Can someone help me with the equation below??

x<-2 or x>5 (the second choice)

Step-by-step explanation:

This is an "or" inequality. Both of the circles on the number line are open, so the signs we will use are going to be either < or >.

The shaded line on the left means x is less than -2 (x<-2). The shaded line on the right means x is greater than 5 (x>5).

If we write this as an "or" inequality, it will be...

x<-2 or x>5

In case you didn't know, if this was an "and" inequality, the shaded lines would intercept. But in this case, the shaded parts went opposite ways, so this is most likely an "or" inequality.

Hope this helped you out!! Have an amazing day c:

  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.

answer: 298143

step-by-step explanation: