I am a new student of Python programming.

Hello, very good! Hope you can learn what you need from python.

Does anyone have any examples to display a python code stating “hello world”.

Maybe consider the built in print function?

```
https://docs.python.org/3/library/functions.html#print
```

Cheers,

Cameron Simpson cs@cskk.id.au

Hi Yilton,

Googling for “python hello world” is pretty easy to do:

https://duckduckgo.com/?q=python+hello+world

If you don’t know how to google, you’re going to struggle to program.

There is a huge amount of tutorials, How To guides, help forums, source

code, etc.

https://duckduckgo.com/?q=python+tutorial

There’s tons of stuff about Python on the web, and knowing how to google

for it is important.

Thanks, Steven

Thanks for the tip. I do know how to use google pretty well. I just posted a simple question on this discussion to see if anyone on here has different ways to print that statement. But ofcourse i know how to use google, i think even my grandma knows hows to google. I’m just trying to use as many different sources of information as possible google is not always the most reliable place to get information some information on there is not always correct and can mislead you.

Hi Yilton,

You ought to fill us in with what you already know and what you are

looking for. If you don’t, we’ll probably think you are one of the many

beginners and newbies here who don’t always think of googling before

asking for help.

`print("Hello world")`

is, of course, the most obvious way. But we can

also do:

```
import sys
sys.stdout.write("Hello world\n")
```

or if you want something even more obfuscated:

```
print(*map(chr, (0x48, 0x65, 0x6c, 0x6c, 0x6f,
0x20, 0x77, 0x6f, 0x72, 0x6c, 0x64)), sep='')
```

And then there’s this:

```
try: 1/0
except Exception as e:
(vars(e.__traceback__.tb_frame.f_globals[
''.join(chr(ord(c)-81) for c in '°°³Æº½Åº¿Ä°°')])
[''.join(chr(ord(c)-62) for c in '®°§¬²')](
*map(chr, (0x48,0x65,0x6c,0x6c,0x6f,0x20,0x77,0x6f,0x72,0x6c,0x64)
), sep=''))
```

You are not supposed to understand that. I wrote it and I barely

understand it *wink*

Enjoy!

Does anyone know how to prompt a user in kilometers and calculates a light year and displays the result? The information i have for the calculation is 90 degrees containing 60 minutes of arc and also a kilometer is 1/10,000 the distance from the north pole to the equator.

A light-year is its own unit of distance, defined as the span light

travels through a vacuum over the course of one terrestrial year,

roughly ten trillion kilometers. I think you’re wanting to calculate

something else, based on your description, but may have chosen the

wrong words?

Hi Yilton,

“90 degrees containing 60 minutes of arc” is a nonsense sentence. It is

literally true, in that 60 minutes of arc is smaller than 90 degrees,

like an inch is smaller than a mile. But 90 degrees has nothing to do

with a lightyear, or a kilometre. It is an angle, not a distance.

The distance from the north pole to the equator is 9944.35 km, not

10000.

If you google for “lightyear” you will find out exactly how many

kilometres are in a lightyear. Then write a program to prompt the user

for some kilometres, and do your conversion. You can start with the

Python function `input`

:

```
distance = input("How many inches? ")
distance = float(distance)
print(f"You asked for {distance} inches.")
converted = distance/12
print(f"That is {converted} feet.")
```

If you have trouble, show us the code you have written. Don’t take a

photo of it, or a screen shot. Copy and paste the code into your

message. Don’t forget to format it as code: if you use the GUI editor,

there’s a button that formats it as code.

This is the question from the book. I actually meant nautical miles, not light-years lol.

Write a program that takes as input a number of kilometers and prints the corresponding number of nautical miles.

Use the following approximations:

- A kilometer represents 1/10,000 of the distance between the North Pole and the equator.
- There are 90 degrees, containing 60 minutes of arc each, between the North Pole and the equator.
- A nautical mile is 1 minute of an arc.

I meant nautical miles. The program should request input for Kilometers and output nautical miles

Ah, well do it on paper first.

Approximation 1 means you can pretend there are 10000 kilometers from

the north pole to the equator.

How many minutes of arc are there in that arc, according to

approximation 1 (which isn’t approximate).

Divide one by the other, since a nautical mile is 1 minute of arc.

Write those down as equations on paper. Solve.

Write the same equations into a Python programme, add an input() call to

get a value for kilometers, apply the equations to compute nautical

miles. print() the result.

If you get stuck, come back here with what you have accomplished and

what’s going wrong.

Cheers,

Cameron Simpson cs@cskk.id.au

This is where I am. I inputted 100 and the answer was supposed to be 54 nautical miles but instead, I’m getting 150

#declare variables and constants

minutes = 60

degrees = 90

#prompt user

kilometer = int(input("Enter the number of kilometers: "))

#calculate nautical miles

arc = minutes / degrees

nautical = kilometer / arc

#display number of nautical miles

print("The number of nautical miles is " + str(nautical))

Assorted remarks:

These are just variable assignments (aka bindings). There pretty much

are not “declarations” in Python.

```
#declare variables and constants
minutes = 60
degrees = 90
```

Longer names will make these easier to use and understand. “minutes per

degree” and “degrees_per_arc” might be clearer.

```
#prompt user
kilometer = int(input("Enter the number of kilometers: "))
#calculate nautical miles
arc = minutes / degrees
```

This does not make sense. There are 60 minutes in a degree, and 90

degrees in the arc. So should this not be a multiplication? Giving

(again for a longer name) minutes_per_arc.

You’re missing a constant for kilometers_per_arc = 10000.

```
nautical = kilometer / arc
```

If you have minutes_per_arc and kilometers_per_arc, and a nautical mile

is 1 minute of arc, then you should now compute the ratio between

kilometers and minutes as a distinct variable eg kilometers_per_minute

or minutes_per_kilometer.

Print it out. As a rough sanity check, a nautical mile is a little like

a land mile (they are *not* equal, though), and a land mile is 1.609344

kilometers. So your ratio should be in this ballpark. Or its

reciprocal.

Then multiply (or divide, depend on which ratio you compute above)

“kilometer” by that ratio.

```
nautical = kilometer * the_ratio_from_above
#display number of nautical miles
print("The number of nautical miles is " + str(nautical))
```

print() will do str for you. So you could write:

```
print("The number of nautical miles is", nautical)
```

for less complication.

Cheers,

Cameron Simpson cs@cskk.id.au

Awsome Cameron. Thanks for the help, I got the program to run correctly with the correct calculations.

#declare variables and constants

minutes = 60

degrees = 90

#prompt user

kilometer = int(input("Enter the number of kilometers: "))

#calculate natical miles

arc = degrees * minutes

nautical = kilometer /10000 * arc

#display number of nautical miles

print("The number of nautical miles is ", nautical)

Im having a hard time making a table with this information. Does anyone have any tips on finishing this code. Below is my current code.

The credit plan at TidBit Computer Store specifies a 10% down payment and an annual interest rate of 12%. Monthly payments are 5% of the listed purchase price, minus the down payment. Write a program that takes the purchase price as input. The program should display a table, with appropriate headers, of a payment schedule for the lifetime of the loan. Each row of the table should contain the following items:

the month number (beginning with 1)

the current total balance owed

the interest owed for that month

the amount of principal owed for that month

the payment for that month

the balance remaining after payment

The amount of interest for a month is equal to balance * rate / 12. The amount of principal for a month is equal to the monthly payment minus the interest owed.

#declare variables

downPaymentRate = 10

monthlyPaymentRate = 5

anaulInterest = 12

month = 0

#Prompt user

purchasePrice = float(input("Enter purchase price: "))

#convert down payment rate to decimal

downPaymentRate = downPaymentRate / 100

monthlyPaymentRate = monthlyPaymentRate / 100

anaulInterest = anaulInterest / 100

#display the table header

print("%5s%16s%14s%15s%16s%26s" %

(“Month”, “Current balance”,

“Interest owed”, “Principal owed”, “Monthly payment”,

“Balance due after payment”))

#compute and display results for each month

for month in range(1, months + 1):

currentBalance = downPaymentRate * purchasePrice

interestOwed = currentBalance * anaulInterest

principalOwed = currentBalance

monthlyPayment = currentBalance + interestOwed

endingBalnace = currentBalance - monthlyPayment

print("%5d%16.2f%14.2f%15.2f%16.2f%26.2f" %

(month, CurrenBalance,

interestOwed, principalOwed, monthlyPayment,

endingBalance))

Hello Everybody, I too

Can anyone help me code this problem.

Qustion?

The `newton`

function can use either the recursive strategy of Project 2 or the iterative strategy of the Approximating Square Roots Case Study. The task of testing for the limit is assigned to a function named `limitReached`

, whereas the task of computing a new approximation is assigned to a function named `improveEstimate`

. Each function expects the relevant arguments and returns an appropriate value.

**Code i have so far**

import math

def limitReached(val,last):

```
return abs(val - last)<1e-2
```

def improvedEstimate(val,n):

```
return (val+n /val)*0.5
```

def newton(n):

```
val = n
while True:
last = val
val = improvedEstimate(val,n)
if limitReached(val,last):
break
return val
```

while True:

```
n=(input("Enter a positive number or enter/return to quit: "))
if n == '':
break
else:
print("The program\'s estimate is: ",newton(eval(n)))
print("Python\'s estimate is: ",math.sqrt(eval(n)))
print()
```