Given n 0’s and n 1’s distributed in any manner whatsoever around a circle (see the following figure), demonstrate, using induction on n, that it is possible to start at some number and proceed clockwise around the circle to the original starting position so that, at any point during the cycle, we have seen at least as many 0’s as 1’s. In the following figure, a possible starting point is marked with an arrow.

nswer: a   y-axis

step-by-step explanation:

x-3y=4

step-by-step e

graph #4 (last graph) depict the function   f(x)= -m(x-d)^1/5 i

step-by-step explanation:

we know that

m=slope

but in this function slope is -m that means slope is decreasing and it is decreasing by 1/5 points down the x-axis.

which is shown in graph four

a decreasing slope down x-axis by 1/5 points.

hence 4th graph is the right option for this.

c.

step-by-step explanation: it's not a. because 1/3 x 1/3 x 1/3 = 1/27 so that's not correct. | b. is also not correct because 7 x 7 1/3 is not actually 3 x 7 1/3 . and for d. multiplications are not additions so that left you with c. ( hope this out : d )