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3 years ago

Factorising quadratics X squared + 5x -6 X squared - 2x15 X squared -7x +12

Mathematics

1 answer:

34kurt3 years ago

5 0

Answer:

(x -1)(x +6)
(x -5)(x +3)
(x -4)(x -3)

Step-by-step explanation:

In each case, you're looking for divisors of the constant that have a sum equal to the x-coefficient. Those divisors are the constants in the binomial factors.

1) -6 = -1·6 = -2·3 . . . . . (-1)+(6) = 5, so these are the constants of interest.

x^2 +5x -6 = (x -1)(x +6)

2) -15 = -1·15 = -3·5 = -5·3 . . . . . (-5)+(3) = -2, so these are the constants of interest

x^2 -2x -15 = (x -5)(x +3)

3) 12 = -1·-12 = -2·-6 = -3·-4 . . . . . (-3) +(-4) = -7, so these are the constants of interest

x^2 -7x +12 = (x -3)(x -4)

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Answer:

Step-by-step explanation:

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3 years ago

Suppose that, in addition to edge capacities, a flow network has vertex capacities. That is each vertex has a limit l./ on how m

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Answer:

See explanation and answer below.

Step-by-step explanation:

The tranformation

For this case we need to construct G' dividing making a division for each vertex v of G into 3 edges that on this case are $v_1, v_2 and l(v)$ .

We assume that the edges from the begin are the incoming edges of $v_1$ and all the outgoing edges from v are outgoing edges from $v_2$

We need to construct $G' = (V', E')$ with capacity function a' and we need to satisfy the follwoing:

For every $v \in V$ we create 2 vertices $v_1, v_2 \in V'$

Now we can add a new edge asscoiated to $v_1, v_2 \in E'$ with the condition $a' (v_1,v_2) = l(v)$

Now for each edges $(u,v)\in E$ we can create the following edge $( u_r, v_1) \in E'$ and the capacity is given by: $a' (u_r, v_1) = a (u,v)$

And for this case we can see this:

$|V'| = 2|V|, |E'|= |E| +|V|$

Now we assume that x is the flow who belongs to G respect vertex capabilities. We can create a flow function x' who belongs to G' with the following steps:

For every edge $(u,v) \in G$ we can assume that $x' (u_r ,v_1) = x(u,v)$

Then for each vertex $u \in V -t$ and we can define $x\(u_1,u_r) = \sum_{v \in V} x(u,v)$ and $x' (t_1,t_2) = \sum_{v \in V} x(v,t)$

And after see that the capacity constraint on this case would be satisfied since for every edge in G' on the form $(u_r, u_1)$ we have a corresponding edge in G because:

$u \in V -(s,t)$ we have that:

$x' (u_1, u_r) = \sum_{v \in V} x(u,v) \leq l(u) = a' (u_1, u_r)$

$x' (t_1,t_2) = \sum_{v \in V} x(v,t) \leq (t) = a' (t_1,t_2)$

And with this we have the maximization problem solved.

We assume that we have K vertices using the max scale algorithm.

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3 years ago