To find the gcf you have to wright down the factors of both the numbers, all of them. Then, find the one that is the greatest number, but both 36 and 27 have, which would be 9
For a quadratic equation ax^2 + bx + c = 0, the discriminant is given by b^2 - 4ac
Thus for a^2 - 2a + 5 = 0,
a = 1, b = -2 and c = 5
b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16
Let the numbers be 1k, 2k , and 3k respectively
Since the sum of their cubes is 4500,
(1k)³+(2k)³+(3k)³=4500
--> k³(36)= 4500
--> k = 5
Hence, the numbers are: 5,10 and 15 respectively
Answer:
The answer is 128 bunches because 1027/8 = 128.375
Then I multiplied 128*8
which is equal to 1024
So the left over is 3 after subtracting the initial.