I think you need to double-check that date -- it should be 1982 -- Mexico was a republic in 1882. Unless the coin says onza plata pura (1 ounce pure silver), it's value is negligable -- maybe a couple cents. An onza is currently worth about $10.00 The coin you have is : = REPUBLICA Mexico 1882 LIBERTAD .8R .Z .1882. J. S. 10D. 20G. COIN. It is 46.5% silver and today is worth about 10US Dollars = = ********* = = Pls reference this web site: = = http://www.identificacion-numismatica.com/otras-incluso-extranjeras-f6/r-mexicana-8-reales-1882-t22318.htm = = According to this site, the 10D 20G refers to the purity, where 12D is pure silver and each D equals 24G. = = So, the coin weighs about 27 grams, or about 2/3 of a troy ounce, and is about 90% pure. =
The date is actually 1992 on this 1 New Peso and it may be worth .10 to maybe $1 depending on amount of wear and collector demand. There were 144 million minted that year of a design used 1992-95. You might also try a library for a copy of the Standard Catalog of World Coins for pictures, values and lots more interesting info.
There was no 1889 issue of the Mexican 1 Peso note. They are dated either 5.5.1823 or 10.1.1920.
.05cents
the value of the exponent n1
the value of the exponent n1
the value of the exponent n1
n1= 25 n2= n1+1 n3= n1-1 n4=n1+2 n5=n1-2
#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; }
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
The N1 is a rocket.
#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; }
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }
#declare value of product product = 0 #while loop in python #while (test condition) : # statements (s) while product < 100: #prompt for a user to enter a number n1 = input ("Enter a number: ") #multiply n1 times 10 product = n1 * 10
The sum of the first 10 positive integers, using the formula N1 + (N1 + 1) + ... + N2 = N2 * (N2 + 1) / 2 - (N1 - 1) * N1 / 2 is: 55
what are the price of a movie ticket at N1 city