Answer:
2 vertical asymptotes occurring at x = 5 and x = -1
Step-by-step explanation:
given
recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)
Equating the denominator to zero,
(x-5)(x+1) = 0
(x-5) =0
x = 5 (first asymptote)
or (x+1) = 0
x = -1 (2nd asymptote)
Answer:
Both functions have one x-intercept each.
Step-by-step explanation:
The first function is
This is a parabola with vertices at the origin and has one x-intercept at t=0.
The transformed function is
The function g(t) is obtained by shifting the graph of f(t) to the left by 3 units.
This graph also has one x-intercept at x=-3.
Therefore both functions has the same number of x-intercepts
Wait are you going to send a picture to fully show us what the question is about
The inclusion/exclusion principle states that
That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).
Therefore,
will have the most elements when the sets
and
are disjoint, i.e.
, which would mean the most we can can in this case would be
(Note that
denotes the cardinality of the set
.)
Answer:
x = 11
Step-by-step explanation:
Given :
AB = x + 5
BC = 2(x - 3)
AB = BC
AB = BC
x + 5 = 2(x - 3)
x + 5 = 2x - 6
5 + 6 = 2x - x
11 = x
x = 11
happy to help :)
