Answer:
1. Rolling a number less than 5
The numbers on a standard dice are 1, 2, 3, 4, 5, and 6. Therefore it is possible to roll a 6, but it is much more likely to be 1-5, simply because there are more of them.
2. Rolling a 2
Again, it is more likely that the result will be one of the other 5 numbers. But obviously a 2 is still an option, so it's not impossible.
3. Rolling a number less than 10
Every option is less than 10 so it is certain.
X^4 - 4x^3 - 9x^2 + 36x = 0
x(x^3 - 4x^2 - 9x + 36) = 0
x(x^2(x - 4) - 9(x - 4) = 0
x(x + 3)(x - 3)(x - 4) = 0
the zeroes are -3, 0, 3, 4 Answer
If he has 6 friends they will all get 8 cards
6x+1 = 12x because the Z angles are the same
Solve this and you get:
1 = 12x - 6x => x = 1/6
To have roots as described, that means we have the following factors: From multiplicity 2 at x=1 has (x-1)^2 as its factor From multiplicity 1 at x=0 has x as a factor From multiplicity 1 at x = -4 has a factor of x+4 Putting these together we get that P(x) = A (x) (x+4) (x-1)^2 Multiply these out and find P(x) = A (x^2 + 4x) (x^2 - 2x + 1) A ( x^4 - 2x^3 + x^2 + 4x^3 - 8x^2 + 4x ) Combine like terms and find P(x) = A (x^4 + 2x^3 - 7x^2 + 4x) To find A, we use the point they gave us (5, 72) P(5) = A [ (5)^4 + 2(5)^3 - 7(5)^2 + 4(5) ] = 72 A [ 625 + 250 - 175 + 20 ] = 72 A [ 720 ] = 72 Divide both sides by 720 and find that A = 0.1 Final answer: P(x) = 0.1 ( x^4 + 2x^3 - 7x^2 + 4x) or P(x) = 0.1 x^4 + 0.2 x^3 - 0.7x^2 + 0.4x
