yes
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.
83.0 x 115.4 cm / 32.7 x 45.4 inches
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)
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 is 1 meter square.
A0
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, ....
A0 paper is 46.8 x 33.1 in.
/* 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
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.
16 pages of A4 fit into an A0
Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.