If f (x) = 2x2 + 3x + 2 and g(a) = x + 1, write an expression for f (x) x g(x). o 2x3 + 5x2 + 5x – 2
o 2x + 5x2 + 5x + 2
o 2x3 + 3x2 + 3x + 2
o 2x3 + 3x2 + 3x – 2

  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.

a .4y4

hope this !

c = 0.429t

step-by-step explanation:

the constant rate at which carlos harvests cassava is 35 minutes per 15 cassavas 15/35 = 0.429 cassava per minute

therefore, an equation that models the number of cassava (c) that harvest carlos in a time (t) is the following:

where t is the time in minutes.

this means that when 35 minutes have elapsed 15 cassava have been harvested, at 70 minutes 30 cassava have been harvested and so on.