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Are there infinitely many primes of the form n2-1 and n2-2 and n2-3 and n2-4?
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
Find LCM and HCF of two given number by using for loop in c?
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
What is Factor 0.04m2 - n2?
.04m2-n2=(.2m+n)(.2m-n)
How do you factor 5n2 10n 20?
Take 5 out. If the missing signs are pluses, it becomes 5(n2 + 2n + 4) If the missing signs are minuses, it becomes 5(n2 - 2n - 4)
What is the factor of n2 - 6n - 40?
(n - 10)(n + 4)
What are the factors of the trinomial n plus 11n plus 30?
Assuming you mean n2 + 11n + 30, the factors are (n + 6)(n + 5).
How do you factor special cases like n2-4 or a2 plus 2a plus 1?
n2 - 4 is the "difference of two squares." That factors to (n - 2)(n+2)a2 + 2a + 1 is "additive squares." That factors to (a + 1)(a + 1) or (a + 1)2
N2 minus 11n minus 26 need to writte the expression in factored form?
n2 - 11n - 26what two factors of - 26 add up to - 11?(n + 2)(n - 13)===========
What are the factors of n to the twelfth?
The factors of n12 are 1, n, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, and n12
Are there infinitely many primes of the form n2-1 and n2-2 and n2-3 and n2-4?
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
What is the answer to the equation 0 equals n2-n?
equation works for n = 0 and n = 1 as factors are n and n - 1
What is the pmf of trinomial distribution?
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
What is the oxidation number for N2?
0 in N2
How do you factor n square minus n minus 56?
The expression n2 - n - 56 factors to (n - 8)(n + 7).
What is oxidation number for N2?
0 in N2
What is n2 in math?
n x n = n2
Codings for coprime number in c?
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
