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4 years ago

What is the maximum number of edges for an undirected graph? A directed graph?

Computers and Technology

1 answer:

siniylev [52]4 years ago

5 0

Answer:

For undirected graph n(n-1) for directed graph $\frac{n(n-1)}{2}$

Explanation:

The maximum number of edges for a undirected graph is given by $n(n-1)$ , where n is number of vertices

The maximum number of edges in directed graph is given by $\frac{n(n-1)}{2}$ , where n is number of vertices

The maximum number of edges is greater in undirected graph than directed graph

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1.Choose a clear central message 2. Embrace conflict 3.Have a clear structure

Explanation:

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2 years ago

What are some tips you can use when taking close up photos

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3 years ago

Read 2 more answers

Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two

Alexeev081 [22]

Answer:

In Python:

cprice= int(input("Current price: "))

lmonth= int(input("Last month's price: "))

print("This house is $"+str(cprice))

print("The change is $"+str(cprice-lmonth)+" since last month")

print("The current mortage $"+str((cprice * 0.051) / 12)+" since last month")

Explanation:

Get current price

cprice= int(input("Current price: "))

Get last month's price

lmonth= int(input("Last month's price: "))

Print the current price

print("This house is $"+str(cprice))

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print("The change is $"+str(cprice-lmonth)+" since last month")

Print the mortgage

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3 years ago

(15) You are a Pascal teacher (a very good programmer using assembly language(i.e., machine language) of your local machine). Yo

ASHA 777 [7]

If you Could out A B C D separate that would help

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3 years ago

A computer cooled by a fan contains eight PCBs, each dissipating 10 W power. The cooling air is supplied by a 25-W fan mounted a

Kay [80]

Given Information:

Qin = 8*10 = 80 W

Win = 25 W

Temperature = T2-T1 = 10°C

Required Information:

(a) flow rate of the air = m = ?

(b) fraction of the temperature rise of air = f = ?

Answer:

(a) flow rate of the air m = 0.01044 kg/s

(b) fraction of the temperature rise of air f = 23.8 %

Solution:

(a) The flow rate of the air that fan needs to deliver can be found by

$m = Q_{in} + W_{in} /C_{p} (T_{2} -T_{1} )$

Where the value of specific heat of air is 1005 j/kg.C

m = (80 + 25)/1005*10

m = 0.01044 kg/s

(b) fraction of the temperature rise of air due to fan

ΔT = Q/mCp

ΔT = 25/0.01044*1005

ΔT = 2.38°C

So the fan causes a temperature rise of 2.38°C

f = ΔT/T2-T1

f = 2.38/10

f = 0.238 = 23.8 %

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3 years ago