Gadfly

The first step is to draw the perpendicular bisector of the segment AB since it would contain centers of the circles that pass through both points. If we call the centers, only one in the image, O(a, b), and the point of tangency H, then the three segments AO, BO, and HO would have to have the same length and that length is b, the radius of the circle. So we can write the following:

\((a-1)^2+(b-9)^2=(a-8)^2+(b-8)^2=b^2\)

This is a system of two equations and two unknowns that can be written as follows:

\(\frac{(a-1)^2}{9}=2b-9\)

\(\frac{(a-8)^2}{8}=2b-8\)

which implies that

\(\frac{(a-1)^2}{9}+9=\frac{(a-8)^2}{8}+8\)

Simplifying, we end up with the quadratic equation \(a^2-128a+496= (a-4)(a-124)=0\) that has solutions

\(a=124\) and \(a=4\). These values of \(a\) result in two values for \(b\) giving the radii of the circles:

\(\frac{(4-1)^2}{9}=2b-9\), implying \(b=5\), and

\(\frac {(124-1)^2}{9}=2b-9\), implying \(b=845\), a rather surprising result.

Since the teacher is concerned with how difficult the words in a whole book would be, we can think of all the words in the book as the population. The sample would then be all the words she actually reads and assesses for difficulty, that is all the words contained in the first sentences of the first page of each chapter of the book. She is sampling using a method called systematic sampling. The three difficult words out of all the words she has read would indicate a level of difficulty that would have to be defined.

\((22.4, 26.5)\)is caalled a confidence interval. The way you form a confidence interval is by adding and subtracting the margin of error from the sample mean. so

\(26.5=sample\ mean+margin\ of\ error\), and

\(22.4=sample\ mean-margin\ of\ error\).

Solve the system of equations by adding the two equations to eliminate margin of error and get

\(2\times sample\ mean=22.4+26.5=48.9\), or

\(sample\ mean=\frac{48.9}{2}=24.45\).

So \(margin\ of\ error=26.5-24.45=2.05\)

As you can see from the underlined table (red for part (1) and blue for part (2)) , the numbers you want, 6 from each category for the first part, are:

Clothing and Jewlery: 011, 106, 135, 138, 109, 071

Food and Services: 157, 163, 154, 197, 174, 185

In the second part, your sample should have \(\frac{1}{4} \times 12 =3\) stores from Food and Services category since one fourth of the population is in that category; the rest, 9 stores, should be in the other category. These are underlined in blue:

Clothing and Jewlery:070, 143, 043, 036, 024, 117, 092, 126, 136

Food and Services:192, 193, 177

It is very easy to make mistakes reading the random digits table, so you should double check my answers. All you do is start at the line suggested and look at three digits at a time from left to right and select the ones that are in the range of IDs assigned to the stores. Read the table as if the digits followed one another in a continuous stream with no gaps for columns and no breaks for lines because in reality that is how it is and the digits are arranged in columns to give your eyes reference points.

\(\frac {X^5-1}{x-1}=x^4+x^3+x^2+x+1\); this polynomial can be written as:

\(1\cdot x^0+1\cdot x^1+1\cdot x^2 +1 \cdot x^3+1\cdot x^4\). So  N= 11,111 (base x)

Ha!!?? the same distortions happened again. I used latex to write three inequalities involving x and r , but all three were somehow Butchered in the process of uploading. Well I don't think images will be distorted, so here they come as pictures:

A number of errors crept in when the solution was published;

line 1: "the decimal part of x , \(0" should read "the decimal part of x, \(0 "

line 10 "Since r cannot be negative, we must choose \(0 and thus \(1 . " should read " since r cannot be negative, we must choose

\(0 and thus \(1 "

let \(x=1 + r\), where r is the decimal part of x , \(0 . Then \(\lfloor x \rfloor=1\), and, therefore,  \(\lfloor x \rfloor ^2=1\) . So we have the left side of the equation and we must look for conditions on r that give you \(\lfloor x^2 \rfloor=1\). But

\(x^2=(1+r)^2=1+2r+r^2\), so that for \(\lfloor x^2 \rfloor=\lfloor 1+2r+r^2\rfloor\) to equal 1, we must have the remainder of the expression, namely, \(2r+r^2\), be a number between 0 and 1, that is, \(0<2r+r^2<1\). We only have to consider, however, the less-than part of the inequality since r is a positive number.

For \(2r+r^2<1\) to hold, however, we can solve \(r^2+2r-1<0\) using a sign chart, but a better approach would be to notice that we are dealing with a parabola that opens up and that has vertex below the x-axis (the vertex is (-1, -2)) and so it would have to be below the x-axis for all values of r between the two x-intercepts. Setting the expression equal to zero and using the quadratic formula we get \(r=-1 \pm \sqrt2\). Since r cannot be negative, we must choose \(0 and thus \(1 .

I tested the result on a few values of x in the above interval and here is the result:

\(\sqrt2\) is approximately 1.414213562373, and I purposely entered some values of x larger than square root of 2 into the table; these are in the shaded rows. Notice that even if x is very slightly larger, say by one unit in the fifth digit after the point (see the last row) the relationship ( FLOOR(x))^2 = FLOOR(X^2) does not hold

The lattice I have given is precisely the set of points we are looking for. But a proof is wanting.

(Notice that 'our friend with the few words of encouragement' who said to my partitions "bits iss wrong" has packed his bags and left the ring. That to me is an indication that he/she has either run out of words completely or that I have happened to hit the jackpot with my poorly dotted diagram of the terminal points. Have we been mean to him/her, perhaps? Is it possible that English is his/her third or fourth tongue, and that that tongue is still developing? I remember a long time ago as an exchange student I developed an intense dislike  for anyone who expected me to talk with more words than the vocabulary of a parrot contains. In fact for about 6 months after arrival I hardly opened my mouth. I guess I was waiting for the moment of glory, when my English would suddenly become bedazzlingly fluent and my voice commanding and irresistable. Of course that never happened, but I manage and yet still have very little to say.)

I am still looking for the overlapping partitions though even though I am becoming convinced that that approach is suffering from a fundamental  flaw; there are too many overlaps and not enough partitions to cover all 221Points.

I don't believe we can assume that a = b = c, since all we are given is that their three moduli are equal and nonzero. So I will provide an answer that does not reqire that assumption.

We are given that \(az^2+bz + c = 0\), so we can assume

\(0= |az^2+bz+c| \geq|az^2+bz|-|c| \geq|az^2|-|bz|-|c|\).

But then

\(|a||z|^2-|b||z|-|c|=k|z|^2-k|z|-k=k(|z|^2-|z|-1)\leq0\),

where \(k=|a|=|b|=|c|\)is nonzero and positive.

This forces the inequality \(|z|^2-|z|-1\leq0\).

From the graph of the real-valued function \(f(|z|)=|z|^2-|z|-1\) given above, one can see that the values of \(|z|\) for the portion of the graph below and touching the \(|z|\) axis range from

|z|= \(\frac{1-\sqrt{5}}{2} \) to |z|=\(\frac{1+\sqrt{5}}{2} \). So the largest value of |z| is \(\frac{1+\sqrt{5}}{2} \).