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exis [7]
3 years ago
6

A telephone number consists of seven digits, the first three representing the exchange. how many different telephone numbers are

possible within the 537 exchange? the answer is .
Mathematics
1 answer:
AveGali [126]3 years ago
8 0
Since the first three digits are 537 (the exchange), we have 7-3=4 digits left. In each of those digits, there are 10 possible numbers, 0 to 9. As there are 10 numbers possible for the first digit, there are 10 numbers for each number for the second digit, and so on. For example, if the first digit was 1, we would have 1___ (3 blanks) with each blank having 10 possibilities. As there are 10 possibilities for the second number if the number 1 is there, there are 10 possibilities for the second digit for each number possible for the digit, getting 10*10=100 possible combinations with only two numbers. The same applies to the third and fourth blank, as if the second number is 2, there are 100 possibilities for 12__ (2 blanks) to be a possibility, and there are 10 blanks for each possible combination of the first two numbers, getting 10*10*10=10^3=1,000 possible combinations. Repeat the process for the fourth digit, and you end up with 10,000 possible combinations.

Feel free to ask further questions!
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3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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