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Vanyuwa [196]
3 years ago
9

Suppose that a "code" consists of 4 digits, none of which is repeated. (A digit is one of the 10 numbers 0, 1, 2, 3, 4, 5, 6, 7,

8, 9.) How many codes are
possible?
Mathematics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

5040

Step-by-step explanation:

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X=30 degrees for each of the congregant angles
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What is 75.4 in expanded form
Shalnov [3]

Answer: 75.4 =

 70  

+ 5  

+   0.4

Step-by-step explanation:

Hope this helps :)

8 0
3 years ago
Three brothers share 2 sandwiches equally.How much of a sandwich does each brothers share
Irina-Kira [14]

Answer:

2/3 per brother

Step-by-step explanation:

2/3 plus 2/3 plus 2/3 is 6

5 0
3 years ago
Rewrite the following equation in logarithmic form.
zloy xaker [14]

Answer:

log2(0.25) = -2

512 = 8^3

Step-by-step explanation:

  • <em>Use of formula: </em><em>loga (b) = c ⇔ b = a^c</em>

Rewrite the following equation in logarithmic form.   0.25 = 2^-2

  • log2(0.25) = -2

Rewrite the following equation in exponential form.  log8 (512) = 3

  • 512 = 8^3
3 0
3 years ago
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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