Graph a line that contains the point (6, -5) and has a
2
slope of
3

  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.

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

step-by-step explanation:

if you want to separate 54 into 6 individual tiers then you need to divide if you want to separate 54 into 6 individual tiers then you need to divide 6 and 54.

if you divide 9 by 6 you will get 54.

- hope i .

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: