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Stereotypical for Asian parents to do what? Whip some asss?

cos theta can be 1/2 and 1.

It just sterotypical for asian parents to do that

Oh? Is an asss whipping more likely or less likely from Asian parents?

depends if parents are asian

An asss whipping?

15 - 4 = 11

11/15  = 0.73333333333

So you got a 73%

The answer is 18.

I understand how the side BC is 4 but how did you get BD or BE? What work did you do, a proportion?

Isn't it an arithmetic series with terms =1, 2, 3, 4...........100 repunits.

The sum(S) = [100  x   101] / 2 =5050

Sum of digits of S =5 + 0 + 5 + 0 =10

I will guess  15

The answer is 35.

thanks

I was careless. I will attempt that again.

I suppose that it depends on your understanding of the word distinct.

In the context of this question, I would take it to mean different.

PLZ HELP!!!!!!!! QUICK!!!!!!!

(WRONG ANSWER: DELETED)

Denote \(r_n\) the n^th repunit.

\(\begin{array}{rcl} S &=& \displaystyle \sum^{100}_{n = 1} r_n\\ &=& \displaystyle \sum^{100}_{n = 1} \sum^{n-1}_{k = 0} 10^{k}\\ &=& \displaystyle \frac{1}{10}\sum_{n = 1}^{100} \sum_{k = 1}^{n} 10^{k}\\ &=& \displaystyle \dfrac{1}{10} \sum_{1 \leq k \leq n \leq 100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \sum_{n=k}^{100} 10^{k}\\ &=& \displaystyle \dfrac{1}{10}\sum_{k = 1}^{100} \left(10^{k} (101 - k)\right)\\ &=& \displaystyle \dfrac{1}{10}\left(101\left(\dfrac{10\left(10^{100} - 1\right)}{10 - 1}\right)-\sum_{k= 1}^{100}k\cdot 10^{k}\right) \end{array}\\ \text{Let }S_{AG} = \displaystyle \sum_{k = 1}^{100} k\cdot 10^{k}\\ \begin{array}{rcl} S &=& \dfrac{101}{9}\left(10^{100} - 1\right) - \dfrac{1}{10}S_{AG}\\ S_{AG} &=& 1\cdot 10^1 + 2\cdot 10^2 + \cdots + 100\cdot 10^{100}---(1)\\ \dfrac{1}{10}S_{AG} &=& 1 + 2\cdot 10^1+\cdots +100\cdot 10^{99}---(2)\\ (1)-(2):\dfrac{9}{10}S_{AG} &=& 100\cdot 10^{100} - (1 + 10^1 + 10^2 +\cdots + 10^{99})\\ &=&10^{102} - \dfrac{10^{100} - 1}{9}\\ S_{AG} &=& \dfrac{10^{103}}{9} - \dfrac{10^{101} - 10}{81}\\ S &=& \dfrac{101}{9} (10^{100} - 1) - \left(\dfrac{10^{102}}{9}-\dfrac{10^{100} - 1}{81}\right)\\ &=& \dfrac{10^{100}}{9} - \dfrac{101}{9}+\dfrac{10^{100} - 1}{81}\\ &=& \dfrac{10^{100} - 1}{9}+\dfrac{10^{100} - 901}{81} \end{array}\\\)