Answer:
Show that if for some where , then by Rolle's Theorem for some . However, no such exists since for all .
Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.
Step-by-step explanation:
The function is continuous and differentiable over . By Rolle's Theorem. if for some where , then there would exist such that .
Assume by contradiction does have more than one roots over . Let and be (two of the) roots, such that . Notice that just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist such that .
However, no such could exist. Notice that , which is a parabola opening upwards. The only zeros of are and .
However, neither nor are included in the open interval . Additionally, , meaning that is a subset of the open interval . Thus, neither zero would be in the subset . In other words, there is no such that . Contradiction.
Hence, has at most one root over the interval .
Answer:
ask 30 5th grade students
Step-by-step explanation:
if you only ask a certain group then you wont get a accurate answer
Answer:
36V²+84V+49
Step-by-step explanation:
since it is (7+6V) in two places,therefore
7(7+6v)+6v(7+6v)
49+42v+42v+36v²
36v²+84v+49
Given is the base area of a solid crate, B = 6 square meters.
Given is the height of the solid crate, H = 4 meters.
We know the formula for volume of any solid is given as follows :-
(Volume of the solid) = (Base area of the solid) x (Height of the solid).
Volume = B x H.
Volume = (6 square meters) x (4 meters).
Volume = 24 cubic meters.
Hence, the volume of the solid crate would be 24 cubic meters.