What is the relationship between .04 and .004
2 answers:
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Answer:
Ix = Iy =
Radius of gyration x = y =
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is . Moment of inertia of quarter disk about its perpendicular is
.
Now using perpendicular axis theorem, Ix = Iy = =
.
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.
Pretend these are coordinates that you can use to find the slope of the line.
(10, 40) and (15, 60). Fit these into the slope formula to find the slope of the line you are looking for:
and the slope is 4. Now use one of the points and the slope of 4 to solve for b, the y-intercept:
40 = 4(10) + b so b = 0. The equation of the line then is y = 4x + 0 or just
y = 4x
(10, 40) and (15, 60). Fit these into the slope formula to find the slope of the line you are looking for:
40 = 4(10) + b so b = 0. The equation of the line then is y = 4x + 0 or just
y = 4x
Answer: option d. x = 3π/2
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).