Answer:
I believe it is Option 1
Step-by-step explanation:
Since the fourth root of 81 will be 3 and of 16 will be 2.
I think the answer to this question is letter A
![\bf \begin{array}{lllll} round(x)&\boxed{1}&2&3&\boxed{4}\\\\ wrestlers[f(x)]&\boxed{64}&32&18&\boxed{9} \end{array} \\\\\\ slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby \begin{array}{llll} average\ rate\\ of\ change \end{array}\\\\ -------------------------------\\\\ f(x)= \qquad \begin{cases} x_1=1\\ x_2=4 \end{cases}\implies \cfrac{f(4)-f(1)}{4-1}\implies \cfrac{9-64}{4-1}\implies \cfrac{-55}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0Around%28x%29%26%5Cboxed%7B1%7D%262%263%26%5Cboxed%7B4%7D%5C%5C%5C%5C%0Awrestlers%5Bf%28x%29%5D%26%5Cboxed%7B64%7D%2632%2618%26%5Cboxed%7B9%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Aslope%20%3D%20%7B%7B%20m%7D%7D%3D%20%5Ccfrac%7Brise%7D%7Brun%7D%20%5Cimplies%20%0A%5Ccfrac%7B%7B%7B%20f%28x_2%29%7D%7D-%7B%7B%20f%28x_1%29%7D%7D%7D%7B%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%7D%5Cimpliedby%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Aaverage%5C%20rate%5C%5C%0Aof%5C%20change%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Af%28x%29%3D%20%20%20%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ax_1%3D1%5C%5C%0Ax_2%3D4%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7Bf%284%29-f%281%29%7D%7B4-1%7D%5Cimplies%20%5Ccfrac%7B9-64%7D%7B4-1%7D%5Cimplies%20%5Ccfrac%7B-55%7D%7B3%7D)
55 over 3, or 55 wrestlers for every 3 rounds, but the wrestlers value is negative, thus 55 "less" wrestlers for every 3 rounds on average.
Answer:

Step-by-step explanation:
<em>Given:</em>
Mn is diameter of circle having centre O
and BD = OD,
<em><u>To prove that:</u></em>
<u>
</u>
<em>Solution:</em>
Join the points O and B and draw OB,
On joining the line,
in ∆OCD and ∆OBD,
OC =OB → (Radius of same circle)
BD =CD → (from given)
OD =OD → (Common side in both the triangles)
Hence ∆OCD and ∆OBD are congruent from SSS property.
so we can say that,

Consider above prove as statement A
Corresponding angles of congruent traingle.
in ∆ OAB,
OA = OB (radius of same circle)
hence ∆OAB is an isosceles traingle.
We know that opposite angle of isosceles traingle are always equal. hence,

Consider above prove as statement B
From Statement A & B we can say that

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