This is not a question, i just wanted to post a detailed explanation of a property. this is a good thing to remember, and i had trouble understanding it at first, so i want to explain to others. i can now understand it and will explain in case someone is unsure. 1. ⁠the first sum is just 1 + 2 + 3 + + 10. which equals 55, times 2. which is 110. easy right? 2. ⁠the second sum is simply 2(1) + 2(2) + 2(3) + + 2(10). this is also equal to 110. 3. ⁠think of it like transferring the 2, from outside the first σ, to inside the second σ. so, logically you would be distributing the 2 to each term and then adding the up, instead of adding them up and then multiplying by 2. 4. ⁠if it , think if it like this: 2 ( 1 + 2 + 3 + + 10 ) = 2(1) + 2(2) + 2(3) + + 2(10). all you’re doing is distributing the 2 to each term. i hope this lets you understand the concept easier. i know this is not a very complicated idea, but i want to people understand the basics. it’s good to have a full understanding of the basic when you’re doing more complex math.

  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.

answer: 10 meter

step-by-step explanation:

need details

what

where is the diagram? shown below where? ?

step-by-step explanation: