Answer:
In order to have a consistent linear system represented by the augmented matrix:
![\left[\begin{array}{ccc}2&-3&h\\-6&9&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%26h%5C%5C-6%269%265%5Cend%7Barray%7D%5Cright%5D)
the value of h must be:
Step-by-step explanation:
A system is consistent if it has a solution, this solution can be unique or a set of infinite solutions.
First, you take the augmented matrix and find the equivalent row echelon form using Gaussian-Jordan elimination:
To do this, you have to multiply the 1st row by 3 and add it to the 2nd row, the resulting matrix is:
![\left[\begin{array}{ccc}2&-3&h\\0&0&5+3h\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-3%26h%5C%5C0%260%265%2B3h%5Cend%7Barray%7D%5Cright%5D)
Now, write the system of equations:
The only way this system has a solution is if 5+3h=0, then, to satisfy this, the value of h must be:
Answer:
z ≥ 2
Step-by-step explanation:
Step 1: Write inequality
-7z - (-5z - 4) ≤ -3z - 2 + 4z
Step 2: Solve for <em>z</em>
- Distribute negative: -7z + 5z + 4 ≤ -3z - 2 + 4z
- Combine like terms: -2z + 4 ≤ z - 2
- Add 2z to both sides: 4 ≤ 3z - 2
- Add 2 to both sides: 6 ≤ 3z
- Divide both sides by 3: 2 ≤ z
- Rewrite: z ≥ 2
For the first one: i don't know...i think it can't be done.
for the second one, it is : 4 × (2 + 3) - 1 = 19
for the third one, it is 4 + ((3² +1) ÷ 5) = 6
for the fourth one, it is : (3 + 2) × 6 - 3 = 27
for the fifth one, it is : 7 + 4 - (9 ÷ 3) = 8
for the sixth one, it is : (6 ÷ 2) + (4 × 2) = 11
for the seventh one, it is : (3 + 1)² ÷ 4 = 4
for the eight one, it is : (12 + 20) ÷ 4² = 2
for the ninth one, it is : 7 + (7 - 18 ÷ 6)² = 23
Answer: n=4 y=3
Step-by-step explanation:
3n - y = 9
2n + y = 11
n=4
3(4) - y = 9
12 - y = 9
y=3
