Answer:
Elimination
Step-by-step explanation:
The following system of equations can be easily solved by elimination as we can simply eliminate x-term.
Substitution will take a lot of time because you need to move to make either x-term or y-term as the subject then substitute in.
Graphing Method — You need to find the intercepts of both equations. You also have to convert into slope-intercept form which takes a lot of time.
D = 1/2hw
2dhw = 1
h = 1/2dw
i am a mathematics teacher. if anything to ask please pm me
Answer:
Domain: (2,7,1,3,8) Range: (4,-9,-6,0)
Step-by-step explanation:
The Karger's algorithm relates to graph theory where G=(V,E) is an undirected graph with |E| edges and |V| vertices. The objective is to find the minimum number of cuts in edges in order to separate G into two disjoint graphs. The algorithm is randomized and will, in some cases, give the minimum number of cuts. The more number of trials, the higher probability that the minimum number of cuts will be obtained.
The Karger's algorithm will succeed in finding the minimum cut if every edge contraction does not involve any of the edge set C of the minimum cut.
The probability of success, i.e. obtaining the minimum cut, can be shown to be ≥ 2/(n(n-1))=1/C(n,2), which roughly equals 2/n^2 given in the question.Given: EACH randomized trial using the Karger's algorithm has a success rate of P(success,1) ≥ 2/n^2.
This means that the probability of failure is P(F,1) ≤ (1-2/n^2) for each single trial.
We need to estimate the number of trials, t, such that the probability that all t trials fail is less than 1/n.
Using the multiplication rule in probability theory, this can be expressed as
P(F,t)= (1-2/n^2)^t < 1/n
We will use a tool derived from calculus that
Lim (1-1/x)^x as x->infinity = 1/e, and
(1-1/x)^x < 1/e for x finite.
Setting t=(1/2)n^2 trials, we have
P(F,n^2) = (1-2/n^2)^((1/2)n^2) < 1/e
Finally, if we set t=(1/2)n^2*log(n), [log(n) is log_e(n)]
P(F,(1/2)n^2*log(n))
= (P(F,(1/2)n^2))^log(n)
< (1/e)^log(n)
= 1/(e^log(n))
= 1/n
Therefore, the minimum number of trials, t, such that P(F,t)< 1/n is t=(1/2)(n^2)*log(n) [note: log(n) is natural log]
domain represents the x values so for example in a diagonal line that continues infinitely, the domain is all real numbers or (-infinity, infinity)
range represents y values so it would also be all real numbers or (-infinity, infinity)
let’s say there is a line (refer to pic) that moves ONLY from point (-3, -1) and (2, 2)
the domain would be [-3, 2]
we use brackets because it’s a real number unlike infinity (also because it’s a closed circle on the graph; if the graph had an open circle you would use a parenthesis)
and the range would be [-1, 2]
if you have any more questions about this explanation feel free to ask!
