Unanswered

Unanswered

Answered

Catija Cotija Girija Full Answer

Answered

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

Answered

In
Cheeses

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

Answered

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

Answered

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

Unanswered

Answered

In
Recipes

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

Answered

In
Algebra

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

Answered

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

Answered

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

Answered

In
Algebra

n-1 = 1/n Full Answer

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

n - 3 - 1 = n - 4 Full Answer

Answered

nn+1n+2n+3n+4 Full Answer

Answered

N+2h=-1 n+2(1)=-1 n+2=-1n=-3 Full Answer

Answered

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

Answered

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

Answered

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

Answered

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

Answered

The GCF of n and 1 is 1. Full Answer

Answered

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

Answered

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

Answered

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

Answered

In
N-Dubz

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

Answered

In
Algebra

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

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

Answered

equation works for n = 0 and n = 1 as factors are n and n - 1 Full Answer

Answered

Answered

Answered

Answered

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

Answered

In
Calculus

It is (x^(n+1))/(n+1) Full Answer

Answered

n*(n+1)/2 Full Answer