Minus 1 both sides
x²-8x=-1
take 1/2 of the linear coefient and square it
-8 is the linear coefient
-8/2=-4, (-4)²=16
add 16 to both sides
x²-8x+16=16-1
factor perfect squaer trinomial
(x-4)²=15
square root both sides, remember to take positive and negative roots
x-4=+/-√15
not exactly sure what to put in the blnaks
but
the solutions are x=4+√15 and x=4-√15
The Angle formed by an arc on the centre of the circle is double the angle made by the same arc on the circumference of the circle.
So,
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NO! 3 points cannot be non-co-planar
Good luck! :)
<3
Answer:
True
the answer to the above question is true
