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>
60 inches because the square has 4 sides so 4x2 is 8 and 52+8 is 60.
Looks like the given curve is
5<em>xy</em> + <em>z</em>⁴<em>x</em> - 3<em>yz</em> = 3
Differentiate both sides with respect to <em>x</em> :
5<em>y</em> + 4<em>z</em>³<em>x</em> ∂<em>z</em>/∂<em>x</em> + <em>z</em>⁴ - 3<em>y</em> ∂<em>z</em>/∂<em>x</em> = 0
Solve for ∂<em>z</em>/∂<em>x</em> :
(4<em>z</em>³<em>x</em> - 3<em>y</em>) ∂<em>z</em>/∂<em>x</em> = - (5<em>y</em> + <em>z</em>⁴)
∂<em>z</em>/∂<em>x</em> = - (5<em>y</em> + <em>z</em>⁴) / (4<em>z</em>³<em>x</em> - 3<em>y</em>)
At the point (1, 1, 1), this derivative is
∂<em>z</em>/∂<em>x</em> (1, 1, 1) = - (5 + 1) / (4 - 3) = -6
Answer: base times height divided by two
Additive Property of Equality, sometimes called the Addition Property of Equality.
You can add or subtract the same value from both sides of the equation and still maintain the equality.
