Answer:
11 feet away.
Step-by-step explanation:
Given:
A water and sprinkle our sins water out in a circular pattern
Area formed by the watering pattern is 379.94 square feet
To find:
How many feet away can it spread water means we have to find distance from the middle point to the last end of circular pattern of water sprinkling (radius of a circle) = ?
solution:
By using ,Area of a circle = 
Area of a circle = 379.94 square feet (given)
= 
Dividing both side by 3.14
Taking square root both side
![r = \sqrt[2]{11\times11}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B2%5D%7B11%5Ctimes11%7D)
(Taking common)
Therefore, radius = 11
Thus, it can spread water 11 feet away.
Answer in pic
Step-by-step explanation:
Z=7 all you really gotta do is subtract 12 from 19 and you get 7
Compute the derivative dy/dx using the power, product, and chain rules. Given
x³ + y³ = 11xy
differentiate both sides with respect to x to get
3x² + 3y² dy/dx = 11y + 11x dy/dx
Solve for dy/dx :
(3y² - 11x) dy/dx = 11y - 3x²
dy/dx = (11y - 3x²)/(3y² - 11x)
The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when
11y - 3x² = 0
or
y = 3/11 x²
(provided that 3y² - 11x ≠ 0)
Substitute y into into the original equation:
x³ + (3/11 x²)³ = 11x (3/11 x²)
x³ + (3/11)³ x⁶ = 3x³
(3/11)³ x⁶ - 2x³ = 0
x³ ((3/11)³ x³ - 2) = 0
One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with
(3/11)³ x³ - 2 = 0
(3/11 x)³ - 2 = 0
(3/11 x)³ = 2
3/11 x = ³√2
x = (11•³√2)/3
Solving for y gives
y = 3/11 x²
y = 3/11 ((11•³√2)/3)²
y = (11•³√4)/3
So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).
Answer:
2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)
Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?
ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol
