Orders: x
Inventory: y
1) First table
x2-x1=6-3→x2-x1=3
y2-y1=1920-1960→y2-y1=-40
x3-x2=9-6→x3-x2=3=x2-x1
y3-y2=1900-1920→y3-y2=-20 different to y2-y1=-40. The table does not represent a linear relationship.
2) Second table
x2-x1=7-5→x2-x1=2
y2-y1=1860-1900→y2-y1=-40
x3-x2=9-7→x3-x2=2=x2-x1
y3-y2=1820-1860→y3-y2=-40=y2-y1
x4-x3=11-9→x4-x3=2=x3-x2
y4-y3=1780-1820→y4-y3=-40=y3-y2
x5-x4=13-11→x5-x4=2=x4-x3
y5-y4=1740-1780→y5-y4=-40=y4-y3
The table represents a linear relationship.
3) Third table
x2-x1=2-1→x2-x1=1
y2-y1=1000-2000→y2-y1=-1000
x3-x2=3-2→x3-x2=1=x2-x1
y3-y2=500-1000→y3-y2=-500 different to y2-y1=-1000. The table does not represent a linear relationship.
4) Fourth table
x2-x1=6-4→x2-x1=2
y2-y1=1640-1840→y2-y1=-200
x3-x2=8-6→x3-x2=2=x2-x1
y3-y2=1360-1640→y3-y2=-280 different to y2-y1=-200. The table does not represent a linear relationship.
Answer: The second <span>table best represents a linear relationship.</span>
Answer:
y=x\\
x=y^{2} + 12y\\
y^{2} + 12y -x = 0\\
Delta = (12^{2}) - 4.1.(-x) = 144 +4X = 36.(4+x)
\sqrt{Delta} = 6 . \sqrt{(4+x)} \\
y' = \frac{-12 + 6.(\sqrt{(4+x)}}{2} = -6 + 3.\sqrt{(4+x)}\\
y" = -6 - 3.\sqrt{(4+x)}\\\\
y' = 3.\sqrt{(4+x)} - 6\\
y''= -3.\sqrt{(4+x)} - 6\\
Look at the tenth's place. (:
Answer:
562.5
Step-by-step explanation:
11.25*40 = 562.5
