It is said that a human child is born with two fears falling and loud noises but over time more fears are added and taken away, those fears will hold you back in life and will take yourself away from you. but if you read the book of psalms it tells us that with the Lord we shall not fear psalms 23 says ¨The lord is my shepard I shall not want. He maketh me to lie down in green pastures he leadeth me beside the still waters. He restoreth my soul He leadeth me in the paths of righteousness for his names sake. Yea though I walk through the valley of the shadow of death I will fear no evil for thou art with me thy rod and thy staff they comfort me. Thou preparest a table before me in the presence of mine enemies thou anointest my head with oil my cup runneth over. Surely goodness and mercy shall follow me all the days of my life and I will dwell in the house of the Lord forever. (KJV) When God is with you ye shall not fear anymore because he declares you one of his own he is your protector and comforter.

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Step-by-step explanation:

What is the question?

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.