79 + 1 78 + 2 77 + 3
76 + 4 75 + 5 74 + 6 73 + 7 72 + 8
(Etc.) I hope you get the picture.
The numbers are 10 and 8
Let's denote the two square numbers as x^2 and y^2. The difference between two square numbers can be expressed as (x^2 - y^2), which can be factored into (x + y)(x - y). Since the difference is 80, we have (x + y)(x - y) = 80. The factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. By testing out different combinations of these factors, we find that the pair of square numbers that make a difference of 80 is 82^2 and 2^2.
There's only one number that's equal to 80. It's 80.There's a good chance that there could be three numbers whose sum is 80, or threedifferent numbers whose joint product is 80, but the question doesn't ask for those.
The two numbers that equal 160 are 80 and 80. This is because when you add 80 to 80, you get 160. Another way to represent this mathematically is 80 + 80 = 160.
80 + 80 can equal 160
They can be: 73+7 = 80
80= 5x5x3+5
The numbers are 10 and 8
100 and 80
2 and 40, for example.
32 x 5/√4 = 32 x 5/2 = 80
The prime numbers (factors) of 80 are: 2 and 5
u culd make 80
Numbers that are divisible by 80 are itself and any of its other multiples
Prime numbers between 71 to 80 = 73 and 79 Therefore there are 2 prime numbers between 71 to 80.
There are an infinite number of sets with mean 80. Here are some: {80, 80, 80}, {80, 80, 80, 80, 80, 80} {79, 80, 81}, {79, 79, 80, 81, 81}, {79, 79, 80, 82} (1, 80, 159}, {-40, 200} To produce a set of n numbers with mean 80, start with any set of n-1 numbers. Suppose their sum is S. Then add the number 80*n-S to the set. You will now have n numbers whose sum is S+80*n-S = 80*n So the mean of this set is 80.
prime numbers between 50 and 80 = 53,59,61,67 ,71,73,79