I’ve actually been struggling with this:

For $n>0$, $\lbrace s_i\rbrace$ is said to simply solve $SRS(n)$ if for all $k=0,1,...,n-1$, it is true that $s_k
$s_k = 1 + \sum_{i=0}^{n-1} \text{Ch}_{k}(s_i)$ where $\text{Ch}_k$ is the indicator function for the singleton set containing $k$.

Prove or disprove that the only simple solution that exists for $n>5$, which aren’t multiples of 3, is

$s_0=s_{n-2}=s_{n-1}=1$

$s_1=n-3$

$s_2=3$

$s_i=2$ for $2
And that there are none for $n$ which are multiples of $3$

109 trust me

step-by-step explanation:

segment bisector

step-by-step explanation:

line segment bisector. definition: a line, ray or segment which cuts another line segment into two equal parts. try this drag one of the orange dots at a or b and note the the line ab always divides the line pq into two equal parts.

mark brainliest!

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation:

hi there, i would be interested to you with your test, may you add the questions and answer options.

- : )