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: 14
= 16 x (7/8)
= 8 goes into 16 2 times
= 2 x 7
= 14
Answer:
Step-by-step explanation:
y = -3x² + 24x - 50
factor out leading coefficient
y = -3(x² - 8x) - 50
complete the square
coefficient of the x term: -8
divide it in half: -4
square it: (-4)² = 4²
use 4² to complete the square and keep the equation balanced:
y = -3(x² - 8x + 4²) + 3(4²) - 50
y = -3(x-4)² - 2
Answer: The vector relies on how the objects of where the magnitude and directions should lay on.
Step-by-step explanation:
So, I’ll help you with these 4 problems. So, for the vector 2/4, it would 2 units up and 4 units to the right. The second would be 5 units down and 6 units to the right. The third one would be 3 units up and 1 unit to the left. The last one would be 10 units downs and 12 units to the left. A simplified way of thinking of this is to just look at the signs of the number and see where the vector’s magnitude and directions would go in that point of view when you visual the graph or count it in your head.
Could you take a better picture please?
