The second one.
3.54 ? 3.44
3.54
N=0.0
Greater
Assuming x is your number, x * 10^n = x moved n decimal places. When n is positive, move the decimal point n places to the right. When n is negative, move the decimal point n places to the left. When n is 0, do nothing.
7.5 percent is equivalent to 75/1000 or 3/40 in fraction or 0.075 in decimal
0.2
If a number called 'N' is greater than 20, you would write N > 20 You can see how the symbol tapers down as it goes from 'N' to 20, because 'N' is bigger and 20 is smaller.
I'm assuming the question should read n greater than 10 and less than 100 and there are 8 numbers that satisfy this,1223344556677889So the answer is dhttp://www.webanswers.com/share-question.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/answer/1566078/education/how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff| http://www.webanswers.com/report-abuse.cfm?q=how-many-integers-n-greater-than-and-less-than-100-are-there-such-that-if-the-digits-of-n-are-the-is-1379ff&p=1566078
1
For simplicity I will assume you're working in base x, for any integer x greater than 1, although the argument extends to integers greater than 1 in absolute value (note that in base -1,1 all decimal numbers are in fact integers and that in base 0 decimals are not very well defined). In base x, x can of course be conveniently denoted as 10, so in the remainder of this answer I will work in base x. It is sufficient to show that there exists a decimal number that is not an integer so take 0.1 or 10^-1. This number has the property that 10*0.1 = 1, it is the multiplicative inverse of 10. I will now prove by induction that no positive integer has this property. Base case: 1*10 = 10 which is greater than 1 by assumption. Suppose n*10 is greater than 1, then (n+1)*10 = n*10+1*10 = n*10 + 10 which is still greater than 1. So we now know that n*10 is always greater than 1 for any n greater than 0, from which it can be deduced that for these n, n*10 is also unequal to -1. Therefore, for no integer n unequal to zero can n*10 be 1. Now assume n=0, then n*10 = 0*10 = 0 which is not equal to 1 either. Thus, no integer n has the property n*10=1, whereas the decimal number 0.1 does. So 0.1 is not an integer and therefore the decimal numbers are not integers.
34 654
Assuming x is your number, x * 10^n = x moved n decimal places. When n is positive, move the decimal point n places to the right. When n is negative, move the decimal point n places to the left. When n is 0, do nothing.
l is greater than n
include <iostream> using namespace std; int main() { int n; // number to convert to binary while (cin >> n) { if (n > 0) { cout << n << " (decimal) = "; while (n > 0) { cout << n%2; n = n/2; } cout << " (binary) in reverse order" << endl; } else { cout << "Please enter a number greater than zero." << endl; } } return 0; }//end main
This is impossible. A two digit number n divided by 345 has a remainder equal to n. If you meant to say divided by 3,4,5 or 6 then the answer is 62.
The decimal form of a fraction is either a terminating or recurring decimal.
No, for any number n greater than zero, the LCM of n and n is n.
N/40 = N*0.025
n + 4 =====
It depends on what you mean by a number. If n is a positive integer (except for 1), then n^2 is greater than n. If n = 0 or 1, then n and n^2 are equal. If n = 1/2, then n is greater than its square. If n is negative, then n is always less than its square.