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Butoxors [25]
3 years ago
13

А,

Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

\displaystyle 35

Explanation:

Complementary Angles sum up to ninety degrees, so do the following:

\displaystyle 90° = 55° + m\angle{DEF} \\ \\ 35° = m\angle{DEF}

I am joyous to assist you at any time.

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A block of metal with volume 15 m3 and density 8,800 kg/m3 is combined with another metal of mass 850 kg and volume 6.2 m3 to fo
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El aluminio es un metal ligero con una densidad de 2.70 g/cm3, y por ello, aunque las aleaciones de aluminio tienen características mecánicas relativamente bajas comparadas con las del acero, su relación resistencia-peso es excelente. Es precisamente debido a esto que el aluminio se utiliza cuando el peso es un factor importante, como ocurre en las aplicaciones aeronáuticas y de automoción.

 

Tabla 13.8. Propiedades mecánicas y aplicaciones de algunas aleaciones comerciales de aluminio.

El aluminio también responde fácilmente a los diferentes mecanismos de endurecimiento, tal como se recoge en la tabla 13.8, donde se observa que el mecanismo más notable es el de endurecimiento por precipitación, donde se consigue una dureza hasta 30 veces superior a la del aluminio puro.

Por otra parte, el aluminio no suele presentar un límite de resistencia a la fatiga bien definido, de modo que la fractura puede suceder incluso a niveles muy bajos. Debido a su bajo punto de fusión, el aluminio no se comporta bien a temperaturas elevadas. Finalmente, las aleaciones de aluminio tienen escasa dureza, lo que origina poca resistencia al desgaste abrasivo en ocasiones.

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english ?

Step-by-step explanation:

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Step-by-step explanation:  

hello :

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3 years ago
How do i do this? i'm so lost. please explain and show your work
elena-s [515]
You want to make sure to factor each equation as much as you can first.
So just focus on the numerator first. x^3-6x^2+8x
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(x)(x^2-6x+8) factor out an x
(x)(x-4)(x-2) factor out the rest completely.

Now for the denominator: -4x^2+4x+24
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I'm not sure what you're supposed to find exactly, so if you need more help you can ask me (:
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