Use the distributive property to write an expression that is equivalent to each expression. If
you get stuck, consider drawing boxes to help
organize your work.

D. 8(-x-1/2)

e. -8(-x-3/4y+7/2

D) -8x-4

E) 8x + 6y - 28

Step-by-step explanation:

you just want to multiply everything in the brackets by what is outside

so for D

8(-x-1/2) = -x * 8 - 1/2 * 8 = -8x-4

for E

-8(-x - 3/4 y + 7/2) = -x * -8 - 3/4 * -8 + 7/2 * -8 = 8x + 6y - 28

pls thank and mark brainliest

Provided by Guathmaths

  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.

you would 18 *0.25 = 4.5