Math 25- Study Guide
... 68) If an object is thrown upward with an initial velocity of 128 ft/sec, its height after t sec is given by h = 128t - 16t2 . Find the maximum height attained by the object. (The object will attain maximum height exactly at the halfway point in terms of the time t, where t = 0 is at the beginning ...
... 68) If an object is thrown upward with an initial velocity of 128 ft/sec, its height after t sec is given by h = 128t - 16t2 . Find the maximum height attained by the object. (The object will attain maximum height exactly at the halfway point in terms of the time t, where t = 0 is at the beginning ...
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... A student makes the following conjecture about the difference of two numbers. Find a counterexample to disprove the student’s conjecture. Conjecture: The difference of any two numbers is always smaller than the larger number. ...
... A student makes the following conjecture about the difference of two numbers. Find a counterexample to disprove the student’s conjecture. Conjecture: The difference of any two numbers is always smaller than the larger number. ...
33rd USAMO 2003
... allowing us to reduce the largest factor. Thus for n > 3, n ↔ n(n - 1) ↔ n(n - 1)(n - 2) ↔ n(n - 1)(n - 2)(n - 3) ↔ 2(n - 1)(n - 2) ↔ (n - 1)(n - 2) ↔ n - 1. But linking n and n1 obviously allows us to link any two integers > 3. That leaves 3 itself, but the question already shows how to link that t ...
... allowing us to reduce the largest factor. Thus for n > 3, n ↔ n(n - 1) ↔ n(n - 1)(n - 2) ↔ n(n - 1)(n - 2)(n - 3) ↔ 2(n - 1)(n - 2) ↔ (n - 1)(n - 2) ↔ n - 1. But linking n and n1 obviously allows us to link any two integers > 3. That leaves 3 itself, but the question already shows how to link that t ...
32(2)
... m = 4; Note that x4 = aif^, x = d^, and x3 = i^. Therefore, applying concatenation to the alignments cdz)d;c and P4 ZDP2;P3 implies that x4 IDX;CX3. Consequently, by Lemma 1,(1) cannot hold for m = 4, since x2 begins with a rf. Similar reasoning shows that (1) is false for m = 9,12,... . To generali ...
... m = 4; Note that x4 = aif^, x = d^, and x3 = i^. Therefore, applying concatenation to the alignments cdz)d;c and P4 ZDP2;P3 implies that x4 IDX;CX3. Consequently, by Lemma 1,(1) cannot hold for m = 4, since x2 begins with a rf. Similar reasoning shows that (1) is false for m = 9,12,... . To generali ...
