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k0ka [10]
3 years ago
7

5 características de la obsolescencia programada...

Computers and Technology
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

Obsolescencia programada es cuando un producto está diseñado deliberadamente para tener un tiempo de vida específico. ... Los productos dejan de funcionar al cabo de un tiempo, no porque estén estropeados, sino por que han sido diseñados para fallar al cabo de ese periodo.

Explanation:

espero y esto te pueda ayudar

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Aiming high means???
Llana [10]

Answer:

in what context is it in? it could be aim high for goals set limits that are hard to reach but not impossible

Explanation:

5 0
3 years ago
What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent
Scrat [10]

Answer:

The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

Then satisfying this theorem the system is consistent and has one single solution.

Explanation:

1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

rank(A)

Then the system is consistent and has a unique solution.

<em>E.g.</em>

\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5  \end{matrix}\right.

2) Writing it as Linear system

A=\begin{pmatrix}1 & -3 &-2 \\  2& -4 &-3 \\ -3 &6  &8 \end{pmatrix} B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}

rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3

3) The Rank (A) is 3 found through Gauss elimination

(A|B)=\begin{pmatrix}1 & -3 &-2  &6 \\  2& -4 &-3  &8 \\  -3&6  &8  &-5 \end{pmatrix}

rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3

4) The rank of (A|B) is also equal to 3, found through Gauss elimination:

So this linear system is consistent and has a unique solution.

8 0
3 years ago
I don't know what to do, anyone there?​
s2008m [1.1K]
I’m here, hello. . . . . . .
3 0
3 years ago
Write a program that plays the popular scissor-rockpaper game. (A scissor can cut a paper, a rock can knock a scissor, and a pap
kolbaska11 [484]

Answer:

import random

computer = random.randint(0, 2)

user = int(input("scissor (0), rock (1), paper (2): "))

if computer == 0:

   if user == 0:

       print("The computer is scissor. You are scissor too. It is a draw")

   elif user == 1:

       print("The computer is scissor. You are rock. You won")

   elif user == 2:

       print("The computer is scissor. You are paper. Computer won")

elif computer == 1:

   if user == 0:

       print("The computer is rock. You are scissor. Computer won")

   elif user == 1:

       print("The computer is rock. You are rock too. It is a draw")

   elif user == 2:

       print("The computer is rock. You are paper. You won")

elif computer == 2:

   if user == 0:

       print("The computer is paper. You are scissor. You won")

   elif user == 1:

       print("The computer is paper. You are rock. Computer won")

   elif user == 2:

       print("The computer is paper. You are paper too. It is a draw")

Explanation:

*The code is in Python.

Import the random to be able to generate number number

Generate a random number between 0 and 2 (inclusive) using randint() method and set it to the computer variable

Ask the user to enter a number and set it to the user variable

Check the value of computer the computer variable:

If it is 0, check the value of user variable. If user is 0, it is a draw. If user is 1, user wins. If user is 2, computer wins

If it is 1, check the value of user variable. If user is 0, computer wins. If user is 1, it is a draw. If user is 2, user wins

If it is 2, check the value of user variable. If user is 0, user wins. If user is 1, computer wins. If user is 2, it is a draw

8 0
2 years ago
Acess labeling windows​
Debora [2.8K]

Answer:

I got the same thing

Explanation:

8 0
3 years ago
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