Segment AB has length a and is divided by points P and Q into AP , PQ , and QB , such that AP = 2PQ = 2QB. A) Find the distance
1 answer:
Answer:
A)
B)
Step-by-step explanation:
AB has length a and is divided by points P and Q into AP , PQ , and QB , such that AP = 2PQ = 2QB
A) Therefore, AP = 2QB
QB = AP/2
The midpoint of QB = QB/2 = (AP/2)/2 = AP/4
AP = 2PQ, Therefore PQ = AP/2
Since the length of AB = a
AB = AP + PQ + QB = a
AP + AP/2 + AP/2 = a
AP + AP = a
2AP = a
AP = a/2
The distance between point A and the midpoint of segment QB = AP + PQ + QB/2 = AP + AP/2 + AP/4 = 7/4(AP)
But AP = a/2
Therefore The distance between point A and the midpoint of segment QB = 7/4(a/2)=
B)
the distance between the midpoints of segments AP and QB = AP/2 + PQ + QB/2 = AP/2 + AP/2 + AP/4 = 5/4(AP)
But AP = a/2
Therefore the distance between the midpoints of segments AP and QB = 5/4(AP) =
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Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
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Explanation:
Recall that tan(x) = sin(x)/cos(x).
The discontinuities occur whenever cos(x) is equal to zero.
Solving cos(x) = 0 will yield the locations when we have discontinuities.
This all applies to tan(x), but we want to work with tan(x/2) instead.
Simply replace x with x/2 and solve for x like so
cos(x/2) = 0
x/2 = arccos(0)
x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k
x = pi + 4pi*k or x = -pi + 4pi*k
Where k is any integer.
If we make a table of some example k values, then we'll find that we could get the following outputs:
- x = -3pi
- x = -pi
- x = pi
- x = 3pi
- x = 5pi
and so on. These are the odd multiples of pi.
So we can effectively condense those x equations into the single equation x = pi+k*pi
That equation is the same as x = (k+1)pi
The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).
