(6 1/3) / (5/6) = (19/3) / (5/6) = 19/3 * 6/5 = 114/15 = 38/5
3 - 7^2 + 3 * 4^2
3 - 49 + 3 * 16
3 - 49 + 48
- 46 + 48
2
(7b - 2) / (-a + 1)....a = -2, b = 3, c = -1/3
where is he c in this equation ? Because if I just use a and b, my answer is not an answer choice
(7b - 10) / (a - 1)...a = -1, b = 5, c = -2/3
same as before...where is the c in this equation
V = (pi) r^2* h
V = (pi)(5^2)(7)
V = (pi)(25)(7)
V = 175(pi)
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:
I don't know the answer to the question but I just wanted to say I love your profile picture. I love Quackity :) <3
Answer:
22.05
Step-by-step explanation:
i used a calculator
