Suppose the sequence is defined by an = a0 + n*d Then a1 = a0 + d = 15 and a13 = a0 + 13d = -57 Subtracting the first from the second: 12d = -72 so that d = -6 and then a0 - 6 = 15 gives a0 = 21 So a32 = 21 - 32*6 = -171
It is not clear what the question requires. Yes, there are plenty of equations that have the same solution. For example, each and every equation of direct proportionality has the solution (0, 0). So what? every polynomial of the form y = anxn + an-1xn-1 + ... + a1x + a0 has the solution (0, a0). Again, so what?
Suppose an (n+1)-digit number, X, is divisible by 3. Let X = a0*100 + a1*101 + a2*102 + ... + an*10n = a0 + a1*(1+9) + a2*(1+99) + ... + an*(1+999..9) = [a0 + a1 + a2 + ... + an] + [a1*9 + a2*99 + ... + an*999..9] 3 divides 9, 99, 999, etc so 3 divides each term in the second brackets (parentheses). Therefore 3 must divide the sum in the first brackets. That is, 3 must divide the sum of digits.
Anything (except zero) to the power of zero is 1. If written as 7a0, this is operated as 7 x (a0) = 7 x 1 = 7. If written as (7a)0, it is simply 1 by the first statement.
That is equal to 500(1 + 500) / 2.That is equal to 500(1 + 500) / 2.That is equal to 500(1 + 500) / 2.That is equal to 500(1 + 500) / 2.
Any value with a 'zero' exponent is equaL TO '1'. A^(0) = 1 proof Let a^(0) =. a^(n - n) = a^(n) / a^(n) Cancel down by a^(n) hence it equals '1'.
a0=(a-1\a-1)=a\a=1
t is equal to = (1/2)ln(A/A0))
A= A0e^-kt A0= A/ e^kt = Ae^kt A0= A+ D* D*= A0- A D*= Ae^kt - A D*= A(e^kt - 1)
A0 is 1 meter square.
A0
Let the inputs be A2 A1 A0 & outputs be S5 S4 S3 S2 S1 S0. Now, make a truth table as follows A2 A1 A0 S5 S4 S3 S2 S1 S0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 and so on....... Finally we'll get S0 = A0 S1 = 0 S2 = A1 A2(bar) S3 = A0 [ A1 XOR A2] S4 = A2 [A1(bar) + A0 ] S5 = A1 A2
one rule would be an+1 = an + 4 ; a0= 4. This gives 4,8,12,16,20,..... This is called an arithmetic sequence. A geometric rule would be an+1 = 2an; a0= 4. This gives 4,8,16,32,64,... Another rule is an+1 = an/2 + 6 ; a0= 4. This gives 4, 8, 10, 11, 11.5,11.75, ....
/* the sequence printed is Fibonacci's sequence, each element is calculated as a sum of two previous elements */#includeint main(){int i;int n;int a0=0;int a1=1;printf("How many elements do you want to print? ");scanf("%d",&n);printf("0 ");if (n > 0)printf("1 ");for (i = 2; i
A0 paper is 46.8 x 33.1 in.
The A series of paper is such that each numbered size of paper has exactly half the area of the previous size. ie A1 is 1/2 the area of A0, A2 is half the area of A1, and so on. Also, A0 has an area of exactly 1 sq m. Thus A1 has an area of 1/2 that of A0, A2 has an area of (1/2)^2 = 1/4 of A0, An has an area of (1/2)^n of that of A0 = (1/2)^n sq m 1 m = 100 cm 1 sq m = 1 m x 1 m = 100 cm x 100 cm = 10000 sq cm → A4 has an area of (1/2)^4 sq m = 1/16 sq m = 0.0625 sq m = 0.0625 x 10000 sq cm = 625 sq cm.
210 × 297 mm. A4 is the standard size for most countries, excluding the US, Canada and a few others.A4 is one of a series of paper sizes, starting with A0. A0 has an area of 1 m². If you fold A0 into two you get A1, and so on.All sizes in the A-series have the same ratio of length/width, which is the square root of 2.