I have an assignment due and I need help. These are the problems! WILL MARK BRRAINLIEST 1. Find the selling price if the cost to the store is $80 and the Markup is 60%. Do not label your answer
2. Find the selling price if the cost to the store is $140 and the Markup is 25%. Do not label your answer
3. Find the selling price if the cost to the store is $50 and the Markup is 10%. Do not label your answer.
4.Find the cost to the store if the store is selling a game for $48 with a markup of 20%. Do not label your answer
5. Find the cost to the store if the store is selling shoes for $54 with a markup of 8%. Do not label your answer
6. Find the cost to the store if the store is selling chips for $3.24 with a markup of 35%. Do not label your answer
7. Find the precent of markup if the cost to the store is $350 and the store is selling the bike for $402.50. Do not label your answer
8. Find the precent of markup if the cost to the store is $250 and the store is selling the TV for $312.50. Do not label your answer
9. Find the precent of markup if the cost to the store is $200 and the store is selling the bike for $376. Do not label your answer
and then there are some problems that I will ask please help ASAP

1. Would  60%×80+80 = $128

(P.S. still thinking of the other ones.)

  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.

5/12, or 5: 12

step-by-step explanation: