1- Diante da expressão, sendo x= 2-5 e y=2-2
. Os alunos do 8 ano do colégio Santana Lima, Nicolas
e Rudinei optam por resolvê-la de maneiras diferentes. Nicolas decide subtrair os valores de x e y na
expressão para em seguida resolvê-la, enquanto Rudinei prefere simplificar ao máximo a expressão,
utilizando conceitos que aprendeu em anos anteriores, para só então fazer a subtração.

2
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6
(.
2
)

4
6
4
Escolha um dos métodos de resolução e responda o problema.

  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.

the correct answer is b) histograms can be used to exhibit the shape of distributions. this is because histograms count the frequency of each possible answer, so one is able to see which results are more common and which are less common.