2.a newspaper started an online version of its paper 14 years ago. in a recent presentation to stockholders, the lead marketing executive states that the revenues for online ads have more than doubled that of the revenues for printed ads since starting the online version of the paper. use the graph below to justify the lead executive’s statement and to determine the approximate year that the two ad revenues were equal.
graph with x axis labeled years. y axis labeled revenue in dollars. the line for printed ad revenue starts at 0, 3 and goes through the10, 2. the line for online ad revenue starts at 0, 0 zero and goes through the 10, 3.

At year 14 online ads revenues  are more than double of printed ads

That is, after 7 and a half years the two ad revenues were equal.

Step-by-step explanation:

Printed ads revenues line:

y-intercept: 3

slope: (2-3)/(10-0) = -1/10

equation: y1 = (-1/10)*x + 3

Online ads revenues line:

y-intercept: 0

slope: (3-0)/(10-0) = 3/10

equation: y2 = (3/10)*x

where x represents years elapsed since the newspaper started an online version of its paper, and y represents revenue in dollars

At x = 14 years:

y1 = (-1/10)*(14) + 3 = 1.6

y2 = (3/10)*(14) = 4.2

Then, online ads revenues  are more than double of printed ads (4.2 > 1.6 *2 = 3.2)

In the year that the two ad revenues were equal y1 = y2, then:

(-1/10)*x + 3 = (3/10)*x

3 = (3/10)*x + (1/10)*x

3 = (2/5)*x

3*(5/2) = x

7 1/2 = x

That is, after 7 and a half years the two ad revenues were equal.

  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.

dont trust me im no expert these losers made me fail a test bro this sucks

step-by-step explanation: