Answer:
y<2/3x-1
Step-by-step explanation:
plotted point is on -1
slope of the line is rise 2, run 3
so its the last option
Answer:
d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)
Step-by-step explanation:
d²y/dx² = (-2y² − x²) / (4y³)
Take the derivative (use quotient rule and chain rule):
d³y/dx³ = [ (4y³) (-4y dy/dx − 2x) − (-2y² − x²) (12y² dy/dx) ] / (4y³)²
d³y/dx³ = [ (-16y⁴ dy/dx − 8xy³ − (-24y⁴ dy/dx − 12x²y² dy/dx) ] / (16y⁶)
d³y/dx³ = (-16y⁴ dy/dx − 8xy³ + 24y⁴ dy/dx + 12x²y² dy/dx) / (16y⁶)
d³y/dx³ = ((8y⁴ + 12x²y²) dy/dx − 8xy³) / (16y⁶)
d³y/dx³ = ((2y² + 3x²) dy/dx − 2xy) / (4y⁴)
Substitute:
d³y/dx³ = ((2y² + 3x²) (-x / (2y)) − 2xy) / (4y⁴)
d³y/dx³ = ((2y² + 3x²) (-x) − 4xy²) / (8y⁵)
d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)
Answer:
The measure of ∠EFG is 52°
Step-by-step explanation:
Given line m is parallel to line p. m∠HEF = 39º and m∠IGF = 13º.we have to find m∠EFG.
In ΔJFG,
By angle sum property of triangle, which states that sum of all angles of triangle is 180°
m∠FJG+m∠JGF+m∠JFG=180°
⇒ 39°+13°+m∠JFG=180°
⇒ m∠JFG=180°-39°-13°=128°
As JFE is a straight line ∴ ∠JFG and ∠EFG forms linear pair
⇒ m∠JFG+m∠EFG=180°
⇒ 128°+m∠EFG=180°
⇒ m∠EFG=52°
The measure of ∠EFG is 52°
Answer:
1 =x-5 = as pytanerosa tery says
23
Step-by-step explanation:
Answer:
Carriage outwards is also referred to as freight-out, transportation-out, or delivery expense. The cost of carriage outwards should be reported on the income statement as an operating expense in the same period as the revenue from the sale of the goods.
