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Irina18 [472]
3 years ago
13

If AACD ~ AABE, find the value of x.

Mathematics
1 answer:
Anastaziya [24]3 years ago
4 0

9514 1404 393

Answer:

  x = 16

Step-by-step explanation:

Corresponding sides are proportional.

  BE/CD = AE/AD

  20/(3x+8) = (x-1)/(x-1+27)

  20(x+26) = (3x+8)(x -1)

In standard form, this is ...

  3x^2 -15x -528 = 0

  x^2 -5x -176 = 0 . . . . . . divide by 3

  (x -16)(x +11) = 0 . . . . . . . factor

  x = 16

__

This is the positive value that makes the product zero. x=-11 will also make the product 0, but gives negative segment lengths in the geometry. It is an extraneous solution.

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Hi can some please help?
Oduvanchick [21]
So I'm assuming that you're taking Calculus.
The first thing you want to do is take the integral of f(x)...
Use the power rule to get:
4X^2-13X+3.
Now solve for X when f(x)=0. This is because when the slope is 0, it is either a minimum or a maximum(I'm assuming you know this)
Now you get X=0.25 and X=3. Since we are working in the interval of (1,4), we can ignore 0.25
Thus our potential X values for max and min are X=1,X=4,X=3(You don't want to forget the ends of the bounds!)
Plugging these value in for f(x), we get
f(1)=2.833
f(3)=-8.5
f(4)=1.667
Thus X=1 is the max and X=3 is the min.
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min:(3,-8.5)

Hope this helps!
3 0
3 years 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

5 0
1 year ago
6.Calcular: 2^3 ; (-2)^3; ; -2^3 *
olga2289 [7]

Answer:

90-8y66697u

Step-by-step explanation

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2 years ago
Which table shows a proportional relationship and the correct amount of each ingredients
ollegr [7]
I am pretty sure that it is table one because each number from the original recipe to the teachers recipe is multiplied by three.
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3 years ago
When there is a positive trend (or positive correlation), do the y-values of the points on the scatter plot tend to increase or
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As the y-axes decreases the x-axes increases 

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3 years ago
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