Consider the game of up-and-down rooks played on a standard chess- board. player i has a set of white rooks initially located in the bottom row, while player ii has a set of black rooks in the top row. in each turn, a player moves one of its rooks up or down a column, without skipping over the other rook or occupying its position. the first player who cannot move loses. this game is not progressively bounded, yet an optimal strategy exists. find such a strategy by relating this game to a nim position with 8 piles.

y=x+9

?

step-by-step explanation:

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

i would say true? if the unit was informational and it "refreshed your understanding", then true.

step-by-step explanation: