*Other mainly includes mid-stream urine, other urinary checks and discussion / counselling.
What keeps people keep coming back to Likealittle is the compulsion to check the discussion around their updates and comments as well as check whether or not someone has mentioned somebody that looks like them.
They begin with background checks and discussions about the clients' needs.
Amref has already identified villages still in need of a safe water source close by and, after various checks and discussions, should be in a position to start the work early in the new year.
Otherwise, participants had to check the discussion results by reviewing the chat history; audio devices and web cameras are better for online deliberation.
Using substitution (y=tx), (tge 0), the last inequality reduces to begin{aligned} |3-2t|le max {6|1-t|,tfrac{11}{2},tfrac{17}{3}t,tfrac{1}{2}[|6-tfrac{1}{3}t|+|6t-tfrac{1}{2}|]}, end{aligned}and can be checked by discussion on possible values for (tge 0).
Using substitution (y=tx), (t>0), the last inequality reduces to begin{aligned} |3-2t|le 5max left{ |1-t|,tfrac{1}{2},tfrac{2}{3}t,tfrac{1}{2}left[|1-tfrac{1}{3}t|+|t-tfrac{1}{2}andght]right}, end{aligned}and can be checked by discussion on possible values for (t>0).
Using substitution y = t x, t ≥ 0, the last inequality reduces to | 3 − 2 t | ≤ max { 6 | 1 − t |, 11 2, 17 3 t, 1 2 | 6 − 1 3 t |, 1 2 | 6 t − 1 2 | }, and can be checked by discussion on possible values for t ≥ 0. Hence, all the conditions of Theorem 4 are satisfied and S, T, ℛ have a common fixed point (which is 0).
Using substitution y = tx, t ≥ 0, the last inequality reduces to 3 - 2 t ≤ max 6 1 - t, 11 2, 17 3 t, 1 2 6 - 1 3 t + 6 t - 1 2, and can be checked by discussion on possible values for t ≥ 0. Hence, all the conditions of Theorem 4 are satisfied and S, T, R have a unique common fixed point in O x 0 ; S, T, R - (which is 0).
Using substitution x = 1 − ξ, y = 1 − ξ t, 0 ≤ ξ ≤ 1, t ≥ 0, the last inequality reduces to | 3 t − 2 | ≤ 5 max { | 1 − t |, 2 3, 1 2 t, 1 2 [ | 1 2 t − 1 | + | t − 1 3 | ] }, and can be checked by discussion on possible values for t ≥ 0. Hence, all the conditions of Theorem 2 are satisfied and S, T have a common fixed point (which is z = 1 ).
