Answer:
Thank you for your help on this I want to change my username
Explanation:
Answer:
MAX_VALUE
Explanation:
Number.MAX_VALUE represents the maximum possible value that javascript can represent. Its value can be determined using the following code segment:
<script>
var x = Number.MAX_VALUE;
alert(x);
</script>
This will display a value of 1.7976931348623157e+308 .
Similarly Number.MIN_VALUE represents the minimum value that can be handled by javascript. Its value is 5e-324.
Answer:
Use a <u>for</u> loop when you know you want to make 10 passes through a loop
Use a <u>while</u> loop when you want to continue passing through a loop until a condition is met
Explanation:
Let us first clear that when do we use for loop and when do we use while loop.
For loop is used when the programmer already knows the number of repetitions.
While loop is used when the number of iterations are not known but the condition has to be used to determine the number of iterations.
Hence,
Use a <u>for</u> loop when you know you want to make 10 passes through a loop
Use a <u>while</u> loop when you want to continue passing through a loop until a condition is met
6. x, y, and z (x is right, z is forward, and y is up)
7. true
8. plane
9. Cartesian grid
10. They describe a location using the angle and distance from the original.
11. effects that alter the look of an object.
12. true
13. true
14. (not sure, but I would go with conceptual)
15. 3-D elements
Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a great day!
Stay Brainy!
Answer:
Check the explanation
Explanation:
#!usr/bin/python
#FileName: sieve_once_again.py
#Python Version: 2.6.2
#Author: Rahul Raj
#Sat May 15 11:41:21 2010 IST
fi=0 #flag index for scaling with big numbers..
n=input('Prime Number(>2) Upto:')
s=range(3,n,2)
def next_non_zero():
"To find the first non zero element of the list s"
global fi,s
while True:
if s[fi]:return s[fi]
fi+=1
def sieve():
primelist=[2]
limit=(s[-1]-3)/2
largest=s[-1]
while True:
m=next_non_zero()
fi=s.index(m)
if m**2>largest:
primelist+=[prime for prime in s if prime] #appending rest of the non zero numbers
break
ind=(m*(m-1)/2)+s.index(m)
primelist.append(m)
while ind<=limit:
s[ind]=0
ind+=m
s[s.index(m)]=0
#print primelist
print 'Number of Primes upto %d: %d'%(n,len(primelist))
if __name__=='__main__':
sieve()