Answer:
If we write the given matrix in a reduced row echelon form, it will have a minimum of one row comprising zeros. In addition, if we try to solve the expression Ax = b, the specific row will have an expression/equation compressing a zero on the left side of the expression/equation and nonzero values of b on the other side of the equation.
Step-by-step explanation:
If we write the given matrix in a reduced row echelon form, it will have a minimum of one row comprising zeros. In addition, if we try to solve the expression Ax = b, the specific row will have an expression/equation compressing a zero on the left side of the expression/equation and nonzero values of b on the other side of the equation.
Answer:
See my question............
Answer:
Step-by-step explanation:
your photo is not clerly
Answer:
13/30
Step-by-step explanation:
Add: -1/
3
+ 2/
5
= -1 · 5/
3 · 5
+ 2 · 3/
5 · 3
= -5/
15
+ 6/
15
= -5 + 6/
15
= 1/
15
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(3, 5) = 15. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 5 = 15. In the next intermediate step the fraction result cannot be further simplified by canceling.
In words - minus one third plus two fifths = one fifteenth.
Subtract: 1/
2
- the result of step No. 1 = 1/
2
- 1/
15
= 1 · 15/
2 · 15
- 1 · 2/
15 · 2
= 15/
30
- 2/
30
= 15 - 2/
30
= 13/
30
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(2, 15) = 30. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 2 × 15 = 30. In the next intermediate step the fraction result cannot be further simplified by canceling.
In words - one half minus one fifteenth = thirteen thirtieths.
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Answer:
C. {x² + 4, x < 2
{-x + 4 x ≥ 2
Step-by-step explanation:
This is a piecewise function, where the two equations are different. They are:
y = x²+ 4
y = -x + 4
The function x² + 4 is graphed where x < 2. (< is used because the circle is open)
The function -x + 4 is graphed where x ≥ 2. (≥ is used since the endpoint is closed)
Therefore, the correct answer is:
C. x² + 4, x < 2
-x + 4 x ≥ 2