Solve the following equation. X cubed minus 6X squared plus 6X equals zero
1 answer:
We have to solve this equation:
Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:
Now we can find the roots of the quadratic polynomial as:
Then, the solutions to the equation are:
x = 0
x = 3 - √3
x = 3 + √3
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#4
White block(s) = 2
Red block(s) = 1
Purple block(s) = 3
Total = 2 +1 + 3 = 6 blocks
a) P(white) =
P(red) =
P(purple) =
b)Not white block:
1 -
OR 
Because, when they say no white blocks, we simply do not count them and add the rest to find that probability without white blocks.
c) The probability stays the same: lets say now we have
4 white blocks, 2 red, and 6 purple, total will be 12
P(white)=
which is still
d) We get two more blocks in the numerator: lets say we have 4 white blocks, 3 red, 5 purple (after adding 2 of each color), total will be 12
P(purple)=
(im not quite sure if my explanation here helps you though)
e) 1 more of white and purple, 5 more of red
white = 3, purple = 4, red = 6, total = 12
(you can either add 2 to white or purple but make sure you add 5 of red)
P(red)=
= 
White block(s) = 2
Red block(s) = 1
Purple block(s) = 3
Total = 2 +1 + 3 = 6 blocks
a) P(white) =
P(red) =
P(purple) =
b)Not white block:
1 -
Because, when they say no white blocks, we simply do not count them and add the rest to find that probability without white blocks.
c) The probability stays the same: lets say now we have
4 white blocks, 2 red, and 6 purple, total will be 12
P(white)=
d) We get two more blocks in the numerator: lets say we have 4 white blocks, 3 red, 5 purple (after adding 2 of each color), total will be 12
P(purple)=
(im not quite sure if my explanation here helps you though)
e) 1 more of white and purple, 5 more of red
white = 3, purple = 4, red = 6, total = 12
(you can either add 2 to white or purple but make sure you add 5 of red)
P(red)=