Let there be a sequence of integers a_1, a_2, a_3, ... For all a_i written as a base 10 number, multiply it by 5^100 and write it in base 10, replace each digit with its remainder when divided by 2, and read that number as if it's binary. Call that number a_(i+1). Prove that if a_1 is any positive integer less than 2^101, the sequence is periodic and has period 2^k for some k.

answer: 20

step-by-step explanation:

step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.