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The circumcenter is the center of the circle which goes through the triangle's vertices, so the circumcenter of the triangle and the center of that circumscribed circle MUST be the same point.
The same goes for the incenter and the center of the inscribed circle, though these will not, in general, be the same point as the circumcenter.
You use long division to divide decimals
Answer:
Second puck travels farther
Step-by-step explanation:
Maximum height of first puck = 3 feet
The height of a second hockey puck is modeled by:


To find maximum height of second puck

Equate the derivative equals to 0
0.15-0.002x=0

75 = x
At x = 75

So, The maximum height of second puck is greater than first puck
So, Second puck travels farther
Answer:
X=5
Step-by-step explanation:
f(1)= -1^2+4(1)+12=15
f(2)= -2^2+4(2)+12=16
f(3)= -3^2+4(3)+12=15
f(4)= -4^2+4(4)+12=12
f(5)= -5^2+4(5)+12=7
g(1)=1+2=3
g(2)=2+2=4
g(3)=3+2=5
g(4)=4+2=6
g(5)=5+2=7
so g(5)=f(5)= X=5