What is the solution set of the quadratic inequality x^2-5=< 0

{x|√5 ≥ x ≥ -√5}

Step-by-step explanation:

Simply take the ±√5 then square it to get rid of the radical [5].

I am joyous to assist you anytime.

Step-by-step

x^2-5<=0 (1)

can be rewritten as

x^2<=5  (2)

which gives the solution set

|x| <=sqrt(5)   (3)

(3) can be rewritten as

x<=sqrt(5), and -x <=sqrt(5) which gives a solution set of

x<=sqrt(5) and x>=-sqrt(5) (4)

The combined conditions of (4) can be rewritten as

-sqrt(5) <= x <= +sqrt(5)

which can also be rewritten in interval notation

[-sqrt(5), sqrt(5)]

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.