Answer:
45.1feet
Step-by-step explanation:
Given the following
∠I=90°
∠G=62°, and
GH = 96 feet = Hypotenuse
Required
IG = Adjacent side
Using the SOH CAH TOA identity
Cos theta = Adj/hyp
Cos 62 =IG/96
IG = 96cos62
IG = 96(0.4695)
IG = 45.1feet
Hence the length of IG to the nearest tenth is 45.1feet
The ANWSER is 2/4 I hope I helped
Answer:
n =24
Step-by-step explanation:
pretend n is x its just a normal fraction lol
Answer:
Step-by-step explanation:
By applying the concept of calculus;
the moment of inertia of the lamina about one corner 
is:
where :
(a and b are the length and the breath of the rectangle respectively )
![I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}](https://tex.z-dn.net/?f=I_%7Bcorner%7D%20%3D%20%20%5Crho%20%5B%5Cfrac%7Bbx%5E3%7D%7B3%7D%2B%20%5Cfrac%7Bb%5E3x%7D%7B3%7D%5D%5E%20%7B%5E%20a%7D%20_%7B_0%7D)
![I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]](https://tex.z-dn.net/?f=I_%7Bcorner%7D%20%3D%20%20%5Crho%20%5B%5Cfrac%7Ba%5E3b%7D%7B3%7D%2B%20%5Cfrac%7Bab%5E3%7D%7B3%7D%5D)
Thus; the moment of inertia of the lamina about one corner is
Answer:
54.
Step-by-step explanation:
(self-explanatory.)
