The teams that finish in first place and in second place in Group A.
There is no such thing as a1 litre cup.
add a1/4 cup of sugar
The mean of the numbers a1, a2, ..., an is equal to (a1 + a2 + ..., + an)/n. This number is used mostly as the average. It is called the arithmetic mean.
satellite A1
it was modified................
(a1+a2+....+am )2
You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.You could mean a cell reference, such as A1. You can also give cells a name to make them easier to remember and then reference.
You need to use the LOWER function and put the cell reference or text in the brackets: =LOWER(A1) =LOWER("I WANT THIS TO BE IN LOWER CASE")
oil change due
=AVERAGE(A1:A34)
There is no cell A1A2. What you may mean is A1:A2 which refers to the range consisting of cell A1 and cell A2.
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.