Answer:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R
2
and \displaystyle {R}_{3}R
3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z=1
Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).
3y - 1 = 28
+ 1 + 1
----------------
3y = 29
---- ----
3 3
y= 9.66
Answer:
12 stores
Step-by-step explanation:
180 * 1/15 = 180/15 = 12
Here is your answer
=
REASON:
On solving the given two terms
-|2-5|= -|-3|
= -3 (since |-3|=3)
and,
(8-11)
= -3
Hence,
-|2-5| <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>=</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>(8-11)
HOPE IT IS USEFUL
Answer:-3
Just simplify the equation and you will x=-3
