14 dollars an hour is your answer
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:
The slope of the line is 8.
Step-by-step explanation:
The points on the straight line are:
A = (-1, -12)
B = (9, 68)
Compute the slope of the line as follows:
Thus, the slope of the line is 8.
A, the -4 x 6 = 24 +5 = 29 so -29/6 and the -3/5 stays the same. You just have to put the mixed fraction into a improper fraction. That should be the answer.
Commutative property of addition