lets say i'm finding the zeros... and my factored equation is -1(x+6)(x-2)
i know x=-6,2. But what about the -1 out front?
or if I have
3(x+2)(x-3) ==>what about the 3 out front?
do those numbers count as a zero?
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Help with Cellular Automata?
In one-dimensional cellular automata, each rule is a static eight-bit number that represents each three-cell combination that produces an "alive" state. In two-dimensional cellular automata, the rule can be a 512-bit number because the nine-cell neighborhood has 512 possible combinations, and each bit in the rule's number determines whether each combination produces an alive (1) or dead (0) state. Because of such complexity and the symmetry of the 2-D rules, the representation is simplified by stating how many cells in a neighborhood cause a cell to be born, to continue living, or to die. In the case of Conway's Game of Life, the rule is 3 neighbors must be alive in order for a cell to come to life, and an alive cell with 2 or 3 alive neighbors will remain alive in the next generation.
I want to find out the 512-bit number that represents this rule. Assuming the middle cell represents the one's place value in each 9-bit combination (it represents the two's place value in one-dimensional CA), what would the rule be? I can't find an online calculator that handles 512-bit numbers, but I believe I've found the formula for it.
3*Sigma(2^(2^a+2^b+2^c)) + Sigma(2^(1+2^d+2^e)),
With 1<=aAnd 1<=d
Multiplying the first half of the equation by 3 is the same as writing it as Sigma(2^(2^a+2^b+2^c) + 2^(1+2^a+2^b+2^c)), which covers three alive neighbors surrounding a cell, whether it is dead or alive. The second half of the equation covers an alive cell with exactly two alive neighbors.
Keep in mind, I've never been formally educated on sigma notation, so I don't know for certain if this covers any combinations I didn't intend or doesn't cover enough. If someone can tell me if it's the correct formula, I'd be very grateful.
View the original article here
I want to find out the 512-bit number that represents this rule. Assuming the middle cell represents the one's place value in each 9-bit combination (it represents the two's place value in one-dimensional CA), what would the rule be? I can't find an online calculator that handles 512-bit numbers, but I believe I've found the formula for it.
3*Sigma(2^(2^a+2^b+2^c)) + Sigma(2^(1+2^d+2^e)),
With 1<=aAnd 1<=d
Multiplying the first half of the equation by 3 is the same as writing it as Sigma(2^(2^a+2^b+2^c) + 2^(1+2^a+2^b+2^c)), which covers three alive neighbors surrounding a cell, whether it is dead or alive. The second half of the equation covers an alive cell with exactly two alive neighbors.
Keep in mind, I've never been formally educated on sigma notation, so I don't know for certain if this covers any combinations I didn't intend or doesn't cover enough. If someone can tell me if it's the correct formula, I'd be very grateful.
View the original article here
A few Math questions!(: 10 POINTS BEST AWNSER!?
9. What is the vertex form of the equation?
y = x^2 – 6x + 8 (2 points)
13. What is the expression in factored form?
6x^2 + 4x (2 points)
14. What is the expression in factored form?
4x^2 + 11x + 6 (2 points)
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y = x^2 – 6x + 8 (2 points)
13. What is the expression in factored form?
6x^2 + 4x (2 points)
14. What is the expression in factored form?
4x^2 + 11x + 6 (2 points)
View the original article here
What does interpret the value of the average rate of change mean?
This is my problem and my math book isn't giving me any information about what this question means..
"The graph shows the population P in a small industrial city from 1950-2000. The variable X represents the number of years since 1950."
Graph below points are on X = 20,25,40,50 ; Y= 40,50,40,20
50-
40-
30-
20-
10-
0 ____________________________
--------------10 20 30 40 50
(a) Determine the intervals on which the function P is increasing and on which it is decreasing.
My answer= Increasing [0,20] [20,25] ; Decreasing [25,40] [40,50]
(b) What was the average rate of change of P between X=20 and X=40 ?
My answer = 40 - 40 / 40-20 = 0/20 = 0
(c) Interpret the value of the average rate of change that you found in part (b)
** This is where I am lost. I'm not sure what the question is actually asking, I've done some searching on the internet about the value of the average rate of change, although I think I'm either finding the wrong thing, or not quite catching the concept. Intuitively I consider the fluctuation between the two intervals, but as the past has shown me, intuition and math don't necessarily make good partners ;X
View the original article here
"The graph shows the population P in a small industrial city from 1950-2000. The variable X represents the number of years since 1950."
Graph below points are on X = 20,25,40,50 ; Y= 40,50,40,20
50-
40-
30-
20-
10-
0 ____________________________
--------------10 20 30 40 50
(a) Determine the intervals on which the function P is increasing and on which it is decreasing.
My answer= Increasing [0,20] [20,25] ; Decreasing [25,40] [40,50]
(b) What was the average rate of change of P between X=20 and X=40 ?
My answer = 40 - 40 / 40-20 = 0/20 = 0
(c) Interpret the value of the average rate of change that you found in part (b)
** This is where I am lost. I'm not sure what the question is actually asking, I've done some searching on the internet about the value of the average rate of change, although I think I'm either finding the wrong thing, or not quite catching the concept. Intuitively I consider the fluctuation between the two intervals, but as the past has shown me, intuition and math don't necessarily make good partners ;X
View the original article here
Why is the answer 2y/x^2 and not 2x^-2*y?
x^(-a) = 1/x^a
x^(-2) = 1/x^2
Therefore, 2y/x^2 and 2x^(-2)y are equal
But often, you will be asked to use positive exponents,
so it's best to answer 2y/x^2
View the original article here
x^(-2) = 1/x^2
Therefore, 2y/x^2 and 2x^(-2)y are equal
But often, you will be asked to use positive exponents,
so it's best to answer 2y/x^2
View the original article here
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