Answer:
110
Step-by-step explanation:
you place 30 in the place of X, and it becomes 4×30-10. in PEMDAS, multiplication goes first and 4×30=120. 120-10=110
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).
Answer:
All of them? Pls reply if so!
Step-by-step explanation:
To convert a fraction to a decimal just the divide the bottom into the top.
To convert a decimal into a fraction just put the decimal over 1 or divide the decimal by preference of tenths, hundredths, or thousandth place.
Yes- though we may say "When will we ever use this?" ever so often in class, the reality is that we use mathematics in everyday life. From simple addition, to factoring, to finding the angles of various components to a building, math is always being used in real life.
You have to add the $51 to $13 and find the answer then divide the sum by $28,500
