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Natasha_Volkova [10]
3 years ago
10

Use the Distributive Property to solve the equation below. Use pencil and paper. Describe what it means to distribute the 2 to e

ach term inside the parentheses.
2(m+4)=20
Mathematics
1 answer:
Bad White [126]3 years ago
8 0

Answer:

Use the distributive property to solve the equation 2(m + 2) = 22. Describe what it means to distribute the 2 to each term inside the parentheses

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Write the numbers in order from least to greatest. 0.3.0.5.-1.6.-1.35​
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-1.6
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Top to bottom, top is the lowest
4 0
2 years ago
After Hillary realized her clock was running 30 minutes fast, she used a knob on the clock to turn the minute hand back 30 minut
iragen [17]

Answer:

Counterclockwise rotation

Step-by-step explanation:

It’s counter clockwise rotation because The clock was 30minutes faster , so she turned the minute hand backwards to correct the clock. Counterclockwise is opposite of clockwise

8 0
3 years ago
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What is the area of the trapezoid with height 11 units?
cricket20 [7]
I hope this helps you



Area =(34+16).11/2


25.11


Area =275
6 0
3 years ago
What is the point-slope form of the equation for the line with a slope of -2 that passes through (1,4)?
morpeh [17]

\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-2(x-1)

6 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
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