Identify the area of the polygon with vertices t(6,5), a(8,−1), s(4,−2), and k(−1,4).
1 answer:
Answer:
36.5 square units
Step-by-step explanation:
There are a number of ways to compute the area of a polygon from the coordinates of its vertices. There are algebraic formulas, geometric methods involving decomposing the figure, and an interesting method described by Pick's theorem.
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<h3>Geometry</h3>The figure can be enclosed by a rectangle that is 7 units high by 9 units wide. The actual figure is obtained by removing the triangular corners of this rectangle. The areas of those are given by the triangle area formula ...
A = 1/2bh
Working clockwise from lower left, the triangle base (horizontal) and height (vertical) dimensions are 5×6, 7×1, 2×6, 4×1. That means the area of the triangles is ...
A = 1/2(5×6 +7×1 +2×6 +4×1) = 1/2(30 +7 +12 +4) = 1/2(53) = 26.5
The area of the bounding rectangle is ...
A = bh = 9×7 = 63
So, the area of the figure is the difference ...
A = 63 -26.5 = 36.5 . . . square units
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The other methods of finding the area have been added as attachments.
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Answer:
50
Step-by-step explanation:
The figure represents a 30-60-90 special triangle
in this special right triangle the measure of sides lengths is represented by
a, a, and 2a
The side length that sees angle measure 30 is represented by a and here we see a = 50m
The height of the building, which sees angle measure 60, is equal to
50
Answer:
a) An alternating series is a whose terms are alternately positive and negative
b) An alternating series where bn = |an|, converges if
for all n, and
0
c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1
Step-by-step explanation:
<em>Part a</em>
An Alternating series is an infinite series given on these three possible general forms given by:
For all
The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.
<em>Part b</em>
An alternating series where bn = |an| converges if
for all n and
=0
Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:
<em>Theorem (Alternating series test)</em>
If a sequence of positive terms {bn} is monotonically decreasing and
<em><em>, then the alternating series
converges if:</em></em>
<em>i) </em>
<em>ii) </em>
then <em> converges</em>
<em>Proof</em>
For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then
for all n, and that implies that the series converges to a value, s.
And this complete the proof.
<em>Part c</em>
An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series
By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:
Rn = s - sn
Where represent the partial sum for the series and s the total for the sum.
Is important to notice that the size of the error is at most by the following theorem:
<em>Theorem (Alternating series sum estimation)</em>
<em>If is the sum of an alternating series that satisfies</em>
<em>i) </em>
<em>ii) </em>
Then then
<em>Proof</em>
In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1.
And this complete the proof.