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1 answer:
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There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/-
x = +/- i
Answer:
- 0.0374
- 0.0010
Step-by-step explanation:
number of cards = 52
number of queen = 4
number of spades = 13
A) probability that the tenth card is a queen
drawn time (r) = 1
position of success(x) = 10th
p = 4/52
P( x,r,p) =
p(10,1,4/52) = 9C0(4/52)^1 * (48/52)^9 = 0.0374
B) probability the twentieth card is a spade
x = 20
r = 1
p = 13 / 52
P(20,1,26/52) = 19C0(26/52)^1 * (26/52)^19 = 0.0010
c) The last five cards been spades
p(last five cards been spades )
p(48..52, 5, 13/52 ) = 47...52C4(13/52)^5 * (39/52)^48..52 - 5