Answer:
The answer to your question is:
x = 4
y = -1
z = -3
Step-by-step explanation:
3 x + 2 y + z = 7
5 x + 5 y + 4 z = 3
3 x + 2 y + 3 z = 1
![\left[\begin{array}{ccc}3&2&1\\5&5&4\\3&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%261%5C%5C5%265%264%5C%5C3%262%263%5Cend%7Barray%7D%5Cright%5D)
= 45 + 10 + 24 - (30 + 24 + 15)
= 79 - 69
Δ = 10
![\left[\begin{array}{ccc}7&2&1\\3&5&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%262%261%5C%5C3%265%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D)
= 105 + 6 + 8 - (18 + 56 + 5)
= 119 - 79
Δx = 40
![\left[\begin{array}{ccc}3&7&1\\5&3&4\\3&1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%267%261%5C%5C5%263%264%5C%5C3%261%263%5Cend%7Barray%7D%5Cright%5D)
= 27 + 5 + 84 - ( 105 + 12 + 9)
= 116 - 126
Δy = -10
![\left[\begin{array}{ccc}3&2&7\\5&5&3\\3&2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%267%5C%5C5%265%263%5C%5C3%262%261%5Cend%7Barray%7D%5Cright%5D)
= 15 + 70 + 18 - (10 + 18 + 105)
= 103 - 133
= -30
Δz = -30
x = Δx /Δ = 40/10 = 4
y = Δy/Δ = -10/10 = -1
z = Δz/Δ = -30/10 = -3
Step-by-step explanation: To solve this absolute value inequality,
our goal is to get the absolute value by itself on one side of the inequality.
So start by adding 2 to both sides and we have 4|x + 5| ≤ 12.
Now divide both sides by 3 and we have |x + 5| ≤ 3.
Now the the absolute value is isolated, we can split this up.
The first inequality will look exactly like the one
we have right now except for the absolute value.
For the second one, we flip the sign and change the 3 to a negative.
So we have x + 5 ≤ 3 or x + 5 ≥ -3.
Solving each inequality from here, we have x ≤ -2 or x ≥ -8.
Answer:
Let's say 50
Step-by-step explanation:
40% of 100 is 40
50% of 117 is about 59
so number should be a somewhere in between.
Answer:
-3q² + 3qp + 2rp - 2rq + Sq - Sp
Step-by-step explanation:
first part
3q(p-q) = 3qp - 3q²
second part
2r(p-q) = 2rp - 2rq
third part
S(q-p) = Sq - Sp
then we put it all together
3qp - 3q² + 2rp - 2rq + Sq - Sp
in the right place possibly
-3q² + 3qp + 2rp - 2rq + Sq - Sp