Math
just answer as many as you can : (

(1) what should you do to both sides of the inequality in order to solve x + (-2) > 3?
a) subtract 2.
b) add 2.
c) add 3.
d) subtract 3.

(2) which of the following statements describes the process for solving 4x ≥ -24?
a) divide both sides of the inequality by -24 and flip the symbol.
b) divide both sides of the inequality by -24 but don't flip the symbol.
c) divide both sides of the inequality by 4 and flip the symbol.
d) divide both sides of the inequality by 4 but don't flip the symbol.

(3) what is the solution for -10 < x - 9?
a) x > -19
b) x < -19
c) x > -1
d) x < -1

(4) what is the solution for -3x ≥ 36?
a) x ≤ -12
b) x ≥ -12
c) x ≤ -108
d) x ≥ -108

(5) how would you graph the solution set of x + 4 > -9?
a) open circle on -13 and shaded to the right
b) closed circle on -13 and shaded to the right
c) open circle on -5 and shaded to the right
d) closed circle on -5 and shaded to the right

(6) how would you graph the solution set of x/6 ≤ -18?
a) closed circle on -3 and shaded to the right
b) closed circle on -3 and shaded to the left
c) closed circle on -108 and shaded to the right
d) closed circle on -108 and shaded to the left

  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.

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step-by-step explanation: