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
The N1 is a rocket.
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
#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