The question is incomplete. The complete question is :
A plate of uniform areal density 
is bounded by the four curves:
where x and y are in meters. Point 
has coordinates 
and 
. What is the moment of inertia 
of the plate about the point ?
Solution :
Given :
and , 
, 
.
So,
, 
![$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%5Cleft%28%20%5Cleft%5B%20%28x-1%29%5E2y%2B%5Cfrac%7B%28y%2B2%29%5E3%7D%7B3%7D%5Cright%5D_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%5Cright%29%20%5C%20dx%24)
So the moment of inertia is 
.
An impulsive force is a force that is acting only during a short time, I mean, for an instant. Impulse is a physics magnitude define by the product of the impulsive force and the time that it was acting.
Is there any mistake in my English? Please, let me know.
Answer:
See below
Explanation:
See attached diagram
280 km east then 190 km north
Use Pythagorean theorem to find the resultant displacement
d^2 = 280^2 + 190^2
d = 338.4 km
Angle will be arctan ( 190/280) = 34.16 °
The answer is c.
Sound, light and heat energy.
Hope this helped :)
