Asked by Aline
1)The function f is defined by the equation f(x)+ x-x^2. Which of the following represents a quadratic with no real zeros?
A)f(x) +1/2
B)f(x)-1/2
C)f(x/2)
D)f(x-1/2)
2) If I^(2k) = 1, and i = radical -1, which of the following must be true about k?
A) k is a multiple of 4
B) k is a positive integer
C)when 2k is divided by 4, the remainder is 1
D) k/2 is an integer
^I got B
3)
For all the numbers x and y, let z be defined by the equation z=I 2^2 - x^2 - y^2I +2^2. What is the smallest possible value of z?
^I got 0
A)f(x) +1/2
B)f(x)-1/2
C)f(x/2)
D)f(x-1/2)
2) If I^(2k) = 1, and i = radical -1, which of the following must be true about k?
A) k is a multiple of 4
B) k is a positive integer
C)when 2k is divided by 4, the remainder is 1
D) k/2 is an integer
^I got B
3)
For all the numbers x and y, let z be defined by the equation z=I 2^2 - x^2 - y^2I +2^2. What is the smallest possible value of z?
^I got 0
Answers
Answered by
Arora
1) f(x) = x - x^2
When the determinant (b^2 - 4ac) of a quadratic is negative, it will have non-real zeros. This is true for f(x) - (1/2)
2) No
In order to get i^x = 1, x must be four, or some multiple of four (could also be negative because 1/1 is equal to 1)
So, if 2k is a multiple of 4, then k is a multiple of 2 hence k/2 is an integer.
(To show that B is untrue, take k = -2 => 2k = -4. i^-4 = 1/1 = 1)
When the determinant (b^2 - 4ac) of a quadratic is negative, it will have non-real zeros. This is true for f(x) - (1/2)
2) No
In order to get i^x = 1, x must be four, or some multiple of four (could also be negative because 1/1 is equal to 1)
So, if 2k is a multiple of 4, then k is a multiple of 2 hence k/2 is an integer.
(To show that B is untrue, take k = -2 => 2k = -4. i^-4 = 1/1 = 1)
Answered by
Aline
I think number 3 is C?
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