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.
Answer:
216
according to 9*mean(24)
Step-by-step explanation:
This is similar to the other question you posted. Follow the same steps as before.
First find g(41).
g(41) = sqrt{x - 5}
g(41) = sqrt{41 - 5}
g(41) = sqrt{36}
g(41) = 6
We now find f(6).
f(6) = -7(6) + 1
f(6) = -42 + 1
f(6) = -41
Answer:
(fºg)(41) = -41
Did you follow?
The correct answer is C
to get 5 to 100, multiply by 20, so multiply 3 by 20 as well.
1/4 I think because you would limit the eights
