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1) Let S be the sum of a finite geometric series with negative common ratio whose first and last terms are 1 and 4, respectively. (For example, one such series is 1-2+4, whose sum is 3.)
There is a real number L such that S must be greater than L, but we can make S as close as we wish to L by choosing the number of terms in the series appropriately. Determine L.
The series clearly has an odd number of terms. Hence
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And since n is odd,
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Thus
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Here I cheated and used WolframAlpha. It turns out that S_n is a decreasing sequence converging to 5/2. Hence
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3) Let ABCD be a square of side length 4. Let M be on side BC such that CM = 1, and let N be on side AD such that DN = 1. We draw the quarter-circle centered at A.
Let x and y denote the areas of the shaded regions, as shown. Find x - y.
Let z be the area of the quarter-circle bounded by DN, MN and the arc. Then
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It should be straightforward to find x−y.
If you don’t need to simplify your polynomial formula, you can write it down immediately.
15, 35, 53, 100, 243, ...
Answer:
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Can you see the trick? If you substitute n=1, all but the first term disappear; if you substitute n=2, all but the second term disappear, etc.
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The sum of the nth column is
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Hence
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Object–subject–verb order Yoda uses. Called anastrophe this is. An example of hyperbaton it is.
24. Under a certain enlargement (3,4) → (0,-2) and (-5,2) → (4,-1). Find the image of (-3,-8) under this enlargement.
The quickest way to solve this is probably to plot a graph, as Bob Bundy did in post #2. However, if you need to solve it algebraically, you can do this.
Let the centre of enlargement be (a,b). We apply the translation (x,y) → (x−a,y−b) to make the centre of enlargement coincide with the origin. Under the new coordinates the enlargement becomes
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Thus if k is the enlargement factor we have the following equations:
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Solve these equations for a, b, k. The image of (−3,−8) under the enlargement is ((−3−a)k+a,(−8−b)k+b).
I found that 3i, -3i, and 2 work so theres one.
Any numbers of the form[list=*]
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I learn that it is possible to make a tetradecahedron with just regular hexagons and squares.
I like Dvořák’s “New World” symphony.
Let T(n) represent the nth triangular number; then the nth row is
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Hence the sum is
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It is well known that
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for any a > 1.
What’s so hard about it?
Suppose both roots are integers. Then a and b must both be integers and the discriminant is a perfect square. We have
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[EDIT: Oops, mistake. Salvaging operation in progress.]
My question: what is a matrix with variables called? That is, not a matrix of coefficients that corresponds to a linear system; rather, what is a matrix that contains variables called?
Random example: Suppose a person were to encounter a matrix that looks something like this:
where x,y are real variables and x1,y1∈R. What kind of matrix is M? Not a matrix of coefficients, nor a matrix that corresponds to a linear system (I don't think?) but a... ? matrix.
That’s just wat it is, a matrix.
While I am here, does a semicolon mean "to solve for"? I have seen it in various contexts and I was wondering if det(M)=0;y would properly mean "to solve for y".
I’ve never seen anything like this before so I don’t know what it means.
I mention this because a lot of people I come across seem to equate atheism with non-belief in a religion. This is not true – as in the case of Buddhism. Yet in debates between science and religion the confusion with the term is all too prevalent. In fact such debates tend to bracket all people into just two cetegories: “Christians” and “atheists”. If you’re religious you’re a “Christian”, if you’re not religious you’re an “atheist”. Ridiculous!
Is a Buddhist an atheist? A Buddhist has a religion but does not believe in deities.