# Results for: cotija-michoac-n-1

### Where country you can find mount Paricutan?

Parícutin is a scoria-cone volcano located in the Mexican state of Michoacán, near the city of Uruapan and about 322 km west of Mexico City. Full Answer

### What is cotija cheese?

Cotija cheese is a Mexican cheese made with raw milk and aged 3-12 months. This is a very salty cheese used as a garnish for re-fried beans, tacos, salads and chili. Full Answer

### How do you say old cheese in spanish?

Queso viejo if you want to be correct. Are you thinking of cheeses like old cheddar? Aging of cheese is not a common practice nor a common term here though cotija has 2 versions, the older one being called cotija… Full Answer

### Cotija de la paz map?

if u type it in English ill be able to tell you!k? Full Answer

### What ingredients are included in a taco soup recipe?

Taco soup generally consists of beef or chicken, Mexican spices such as ground cumin, chili powder, and garlic, and optional beans, corn, onions, and peppers. Toppings may include cilantro, sour cream, tortilla chips, hot sauce, and grated block cheese or… Full Answer

### What is the factorial of 0?

A recursive formula for the factorial is n! = n(n - 1)!. Rearranging gives (n - 1)! = n!/n, Substituting 'n - 1' as 0 -- i.e. n = 1 -- then 0! = 1!/1, which is 1/1 = 1. Full Answer

### Solve 2xsquared plus 5x equals 5?

n! = n * (n-1) !n!/n = n*(n-1)!/n //divid by n both sidesn!/n = (n-1)!let n = 11!/1 = (1-1)!1 = 0!Hence proved that 0! = 1. Full Answer

### What is the sum of the first n numbers?

Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n… Full Answer

### What is the sum of successor of n and predecessor of n is?

Successor of n: n + 1Predecessor of n: n - 1Add them together, and you get: n + 1 + n - 1 = 2n Full Answer

### How to simplify n plus 1 factorial divided by n factorial?

Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■ Full Answer

### Prove that nCr plus nCr minus 1 equals n plus 1Cr?

nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!]= n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1}= n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]}= n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]}= (n+1)!/[r!(n+1-r)!]= n+1Cr Full Answer

### How do you solve sin x equals n over 1?

sinx=n/1 (1)sinx=n/1(1) sin(-n)x=n(-n) six=6 Full Answer

### What is the answer to n factorial?

n! = 1*2* ... * (n-1)*n Full Answer

### What is n3 plus 1?

n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - n + 1) Full Answer

### Use S n(n plus 1)2 to find the sum of 1 plus 2 plus 3 plus . . . plus 825?

Plug into the formula the value of n = 825 as there are 825 numbers in {1 2, ..., 825} and see what comes out: S = n(n + 1)/2 = 825(825 + 1)/2 = 825 × 826 / 2… Full Answer

### What is 52 equal 4 plus (n-1)3?

52 = 4 + (n - 1)³ → (n - 1)³ = 48 → n - 1 = ³√48 → n = 1 + ³√48 ≈ 4.634 Full Answer

### How to solve why any number raise to 0 is equal to 1?

Let n be any number and n/n = 1 and n1/n1 = n1-1 which is n0 that must equal 1 Full Answer

### What is n-3 -1?

n - 3 - 1 = n - 4 Full Answer

### What is the answer to 4 divided by 8 over n?

(4/8)/n = (1/2)/(n/1) = 1/2 x 1/n = 1/(2n) Full Answer

### How is it that 3 is the only prime that is before a square number?

Take any positive integer n. If you square it, and subtract 1, you get (x squared - 1). If you take (n - 1) and (n + 1), and multiply them together, you get n squared - n + n… Full Answer

### What is n 1 if n is 4?

if n=4 then n+1 would be 4+1 which equals 5 Full Answer

### What is the sum 1 to 200?

1 to 200 sum to 20100. 1 + ... n = n(n+1)/2. Full Answer

### Is the greatest common factor of any n and 1 n?

The GCF of n and 1 is 1. Full Answer

### How do you calculate the time complexity of selection sort?

T(n)=n+(n-1)+(n-2)+.............+2+1 make pair the first and last terms, the second and next to last terms, and so forth. it ends up looking like this:- T(n)=[n+1]+[(n-1)+2]+[(n-2)+3]+......... =(n+1)+(n+1)+(n+1)+............ In an algebra course, the sum of first N integers is given by the… Full Answer

### What is 52 equal 4 plus (n-1) x 3?

52 = 4 + (n-1)3 → 3(n-1) = 48 → n - 1 = 16 → n = 17 Full Answer

### What is the product of a number n-1 and the number n plus 1 is always equal?

Difficult to tell without brackets, but (n-1)*(n+1) = n^2 - 1 [in other words, n squared minus 1] is a useful algebraic identity. Full Answer

### Suppose that X1 1 and thatXn 1 1 for n NProve by induction that xn for n N?

Suppose that X1 = 1 and that Xn+1 = 1+ , for n > N Prove by induction that xn for n N Full Answer

### Which expression shows the multiplicative inverse of n?

The multiplicative inverse of the expression n is 1/n because when n is multiplied by 1/n, the result is 1. Full Answer

### What are the 3 numbers of the three consecutive?

n-1, n and n+1 where n is an integer. Full Answer

### Does the series 1 divided by ln x converge?

Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges. For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in… Full Answer

### What is the formula of the nth term of Fibonacci sequence?

x(n)=x(n-1)+x(n-2) n, n-1 and n-1 are subscript. Full Answer

### What is first difference?

In a sequence of numbers, a(1), a(2), a(3), ... , a(n), a(n+1), ... he first differences are a(2) - a(1), a(3) - a(2), ... , a(n+1) - a(n) , ...Alternatively, d the sequence of first differences is given byd(n) =… Full Answer

### How do I find the sum of 1 to 20?

The sum from 1 to n is n*(n+1)/2 In this case that mean 20*21/2 = 210 Full Answer

### What is the sum of the series 4 12 24 40 60 upto n terms?

The kth term, t(k) is given by t(k) = 2k2 + 2k So the sum of the first n terms is 2*(12+22+32+...+n2) + 2*(1+2+3+...+n) = 2*n(n+1)(2n+1)/6 + 2*n(n+1)/2 = n(n+1)*(2n+1)/3 + n(n+1) = n(n+1)*(2n+1+3)/3 = 2*n(n+1)(n+2)/3 Full Answer

### Who Do solve Fermat in one page?

PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE… Full Answer

### What is the 34th number in this sequence 2 6 12 20?

The 34th number is 1190. First number is 2 = 2n where n = 1 Second number is 4 + 2 = 2n + 2(n-1) where n = 2 Third number is 6 + 4 + 2 = 2n +… Full Answer

### What is the product of an non-zero number n and its reciprocal?

n x 1/n =n/n = 1 Full Answer

### 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 Full Answer

### Who is a genius?

PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE… Full Answer

### Can you tell how to submit an attempted proof of Fermat's Last Theorem?

PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE… Full Answer

PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE… Full Answer

### Sum of 1 plus 2 plus 3 plus 4 plus .. plus n?

n(n+1)/2You can see this from the following:Let x=1+2+3+...+nThis is the same as x=n+(n-1)+...+1x=1+2+3+...+nx=n+(n-1)+...+1If you add the corresponding terms on the right-hand side of the two equations together, they each equal n+1 (e.g., 1+n=n+1, 2+n-1=n+1, ..., n+1=n+1). There are n such… Full Answer