two to the third power, minus four (It's hard to type this, the previous answer had -4 as an exponent as well) What's below is the best I can do
23 - 4
So we have: the mean; (a + b + c)/3 = 10 yield that a + b + c = 30 (numbers in order) the range; c - a = 2 yield that c = a + 2 the median; b = 10.3 Substituting this information, we have: a + 10.3 + a + 2 = 30 2a = 17.7 a = 17.7/2 a = 8.85 yield that c = 10.85 So the numbers are 8,85, 10.3, and 10.85. Or, since b = 10.3 yield that a + c = 19.7 a + c = 19.7 -a + c = 2 add both equations 2c = 21.7 c = 18.85 which yield a = 8.85
By using: (3*6)+(3*2) = 24
8 divided by 3 = 2 and 2/3
7 × 3 + 6 ÷ 2
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
So we have: the mean; (a + b + c)/3 = 10 yield that a + b + c = 30 (numbers in order) the range; c - a = 2 yield that c = a + 2 the median; b = 10.3 Substituting this information, we have: a + 10.3 + a + 2 = 30 2a = 17.7 a = 17.7/2 a = 8.85 yield that c = 10.85 So the numbers are 8,85, 10.3, and 10.85. Or, since b = 10.3 yield that a + c = 19.7 a + c = 19.7 -a + c = 2 add both equations 2c = 21.7 c = 18.85 which yield a = 8.85
By using: (3*6)+(3*2) = 24
21+3 or using all the numbers: 21 + (3*(2-1))
8 divided by 3 = 2 and 2/3
(5 - 3)/2
7 × 3 + 6 ÷ 2
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
Using 1, 2, and 3, you can make 27 whole numbers.
There are no two real numbers that do. Using complex numbers, these two do: (-3/2 + i√151/2) & (-3/2 - i√151/2) Two numbers that add to -3 and multiply to -40 are -8 & 5 Two numbers that add to 3 and multiply to -40 are 8 & -5 Two complex numbers that add to 3 and multiply to 40 are (3/2 + i√151/2) & (3/2 - i√151/2)
(9-(2*3))8
5+4-3+2
9+3+(-2*-6) = 24