Irrational because 3.14... is pie.
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
Answer:
16/3 because you put a 1 under 8 since it's not a fraction hen multiple across
Step-by-step explanation:
pls give brainly
Answer:
1. 1
2. 10
3. 12
Step-by-step explanation:
