No solution but the equation is true
Answer: x=3
Step-by-step explanation:
To find the zeros, you want to first factor the expression.
x³+x²-36
(x-3)(x²+4x+12)
Now that we have found the factors, we set each to 0.
x-3=0
x=3
Since x²+4x+12 cannot be factored, we can forget about this part.
Therefore, the zeros are x=3. You can check this by plugging the expression into a graphing calculator to see the zeros.
Solution:
The total number of possiblities rolling a number cube with faces numbered 1 to 6 is;
And the total number of possibilities spining the spinner is;
The number of possibility where outcome on the cube is greater than 2 is;
And the number on the spinner less than 9 is;
Hence, the unique combination is;
CORRECT OPTION: A
I dont remember that one sorry
Step-by-step explanation:
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).
