Answer:91 i think
Step-by-step explanation:
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.
What’s angle 1? U need to post a pic
this would look like this: 0.0000001^10
If we were to do 0.0000001*0.0000001 it would be 0.00000000000010
We doubled the zeroes and moved the one up. Continue on until you reach the tenth power.
