Is the initial graph a function ? Is the inverse of the graph a function ?
1 answer:
You might be interested in
Given that formula
((4-(-2))^2- (5-5))^1/2
((4+2)^2 -(0))^1/2
((6)^2 -0)^1/2
(36)^1/2
The answer is 6
(Note the 1/2 is also a substitute for the square root sign)
((4-(-2))^2- (5-5))^1/2
((4+2)^2 -(0))^1/2
((6)^2 -0)^1/2
(36)^1/2
The answer is 6
(Note the 1/2 is also a substitute for the square root sign)
Answer:
V = π /2 volume units
Step-by-step explanation:
Volume of y = √x bounded by y = x and y = 1
y = √x
limits of integration ( see Annex )
By simple inspection limits of integration are x = 0 to x = 1
or y = √x and y = x
solving these two equations
x = 0 y = 0 x = 1 y = ± 1
V = ∫ π*f(x)²*dx
V = π * ∫₀¹ x*dx = π * x²/2 |₀¹
V = π * (1/2) - 0
V = π /2 volume units
