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Arte-miy333 [17]
3 years ago
5

(8 + 5r3 - 2r2) - (8r - 8 - 6r2) Simplify expression

Mathematics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

5r^3 + 4r^2 - 8r + 16

Step-by-step explanation:

(8 + 5r^3 - 2r^2) - (8r - 8 - 6r^2) =

The first set of parentheses is there just to show you that what is inside is a polynomial. The second set of parentheses has a second polynomial inside. The subtraction sign just to the left of the second set of parentheses shows that you are subtracting the second polynomial from the first one.

The first set of parentheses is not needed and can be dropped.

You are subtracting the second polynomial fromt he first one, so you can think of the the subtraction sign as a -1, and you need to distribute the -1 by the second polynomial, That will result in all signs inside the second set of parentheses changing.

Below, just the first set of parentheses is removed.

= 8 + 5r^3 - 2r^2 - (8r - 8 - 6r^2)

Now, we change every sign inside the second set of parentheses by distributing -1.

= 8 + 5r^3 - 2r^2 - 8r + 8 + 6r^2

Now we need to combine like terms. Like terms have the same variable part. We can rearrange the terms grouping like terms together before combining them. Also, it is customary to list the terms in descending order of degree.

= 5r^3 - 2r^2 + 6r^2 - 8r + 8 + 8

Now we combine like terms.

= 5r^3 + 4r^2 - 8r + 16

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<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)

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σ = 25 N / [π · (0.02 m)² / 4]

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<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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