Apply to Row 2 : Row 2 + Row 1
2x + 2y + 3z = 0
y + 4z = -3
2x + 3y + 3z = 5
Apply to Row 3: Row 3 - Row 1
2x + 2y + 3z = 0
y + 4z = -3
y = 5
Apply to Row 3: Row 3 - Row 2
2x + 2y + 3z = 0
y + 4z = -3
-4z = 8
Simplify rows
2x + 2y + 3z = 0
y + 4z = -3
z = -2
<em>Note that the matrix is in echelon form now. The next steps are for back substitution.</em>
Apply to Row 2: Row 2 - 4 Row 3
2x + 2y + 3z = 0
y = 5
z = -2
Apply to Row 1: Row 1 - 3 Row 3
2x + 2y = 6
y = 5
z = -2
Apply to Row 1: Row 1 - 2 Row 2
2x = -4
y = 5
z = 2
Simplify the rows
<u>x = -2</u>
<u>y = 5</u>
<u>z = -2</u>
Answer:
D -13
Step-by-step explanation:
5x - 3y - z if x = -2, y = 2, and z = -3
5x - 3y - z =
= 5(-2) - 3(2) - (-3)
= -10 - 6 + 3
= -16 + 3
= -13
The temperature got colder by 11 regress
Answer: 63lbs
Step-by-step explanation:
The truck can only hold 16crates and each crate weigh 12lbs, the total weight of the 16 crates will be 12×16= 192lbs
This shows that the truck can only contain extra load of 1200lbs - 192lbs = 1008lbs (excluding weight of crates).
To get the shipment weigh close to the total of 1200lbs, the truck must be loaded with engine components not more than 1008lbs.
Since we have 16 crates to fill with engine components not more than 1008lb, each crates will therefore must not exceed 1008/16 pounds of engine components which is equivalent to 63lbs.
Therefore, the manager should instruct the workers to put 63lbs of machine components in "each crate" in order to get the shipment weight as close as possible to 1200 lbs.
Answer:
0 slope
Step-by-step explanation:
flat horizantal line
no increase in y-value
