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olga2289 [7]
3 years ago
8

Which would be the better value? A) 12 pens for $2.70 B) 16 pens for $3.40

Mathematics
1 answer:
gregori [183]3 years ago
8 0

Answer:

a

Step-by-step explanation:

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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
Find the value of x.
maxonik [38]

180 = 132 + x + x

180 - 132 = 2x

48 = 2x

x = 24⁰

5 0
2 years ago
Read 2 more answers
This quiz needs a reason and an answer
Len [333]

y = -13, 13

There are two answers because you are taking the square root of a number. If we go the other way, to check our work, (-13)^2 = 169 and (13)^2 = 169.

Hope this helps! :)

7 0
3 years ago
Read 2 more answers
A. -2/9<br><br> B. -1/3<br><br> C. 1/3<br><br> D. 2/9
gogolik [260]

Answer:

C. 1/3

Hope this helps!:)

4 0
3 years ago
Read 2 more answers
Workings please......
attashe74 [19]

Answer:

21.142

Step-by-step explanation:

not sure it's correct

3 0
3 years ago
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