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lapo4ka [179]
2 years ago
9

Solve pleaseeeeeeeeeeeee​

Mathematics
2 answers:
lesya692 [45]2 years ago
7 0

Answer: 136 degrees

Step-by-step explanation:

Angle 136 and x are corresponding angles so that’s why x is 136

frozen [14]2 years ago
4 0

Answer:

x= 136

Step-by-step explanation:

The two angles are corresponding angles, therefore they are equal to each other.

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Please urgent answer needed
Daniel [21]

Answer:

8

Step-by-step explanation:

5 0
3 years ago
Solve <br> x2 + 11x + 30 = 0
Zarrin [17]

Answer:

x = - 5, - 6

Step-by-step explanation:

x^2 + 11x + 30 = 0

x^2 + 5x + 6x + 30 = 0

x ( x + 5 ) + 6 ( x + 5 ) = 0

( x + 6 ) ( x + 5 ) = 0

x + 6 = 0

x = - 6

x + 5 = 0

x = - 5

7 0
2 years ago
Aubree has a coin collection. She keeps 5 of the coins in her box, which is 2% of the
Ronch [10]

Answer:

250

Step-by-step explanation:

you enter into a calculator 5x100/2

8 0
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
storchak [24]

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.

6 0
3 years ago
Solve the equation.
expeople1 [14]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

z = 6

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