2 + 4 + 6 + 8 + 10 = 30 In general, the sum of the first n even numbers is n(n+1). Replacing n with 5 we have 5x6 which equals 30 as we found by direct calcuation. The formula is nice when you have many numbers. For example, finding the sum of the first 100 positive even numbers would be a PAIN! But using the formula is it simply 100x101.
no, for example 25 cannot be divided by 2 Half of the multiples of 5 are even numbers. For instance, out of the first ten (5, 10, 15, 20, 25, 30, 35, 40, 45, 50) five of them are even numbers.
5x - x = 116so4x = 116sox = 29 and 5x = 145
Just multiply the first five prime numbers! 2x3x5x7x11 = 6x35x11 = 210x11 = 2310. Any multiple of this number "has the first five prime numbers as factors", but 2310 is the smallest such number.
The first five prime numbers are 2, 3, 5, 7, 11. The product of these five numbers is equal to 2 x 3 x 5 x 7 x 11 = 2310.
0.01, 0.001, 0.0001, 0.0000001 and 2.4542
the first positive even numbers are 2,4,6,8,10
1,2,3,4,5
Assuming the first five numbers is meant to refer, not to the first five real numbers but to the first five positive integers, the answer is 1*2*3*4*5 = 120
Even numbers are numbers that are evenly divisible by 2. The first five even numbers would be 2, 4, 6, 8, and 10.
30
The sum of the first 500 even numbers, excluding zero, is 250,500.
The first five whole numbers are: 01234. The first five natural numbers are: 12345.
(2,1), (4,3), (6,5), (8,7) and (10,9).
No, it is not. Counting numbers are positive whole numbers.
In the context of numbers on a number line, positive five is indeed greater than negative five. This is because numbers increase in value as you move to the right on a number line and decrease as you move to the left. Positive five is to the right of zero, while negative five is to the left of zero, making positive five greater than negative five.
weather you would count '0' or not, i dont know but em... 1,2,3,4,5 or 0,1,2,3,4
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.