The center is at origin O(0,0).
If it contains the point, P(-8,6), then the radius r is, by Pythagoras theorem,
r=sqrt((-8)^2+6^2)=10
The general equation of a circle at centre (xc,yc) with radius r is given by
(x-xc)^2+(y-yc)^2=r^2
Substituting r=10, (xc,yc)=(0,0)
the resulting equation is therefore
(x-0)^2+(y-0)^2=10^2
or simply
x^2+y^2=100
If RS is the hypotenuse of the triangle RST and point T is in Quadrant 3, then point T must be the intersection of the lines: x = - 4 and y = - 5.
Therefore, the coordinates of point T are ( x, y ) = ( - 4, - 5 )
Answer:
T ( - 4, - 5 )
Answer:
10
Σ [n^2]
i=1
Step-by-step explanation:
Hope I helped? I inserted it into a calc and I also got 385 for something if its useful))
Answer:
Remembering the properties of numbers is important because you use them consistently in pre-calculus. The properties aren't often used by name in pre-calculus, but you're supposed to know when you need to utilize them.
5 because it appears most
Mode is most
