Math 2602 Finite and Linear Math Fall `14 Homework 7
... 3. Define a sequence recursively by a1 = 1 and an = 3an−1 for n ≥ 2. Solve the recurrence relation. That is, find an explicit formula for an that does not involve any of the previous terms. Solution: The solution is an = 3n−1 . This is proved by induction and the proof is left to the reader. 12. Def ...
... 3. Define a sequence recursively by a1 = 1 and an = 3an−1 for n ≥ 2. Solve the recurrence relation. That is, find an explicit formula for an that does not involve any of the previous terms. Solution: The solution is an = 3n−1 . This is proved by induction and the proof is left to the reader. 12. Def ...
Chapter 10: Writing and solving systems of linear functions
... 31.Shanaya graphed the line represented by the equation y = x - 6. An equation for a line that is identical to the given line but has different coefficients is ...
... 31.Shanaya graphed the line represented by the equation y = x - 6. An equation for a line that is identical to the given line but has different coefficients is ...
Review 5
... Multiply: Now we multiply the piece we just put as part of the answer (-3) by the entire binomial (x+2).3(x + 2) = -3x -6. This is written underneath the original polynomial (just like we would in an arithmetic long division problem. ...
... Multiply: Now we multiply the piece we just put as part of the answer (-3) by the entire binomial (x+2).3(x + 2) = -3x -6. This is written underneath the original polynomial (just like we would in an arithmetic long division problem. ...
Quadratic Equations
... quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Martin-Gay, Developmental Mathematics ...
... quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Martin-Gay, Developmental Mathematics ...
PDF
... Now what do zero divisors, say g, in Fp [X]/(f ) look like? When you take the image under the chinese remaindering, it should go to some tuple (a1 , a2 , · · · , am ) where some ai = 0. Further, if this zero divisor is nontrivial (0 is a trivial zero-divisor, useless), some other aj 6= 0. What does ...
... Now what do zero divisors, say g, in Fp [X]/(f ) look like? When you take the image under the chinese remaindering, it should go to some tuple (a1 , a2 , · · · , am ) where some ai = 0. Further, if this zero divisor is nontrivial (0 is a trivial zero-divisor, useless), some other aj 6= 0. What does ...
