we have:

From the above graph, it is clear that point P(2,-5) is a point on the circle.

8 0
11 months ago
Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

  • W - Peso de la caja, en newtons.
  • D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

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1 year ago
If you choose one card from a standard deck, what is the probability that the card is either red or a king?
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