Task #1: comparing phone plans plus telecommunications offers a plan of $20 per month for an unlimited calling and data plan and 10 cents per text message sent. talkmore, a competing company, offers a plan of $55.00 per month for an identical unlimited calling and data plan and five cents per text message. how can you determine which plan will be cheaper for you?

Step-by-step explanation:

First company offers a plan of $20 per month for unlimited calling and data plan plus 10 cents per text message.

Second company offers a plan of $55 per month for unlimited calling and data plan plus 5 cents per text message.

Let the messages to be done in a month = x

therefore, charges for a month = 0.10x

for plan B cost of x messages in a month will be = 0.05x

so, monthly charges for plan A with x messages = 20 + 0.10x

similarly for plan B charges = 55 + 0.05x

Both the plans will be same when

20 + 0.10x = 55 + 0.05x

0.10x - 0.05x = 55 - 20

0.05x = 35

x = = 700

For 700 messages both the plans will cost the same.

Below 700 messages (600 messages) we have to pay for plan A = 20 + 0.10×600 = 20 + 60 = $80

And For plan B charges = 55 + 0.05×(600) = 55 + 30

                                       = $85

This reveals that below 700 messages Plan A will be cheaper than Plan B but above 700 messages Plan B will Be economical.

  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.

h df tyjgtyj ttd y ytkgfty ty

step-by-step explanation:

tyty jry j y yj yttj yhgtyj y ghyj ytthj