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RSB [31]
3 years ago
13

What is 400 divided by 200

Mathematics
2 answers:
Butoxors [25]3 years ago
8 0

Answer:

2

Step-by-step explanation:

Andrei [34K]3 years ago
4 0

Answer:(:

Step-by-step explanation:  2

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Zalon walk 3/4 of a mile in 3/10 of an hour what id his speed in miles per hour
mart [117]
Speed in miles per hour is calculated as:

distance (in miles) ÷ time (in hours)

So we have:

(3/4) ÷ (3/10) = (3×10)/(4×3) = 30/12 = 15/6 = 5/2 = 2.5

So his speed is 2.5 mph (miles per hour).
7 0
3 years ago
What is the Complement of 66°?
Romashka-Z-Leto [24]

Answer:

24°

Step-by-step explanation:

Complementary angles are the angles whose measurements equal to 90°

To find the complement of 66 you subtract it from 90

90 - 66 = 24

7 0
3 years ago
The area of a 25 inch TV screen is 300 in.². The area of a 40 inch TV screen is 768 in.². The area of the smaller screen is what
Rina8888 [55]

Answer:

39.0625%

Step-by-step explanation:

The area of a 25 inc TV screen is 300in²

The area of a 40 inch TV screen is 768in²

Percentage of area of the smaller screen to larger screen is;

300/768 × 100 = 39.0625%


5 0
4 years ago
Which expression is equivalent to cos 150°?
Andreas93 [3]
cos150^o=cos(150^o-360^o)=cos(-210^o)


Answer:D
3 0
4 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).

4 0
3 years ago
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