26 for ad and bd 13 I believe
(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous
Answer:
B) 120
Step-by-step explanation:
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.