Answer:
Find the Roots (Zeros) f(x)=x^3-6x^2+13x-20. f(x)=x3−6x2+13x−20 f ( x ) = x 3 - 6 x 2 + 13 x - 20. Set x3−6x2+13x−20 x 3 - 6 x 2 + 13 x - 20 equal to 0 0 .
Step-by-step explanation:
hope this helps
If the negative square root is found to be one of your solutions, then that is indicative of a pair of imaginary roots (the imaginary i). According to the conjugate rule, if you have one solution that is imaginary, you will have another but with the opposite sign. For example, if a solution to a quadratic is found to be 2 - i, then its conjugate, 2 + i is also a solution. They will ALWAYS go in pairs. Same thing with radical solutions. If one solution is found to be 
then 
will also be a solution.
First u do a squared +b squared =c squared aka pythagorean theorem then so u already have c squared so you would do 5x5=25 and 13x13=169 so then the equation would be 25+?=169 then do 169-25 which equals 144 and then do the square root of that which is 12 THE ANSWER IS 12
Answer:
False
Step-by-step explanation:
f(x) = 4x³ - 12x² - x + 15
Set output to 0.
Factor the function.
0 = (x + 1)(2x - 3)(2x - 5)
Set factors equal to 0.
x + 1 = 0
x = -1
2x - 3 = 0
2x = 3
x = 3/2
2x - 5 = 0
2x = 5
x = 5/2
-2 is not a lower bound for the zeros of the function.
15 cookies burned because if you Turn
into a decimal then multiply it by 20 you get 15 and thats your answer.
