Explain how (13 + 10) + 5 is solved differently from 13 + (10 + 5). a) for (13 + 10) + 5, you add 13 + 5 first, then add 10. for 13 + (10 + 5), you add the 10 + 13 first, then add 5. eliminate b) for (13 + 10) + 5, you add 13 + 10 first, then add 5. for 13 + (10 + 5), you add the 10 + 5 first, then add 13. c) for (13 + 10) + 5, you add 13 - 10 first, then add 5. for 13 + (10 + 5), you add the 10 - 5 first, then add 13. d) for (13 + 10) + 5, you add 13 + 10 first, then subtract 5. for 13 + (10 + 5), you add the 10 + 5 first, then subtract 13.

B)For (13 + 10) + 5, you add 13 + 10 first, then add 5. For 13 + (10 + 5), you add the 10 + 5 first, then add 13.

Step-by-step explanation:

Apply the associative property of addition.

Let , then

In this case a=13,b=10,c=5.

When we substitute into the above relation, we get:

For the LHS, you add 13+5 first, then add 5

For the RHS, you add 10+5 first, then add 13

The correct answer is B

  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.

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.