He base of a pentagonal prism has an area of 2.4 m2. The prism has a volume of 4.8 m3. What is its height? Use sketches, formula
1 answer:
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Answer:
5.25 m
Step-by-step explanation:
A diagram can help you understand the question, and can give you a clue as to how to find the answer. A diagram is attached. The problem can be described as finding the sum of two vectors whose magnitude and direction are known.
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<h3>understanding the direction</h3>In navigation problems, direction angles are specified a couple of different ways. A <em>bearing</em> is usually an angle in the range [0°, 360°), <em>measured clockwise from north</em>. In land surveying and some other applications, a bearing may be specified as an angle east or west of a north-south line. In this problem we are given the bearing of the second leg of the walk as ...
N 35° E . . . . . . . 35° east of north
Occasionally, a non-standard bearing will be given in terms of an angle north or south of an east-west line. The same bearing could be specified as E 55° N, for example.
<h3>the two vectors</h3>A vector is a mathematical object that has both magnitude and direction. It is sometimes expressed as an ordered pair: (magnitude; direction angle). It can also be expressed using some other notations;
- magnitude∠direction
- magnitude <em>cis</em> direction
In the latter case, "cis" is an abbreviation for the sum cos(θ)+i·sin(θ), where θ is the direction angle.
Sometimes a semicolon is used in the polar coordinate ordered pair to distinguish the coordinates from (x, y) rectangular coordinates.
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The first leg of the walk is 3 meters due north. The angle from north is 0°, and the magnitude of the distance is 3 meters. We can express this vector in any of the ways described above. One convenient way is 3∠0°.
The second leg of the walk is 2.5 meters on a bearing 35° clockwise from north. This leg can be described by the vector 2.5∠35°.
<h3>vector sum</h3>The final position is the sum of these two changes in position:
3∠0° +2.5∠35°
Some calculators can compute this sum directly. The result from one such calculator is shown in the second attachment:
= 5.24760∠15.8582°
This tells you the magnitude of the distance from the original position is about 5.25 meters. (This value is also shown in the first attachment.)
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You may have noticed that adding two vectors often results in a triangle. The magnitude of the vector sum can also be found using the Law of Cosines to solve the triangle. For the triangle shown in the first attachment, the Law of Cosines formula can be written as ...
a² = b² +o² -2bo·cos(A) . . . . where A is the internal angle at A, 145°
Using the values we know, this becomes ...
a² = 3² +2.5² -2(3)(2.5)cos(145°) ≈ 27.5373
a = √27.5373 = 5.24760 . . . . meters
The distance from the original position is about 5.25 meters.
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<em>Additional comment</em>
The vector sum can also be calculated in terms of rectangular coordinates. Position A has rectangular coordinates (0, 3). The change in coordinates from A to B can be represented as 2.5(sin(35°), cos(35°)) ≈ (1.434, 2.048). Then the coordinates of B are ...
(0, 3) +(1.434, 2.048) = (1.434, 5.048)
The distance can be found using the Pythagorean theorem:
OB = √(1.434² +5.048²) ≈ 5.248
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that . Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that . In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and . So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.