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Ghella [55]
3 years ago
10

Michael arranged his pennies in the following display what is the answer

Mathematics
1 answer:
Vinil7 [7]3 years ago
4 0
You did not show the display im so sorry :(
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Solve for x. See image below
Anastasy [175]

Answer:

Step-by-step explanation:

Tangents from the point outside the circle are equal.

x + 7 = 13 +8

x + 7 = 21      {Subtract 7 from both sides}

     x = 21 - 7

x  = 14

6 0
3 years ago
An equation of a circle is given as (x + 6)² + (y − 7)² = 81. Find the center and radius of the circle. Show all work to receive
Salsk061 [2.6K]
Center is (-6,7) and radius is 9. A circle with a center at (h,k) and a radius of r has the equation (x-h)^2 + (y-k)^2=r^2 so h= -6 k= 7 r is 9 you just work back wards, what two numbers multiplied gives you 81? Well 9
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3 years ago
In a particular flow network G = (V,E) with integer edge capacities ce, we have already found the maximum s-t flow. However, we
yulyashka [42]

Answer:

Check the explanation

Step-by-step explanation:

1) Algorithm for finding the new optimal flux: 1. Let E' be the edges eh E for which f(e)>O, and let G = (V,E). Find in Gi a path Pi from s to u and a path P_1, from v to t.  

2) [Special case: If P_1, and P_2 have some edge e in common, then Piu[(u,v)}uPx has a directed cycle containing (u,v). In this instance, the flow along this cycle can be reduced by a single unit without any need to change the size of the overall flow. Return the resulting flow.]

3) Reduce flow by one unit along P_1U{(u,v)}UP_2

4) Run Ford-Fulkerson with this sterling flow.

Justification and running time: Say the original flow has see F. Lees ignore the special case (4 After step (3) Of the elgorithuk we have a legal flaw that satisfies the new capacity constraint and has see F-1. Step (4). FOrd-Fueerson, then gives us the optimal flow under the new cePacie co mint. However. we know this flow is at most F, end thus Ford-Fulkerson runs for just one iteration. Since each of the steps is linear, the total running time is linear, that is, O(lVl + lEl).

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3 years ago
How many unique segments can be named using M L D and R as endpoints
Harlamova29_29 [7]

ML LD DR MD LR MR

It 6

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3 years ago
Which statement is not a step used when constructing an inscribed square?
V125BC [204]
Where is the multiple choice answers ?
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3 years ago
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