Write two different pairs of decimals whose sums are 14.1. one pair ahold involve regrouping

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

step-by-step explanation:

let pq = qr = rs = x, and xz = y and zu = z.

if the horizontal lines are parallel, then corresponding segments are in proportion:

substitute:

therefore 3 = y = z.

xu = y + z → xu = 3 + 3 = 6

on the middle line

step-by-step explanation:

1/5=50

6 cars because if you have 8 and subtract 2 youll have 6 left over