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atroni [7]
3 years ago
13

6+7=13 is an example of a

Mathematics
2 answers:
Reika [66]3 years ago
5 0
Addition problem lol
Goryan [66]3 years ago
4 0
An expression. A numerical expression.
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Please answer will mark brainliesttt
yanalaym [24]

Answer:

C

Step-by-step explanation:

8 0
2 years ago
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Dada a Matriz A= [1 -2 1 ][-3 1 4] [0 6 8] , escreva uma matriz B diferente de A, tal que detA =detB.
Mariulka [41]

Use o fato de que o determinante de qualquer matriz quadrada é o mesmo da sua transposta.

\mathbf A=\begin{bmatrix}1&-2&1\\-3&1&4\\0&6&8\end{bmatrix}

\mathbf B=\mathbf A^\top=\begin{bmatrix}1&-3&0\\-2&1&6\\1&4&8\end{bmatrix}

3 0
4 years ago
This is my problem. I need help ​
koban [17]

Answer:

1: -3

2: She gives him $3 per week

3: 18

4: How much money she initially borrowed

5: y = -3x + 18

7 0
3 years ago
The equation y=2.54x gives the length y in centimeters corresponding to a length x in inches. What is the length in centimeters
bija089 [108]
The equation given in the question is
y = 2.54 x
In this equation, x is in cm and y is in inches
Size of the ruler = 1 foot 
                          = 12 inches.
Putting the value of 12 inches in the equation, we get
y = 2.54x
   = 2.54 * 12
   = 30.48 cm
From the above deduction, we can conclude that the length of the ruler is 30.48 cm. I hope that this is answer that you were looking for and the answer has come to your desired help.
6 0
3 years ago
A spring exerts a force of 6 N when stretched 3 m beyond its natural length. How much work is required to stretch the spring 2 m
lys-0071 [83]

Answer:

4Joules

Step-by-step explanation:

According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

K is the elastic constant

e is the extension

If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'

can be gotten using k = f/e where

F = 6N, e = 3m

K = 6N/3m

K = 2N/m

Work done on an elastic string is calculated using 1/2ke².

If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;

1/2× 2× (2)²

= 4Joules

3 0
3 years ago
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