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vivado [14]
3 years ago
8

John has two open packs of paper. One pack has 31 pieces of paper. The second pack has 45 pieces of paper. How many pieces of pa

per does John have all together? A) 72 B) 73 C) 74 D) 76
Mathematics
1 answer:
Tems11 [23]3 years ago
8 0
D 76 pieces of paper. Just add
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I really need help and I have to show my work
Amanda [17]
10 lbs/$12= 1 lb/x dollars.
If you rewrite it up and down and not slanted, it would be clearer to understand, but anyways, you cross multiply:
12*1=12 
12/10= 1.2
So the answer is 1.2, if you put it on that bubble sheet, put 1, then a decimal, after the decimal 2, and then all 0's,.

8 0
3 years ago
I need help with Function Operations
Leni [432]

Answer:

490

Step-by-step explanation:

d(x)=700-7(30)

7x 30=210

700-210=490

therefore,

r(x)=490

4 0
2 years ago
Help with #5, 6, 7, and 8 <br> pleasee
lesya [120]
5. It is a 5/9 chance he will pick 2 pair of pants. There are 5 pants and 9 in total (clothes and pants)
7. It is a 3/9 chance he will pick 2 items that are black. There are only 3 things that are black and the rest are other colors.
8. It is a 2/9 chance he will pick a pant and a shirt that is white. There is 1 white pants and 1 white shirt.


That is all I know. Number 6 kind of confuses me. Hope I helped!!!
8 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
Which of the following best describes a function? Select all that apply
enot [183]

Answer:

Where are the answer choice's??

Step-by-step explanation:

3 0
2 years ago
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