Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
Answer:
x = 60°
Step-by-step explanation:
let the missing angle be = x
Cos x = 8/16
x = Cos-¹ 0.5
x = 60°
This is an expression equivalent to the problem (f g) (10)
Solution:
The given Polynomial is :
By Rational Root theorem the of Zeroes of the Polynopmial are:
But , 
So, no root of this polynomial is real.
Therefore, All the four roots of Polynomial are imaginary.
So, we can't say whether the number k=2, is an upper or lower bound of the polynomial .
X+2.6=1.7
x+12=7
x+12-12=7-12 (subtract 12 from both sides)
x=-5
