mn = 80
m -n = -79
Substitute
m - 80/m = -79
m^2 - 80 = -79m
m^2 + 79m = 80 (NB Notice change of signs)
Quadratic Eq'n
m = { - 79 +/- sqrt[)79)^2 - 4(1)(-80)}] / 2(1)
m = { - 79 +/- sqrt[6241 + 320}]/ 2
m = { -79 +/- sqrt[6561]} / 2
m = { - 79 +/- 81}/2
m = -160 / 2 = -80 or
m = 2/2 = 1
Hence n = 1 + 79 = 80
80 + 0 = 80. 79 + 1 = 80 78 + 2 = 80. Just continue with that pattern, subtract one from the first number and add one to the second number. 77 + 3 = 80.
80
x = 79 + 1 is one of the infinite number of such equations.
80
1
80 + 0 = 80. 79 + 1 = 80 78 + 2 = 80. Just continue with that pattern, subtract one from the first number and add one to the second number. 77 + 3 = 80.
80
80 is equal to 80 because it is the same number....... :)ex. (79=79) (100=100)
61, 67, 71, 73, 79, 83, 89, 97, so 79 is nearest
83 is the first prime number after 79.
It is: 79
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.
The number of neutrons is 45 because you subtract 80 to 35.
If three of the four numbers are 79, 80, and 65, and the average is 85, then the final number will equal (85 x 4) - 79 - 80 - 65 = 116.
71, 73, 79
It is 80 and not 79