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Solnce55 [7]
3 years ago
5

Solve two-step equations. -5/2 a + 5 = 25

Mathematics
1 answer:
Goryan [66]3 years ago
6 0

Answer:

a = -8

Step-by-step explanation:

-5/2 a + 5 = 25

Subtract 5 from each side

-5/2 a + 5-5 = 25-5

-5/2 a = 20

Multiply each side by -2/5

-2/5 *-5/2 a = 20*-2/5

a = -8

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B.

Step-by-step explanation:

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3 years ago
On a coordinate plane, 2 exponential fuctions are shown. Function f (x) decreases from quadrant 2 into quadrant 1 and approaches
Gala2k [10]

Answer:

g(x)=6(3)^x

Step-by-step explanation:

We are given  that

f(x)=6(\frac{1}{3})^x

Function f decreases from quadrant  2 to quadrant 1 and approaches  y=0

It cut the y- axis at (0,6) and passing through the point (1,2).

Function g(x) approaches y=0 in quadrant 2 and increases into quadrant 1.

It passing through the point (-1,2) and cut the y-axis at point (0,6).

Reflection across y- axis:

Rule of transformation is given by

(x,y)\rightarrow (-x,y)

Using the rule then we get

g(x)=6(\frac{1}{3})^{-x}=6(3)^x

By using

x^{-a}=\frac{1}{x^a}

Substitute x=-1

g(-1)=6\times (\frac{1}{3})=2

Substitute x=0

g(0)=6

Therefore,g(x)=6(3)^x is true.

8 0
3 years ago
Read 2 more answers
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
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