The answer is D
Correct me if I am wrong
Answer:
B. x ≈ 13/8
Step-by-step explanation:
We assume that one iteration consists of determining the midpoint of the interval known to contain the root.
The graph shows the functions intersect between x=1 and x=2, hence our first guess is x = 3/2.
Evaluation of the difference between the left side expression and the right side expression for x = 3/2 shows that difference to be negative, so we can narrow the interval to (3/2, 2). Our 2nd guess is the midpoint of this interval, so is x = 7/4.
Evaluation of the difference between the left side expression and the right side expression for x = 3/4 shows that difference to be positive, so we can narrow the interval to (3/2, 7/4). Our 3rd guess is the midpoint of this interval, so is x = 13/8.
_____
The sign of the difference at this value of x is still negative, so the next guess would be 27/16. It is a little hard to tell what the question means by "3 iterations." Evaluating the function for x=13/8 will be the third evaluation, so the determination that x=27/16 will be the next guess might be considered to be the result of the 3rd iteration.
Answer:
a^2*c^2
Step-by-step explanation: im pretty sure this is it
Answer:
x = -2, -5 where -2 is a double root
Step-by-step explanation:
The real roots are the solutions to the cubic equation. You find them by setting each factor equal to 0 and solving for x.


Since -2 appears twice, it is a double root.
Answer:
−π
----
4
Step-by-step explanation:
Alright, archtan /
tan
−
1
(
x
)
is the inverse of tangent. Tan is
sin
cos
. Like the inverse of sin, the inverse of tan is also restricted to quadrants 1 and 4.
Knowing this we are solving for the inverse of tan -1. We are basically being asked the question what angle/radian does tan(-1) equal. Using the unit circle we can see that tan(1)= pi/4.
Since the "Odds and Evens Identity" states that tan(-x) = -tan(x). Tan(-1)= -pi/4.
Knowing that tan is negative in quadrants 2 and 4. the answer is in either of those two quadrants. BUT!!! since inverse of tan is restricted to quadrants 1 and 4 we are left with the only answer -pi/4.