Q: Divide M2n3 over p3 by mp over n2?

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y = kn / (1 + n2)First, let's express it with negative exponents:y = kn * (1 + n2)-1Then apply the chain rule:dy|dn = k(1 + n2)-1 - kn(1 + n2)-2 * 2nAnd simplify:dy|dn = k / (1 + n2) - 2n / kn(1 + n2)

4*n2 = 16 Divide both sides by 4: n2 = 4 Take square roots: n = -2 or n = 2

If you mean: 6/n times 5/n-1 = 1/3 Then: 30/n2-n = 1/3 Multiplying both sides by n2-n: 30 = n2-n/3 Multiplying both sides by 3: 90 = n2-n Subtracting 90 from both sides: 0 = n2-n-90 or n2-n-90 = 0 Solving the above quadratic equation: n = -9 or n =10 If n is of a material value its more likely to be 10 Note that n2 means n squared

Add 7 to each side: 3n2 = 7 Divide each side by 3: n2 = 7/3 n = sqrt 7/sqrt 3, ie just over 1.5

#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; }

Related questions

N3/p4

y = kn / (1 + n2)First, let's express it with negative exponents:y = kn * (1 + n2)-1Then apply the chain rule:dy|dn = k(1 + n2)-1 - kn(1 + n2)-2 * 2nAnd simplify:dy|dn = k / (1 + n2) - 2n / kn(1 + n2)

4*n2 = 16 Divide both sides by 4: n2 = 4 Take square roots: n = -2 or n = 2

Nitrogen (N2) at a little over 78%

If you mean: 6/n times 5/n-1 = 1/3 Then: 30/n2-n = 1/3 Multiplying both sides by n2-n: 30 = n2-n/3 Multiplying both sides by 3: 90 = n2-n Subtracting 90 from both sides: 0 = n2-n-90 or n2-n-90 = 0 Solving the above quadratic equation: n = -9 or n =10 If n is of a material value its more likely to be 10 Note that n2 means n squared

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.

Add 7 to each side: 3n2 = 7 Divide each side by 3: n2 = 7/3 n = sqrt 7/sqrt 3, ie just over 1.5

N2+ and N2- I just did it on mastering chem and it worked I'm pretty sure its because when you count the valence electrons in N2+ and N2- you get 9 and 11 respectively because these are odd there has to be an unpaired electron in each

n2 + n2 = 2 n2

1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2

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

0 in N2