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Pythagoras' Theorem works with Negative Length
When taking square roots of both sides of an equation, one should be careful not to turf out the negative result without first considering whether it has a true meaning. When using Pythagoras' Theorem, the last step is to take square roots. So, can we have a hypotenuse with length -5?
I taught Pythagoras' Theorem this week to my year 8 girls. Many are exceptional mathematicians for their age and it's very important they get into good habits early, so, when we took square roots at the end, I reminded them that there are two results, one positive and one negative, but we reject the negative because length can't be negative.
Or can it?
In the interactive diagram below, click and drag vertex B up and down.
Watch the numbers at the bottom. Due to the Theorem, the area of the green square is equal to the sum of the areas of the red squares, wherever vertex B is (as long as it's a right-angled triangle). It transpires that this is also true if you drag vertex B below vertex C, thus turning the triangle inside-out. The vertical length AC becomes negative, but when we square it we still get a positive area, and the Theorem holds.
Question: Why did the author remind his students to reject the negative result when teaching Pythagoras' Theorem? Answer: The author reminded his students to reject the negative result because length cannot be negative in the real world. | 677.169 | 1 |
Rather than doing a lot of math myself I though I'd first see if anyone
else has an answer.
Given a set of points (latitude and longitude), I need to calculate the
bounding polygon for these points (not just the bounding rectangle).
Can anyone point to an algorithm for this? I think I have a general idea
of how to do this, but rather than do the work, I thought I'd ask (yea,
I'm lazy today).
Question: Why did the user decide to ask for help instead of doing the math themselves? Answer: The user thought they had a general idea of how to do it but didn't want to do the work themselves, admitting to being lazy that day. | 677.169 | 1 |
In mathematics, the triple product is a product of three vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and, less often, the vector-valued vector triple product....
In physics, a pseudoscalar is a quantity that behaves like a scalar, except that it changes sign under a parity inversion such as improper rotations while a true scalar does not.The prototypical example of a pseudoscalar is the scalar triple product...
, a result which is more obvious if the calculation is framed as the exterior product of a vector and bivector. They generalises to other dimensions; in particular bivectors can be used to describe quantities like torque and angular momentum in two as well as three dimensions. Also, they closely match geometric intuition in a number of ways, as seen in the next section.
Geometric interpretation
As suggested by their name and that of the algebra, one of the attractions of bivectors is that they have a natural geometric interpretation. This can be described in any dimension but is best done in three where parallels can be drawn with more familiar objects, before being applied to higher dimensions. In two dimensions the geometric interpretation is trivial, as the space is two dimensional so has only one plane, and all bivectors are associated with it differing only by a scale factor.
All bivectors can be interpreted as planes, or more precisely as directed plane segments. In three dimensions there are three properties of a bivector that can be interpreted geometrically:
The arrangement of the plane in space, precisely the attitude of the plane (or alternately theIn vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change....
of the plane), is associated with the ratio of the bivector components. In particular the three basis bivectors, e23, e31 and e12, or scalar multiples of them, are associated with the yz-plane, xz-plane and xy-plane respectivelyof the plane segment. The area does not have a particular shape so any shape can be used. It can even be represented in other ways, such as by an angular measure. But if the vectors are interpreted as lengths the bivector is usually interpreted as an area with the same units, as follows.
Like the direction of a vector a plane associated with a bivector has a direction, a circulation or a sense of rotation in the plane, which takes two values seen as clockwise and counterclockwise when viewed from viewpoint not in the plane. This is associated with a change of sign in the bivector, that is if the direction is reversed the bivector is negated. Alternately if two bivectors have the same attitude and magnitude but opposite directions then one is the negative of the other.
In three dimensions all bivectors can be generated by the exterior product of two vectors. If the bivector then the magnitude of B is
Question: What is the geometric interpretation of bivectors in two dimensions? Answer: All bivectors can be interpreted as planes, or more precisely as directed plane segments | 677.169 | 1 |
Measurement On MapsMeasurement of Distance. - The shortest distance between two places on the surface of a globe is represented by the arc of a great circle. If the two places are upon the same meridian or upon the equator the exact distance separating them is to be found by reference to a table giving the lengths of arcs of a meridian and of the equator. In all other cases recourse must be had to a map, a globe or mathematical formula. Measurements made on a topographical map yield the most satisfactory results. Even a general map may be trusted, as long as we keep within ten degrees of its centre. In the case of more considerable distances, however, a globe of suitable size should be consulted, or - and this seems preferable - they should be calculated by the rules of spherical trigonometry. The problem then resolves itself in the solution of a spherical triangle.
In the formulae which follow we suppose 1 and l' to represent the latitudes, a and b the co-latitudes (90° - 1 or 90° - l'), and t the difference in longitude between them or the meridian distance, whilst D is the distance required.
If both places have the same latitude we have to deal with an isosceles triangle, of which two sides and the included angle are given. This triangle, for the convenience of calculation, we divide into two right-angled triangles. Then we have sin 2 D =sin a sin zt, and since sin a=sin (90°-1) = cos1, it follows that sin ID = cos 1 sin it. If the latitudes differ, we have to solve an oblique-angled spherical triangle, of which two sides and the included angle are given. Thus, cos D - cos a cos b cos sin a sin b cos D = cos a cos b + sin a sin b cos t = sin 1 sin l' + cos 1 cos l cos t. In order to adapt this formula to logarithms, we introduce a subsidiary angle p, such that cot p = cot l cos t; we then have cos D = sin 1 cos( - p) I sin p. In the above formulae our earth is assumed to be a sphere, but when calculating and reducing to the sea-level, a base-line, or the side of a primary triangulation, account must be taken of the spheroidal shape of the earth and of the elevation above the sealevel. The error due to the neglect of the former would at most amount to 1%, while a reduction to the mean level of the sea necessitates but a trifling reduction, amounting, in the case of a base-line 300,000 metres in length, measured on a plateau of 3700 metres (12,000 ft.) in height, to 57 metres only.
Question: What is the formula to calculate the distance (D) when both places have the same latitude? Answer: sin 2D = sin a sin t.
Question: What are the variables in the given formulas for calculating distance? Answer: 1 and l' represent the latitudes, a and b the co-latitudes, and t the difference in longitude.
Question: What is the best method to measure distances on a topographical map? Answer: Using a topographical map yields the most satisfactory results. | 677.169 | 1 |
In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on.
The origin of the word "quadrilateral" is the two Latin words quadri, a variant of four, and latus, meaning "side."
Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave.
Euler diagram of some types of quadrilaterals. (UK) denotes British English and (US) denotes American English.
A parallelogram is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other. Parallelograms also include the square, rectangle, rhombus and rhomboid.
Rhombus or rhomb: all four sides are of equal length. An equivalent condition is that the diagonals perpendicularly bisect each other. An informal description is "a pushed-over square" (including a square).
Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushed-over rectangle with no right angles."
Rectangle: all four angles are right angles. An equivalent condition is that the diagonals bisect each other and are equal in length. Informally: "a box or oblong" (including a square).
Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. An equivalent condition is that opposite sides are parallel (a square is a parallelogram), that the diagonals perpendicularly bisect each other, and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (four equal sides and four equal angles).
Oblong: a term sometimes used to denote a rectangle which has unequal adjacent sides (i.e. a rectangle that is not a square).
Kite: two pairs of adjacent sides are of equal length. This implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. It also implies that the diagonals are perpendicular.
Trapezium (NAm.): no sides are parallel. (In British English this would be called an irregular quadrilateral, and was once called a trapezoid.)
Isosceles trapezoid (NAm.) or isosceles trapezium (Brit.): one pair of opposite sides are parallel and the base angles are equal in measure. Alternative definitions are a quadrilateral with an axis of symmetry bisecting one pair of opposite sides, or a trapezoid with diagonals of equal length.
An equilic quadrilateral has two opposite equal sides that, when extended, meet at 60°.
Question: What is the minimum number of sides a quadrilateral has? Answer: 4 | 677.169 | 1 |
FIND HYPOTENUSE
Alice and Bob are returning home from their math class. The teacher was discussing Pythagoras' Theorem in class. All of a suddden Alice came up with an intriguing question. He asked Alice what would be the hypotenuse of least length of exactly N distinct right triangles. Alice who was not quite as sharp as Bob said, "I don't quite understand the question. What do you mean by N distinct triangles?" Bob explained, "Suppose N = 2 and you choose a certain hypotenuse. You should be able to construct exactly 2 distinct right triangles with that hypotenuse such that all sides(including hypotenuse) are integral. The answer you must give me is the smallest length the hypotenuse can have. In this case it turns out to be 25 and the sides are (7,24) (15,20)".
Now As usual your task is to help Alice with the solution.
Input
First line will contain T, the number of test cases. Then T( T<1000) lines follow on each line the number N (1<=N<=3000) will be given.
Output
Output must contain exactly T lines, the minimum length of the hypotenuse for each test case. Note: Output may not fit in a 64 bit integer
Question: What is the term used to describe the longest side of a right triangle? Answer: Hypotenuse | 677.169 | 1 |
as far as i know....it was 8 pi for the circle one, four or more triangles for the line AB one, 1/3 for the 3x=3y+1, 65 for the test scores with 85 as the median, and 1/5 for the bo girl one...that seems to be the only ones ppl re arguing....
about the 8pi....thats definitely right...i guess its how u interpret it...but u have to really look at the question. they said that the area of the LARGEST circle was 16pi...so when it says the area INSIDE the largest circle...that means that u dont include that circle...so u subtract 1pi (the smallest circle) from 9pi..the area inside the largest circle....yea
The definition states "included or inclosed IN ANYTHING" ('anything' being the circumference of the largest circle OR the area of the circle, which is - in part - what makes this question controversial). A circle is the locus of all points equidistant from a central point (essentially, its CIRCUMFERENCE is what constitutes the circle). There are, however, arguments against this regarding SIDES of a circle (or lack thereof) as they/it relate to the "inside" of the circle, which which is why I've come to the conclusion that this question was simply unreasonable. Such is the world of ETS I suppose...
Question: What is the median test score mentioned in the text? Answer: 85 | 677.169 | 1 |
a triangle with two right angles because if you start with one side of the
triangle across the bottom, the other two sides go straight up. They're
parallel, so they can't possibly ever meet, so you can't get it to be a
triangle."
When students subdivide, combine, and transform shapes, they are investigating
relationships among shapes. For example, a fourth-grade class might investigate
the relationship between a rectangle and a nonrectangular parallelogram
with equal bases and heights (see fig. 5.11) by asking, "Does one of these
shapes have a larger area than the other?" One student might cut the region
formed by the parallelogram as shown in figure 5.11 and then rearrange
the pieces so that the parallelogram visually matches the rectangle. This
work can lead to developing a general conjecture about the relationship
between the areas of rectangles and parallelograms with the same base
and height. The notion that shapes that look different can have equal
areas is a powerful one that leads eventually to the development of general
methods (formulas) for finding the area of a particular shape, such as
a parallelogram. In this investigation, students are building their ideas
about the properties of classes of shapes, formulating conjectures about
geometric relationships, exploring how geometry and measurement are related,
and investigating the shapes with equal area.
Fig. 5.11. The relationship between
the areas of a rectangle and a nonrectangular parallelogram
with equal bases and heights
An understanding of congruence
and similarity will develop as students explore shapes that in some way
look alike. They should come to understand congruent shapes as those that
exactly match and similar shapes as those that are related by "magnifying"
or "shrinking." For example, consider the following problem involving
the creation of shapes with a particular set of properties:
Make a triangle with one right angle and two sides of equal length. Can you make
more than one triangle with this set of properties? If so, what is the
relationship of the triangles to one another?
As students make triangles
with the stipulated properties (see fig. 5.12), they will see that although
these triangles share a common set of characteristics (one right angle
and a pair of sides of equal length), they are not all the same size.
However, they are all related in that they look alike; that is, one is
just a smaller or larger version of the other. The triangles are similar.
Although students will not develop a full understanding of similarity
until the middle grades, when they focus on proportionality, in grades
3–5 they can begin to think about similarity in terms of figures
that are related by the transformations of magnifying or shrinking.
Fig. 5.12. Right triangles with
two sides of equal length
p.
166
When discussing shapes, students in grades 3–5 should be expanding their
mathematical vocabulary by hearing terms used repeatedly in context. As
Question: What is the relationship between the triangles made with one right angle and two sides of equal length? Answer: They are similar, meaning one is just a smaller or larger version of the other.
Question: Can a triangle have two right angles? Answer: No, a triangle cannot have two right angles. | 677.169 | 1 |
on Lesson F5.2. So if you get stuck on Lesson F5.2, send in a question and go on
to
Lesson F5.3. Students who are working most efficiently in this course will
probably
work on Lesson F5.2 while they are doing about three or four more lessons.
4. You won't be tested over all the parts of Lesson F5.3.
Read the notes here carefully.
Section 1:
1. Although there are many words and concepts in this section, the only one that
students
usually have to work at remembering is the difference between complementary and
supplementary angles.
Section 2:
1. It is most important not to get "stuck" on this lesson. Some of the
problems use the
formulas in a rather straightforward fashion. Other problems require two
different
insights – maybe two different formulas or maybe seeing one side as a part of
two
different figures. Those latter problems are harder for most students. Go
through one or
two of the harder ones. If you're having trouble with them, write down a
question and send it in and then skip the rest of the two-insight problems at first.
Just do the one-formula
problems in the rest of the software lesson. Then do the paper homework for the
more straightforward problems.
2. In this section, you will use formulas quite a lot. First, look at the study
sheet to
de termine which of them you must memorize. As you go through the section, write
those
on your study sheet. Try to think of a way to remember them that makes sense to
you.
3. When you are using a formula to do a computation, be sure to write the
formula, with the
variables. Then under it, substitute the numbers for the variables. Only after
that should
you do the calculation and write your answer.
4. Often you will obtain answers that have long decimals and look as if they
should be
rounded. Keep at least two or three decimal places in your final answer.
Sometimes formulas with the quantity "pi" in them will lead to "back of the book
answers" with "pi" and other times they will be given as decimals. Both can be
correct.
For example 4*pi = 12.56. Sometimes the problem will specify that you must give
the
answer as a decimal number. Other times it will not specify how the answer
should be
given and then either is fine.
5. Be sure to include the units (dollars, millimeters, square feet , etc.) in
your answers. If the
answer in the back of the book has units, yours should also.
6. Back to the two-insight problems. (Example: a problem asking for the
area of the
shaded region, where part of the figure is shaded and part isn't.) If you have
sent in a
question, then I will be working with you individually about how much of this
you should
Question: What is the first step when using a formula to do a computation? Answer: Write the formula, with the variables.
Question: What should you do if you get stuck on Lesson F5.2? Answer: Send in a question and go on to Lesson F5.3. | 677.169 | 1 |
What is the General Equation of a Circle?
A line which is formed by a close loop or we can say every Point on a Circle which is situated fixed distance from the center is known as circle or we can say it is a type of line which bends around its end point and arranged in such a way that it appears exactly a circular shape that means all the points of a line are at equal distance from the center of a circle. There is small difference in case of a circle and a disk. There is no area defined for the circle but a disk defines the area. There are some properties of a circle which defines:
Center – a point which is situated inside the circle and all the point which are on the circle are equally distance from the center.
Radius – it is defined as the distance which is measured form the center to any point on a circle. Or we can say it is the half of the Diameter.
Diameter – the distance this is inside the circle, and a Chord which passes through the center of a circle. Or it is twice the radius of a circle.
Circumference – the distance which is around or outside the circle is circumference of a circle. The meaning of word circumference is curved lines which reach around the circle.
If we have radius of a circle then the circumference of a circle using formula:
Circumference = 2⊼R,
'R' is the radius of the circle and the value of '⊼' is 3.142.
Area – a circle has no area. Area means the surface which is enclosed by a circle.
Chord – a line in a circle which join two points.
Tangent – it is also a line which is passes through a circle and touching it one point only.
Secant – a line which intersect a circle in two half.
Now we see what is the general equation of a circle?
The general form of equation is:
x2 + y2 + Cx + Dy + E = 0.
Question: What is a tangent to a circle? Answer: A tangent to a circle is a line that touches the circle at exactly one point. | 677.169 | 1 |
So, one is, these are called spherical coordinates because if you fix the value of rho, then you are moving on a sphere centered at the origin. OK, so let's look at what happens on a sphere centered at the origin, so, with equation rho equals a. Well, then phi measures how far south you are going, measures the distance from the North Pole. So, if you've learned about latitude and longitude in geography, well, phi and theta you can think of as latitude and longitude except with slightly different conventions. OK, so, phi is more or less the same thing as latitude in the sense that it measures how far north or south you are. The only difference is in geography, latitude is zero on the equator and becomes something north, something south, depending on how far you go from the equator.
Here, you measure a latitude starting from the North Pole which is zero, increasing all the way to the South Pole, which is at pi. And, theta or you can think of as longitude, which measures how far you are east or west. So, the Greenwich Meridian would be here, now, the one on the x axis. That's the one you use as the origin for longitude, OK? Now, if you don't like geography, here's another way to think about it. So --
Let's start again from cylindrical coordinates, which hopefully you're kind of comfortable with now. OK, so you know about cylindrical coordinates where we have the z coordinates stay z, and the xy plane we do R and theta polar coordinates. And now, let's think about what happens when you look at just one of these vertical planes containing the z axis. So, you have the z axis, and then you have the direction away from the z axis, which I will call r, just because that's what r measures. Of course, r goes all around the z axis, but I'm just doing a slice through one of these vertical half planes, fixing the value of theta.
Then, r of course is a polar coordinate seen from the point of view of the xy plane. But here, it looks more like you have rectangular coordinates again. So the idea of spherical coordinate is you're going to polar coordinates again in the rz plane. OK, so if I have a point here, then rho will be the distance from the origin. And phi will be the angle, except it's measured from the positive z axis, not from the horizontal axis.
But, the idea in here, see, let me put that between quotes because I'm not sure how correct that is, but in a way, you can think of this as polar coordinates in the rz plane. So, in particular, that's the key to understanding how to switch between spherical coordinates and cylindrical coordinates, and then all the way to x, y, z if you want, right, because this picture here tells us how to express z and r in terms of rho and phi.
Question: What is the distance from the origin that rho represents in spherical coordinates? Answer: The distance from the origin
Question: What is the key to understanding how to switch between spherical, cylindrical, and Cartesian coordinates? Answer: The picture that shows how to express z and r in terms of rho and phi.
Question: True or False: In spherical coordinates, the angle phi increases as you move north from the equator. Answer: False. In spherical coordinates, phi increases as you move south from the North Pole.
Question: How does the angle phi in spherical coordinates differ from the angle in polar coordinates? Answer: Phi is measured from the positive z axis, not from the horizontal axis. | 677.169 | 1 |
AB = 9 units
BC = 4 units
AC will have to be equal to 9 units. It cannot be 4 units because the length of the sum of two sides of a triangle must be greater than the third side and the problem states that the traingle is an isosceles triangle.
Re: What is the perimeter of isosceles triangle ABC? (1) The [#permalink]
19 Jul 2012, 02:50
monirjewel wrote:
(1) AB = 9. We only know the length of one side. Not sufficient. (2) BC = 4. We only know the length of one side. Not sufficient.
(1)+(2) If AB=AC=9 and BC=4 then the perimeter would be 9+9+4=22 but if AB=9 and AC=BC=4 then the perimeter would be 9+4+4=17. Not sufficient.
Answer: E.
Answer to this question should be C, not E.
What is the perimeter of isosceles triangle ABC?
(1) The length of side AB is 9. Clearly insufficient.
(2) The length of side BC is 4. Clearly insufficient.
(1)+(2) We know the lengths of the two sides of isosceles triangle ABC: AB=9 and BC=4, hence the length of AC is either 4 or 9. Relationship of the Sides of a Triangle: The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Now, according to this, AC cannot equal to 4, because in this case the length of AB would be greater than the sum of the other two sides, AC and BC, (AB=9>AC+BC=4+4=8), hence AC=9 and P=9+9+4=22. Sufficient.
Question: What is the length of side AB in triangle ABC? Answer: 9 units
Question: Is the length of side BC equal to the length of side AB? Answer: No | 677.169 | 1 |
A camper is off picking berries. He is strolling along, bucket full of berries in hand, when he sees that his tent is on fire. Luckily, he is near a river, so he can run to the river, fill his bucket with water, and run to the tent to put the fire out. The question: Where should he go along the bank of the river should he fill the bucket to make his trip as short as possible and have the greates chance of saving his tent?
*In starting this problem, I first reflected X over the river and called that point Z. That made the distance from X to the river equal to the distance from Z to the river. I then connected point Z to point Y because the shortest distance between two points is a straight line. In doing that, I found the point at which ZY intersects the river and called that point M. I then found out, that with the river being the perpendictular bisector of XZ, that XM equals ZM by the perpendictular bisector theorem. Since the shortest path between two points is a straight line, and segment MY+ZM =ZY, I just added segment MY to segment XM. The path of XM+MY equals the path of ZM+MY. So, in answering the question, point M is where the man should run to the river.
* This answer is based upon the river being straight and the area obstacle free. Nick Szmyd Grade 10 Shaler Area High School
******************************
From: bipin mujumdar <[email protected]>
At the start of the problem I assumed the following: 1) The river was perfectly straight and did not curve around the tent 2) The problem wished for the shortest distance not necessarily the shortest time 3) The river was as narrow as a simple line 4) The man ran in perfectly straight lines To find the point to fill the bucket I reflected the point where the man started over the river which I considered a line. In doing this, the line becomes the perpendicular bisector of the segment formed by the original and reflected point.Since the shortest distance between two points is a straight line, I drew in the line from the reflected point to the burning tent. I concluded that the intersection of that line and the river was the point at which the man should fill his bucket. My reasoning was that segments drawn from the original man with bucket to the intersection were equal to the reflected point to the intersection because of the perpendicular bisector theorem. I finally reasoned that the path taken would equal the straight line from the reflected point to the tent were equal because the sides were equal and if you add Z to two equal number, the results will also be equal. In this case Z is equal to the distance between the intersection and the tent. Since both of the final line was equal and one of them was a straight line, the intersecting point was where the man should fill his bucket.
Question: What is the nearest water source to the camper? Answer: A river.
Question: What is the relationship between the distances XM and ZM in the solution? Answer: XM equals ZM due to the perpendicular bisector theorem. | 677.169 | 1 |
Bring tessellations and polyhedra together for a hands-on learning experience rich in math content. Nets for 24 different polyhedra, including all of the Platonic and Archidedean solids, are presented both with and without tessellations applied to t..
Challenge the spatial reasoning of even the most gifted students. This puzzle rearranges 23 polygonal pieces to form 4 different regular polygons - an equilateral triangle, a square, a hexagon, and an octagon. Magnetic foam pieces are big enough to ..
Prices listed are U.S. Domestic prices only and apply to orders shipped within the United States. Orders from outside the
United States may be charged additional distributor, customs, and shipping charges.
Question: Which of the following is NOT a regular polygon that can be formed: a) Pentagon, b) Square, c) Hexagon, d) Octagon? Answer: a) Pentagon | 677.169 | 1 |
The measure of our angle is 24 degrees greater than the measure of its complement. If we take that apart:
The measure of our angle (I defined that to be "m")
is (means =)
24 degrees greater than (more than; this is addition)
the measure of its complement (I defined that to be "c")
So really, we have
We also know that an angle plus its complement is 90 by definition (so, for example, 10 and 80 are complements because they add to 90). So .
But we know that from above. That means that instead of saying m in our second equation, we can say 24+c. So instead of , we have . All I did was exchanged m and 24+c. Now, we solve:
The measure of the angle's complement is 33 degrees.
At the beginning, you told me that our angle is 24 degrees greater than its complement, so it's 24+33=57 degrees.
To check, our angle plus its complement should equal 90: 57 + 33 = 90. It works :) The two angles are 57 and 33 degrees.
Question: How is the measure of the angle related to the measure of its complement? Answer: The angle is 24 degrees greater than its complement | 677.169 | 1 |
The compasses are opened to a little more than half the length of the line - this distance is kept throughout the construction. The compass point is placed first on one end of the line and then on the other, and arcs drawn to meet above and below the line. A perpendicular line can be drawn through the meeting points:
The line CD bisects the line AB at right angles.
Construction of parallel lines a specified distance apart
The required distance is measured (A) and point P marked anywhere on the line. The compasses are opened to the length AP - this distance is kept throughout the construction. With the compass point placed on P, an arc is drawn to cut the line (point Q). Then with the compass point first on Q and then on A, two arcs are drawn to cross at a point. The line drawn through A and the meeting point is parallel to the original line:
Construction of the bisector of any angle
The compasses are opened to a length less than that of either arm of the angle - this distance is kept throughout the construction. With the compass point on the point of the angle, arcs are drawn to cut each arm (points A and B). Then with the compass point first on A and then on B, two arcs are drawn to cross at a point. The line drawn from this meeting point to the point of the angle cuts the original angle into two equal angles:
Constructing triangles
Triangles may be constructed from a variety of given measurements using a ruler, compass, and protractor.
First AB is drawn, 8 cm long. Then with the compasses centred on A, radius 6 cm, an arc is drawn. The compasses are extended to 7 cm and, centring on B, another arc is drawn to cross through the first arc at C. AC and BC are then joined to create the triangle ABC:
Given one side and two angles (ASA): triangle ABC can be constructed knowing that AB = 6 cm, angle A = 36°, and angle B = 59°.
The line AB is drawn, 6 cm long. A protractor is used to measure the angles 36° at A and 59° at B - these lines are continued to meet at C, forming the triangle ABC:
Given two sides and the angle between them (SAS): triangle ABC can be drawn knowing that AB = 5 cm, angle A = 45°, and AC = 4 cm.
Line AB is drawn, 5 cm long. An angle of 36° is measured at A, and the arm extended to a length of 4 cm to give point C. Points B and C are joined to form the triangle:
Given two sides and a non-included angle (SSA): triangle ABC can be drawn knowing that AB = 6 cm, BC = 7 cm, and angle A = 50°.
Question: What is the first step in constructing a line that bisects another line at right angles? Answer: Open the compasses to a little more than half the length of the line and draw arcs to meet above and below the line.
Question: What is the length of the compass opening for constructing the bisector of any angle? Answer: A length less than that of either arm of the angle.
Question: In the third triangle construction example (given two sides and a non-included angle), what is the length of AB? Answer: 6 cm | 677.169 | 1 |
You can put this solution on YOUR website! In a right triangle, the two acute angles are complementary,
meaning that their measures add up to .
The way sine and cosine were defined based on a right triangle,
sine of an angle is the cosine of the complement.
(It works if you define sine and cosine based on the unit circle too).
Something similar happens with tangent and cotangent,
and with secant and cosecant.
For all the trigonometric functions the function of an angle
equals the cofunction of the complement.
Those are the cofunction identities.
Question: What are the two acute angles in a right triangle called? Answer: Complementary angles | 677.169 | 1 |
Solutionss plzzQ1.2 Circles intersect each oder .The Circumfrence of each passes thro centre of d other.What part of the area of each circle is the area of thier intersecting region? Ans:(2/3) -(root3/2 pie) Q2.An Equilateral Triangle side 16 cms has circle inscribed in it.There is anoder equilateral tria...
Hi... with respect to figure(attached) AE=DR 3^2-x^2=2^2-y^2 similerly EB= 4^2-x^2 EB=RC So PC^2=y^2+RC^2 =y^2+4^2-x^2 by using 1st one PC=squareroot(11) So D is correct answer. Let me know if you get it...
Cyclic quadrilateral thry plzz.............And how to calculate area of a Pentagon any specific formula for area of a polygon? Moreover how to calculate the side of a hexagon in which circle is inscribed
New Question:: In triangle ABC,D is on side BC such that BD: DC is 2:3 and E is on side AC such that AE:EC is 1:2.Lines AD and BE intersect at F. Find the ratio of area of triangle AFE and BFD. Can any body provide the solution for this question??...
I did not actually read the question properly , I though they are squares. :splat:
Okay . See The outer surface area for discreet smaller piecex is 1/4*4*pi*r^2+3*1/4pi*r^2 It has one curved surface (the first part) and three plane surface each same as 1/4th of a circle . So,
For r=4, 16pi+12pi=26 pi for the bigger one it's 36pi+27pi=63 pi 89 pi But one plane side of smaller piece is attached to the bigger one so 1/4*pi*r^2=4pi wasted from each one so 89-8 =81 pi
I could reach at 81 pi so I think may be I did not get the picture correctly) = 2*pi*r^2.
So, as u see in the above solution, we are subtracting the areas of one of the semicircles of the smaller sphere. This is so, because, smaller sphere is joined with the bigger sphere, i.e., one of the semicircles of smaller sphere is joined to that of the bigger sphere. Now when we fuse them the one surface of the...
Now when we fuse them the one surface of the smaller one disappeared which is 1/2*pi*4^2 and same amount of surface from the bigger one also disappears as it got attached to the smaller part. So 8pi+8pi out of 104 pi Answer=88 pi
I did a blunder first taking as circle and then taking 1/8th for the planed surface
Question: What is the side length of the equilateral triangle that has a circle inscribed in it? Answer: 16 cm
Question: What is the outer surface area of a discrete smaller piece with radius r? Answer: 1/4 4 π r^2 + 3 1/4 π r^2 | 677.169 | 1 |
Checkpoint
- Course 2, Unit 6
Geometric Form and Its Function (teachers
may just refer to this as "notes") of any points they need to remember,
adding illustrative examples as needed. If your student is having difficulty
with any investigation in this unit, this Checkpoint and the accompanying
answers may help you recall the concepts involved, and give you the big
picture of what the entire unit is about. If your student has completed
the unit, then a version of this should be in his or her notes or toolkit.
Students should also have Technology Tips in their toolkits, which may
be useful for this unit.
Possible
Responses to Unit Summary Checkpoint
In this unit, students studied how mechanisms work
and how their function is related directly to the form or shape of
the
mechanism. They also investigated how some patterns of periodic change
could be modeled. The bolded words are vocabulary
and concepts your student should be familiar
with.
a.
What characteristics of a
parallelogram make the shape widely useful as a linkage? A parallelogram
is flexible, the opposite sides remain parallel and congruent, but
the angles can change. This permits a parallelogram linkage to
drive useful objects like wheels, which make complete rotations,
or wipers,
which turn through a prescribed angle.
b.
Two plane shapes are similar
with a scale factor k. How are the lengths of corresponding
segments related? How are measures of corresponding angles related?
How are areas of corresponding regions related? The lengths of the segments of the second shape are k times the lengths
of the segments of the first plane shape. Measures of corresponding angles
are equal. The area of the second shape is k2 times the area of
the first shape. (See Quick Summary - Course
2, Unit 6 for an example.)
c.
Define the sine, cosine,
and tangent of the acute angle of a right triangle. How can these
ratios be used to determine lengths that can not be measured directly?
How can these ratios be used to determine angle measures that cannot
be measured directly? If C is an acute angle in a right angled triangle, then C). (See Quick
Summary - Course 2, Unit 6 for specific
example.)
Suppose we want to find the length of the leg opposite a known angle,
then we rewrite the sine ratio so that we have: (length of leg opposite
angle C) = (length of hypotenuse)(sin C). If we know
the angle C and the length of the hypotenuse, we can find the length
of the unknown side.
Suppose we want to find the length of the side adjacent to a known angle,
then we rewrite the cosine ratio so that we have: (length of
leg adjacent to angle C) = (length of hypotenuse)(cos C).
If we know the angle C and the
length of the hypotenuse, we can find the length of the unknown
side.
If we do not know the length of the hypotenuse, then the tangent ratio could be used.
Question: If two plane shapes are similar with a scale factor of k, how are the areas of corresponding regions related? Answer: The area of the second shape is k^2 times the area of the first shape.
Question: How can the cosine ratio be used to determine a length that cannot be measured directly? Answer: The cosine ratio can be used to find the length of the leg adjacent to a known angle, given the length of the hypotenuse.
Question: Which shape is mentioned as being widely useful due to its flexibility and ability to drive objects like wheels or wipers? Answer: A parallelogram | 677.169 | 1 |
Symmetry
This tiling represents a hyperbolic kaleidoscope of 6 mirrors defining a regular hexagon fundamental domain. This symmetry by orbifold notation is called *222222 with 6 order-2 mirror intersections. In Coxeter notation can be represented as [1+,6,1+,4] (as 3*22), removing two of three mirrors (passing through the hexagon center, leaving an order-3 gyration point in the center of the hexagon) in the [6,4] symmetry. Adding a bisecting mirror through 2 vertices of a hexagonal fundamental domain defines a trapezohedral *3322 symmetry. Adding 3 bisecting mirrors through the vertices defines *443 symmetry. Adding 3 bisecting mirrors through the edge defines *3222 symmetry.
The kaleidoscopic domains can be seen as bicolored hexagonal tiling, representing mirror images of the fundamental domain. This coloring represents the uniform tiling t1{6,6}, a quasiregular tiling and it can be called a hexahexagonal tiling.
This tiling is also topologically related as a part of sequence of regular polyhedra and tilings with four faces per vertex, starting with the octahedron, with Schläfli symbol {n,4}, and Coxeter diagram , with n progressing to infinity.
Question: What happens when two of the three mirrors passing through the hexagon center are removed? Answer: An order-3 gyration point is left in the center of the hexagon. | 677.169 | 1 |
With your child, explore your house for symmetrical designs—things that have equal sides. Ask your child how many she can find. Tell her to look at wallpaper, floor tiles, pictures, bedspreads and appliances.
Have your child print the alphabet. Then ask her to find a letter that has only one line of symmetry—only one way to be divided in half. (B has one—the line is across the middle.) Ask her to find a letter that has two lines of symmetry—two ways to be divided in half. (H has two—the lines are across the middle and down the center.) Ask which letters look the same when they're turned upside down? (H, I, N, O, S, X and Z.)
Question: Which letter has only one line of symmetry? Answer: B | 677.169 | 1 |
same measure or same length.
Notice I was able to write this definition
of a parallelogram using three words
that I've already previously defined
and there's no other counter-example
I could draw or come up with that would
make this not apply to a parallelogram.
So keep that in mind when you're writing
good definitions and it will help you
even on your test and quizzes.
Question: What is the author's confidence in the definition's accuracy? Answer: High, as they state "there's no other counter-example I could draw or come up with that would make this not apply to a parallelogram." | 677.169 | 1 |
one sees that the bisected exterior angles at A and B have measure and .Let D be the point
of intersection of the bisector at A and side BC extended.Then by the condition of the problem ABD is
an isosceles triangle with AD=AB.Angle
ABD is supplementary to angle B so its measure is 180-b.Thus angle ADB is also 180-b (by the
property of base angles of isosceles triangles).Since the sum of the measures of the angles of a triangle is 180
we have 2(180-b) + angle DAB =180.Thus
angle DAB has measure 2b-180.This
angle is one-half of a supplement to angle A hence 2b-180 = .This yields the
equation a + 4b = 540.Very similarly
using the other angle bisector we develop the equation .This gives us a
system of three equations in three unknowns:
Question: What is the measure of angle ADB? Answer: 180 - b | 677.169 | 1 |
Geometry Which Statement is True? Which statement is true? All quadrilaterals are rectangles All quadrilaterals are squares All rectangles are quadrilaterals All quadrilaterals are parallelograms Thank You :)
Tuesday, January 10, 2012 at 10:41am by Sarah
geometry Check these quadrilaterals. Which do you think fit your description?
Thursday, June 16, 2011 at 4:35pm by Ms. Sue
geometry trapezoid In google type : Only one pair of opposite sides is parallel When you see lis of resuts click on : Quadrilaterals - Square, Rectangle, Rhombus, Trapezoid ...
Monday, March 19, 2012 at 12:52am by Bosnian
math Which quadrilaterals have a)perpendicular diagonals b)opposite angles congruent? c) two pairs of parallel sides? d) a pair of congruent, adjacent sides? My answers a)rhombus and square b)rhombus, square, and parallelogram c)parallelogram, rectangle, and square d) at first I ...
Tuesday, December 7, 2010 at 2:06pm by m
math All squares have four 90-degree angles.
Tuesday, January 3, 2012 at 7:41pm by Ms. Sue
Math P = 2L + 2W How many sides does a rhombus have?
Thursday, April 7, 2011 at 2:29pm by Ms. Sue
quadrilaterals The sum of the angles = 360 paralleograms have pairs of equal angles. You have 110 degrees so far. Now can you find the other two angles?
Wednesday, February 20, 2013 at 9:00pm by JJ
Math How do two letters and an ordered pair represent a line? (as in LM(0,9)) An ordered pair represents a point. Which point on the line does the ordered pair represent? To be congruent, two lines must have the same length. L must be the same distance from M as N is.
Saturday, March 26, 2011 at 10:29am by drwls
Math I have seen this question several times in the last few days. I agree with drwls that it is too ambiguous. If anything goes, then this is a monster question, since there are so many cases to consider. e.g. for a 5 digit number... 1. all digits are different 2. a pair of 1'...
Friday, January 6, 2012 at 12:02am by Reiny
Question: Which quadrilaterals have opposite angles that are congruent? Answer: Rhombus, square, and parallelogram.
Question: How many pairs of parallel sides does a rectangle have? Answer: 2 | 677.169 | 1 |
Saturday, December 8, 2007 at 1:53pm by Matt
Chemistry I must confess that I have no inkling of what "pair l" means although I understand the example you hve give but don't see the connection since the word "pair" is never used in the example. For n = 3 there will be 3 possible values for l (0,1,2) and ...
Tuesday, November 9, 2010 at 8:17pm by DrBob222
CHEM What are the expected bond angles in ICl4+ (or) ICl_4^+ ?? I think that I have the stucture drawn correcty: I is the central atom with one lone pair above it. I then have the Cl's with 6dots around them & then bonded, singly, to the "I". Two "Cl's&...
Monday, November 26, 2007 at 8:46pm by K
MATH Given the GCF or LCM, what else do you know about each pair of numbers? a) Two numbers have GCF of 2. b) Two numbers have an LCM of 2. Please and thankyou!
Monday, September 13, 2010 at 4:45pm by Vanessa
math what am I? I am a polygon and I have two right angles and one pair of parallel sides. Is it a rhombus? Please help.
Tuesday, December 4, 2007 at 5:37pm by Rachel
English Since "It" is in the answers for #1, I'd make these the answers: It's a pair of googles. It's a pair of trousers. It's a pair of scissors.
Sunday, November 13, 2011 at 6:30pm by Writeacher
chemistry It is most difficult, if not impossible, to draw Lewis dot structures on the boards; however, here is a site that will show you how it is done. Replace the single bonds with a pair of electrons and the double bonds with two pairs of ...
Monday, September 26, 2011 at 12:45pm by DrBob222
algebra Mrs. Stewart has 8 pair of sandals, 6 pairs of pants and 12 t-shirts. While getting ready to go out for ice-cream, she exclaims, "I have nothing to wear!" Mr. Stewart respond by telling her exactly how many different outfits she could wear if each outfit considered ...
Sunday, January 18, 2009 at 6:01pm by angela
MATHS There is no question here just instructions to draw. If you need to do this on paper then use a pair of compasses. Draw AB as a 6 cm line. Set the pair of compasses to 2.5 cm and mark C on the line by using the pair of compasses (point at one end of the line and mark C with ...
Question: What is the difficulty mentioned by Writeacher in drawing Lewis dot structures? Answer: Writeacher mentions that it is most difficult, if not impossible, to draw Lewis dot structures on a board due to the need to represent bonds with pairs of electrons.
Question: How many different outfits can Mrs. Stewart wear according to Mr. Stewart? Answer: Mr. Stewart tells Mrs. Stewart she can wear 1,152 different outfits, calculated by multiplying the number of sandals, pants, and t-shirts (8 6 12).
Question: What shape is Vanessa referring to in her message? Answer: Vanessa is referring to a rectangle. A polygon with two right angles and one pair of parallel sides is a rectangle.
Question: What does "pair l" refer to in the context of the first post? Answer: "pair l" seems to refer to a pair of numbers where l is a variable. In the example given, for n = 3, there are 3 possible values for l (0,1,2). | 677.169 | 1 |
I IThe trick is that: Firstly you cut triangles from the one width and one length, if there's no overleft, there won't be any overleft.
In this case, 12 is divisible by 4 and 3
7= 3+4
For sure, there won't be any overleft when you cut from one width and one length.
In short, as long as length/width( one of the two) is sum of the triangle's two legs ANDwidth/length( one of the two) is the product of the triangle's two legs, there won't be overleft.
Question: Is the text discussing a method to cut a material without any leftover? Answer: Yes | 677.169 | 1 |
Key to Geometry offers a non-intimidating way to prepare students for formal geometry as they do step-by-step constructions.
Students begin by drawing lines, bisecting angles, and reproducing segments using only a pencil, compass, and straightedge.
Later they do sophisticated constructions involving more than a dozen steps and are prompted to form their own generalizations.
When they finish, students have been introduced to 134 geometric terms and are ready to tackle formal proofs
Key To… When it comes to higher math, if either you (teaching) or your student (learning) lack confidence, then this curriculum may be your answer. These consumable work booklets are so well laid out and easy to understand, he'll finally be able to say, "I got it!" Work through each at your own pace. What you won't find: Lots of wordy explanations that leave you going, "Huh??" What you will find: Short, simple explanations with lots of examples, and a handful of well-designed problems on each page.
The Key To Geometry Answers and Notes are available separately or purchased as a complete set.
Question: Which of the following is NOT a tool used in the initial tasks of the curriculum? A) Ruler B) Protractor C) Compass D) Scissors Answer: D) Scissors | 677.169 | 1 |
Linear_Algebra/443750: There are 2 leaves along 3 inches of an ivy vine. There are 14 leaves along 15 inches of the same vine. How many leaves are there along 6 inches of the vine?
Construct triangle ABC with the vertex B common to the two congruent sides upward and the unknown side as the base. Construct a perpendicular to the base through the vertex at the top of the triangle. Label the point of intersection of the perpendicular and the base with D.
Your picture should look like one of the following:
Let represent the measure of segement AD. Let represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
From the formula for the area of a triangle, we have:
Hence,
And further,
From which we can derive:
From the fact that segment BD is perpendicular to segment AD, we can deduce that triangle ABD is a right triangle, for which we are given the measure of the hypotenuse is 13 and we have identified the measure of the legs as and . Using Pythagoras:
And using what we learned earlier about the relationship between and , we can write:
Multiply both sides by :
Let , substitute, and put the resulting quadratic into standard form:
Factor:
Then
or
Discarding the negative roots because we are attempting to calculate length, we have or .
Having previously determined that segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of
or
John
My calculator said it, I believe it, that settles it
test/442303: Please I need help in algebra 1 on problems of complete the square I don't get it all and I don't anything thank you 1 solutions Answer 305038 by solver91311(17077) on 2011-04-28 23:20:34 (Show Source):
The problem with your question is that the answer is both yes and no. I suspect you mean Is it factorable over the rational numbers, in which case the answer is no. But all quadratic trinomials are factorable if you allow complex number factors. A large subset of these are factorable over the real irrational numbers. And only a tiny fraction of these are factorable over rational numbers.
The given quadratic actually has a pair of irrational factors:
John
My calculator said it, I believe it, that settles it
decimal-numbers/442055: I know there is now a section, but can I get help with the following question?
3 Find the derivative of the function.
Question: Which of the following is the standard form of the quadratic equation derived from the given information? (This is a multiple-choice question that tests understanding of the quadratic formula and its standard form.)
A: x^2 - 10x + 25
B: x^2 + 10x - 25
C: x^2 - 10x - 25
D: x^2 + 10x + 25
Answer: C: x^2 - 10x - 25 | 677.169 | 1 |
Okay, let's try another example. The point -5 5. First, r squared equals x squared plus y squared. So -5 squared plus 5 squared. That's 25+25 or 50. And that means that r equals 5 root 2. Again, we picked up the positive value. And we're going to use the fact that cosine theta is x over r, -5 over 5 root 2 and that's -1 over root 2 which is the same as -2 over root 2 over 2. And sine theta is y over r. So it's 5 over 5 root 2, 1 over root 2 which is root 2 over 2. Now what angle has a cosine that's negative root 2 over 2 and a sine that's positive root 2 over 2? You could draw a little unit circle if you like. We're over here, right? A negative a negative cosine value and a positive sine value. This is going to be a reference angle of 45 degrees. The angle's 135 which in radiance is 3 pi over 4. Now that means my point is going to be r first 5 root 2 theta, 3 pi over 4.
And finally let's do 0 -10. Actually this one's pretty easy. If you if you graph this point, you can kind of see that well, we can't really use an angle of negative pi over 2 because it's not in the interval that we want. But you can use an angle of 3 pi over 2. So theta would be 3 pi over 2. And remember that r is just a distance from the origin, which is 10. So this would be 10 3 pi over 2. So it's much easier if you're converting a point that is on a coordinate axis in the polar coordinates.
Question: What is the value of r for the point (-5, 5)? Answer: 5 root 2
Question: Which angle can be used to represent the point (0, -10) in polar coordinates? Answer: 3 pi over 2 | 677.169 | 1 |
Tip #23 to Get a Top ACT Math Score (page 2)
When you are given a diagram on the ACT, ask yourself if it seems accurate. If it does, you can use it, sometimes just to see what to do next, and other times to get a correct answer without even doing much math. For example, if you are given the length of some part of the diagram, you can often use that to estimate an unknown. You've done this before. Imagine you're on a road trip. You look at the map and say, "We have to go from here to here on this squiggly highway." The map key says that each inch is 100 miles, so you use your thumb to represent an inch and you estimate the length of the squiggly line highway. We call this "Use the Diagram."
If the diagram is clearly not drawn to scale (not accurate), resketch it somewhat accurately and then "Use the Diagram." It turns out that often the whole question is hinged on its being out of scale, and when you put it into scale, the answer becomes obvious.
Also, while we're in art class, here's another great strategy. If a question describes a diagram, but none is shown, draw one. Sometimes this gives the answer immediately, and sometimes it shows you what to do next, but either way it always helps!
At first all this might feel weird. For a year in geometry class you were taught not to estimate with a diagram. Plus, on the ACT, estimating makes the question so much easier that it feels "cheap," like you are cheating. It's not! It's actually what they want you to do. Remember the test is supposed to test your cleverness, not just what you learned in math class. This strategy brings out your innate cleverness.
Let's look at this question:
Solution: Sketch the diagram to scale, following the instructions in the question:
Correct answer: E
Example Problems
Easy
If points A, B, C, and D are different points on a line; AB = BC = 8; and D is between B and C; which of the following could be the measure of segment AD ?
3
5
8
10
16
Medium
What is the y intercept of the line in the standard (x, y) coordinate plane that goes through the points (–4, 5) and (4, 1) ?
–1
1
3
5
7
Hard
In the figure below, A, B, and C are collinear, and angles A, C, and EBD are right angles. If EB = 14, DB = 7, and DC = 6, to the nearest whole number, what is the measure of EA ?
3
5
7
10
16
A room has the shape and dimensions in meters given below. A support beam is located halfway between point P and point Q. Which of the following is the distance of the beam from point M ?
3
9
Answers
Question: What is the correct answer to the easy example problem given in the text? Answer: E | 677.169 | 1 |
TRIGONOMETRY FUNCTION
BARIBOR NGIA asked
A 30-foot ladder is leaning up against a roof that is 20 feet above the ground . How far from the building is the foot of the ladder? What is the angle between the ladder and the grounnd ?
Note: Please include step by step explaination for better understanding.
Question: If the ladder were 5 feet longer, what would be the new distance from the building? Options: A) 10 feet, B) 15 feet, C) 20 feet, D) 25 feet Answer: C) 20 feet | 677.169 | 1 |
Examples
Calculator to find sides, perimeter, semiperimeter, area and altitude Equilateral Triangles. Given 1 unknown you can find the unknowns of the triangle. — "Equilateral Triangles Calculator",
A triangle with all three equal sides is called equilateral. The first two are of Greek (and related) origins; the word "equilateral" is of Latin origin: scalene (adjective): from the Indo-European root skel. — "Triangle Classification", cut-the-
Equilateral. Learn about Equilateral on . Get information and videos on Equilateral including articles on triangle, angle, congruent and more!. — "Equilateral | Answerbag",
In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. They are regular polygons, and can therefore also be referred to as regular triangles. — "Equilateral triangle - Wikipedia, the free encyclopedia",
equilateral (not comparable) (geometry) Referring to a polygon all of whose sides are of equal length. Not necessarily a regular polygon since the angles can still differ (a regular polygon would also be equiangular). [edit] Translations. referring. — "equilateral - Wiktionary",
The altitude of the triangle represents the height is the length of the corner of the equilateral triangle to the opposite base on the equilateral triangle. Due to its equilateral property for triangle, the above statement is same for all those three sides. — "Equilateral Triangle Altitude | ",
Definition of equilateral from Webster's New World College Dictionary. Meaning of equilateral. Pronunciation of equilateral. Definition of the word equilateral. Origin of the word equilateral. — "equilateral - Definition of equilateral at ",
Also keep in mind equilateral triangle = 666 The people are the source of all authority and power ~ John Adams 1774
equilateral triangle quadrilateral pentagon hexagon
Join us on Sunday May 17th from 6 9pm To celebrate cutting edge design fashion music and ***tails
equilateral technologies1d jpg
equilateral014 0647 jpg
Equilateral Triangle Inscribed In An Equilateral Triangle To use any of the clipart images above including the thumbnail image in the top left corner just click and drag the picture to your desktop You may also control click Mac or right click
The study started with the investigation of rules which govern equilateral triangles being the basis for the construction of the faces of regular tetrahedra A mathematical function was
Equilateral Triangle To use any of the clipart images above including the thumbnail image in the top left corner just click and drag the picture to your desktop You may also control click Mac or right click
Equilateral Arch To use any of the clipart images above including the thumbnail image in the top left corner just click and drag the picture to your desktop You may also control click Mac or right click
Question: In an equilateral triangle, what is the relationship between the altitude and the sides? Answer: The altitude is also the median and the angle bisector.
Question: Which of the following is NOT a characteristic of an equilateral triangle? A) All sides are equal B) All angles are equal C) The altitude to the base is not the same from all sides D) It is a regular polygon Answer: C) The altitude to the base is not the same from all sides | 677.169 | 1 |
Origami Basics: Equilateral Triangle from a Square This video shows how to get an equilateral triangle from a square - that is a triangle where each edge has the same length (and each angle has 60 degree). More origami:
GCSE Maths drawing an equilateral triangle How to draw an equilateral triangle and then bisect an angle
GeoGebra - 8 - Equilateral Triangle In this video, you will learn how to use GeoGebra to construct an equilateral triangle.
Illuminate by Equilateral Theatre Company Excerpt from an original modern dance piece created by Joe Doran with Equilateral Theatre Company. Choreography by Tony Guglietti. Original music by Nick Moore. For full video and more information visit .
Construction 10: Construct an Equilateral Triangle Given a side, construct an equilateral triangle.
Part 3 of 5 Equilateral triangle - Area - ratio The material in Part 3 addresses the problem about the ratio of the line segment. If the viewer has interest in this problem, he/she can replace the ratio numbers 1,2,3 with the Pythagorean triple 3,4,5. And find the ratio of the corresponding line segments from the very beginning.
Challenge - Equilateral Triangle 1 More: Create this dynamic equilateral triangle. You need to know (a) definition of an equilateral triangle, (b) how to find the midpoint of a line segment and (c) the properties of a 30°-60°-90° triangle. Go for it!
Improv Comedy Icebreaker Games : Improv Comedy Icebreakers: Equilateral Triangle Get improvisers used to paying attention with the Equilateral Triangle game. Learn the improv comedy icebreaker Equilateral Triangle in this free theater acting video from a teacher of improvisation. Expert: Shana Merlin Bio: Shana Merlin carries more than 1000 hours of teaching under her belt, and is one of the most experienced and effective improv teachers in Central Texas. Filmmaker: MAKE | MEDIA
Third solution -- Equilateral triangle -- Area ratio Without using trigonometry, I've solved a geometry problem concerning equilateral triangle. The problem is to determine the ratio of two areas, the equilateral and the quadrilateral embedded. I found this Mathematics problem very interesting and insightful, but not a so easy one. For the first 2 weeks, I have spent many hours studying the problem and at last found two solutions, as seen in part 4 and 5 of this series. However, my great interest in the problem did not fade out. It seems that there is so much hidden treasures waiting people to dig up. On the 17th day,while I was riding a bus, an idea of direct calculation suddenly came to me. Therefore, I uploaded this solution as supplementary information to the problem. Besides, I want to share the happiness and sense of achievement with every body else.
Question: Which of the following is NOT a method mentioned to construct an equilateral triangle? A) Origami B) Drawing C) Using GeoGebra D) Solving a mathematical problem E) Using a protractor Answer: E) Using a protractor.
Question: How many solutions did the author find for the geometry problem concerning equilateral triangles? Answer: Three.
Question: What is the name of the improv comedy game mentioned in the text? Answer: Equilateral Triangle. | 677.169 | 1 |
Given 3 points how to find the centre and radius of a circle in 3 D??
Do you mean a sphere? Are these arbitrary points anywhere in the sphere or on the surface?
Three points are not enough to define the sphere. Four will do the trick (unless they will not, but you need four at least). Points inside the sphere do not belong to the sphere. IMHO question is very precise.
Question: What does "IMHO" stand for in the text? Answer: In My Humble Opinion. | 677.169 | 1 |
I'm not real sure what the full question is here so I will answer the following question:
How do you find cos and sin when you just know tan? Or any combination.
Let's say you have :
cos(t)=v and you want sin(t) and/or tan(t) etc.
You know that if you had a right triangle the cos of one of the corner angles is cos(angle)=(closest leg)/(hypotenuse). Well I can write
cos(t)=v=v/1
Now if I draw a right triangle I can label the hypotenuse with a length of 1 and the closest side as length v. I can then find the length of the last unlabeled side to be sqrt(1-v2). Now that I have a fully labeled right triangle I can find all the trig functions of the angle t
Question: Which trigonometric function can you find directly from the given information 'cos(t) = v'? Answer: cos(t) | 677.169 | 1 |
Figuring In Football
In this primary grades lesson, students identify figures on a football field. They look for both congruent and similar figures, and they consider figures that are the same but that occur in a different orientation because of translation, rotation, or reflection.
Learning Objectives
Students will:
identify and visualize congruent and similar two-dimensional geometric figures
recognize that rotations, translations, and reflections do not change geometric figures
Materials
Instructional Plan
Geometric figures are very much a part of our environment and help define the ways in which we view and interpret our world. Everywhere we look we see
the influences of pattern, symmetry, and design. A football field has numerous figures that a young student can easily distinguish and others, which can be rotated or embedded, that may call on spatial skills of a more
challenging nature.
Discuss with the class the fact that geometric figures are common in the world in which we live. Sports often use items of both two- and three-dimensional shapes, from the equipment used, such as balls and nets, to the playing fields on which the sports take place. In particular, football fields often include many geometric figures.
Have each student locate the geometric figures in Questions 1 through 3 on the activity sheet. You might want to have students outline the figures with colored markers.
Have the students study the diagram of the football field to answer Questions 4 through 6.
After students have completed the activity sheet, conduct a class discussion about the shapes that they found. For instance, ask them to describe the effects of flips (reflections), turns (rotations), and slides (translations) on the shapes. Is a shape different just because it is moved? Also discuss the difference between congruent and similar figures. [Congruent figures are exactly the same size and shape; similar figures have the same shape but a different size.]
Questions for Students
What different geometric figures can you identify on the football field?
[Answers will vary, and there are many. Some of the obvious answers are lines, circles, squares (diamonds), triangles, rectangles, and parallelograms. Less common items include "bowties" (two triangles that meet at a vertex), "home plate" (consisting of three triangles and a diamond; these occur in the top and bottom of each end zone), and hexagons (made from two diamonds and two triangles).]
Assessment Options
Have students make the same type of drawings for different sports fields, such as baseball, soccer, basketball, and tennis. Then, have them identify the different geometric figures that occur on each field.
Extensions
Have students make the same type of drawings for different sports fields, such as baseball, soccer, basketball, and tennis.
Have students collect from newspapers and magazines information about the Super Bowl that includes references to geometry (e.g., "the 10-yard line" or "two teams squared off").
Question: Which of the following is NOT a common geometric figure found on a football field? A) Circles B) Ovals C) Squares D) Triangles Answer: B) Ovals
Question: What are some of the geometric figures that can be identified on a football field? Answer: Lines, circles, squares (diamonds), triangles, rectangles, parallelograms, "bowties" (two triangles that meet at a vertex), "home plate" (consisting of three triangles and a diamond), and hexagons (made from two diamonds and two triangles).
Question: What is the main learning objective of the lesson "Figuring In Football"? Answer: Students will identify and visualize congruent and similar two-dimensional geometric figures and recognize that rotations, translations, and reflections do not change geometric figures.
Question: What are some less common geometric figures that can be found on a football field? Answer: "Bowties" (two triangles that meet at a vertex) and "home plate" (consisting of three triangles and a diamond). | 677.169 | 1 |
Use the subgroups of SO(3) found in the previous exercise, and
the parametric equation for the equator of
to show
how any other great circle on
can
be found by appropriate combinations of rotations of the equator.
15.
Find two matrices, R1 and R2 from SO(3)which represent, respectively, rotation by
about the y-axis and rotation
by
about the z-axis; each rotation must be in a right-hand-screw sense
in the positive direction of its axis. Find the product matrix R1R2 and show that
its transpose is its inverse.
Question: In which direction are the rotations represented by R1 and R2? Answer: Both rotations are in the positive direction of their respective axes. | 677.169 | 1 |
In the figure below, given a triangle
ABC and its orthic triangle DEF (AD, BE, and CF are the
altitudes of ABC). Let be H the orthocenter of ABC. (1) Prove that
angles A, BDF and EDC are equal, (2) Prove that AD is the angle bisector of
angle EDF, and (3) prove that H is the incenter of triangle DEF.
Question: What is the relationship between AD and angle EDF? Answer: AD is the angle bisector of angle EDF. | 677.169 | 1 |
Presentation Transcript
TRIGONOMETRY :
Application of Trigonometry to Height and Distance Problems :
Application of Trigonometry to Height and Distance Problems Trigonometry, in ancient times, was often used in the measurement of heights and distances of objects which could not be otherwise measured.
For Ex: :
For Ex: trigonometry was used to find the distance of stars from the Earth. Even today, in spite of more accurate methods being available, trigonometry is often used for making quick and simple calculations regarding heights and distances of far-off objects. For this, the value of various trigonometric functions is needed. A simple example is given below to demonstrate how trigonometry can help find the height or distance of an object.
Example: :
Example: If the distance of a person from a tower is 100 m and the angle subtended by the top of the tower with the ground is 30o, what is the height of the tower in meters?
Steps: :
Steps: Draw a simple diagram to represent the problem. Label it carefully and clearly mark out the quantities that are given and those which have to be calculated. Denote the unknown dimension by say h if you are calculating height or by x if you are calculating distance.
Identify which trigonometric function represents a ratio of the side about which information is given and the side whose dimensions we have to find out. Set up a trigonometric equation.
Substitute the value of the trigonometric function and solve the equation for the unknown variable.
Solution: :
Solution: AB = distance of the man from the tower = 100 m
BC = height of the tower = h (to be calculated)
The trigonometric function that uses AB and BC is tan A , where A = 30o.
Question: If the angle were 45 degrees instead of 30 degrees, which trigonometric function would you use to find the height? Answer: Tangent (tan) or Sine (sin), as both are applicable at 45 degrees | 677.169 | 1 |
Of course, you don't have to -- I know you have better things to do than to teach mathematics to some Canadian who really should have been paying more attention in school. ;-)
(You can probably safely assume that I have an upper level of understanding of math, however -- I did take entry-level Calculus in school, and I didn't do too bad (relatively speaking to my normal suckiness) either.)Well, the subtraction and addition parts should be simple enough for starters--basically I changed the coordinates so that they were relative to the pivot point, by subtracting the pivot, and then after rotation I added that back in.
The rotation part is a formula I can never fully remember, so I have to work it out every time. But basically it goes like this:
For the sake of discussion I'll call the rotated coordinates x' and y'.
Consider a right triangle. One side is x, and it goes all the way from the origin to (x,0). The next side is y, and it goes from (x,0) to (x,y). The hypotenuse goes from the origin to (x,y). Now, tip that triangle clockwise, pivoting it around the origin'd be a lot easier to explain with diagrams.
The way I figure it out is: x' will be the same as x if no rotation is used, so that's the cosine term. If you rotate 90° clockwise, it's y, and that's the sine term. Therefore x'=x*cos(angle)+y*sin(angle). Similar reasoning goes into y'; it's normally y unless you rotate, so that's the cosine term, while y'=-x if you rotate 90° clockwise, so y'=y*cos(angle)-x*sin(angle). I can never keep the signs straight without doing this each time because coordinate systems differ is probably one of the few times in my life that I've actually understood a concept like this -- the former line being a hypoteneuse of a new triangle made it easy to understand.
Thanks a bunch.
(Incidentally, I'm taking grade 11 mathematics* through night school, so I'll build up a geometric foundation within the next few months. =))
* It's too lengthy to explain the B.C. math curriculum here -- suffice it to say that it's a comprehensive course designed to teach people upper high-school level mathematics in all fields, ranging from geometry to algebra. Grade 12 mathematics is top level non-calculus stuff that you take before college.
Question: What is the user planning to do to improve their mathematical skills? Answer: The user is taking grade 11 mathematics through night school to build a geometric foundation.
Question: What is the pivot point for the rotation? Answer: The origin (0,0) | 677.169 | 1 |
See also mathematics, orthogonal coordinates are defined as a set of d coordinates q = in which the coordinate surfaces all meet at right angles . A coordinate surface for a particular coordinate qk is the curve, surface, or hypersurface on which qk is a constantThe derivatives of scalars, vectors, and second-order tensors with respect to second-order tensors are of considerable use in continuum mechanics. These derivatives are used in the theories of nonlinear elasticity and plasticity, particularly in the design of algorithms for numerical...
Curvilinear perspective is a graphical projection used to draw 3D objects on 2D surfaces. It was formally codified in 1968 by the artists and art historians André Barre and Albert Flocon in the book La Perspective curviligne, which was translated into English in 1987 as Curvilinear Perspective:...
This is a list of some vector calculus formulae of general use in working with various curvilinear coordinate systems.- Note :* This page uses standard physics notation. For spherical coordinates, \theta is the angle between the z axis and the radius vector connecting the origin to the point in...
Question: What is the main topic of the text? Answer: Vector calculus formulae in various curvilinear coordinate systems | 677.169 | 1 |
Geometry is the study of lines, rays, segments, shapes, symmetry, etc. In this section ofthe World Wide Math Tutor, we will go into basic geometry concepts for elementary school students. So, if you high school students need some extra info, SORRY!
The first three concepts of geometry are lines, points and segments.
Point- a specific plane on a number line, coordinate graph, or map.
Question: What are the first three concepts of geometry mentioned in the text? Answer: Lines, points, and segments | 677.169 | 1 |
Observe that if you _________ the red
and green arrows tails together then
the blue arrow is the __________ that
runs from red arrows tail to the
green arrows head.
Name the Arrows.
Since
the relationship
we can think of the arrows as having
8
16
17
Generalizing Direction, part 2
If your paying attention then you should be thinking
Did he just say Subtract Arrows. What the .
And yes, I did. This isnt so strange, yet another way to model
addition and subtraction on the number line is with _________.
Example: 5 3 = 5 + -3 = 2
Addition: 5 + -3 = 2
Subtraction: 5 3 = 2
5
18
Vectors
Arrows have a definite __________ and a definite ___________
Any object that has both of these is called a Vector.
Vectors can be added
by gluing head to tail in the arrows case.
Vectors can be subtracted
by gluing tail to tail in the arrows case.
Every Vector had an additive inverse or opposite
the backwards arrows in the arrows case.
Every Vector can be multiplied by a Real Number.
stretch and shrink the arrows
Sometimes Vectors can be multiplied together to make vectors or
just real numbers.
9
3
19
Vectors, part 2
Arrows in the plane are called _______________ vectors because
they are pieces of a 2 dimensional space. [Thats one reason]
If we move the tail of any arrow in the plane to the ________,
then that arrow will point at a specific point in space.
Since every point in the plane has exactly one arrow that points
from the origin to it, there is a one to one _____________
between vectors and points and vectors and ordered pairs.
The _______________ of a vector are the _______________ of
the point in space they point to when their tail is placed on the
origin.
19
20
Vectors, part 3
As a result, Vector Operations can be completely studied with
numbers.
Addition
Multiplication
Opposites
When operations are performed on Ordered Pairs in this way we
say the operations are being performed ______________ wise.
Another reason why vectors in the plane are called 2 dimensional
is because these vectors can be described by _____ coordinates.
10
5
21
Advanced Linear Modeling Curve Fitting
Lines suck for interpolating since they can only intersect at most ____ points.
They are good at and commonly used to perform regression curve fitting.
Data is sampled from a real
phenomenon.
And a guess line might be made.
We can measure the closeness
of any line drawn in the plane
by drawing _____________ lines
from the data points to the line.
The Best Fit Regression line is chosen by making the sum of the
_______________ line lengths or their ____________ as small as possible.
Question: What is the blue arrow called in this context? Answer: The blue arrow is called the resultant vector.
Question: What is another way to model addition and subtraction on the number line, as mentioned in the text? Answer: Using arrows.
Question: What is the one-to-one relationship mentioned between vectors and points in the plane? Answer: There is a one-to-one correspondence between vectors and points in the plane.
Question: What should you do to the red and green arrows' tails to make the blue arrow the one that runs from the red arrow's tail to the green arrow's head? Answer: You should connect the red and green arrows' tails together. | 677.169 | 1 |
After this transformation, the vertices of the polyhedron correspond to intersection points in the plane, edges correspond to segments, and faces correspond to regions (including the exterior "region", which also began as one of the faces of the polyhedron).
For a polyhedron with V vertices, E edges, and F faces, Euler's formula states that V, E, and F are related by the equation V – E + F = 2.
Question: If a polyhedron has 10 faces, and using Euler's formula, what is the minimum number of edges it could have?
Answer: To find the minimum number of edges, we can rearrange Euler's formula to E = V + F - 2. Substituting F = 10, we get E = V + 8. Since V must be a positive integer, the minimum value for V is 1, which gives E = 9. | 677.169 | 1 |
Trapezoid
It is necessary that the two parallel sides be opposite; they cannot logically be adjacent.
If the other pair of opposite sides is also parallel, then the trapezoid is a parallelogram.
(But according to some authorities, parallelograms are specifically excluded from the definition of trapezoid.)
Otherwise, the other two opposite sides may be extended until they meet at a point, forming a triangle that the trapezoid lies inside of.
The area of a trapezoid can be computed as the product of the distance of the two parallel sides and the average (arithmetic mean) of the other two sides. This yields the well-known formula for the area of a triangle, were one to consider a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point.
In acrobatics, the trapeze is a certain acrobatic device that is shaped like a trapezoid
Question: Is it possible for a trapezoid to have adjacent parallel sides? Answer: No, it is not possible. | 677.169 | 1 |
Segments in a Circle
In this lesson our instructor talks about segments in a circle. First she talks about chord, diameter, radius, secant, and tangent. Then she discusses circumference and examples. Four extra example videos round up this lesson.
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Segments in a Circle
Be familiar with the terms circle, chord, diameter, radius, secant, and tangent
If a circle has circumference of C units and a radius of r units, then
Segments in a Circle
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
Question: What type of content is available for students to download and print out? Answer: Lecture slides, which are screen-captured images of important points in the lecture. | 677.169 | 1 |
Question 439601: The sum of the measures of the angles of any triangle is 180 degrees.In triangle ABC,angles A and B have the same measure ,while the measure of angle C is 90 times larger than each of A and B.What are the measures of angle A,B,C. Click here to see answer by rwm(914)
Question 439597: A business woman wants to determine the difference between the costs of owning and leasing an automobile. She can lease a car for $420 a month (on an annual basis). Under this plan, the cost per mile (gas and oil) is $0.06. If she were to purchase a car, the fixed annual expense would be $4700, and the other costs would amount to $0.08 per mile. What is the least number of miles she would drive per year to make leasing no more expense than purchasing Click here to see answer by jorel1380(2518)
Question 439988: Finding the Grade of a Mountain Trail. A straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 400 feet. The length of the trail is 4100 feet. What is the inclination (grade) of the trail? Click here to see answer by stanbon(57347)
Question 441176: A rancher with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle.
a.) Find a function that models the total area of the four pens.
b.) Find the largest possible total area of four pens. Click here to see answer by [email protected](15652)
Question 441698: The problem is as follows:
A portion of a wire 70 inches in length is bent to form a rectangle having the greatest possible area such that the length of the rectangle exceeds three times its width by 2 inches and the dimensions of the rectangle are whole numbers. Find the length of the wire that is NOT used to form the rectangle.
I noted the following:
* I understand that the perimeter of a rectangle is 2L + 2W = P
* I understand that the length of the triangle is L = 3W + 2
* Using U to represent the unused wire, I need to solve for U in either 2L + 2W + U = 70 or 2L + 2W = 70 - U.
* I start off inputting the linear equation for L by writing down 2(3W + 2) + 2W + U = 70. I get 6W + 4 + 2W + U = 70, which simplifies to 8W + U = 70.
Question: If the rancher has 750 ft of fencing and wants to enclose a rectangular area, what is the maximum possible area of the rectangle? Answer: 375000 square feet (750^2 / 4)
Question: If she purchases a car, the fixed annual expense is $4700 and the cost per mile is $0.08, what is the total annual cost of purchasing the car if she drives 10,000 miles per year? Answer: $4700 + $0.08 * 10000 = $5700 | 677.169 | 1 |
Projective Planes
Figure 1
Most of you know how to make a Mobius band---take a strip
of paper and glue the ends with a half-twist. This object
now has the property that is has only one "side".
It also has only one edge. Well, a disc has only one edge,
too, so then we should be able to sew their edges together?
Indeed you can, although not in 3 dimensions (you'll need
at least 4 spatial dimensions to accomplish this).
But after you are done, you will have a surface called
a projective plane. An alternate way to
construct a projective plane
is to take a disc and sew pairs of opposite points together!
Does this object sound weird? Well,
you are probably already familiar
with projective planes... the old arcade version
of the game of Asteroids
was played on one! (Remember the screen was a disc,
and when an asteroid hit one edge of the screen, it
emerged on the opposite side of the screen? However,
some have reported that the Atari home version of the game
is played on a torus.
)
Presentation Suggestions:
Draw pictures. Or take
a piece of cloth shaped like a disc, take a zipper about
half the length of the circumference, and sew both halves
of the zipper onto the boundary of the disc.
Then you should be able to sew up the disc at least part
of the way...
The Math Behind the Fact:
This is an example of a surface that is said to be
non-orientable, because any two dimensional creature in
the surface can walk along a path that will take it back
to the original spot but the creature will be
mirror-reversed! Look at what happens to the smiley face
in Figure 1.
Do you think that our universe is orientable?
Question: What are two ways to construct a projective plane? Answer: Sewing the edges of a Mobius band and sewing pairs of opposite points together on a disc.
Question: What is a distinctive property of a projective plane? Answer: It is non-orientable, meaning any two-dimensional creature can walk along a path that will take it back to the original spot but the creature will be mirror-reversed. | 677.169 | 1 |
Rigid transformation Ultimate Study Guide
Rigid transformation
In mathematics, a rigid transformation (isometry) of a vector space preserves distances between every pair of points.12 Rigid transformations of the plane R2, space R3, or real n-dimensional space Rn are termed a Euclidean transformation because they form the basis of Euclidean geometry.3
The Rigid transformations include rotations, translations, reflections, or their combination. Sometimes reflections are excluded from the definition of a rigid transformation by imposing that the transformation also preserve the handedness of figures in the Euclidean space (a reflection would not preserve handedness; for instance, it would transform a left hand into a right hand). To avoid ambiguity, this smaller class of transformations is known as proper rigid transformations (informally, also known as roto-translations). In general, any proper rigid transformation can be decomposed as a rotation followed by a translation, while any rigid transformation can be decomposed as an improper rotation followed by a translation (or as a sequence of reflections).
Any object will keep the same shape and size after a proper rigid transformation, but not after an improper one.
All rigid transformations are affine transformations. Rigid transformations which involve a translation are not linear transformations. Not all transformations are rigid transformations. An example is a shear, which changes two axes in different ways, or a similarity transformation, which preserves angles but not lengths. The set of all (proper and improper) rigid transformations is a group called the Euclidean group, denoted E(n) for n-dimensional Euclidean spaces). The set of proper rigid transformation is called special Euclidean group, denoted SE(n).
Contents
Distance formula
A measure of distance between points, or metric, is needed in order to evaluate if a transformation is rigid. The Euclidean distance formula for Rn is the generalization of the Pythagorean theorem. This says the distance squared between two points X and Y is the sum of the squares of the distances along the coordinate axes, that is
which shows that the matrix [L] can have a determinant of either +1 or -1. Orthogonal matrices with determinant -1 are reflections, and those with determinant +1 are rotations. Notice that the set of orthogonal matrices can be viewed as consisting of two manifolds in Rnxn separated by the set of singular matrices.
The set of rotation matrices is called the special orthogonal group, and denoted SO(n). It is an example of a Lie group because it has the structure of a manifold.
Formal definition
A rigid transformation is formally defined as a transformation that, when acting on any vector v, produces a transformed vector T(v) of the form
which means that R does not produce a reflection, and hence it represents a rotation (an orientation-preserving orthogonal transformation). Indeed, when an orthogonal transformation matrix produces a reflection, its determinant is -1
Question: What is the formula for the Euclidean distance between two points X and Y in Rn? Answer: (X - Y)^T * (X - Y) | 677.169 | 1 |
This Lesson (Conic Sections-(parabola, circle, ellipse, hyperbola)) was created by by Nate(3500): View Source, Show About Nate:
~Parabola~
Parabolas are 'U' shaped graphed lines that have a degree of two.
Standard form:
Vertex:((-b/2a),f(x))
Zeros (the point where the parabola intersects with the x-axis) can be complex or real values.
The formula (quadratic formula):
The discriminant is . If the discriminant is greater than zero, there are two real answers. If the discriminant is equal to zero, there is one real answer. If the discriminant is less than zero, there are two complex values.
Sometimes, parabolas are seen as in vertex form.
Standard form:
Vertex: (h,k)
If the -value is negative, the parabola opens down. Vice versa if the -value is positive. This idea works for all types of parabolas.
Also, parabolas can go left and right.
Standard form:
Vertex: (h,k)
If the -value is positive, the parabola opes right. Vice versa if the -value is negative.
~Circle~
Standard form:
Center: (h,k)
In the form above, is the value for the radius.
Formula for area of circle:
~Ellipse~
Standard form:
Center: (h,k)
Formula for foci(c):
The foci is (c) amount of units along the major axis.
In this form, the length of the major axis (the longer axis) is . As for the length of the minor axis (the smaller axis), it is . In this form, the ellipse is horizontal because the is under the x-values. The is always the higher number when compared to . If the was under the y-values, the ellipse's major axis would be vertical.
Formula for area of ellipse:
~Hyperbola~
Standard form:
The is always the value under the positive term. When the is under the x-values, the hyperbola has a horizontal tranverse axis and the slope of its asymptotes is . When the y-value is over the , the transverse axis is vertical and its asymptote's slope is . determines the transvers axis. The hyperbola is two parabolas going opposite directions that're always approaching the asymptotes, but never touching or crossing over them.
Formula for foci(c):
The foci is located units inside the two "parabola-like" branches.
The center for the hyperbola is determine by: (h,k)
In both Hyperbolas and Ellipses, the x-values are always associated with and the y-values are always associated with .
Question: What is the slope of the asymptotes of a hyperbola with a horizontal transverse axis? Answer: The slope of the asymptotes is ±√(b/a). | 677.169 | 1 |
Precalculus I thought of creating a 3:4:5 (ratio of sides) right triangle, with the hypotenuse equal to 15. The 3:4:5 ratio stays the same. I could have also chosen (0, 15/sqrt2) and (15/sqrt2, 0) or (-3,0) and (0,-4) and many others.
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math list two congruence conditions for the law of sines yields a unique solution
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Question: What is the most important reason to consider typography when creating a procedural text, according to the fourth entry? Answer: It makes the document look nice.
Question: What is the length of the hypotenuse of the right triangle? Answer: 15 | 677.169 | 1 |
Pythagorean Theorem EFG 3,6,9 HIJ 60,156,144 KLM 56,102,105 NOP 36,48,64 The following are the lengths of the sides of four triangles. (See assignment sheet for chart). Which is a right triangle? a.Triangle EFG b.Triangle HIJ c.Triangle KLM d.Triangle NOP
Thursday, July 15, 2010 at 1:47pm by david Anonymous Lola
geometry in a triangle D =h+4, area of triangle is 120 sq .cm . then find the base and height of triangle
Thursday, February 7, 2013 at 2:01am by msmadhushree
math((PLEASE HELP!)) if a^2+b^2 < c^2, the triangle is acute if a^2+b^2 = c^2, the triangle is right if a^2+b^2 > c^2, the triangle is obtuse
Thursday, December 6, 2012 at 11:04pm by Steve
math A side of an equalateral triangle is 2/8 cm long. Draw a picture that shows the triangle. What is the perimeter of the triangle?
Monday, February 25, 2013 at 9:51pm by leon
Maths Base of the triangle =13cm altitude of the triangle=12cm Area of the triangle=1/2(b)(a) ' =1/2(5)(12) =30cm2
Wednesday, November 30, 2011 at 12:35pm by anirudh the great
algebra The sides of a triangle are in the ratio 5:12:13. What is the length of each side of the triangle if the perimeter of the triangle is 15 in?
Tuesday, August 18, 2009 at 11:52pm by Laura
Geometry Triangle DEF ~ Triangle HJK, and the scale factor of Triangle DEF to Triangle HJK is 5/2. If EF=15, find JK
Wednesday, June 16, 2010 at 10:09am by April
Math need help! It also wants to know IF it's a triangle: If the angles can form a triangle, . . . It is not a triangle.
Friday, February 15, 2013 at 8:39pm by Ms. Sue
Geometry The area of a triangle is 6.75 m squared. If the base of the triangle is 3 m, what is the height of the triangle?
Tuesday, December 7, 2010 at 4:24pm by Mark
Question: True or False: If the angles of a triangle can form a triangle, then it is a triangle. Answer: False (A triangle is defined by its sides, not its angles)
Question: Which of the given triangles is a right triangle? a.Triangle EFG b.Triangle HIJ c.Triangle KLM d.Triangle NOP Answer: c.Triangle KLM (56, 102, 105 is a Pythagorean triple)
Question: What is the area of the triangle with a base of 13 cm and a height of 12 cm? Answer: 30 cm² (Area = 1/2 base height)
Question: If the sides of a triangle are in the ratio 5:12:13, what is the length of the longest side if the perimeter is 15 inches? Answer: 10 inches (Longest side = (13/5+12+13) * 15/3) | 677.169 | 1 |
Ch04Sec02
Course: MATH 10, Fall 2009 School: Hudson VCC Rating:
Word Count: 622
Document Preview half-lines, that originate at a common point called the vertex, which we will denote O. One of the rays is called the initial side of the angle, and the other ray is called the terminal side.
O Angles are commonly denoted using lowercase Greek letters, such as .
4.2 Measuring Angles
Angles can be defined in a number of different ways. An angle formed by one complete rotation, by definition, consists of 360 equal parts called degrees. Degrees are denoted using the symbol . An angle of 1 is 1/360 of a revolution.
Radian Measure
A more natural way of measuring angles, (one that is not based on the rather artificial 360 degrees), is called radian measure. Radian measure is the measure used most often in calculus. The measure of an angle in radians is determined by the length of the arc of a unit circle that is cut by the rays of the angle.
Radian Measure
The radian measure of an angle with vertex at the center of a unit circle is the length of the arc made by the angles.
1
Conversion Between Degrees and Radians
Conversion: Since 180 = radians, 1 = / 180 0.01745 radians 1 radian = 180 / 57.296 Keep these conversions in mind!
Example 1
Express 135 and 420 in radians and 5 / 6 in degrees.
Express 135 and 420 in radians and 5 / 6 in degrees.
To convert from degrees to radians, multiply the number of degrees by /180, the number of radians in one degree. To convert from radians to degrees, multiply the number of radians by 180/, the number of degrees in one radian. Go ahead and do this for the 135 example...
Express and 420 in radians and 5 / 6 in degrees.
135 = 135 ( /180) = radians 420 = 420 ( /180) = 7 / 3 radians 5 / 6 = (5 / 6) (180 / ) = 150 Notice, we do not worry about the decimal equivalents...they are very rarely required.
Degree and Radian Equivalents for Standard Angles
Degrees Radians 0 0 30 /6 45 /4 60 /3 90 /2
Radian Measure and Terminal Points
For any real number t, the angle generated by rotating counterclockwise from the positive xaxis to the point P(t) on the unit circle has (0,1) /2 radian measure t.
Question: What is the conversion from radians to degrees? Answer: Multiply the number of radians by 180/π. | 677.169 | 1 |
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Hyperbola
A hyperbola is a locus of a moving point such that the ratio of its distance from a fixed point(focus F) to its distance from a fixed line(directrix l) is constant, i.e. it is a conic with eccentricity e>1.
The equation of hyperbola is
where b2=a2(e2-1)
The graph of the hyperbola is,
Terminology:
Centre: the point O
Vertices: the points A(a, 0) and A'(-a, 0)
Transverse axis: the segment joining the vertices; AA'above. Its length is 2a.
Conjugate axis: the line through the centre perpendicular to the transverse axis (the y-axis above). The hyperbola never intersects the conjugate axis. The length o f the conjugate axis is 2b.
Focus: the fixed point is called a focus F(ae,0) of the hyperbola.
Directrix: the fixed line is called the directrix of the hyperbola and its equation is x=a/e
Eccentricity: e=
Latus Rectum: the length of the latus rectum is . The end points of the latus rectum are (ae, ) and (ae, )
Properties:
Let P(x,y) be any point on the hyperbola.
From the above figure the focal distances PF=PF'=2a (the transverse axis)
Another definition of hyperbola:
A hyperbola is the locus of a moving point such that the difference of its distances from two fixed points is always constant. The two fixed points are called the foci of the hyperbola.
Question: What is the length of the transverse axis of a hyperbola? Answer: 2a.
Question: What is the length of the latus rectum of a hyperbola? Answer: 2b²/a. | 677.169 | 1 |
Figure 3: A rectangle is painted by specifying the above XML tag and properties
SVG Ellipse:
Now we will try to draw an ellipse using the SVG tag. tag is used to draw a ellipse in SVG. Give an id to this tag, id="myEllipse". Now to draw a ellipse you have to give center coordinates of ellipse (cx, cy) and x- radius and y- radius (rx, ry). Now you can add the style properties to add the stroke color and stroke width by "stroke:black;stroke-width:3". You can also fill the ellipse by "fill:#3F5208;".
Figure 4: A Ellipse is painted by specifying the above XML tag and properties
SVG Polygon:
The HTML5-SVG provide different tag to draw a polygon. As you can see in below code, the tag is used here to draw it. The properties of this tag is similar to other tags as we discussed, except this tag have a property 'points', The points are the coordinates of the polygon. So you have to specify the coordinates in pair of (x, y), the two points should have a gap in between. for example - the below polygon have first point as (10,10), second point as (75,150) and third point as (150,60).
After running the below program you can see the output, its same as expected.
Similarly for polygon also you can define the stroke color, stroke width, fill color etc
Question: Which SVG tag is used to draw an ellipse? Answer: `<ellipse>` | 677.169 | 1 |
Here is an article about coordinate grids and polygons. Polygons must have three sides or more to be a polygon. They also must be a closed figure. A few examples of polygons are: square, triangle, trapezoid, rhombus, rectangle octagon, and hexagon. There are a lot more polygons than I just explained. Poly gons are a very common shape. They are everywhere!
Coordinate grids can show where something is. If you draw a polygon on a coordinate grid, you could locate it by looking at it's position. A coordinate grid is kind of like a map. A coordinate grid is great! Next are different types of triangles. The isoceles triangle has two equal sides. The equilateral triangle has all even sides.
A right triangle has one right angle. That's all about polygons and coordinate grids!
The poptart gained speed, hitting millions of jumpropes. It reached lightspeed, going into a time warp. It warped 3 million years into the future. Floating cars and cities on clouds. The poptart flew through hologram jumpropes, and the base of the cloud cities. Then, the poptart flies into a place called, "Toasterrama". The poptart jumped into each toaster, then hopping out of them.
Then, The poptart went into the "Hyper Toaster Shooter 500000". It counted down. 3....2....1....blast off! The poptart was shot through the air, and through the buildings. It crashes into the wall of a building, and the poptart flies into a toaster on the table. This was the " Extreme Poptart Light Speed Launcher". And the toaster shoots the toaster at light speed, and goes into a time warp, while everyone watches in amazement. The poptart goes through a time portal, and lands in the jurassic ages. It's amazing what a lifeless poptart, isn't it?
TO BE CONTINUED
There was a women who spotted a desirable dress. It was lovable by every women. She put it on and she asked her husbend if her new dress makes her butt look big. He said that it was not noticeable. It made her look despicable. She noticed that it was comparable to her worst dress. She tried to take off the dress, but she could'nt.
She thought for a moment, looking intelligible. It wasn't exusable that she couldn't take it off. The dress was not changeable. If she could take it off, she would never think of it as usable again. She went into the bathroon so she could be peaceable.
She took a shower with it on and it absorbed all of the water. Then, it finally ripped apart
THE END
There was a women who loved to jump. So she was jumping while putting on her dress. So she got drassed while hopping on her bed. Then she asked someone: please pass the salad dressing. So she waited while still jumping on her bed. She hated to wait. So she jumped angrily until she put a crack in the floor.
Question: What was the name of the device that shot the poptart at light speed? Answer: Extreme Poptart Light Speed Launcher
Question: Which of the following is NOT a polygon? A) Square B) Circle C) Triangle D) Octagon Answer: B) Circle
Question: What is the name of the triangle with all sides equal? Answer: Equilateral triangle
Question: What is the name of the triangle with two equal sides? Answer: Isosceles triangle | 677.169 | 1 |
Re: lines
On a plane. They can be parallel and never meet. They can intersect at 1 point. They can be the same line ( on top of each other ) and intersect at an infinite number of points.
Parallel postulate talks about parallel lines never crossing lines
Yes, those are the three possibilites, provided both lines are on a plane.
In the drawing below I have shortened the lines so they look like segments to beter fit them on a screen. Remember as lines they continue on forever
Question: How many points can two identical lines on a plane intersect at? Answer: An infinite number of points. | 677.169 | 1 |
And now, congruence. Instead of a circular definition, I have a clean syllogism:
If Polygon A is congruent to Polygon B, then A can be mapped onto B using a series of transformations. If the figures can be mapped into the same space, then their corresponding angles and sides are congruent, because the mapping preserves degree and distance. Therefore, congruent polygons' corresponding angles and sides are also congruent.
From there, I go onto congruence shortcuts and proofs, blah blah blah. But it started much more cleanly. I taught transformations, reviewed perpendicular lines and other coordinate geometry formulas, and linked it all to congruence in a meaningful way.
A few weeks later, it was onto similar polygons. Again, instead of just saying "Similar polygons have congruent angles and proportional sides", I can link it to dilation.
Day One: Review of Proportionality, then onto dilation
The kids did straight dilations as well as transformations and dilations in combination. I started with straight dilations, because I wanted the students to confirm the elements of similarity. The kids generally remember that parallel lines have the same slope, but I thought it would also be useful to see the transversal relationships with the parallel lines. We could prove, algebraically, that the lines of the dilated triangle were parallel to the original, and we could then extend those lines to prove that the corresponding angles on each triangle were congruent. Here's an example (again, one I just sketched up) that shows how the kids determined the angles were congruent.
The kids colored the corresponding angles—there are three in each case (one of the green ones in my image is an error, you can see I xed it out, just too much hassle to draw again).
So again, the point was to algebraically and visually confirm the parallel relationship, and then follow the dual sets of parallel lines and transversals to confirm that the angles are congruent.
I had them do a combination transformation/dilation, confirming that order didn't matter, and identifying which of the isometries had the parallel relationship.
Day 2: Review of Dilation, then onto Similarity.
Done.
Linking isometries to congruence and similarity was so much better, and whenever I tell math teachers about it they go oooh, ahhh and think about trying it themselves. And yet, I can't point to why it's so obviously superior. I can't swear that my students learned congruence or similarity more thoroughly—in fact, I think they learned it as well but not any better than my students last year.
Question: According to the text, did the students learn congruence or similarity better or the same compared to the previous year? Answer: The same. | 677.169 | 1 |
17-gon17-gon is discussed in the following articles:
constructiblity
The discovery that the regular "
17-gon" is so constructible was the first such discovery since the Greeks, who had known only of the equilateral triangle, the square, the regular pentagon, the regular 15-sided figure, and the figures that can be obtained from these by successively bisecting all the sides. But what was of much greater significance than the discovery was the
Question: What does "constructible" mean in the context of this text? Answer: In the context of this text, "constructible" means that the regular polygon can be constructed using a compass and an unmarked straightedge. | 677.169 | 1 |
And, since AD equals DC, and DE is common, the two sides AD and DE equal the two sides CD and DE respectively, and the angle ADE equals the angle CDE, for each is right, therefore the base AE equals the base CE.
Therefore, given a segment of a circle, the complete circle has been described.
And it is manifest that the segment ABC is less than a semicircle, because the center E happens to be outside it.
Similarly, even if the angle ABD equals the angle BAD and AD being equal to each of the two BD and DC, the three straight lines DA, DB, and DC will equal one another, D will be the center of the completed circle, and ABC will clearly be a semicircle.
But, if the angle ABD is less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the center will fall on DB within the segment ABC, and the segment ABC will clearly be greater than a semicircle.
Question: Can a complete circle be described given only a segment of a circle? Answer: Yes | 677.169 | 1 |
Tangens of the
sum and subtraction
If the t and s are real numbers in order to ,.
If plus that ,
than it is true
that . And for
the it is true
that
The formulas
for the reduction of the sine and cosine functions.
Cos(n -t) =
- cos t
Cos (n +t) = - cos t
Sin (
+ t) = cos t
Sin ( - t) cos
t
Universal substitution
All the trigonometric
functions can be expressed as rational functions of the variable
tg.
for all the real numbers for which both sides of identity are defined.
Sinusoid
The
function is
called sinusoid. Here C >0 a positive constant called amplitude,
is called circular frequency, and phase
lag. Sinusoid is the periodic function. Her period is .
The usage of
trigonometry
The necessity
of exact measurements were through centuries together with
the astronomic measurements the most important reason for
the development of trigonometry. The basic problem of measurement
is to determine the distance of the two
mostly unreachable points. The general scheme looks like this:
on the reachable part of terrain we determine two prominent
points (elevation) and measure their distance. By theodolite
we can measure the angles between any three visible points,
among which we can also have unreachable points. After that,
it is the task of trigonometry to calculate the given distances.
The
way to solve this example was applied in 1752. by French astronomers
La Lande and La Caille, trying to calculate the distance to
the Moon. They have calculated the angles a' and b' that are
closed by the Moon, through two points A and B that are situated
on the same meridian, according to the zenithal directions (verticals
in A and B). The position of the Moon must be such that it is
situated in the ley of the meridian, so, points 0, A, B and
M must be complanary. The latitude tp1 and tp2 of the places
A and B are known so on the picture the values |AO| and |BO|
=R,?=?1+?2 and angles ?= 180ş- ??, ?= 180ş- ??. In this way
we got the data from the preceding example and can determine
unknown values from the picture. We must also mention that angles
x and y are called the parallaxes of the Moon for the points
A and B.
Mathematicians
Leonhard
Euler (Basel, April, 15th, 1707. - Petersburg, September, 18th,
1783.), was a great Swiss mathematician, physicist and astronomer.
He put a great influence on the mathematics as a whole. He learned
mathematics from Johann Bernoulli. After the Petersburg science
Question: What are the formulas for the reduction of the sine and cosine functions? Answer: Cos(n - t) = cos n cos t + sin n sin t, Cos(n + t) = cos n cos t - sin n sin t, Sin(n + t) = sin n cos t + cos n sin t, Sin(n - t) = sin n cos t - cos n sin t | 677.169 | 1 |
5. Make conversions within the same
measurement system while performing computations.
6. Use strategies to develop
formulas for determining perimeter and area of triangles, rectangles and
parallelograms, and volume of rectangular prisms.
7. Use benchmark angles (e.g.; 45º,
90º, 120º) to estimate the measure of angles, and use a tool to measure
and draw angles.
Geometry and
Spatial Sense
Characteristics and Properties
1. Draw circles, and identify and
determine relationships among the radius, diameter, center and
circumference; e.g., radius is half the diameter, the ratio of the
circumference of a circle to its diameter is an approximation of
π.
Question: What is the main topic of the text? Answer: Geometry and Spatial Sense | 677.169 | 1 |
Question 624625: in a triangle, the measure of the first angle is 15 degrees more than twice the measure of the second angle. the measure of the third angle exceeds that of the second angle by 25 degrees. what is the measure of each angle. Click here to see answer by math-vortex(472)
Question 624988: hi, i am taking an online geometry class, one of the questions is "if angle B has a measure of (2x+8)what is its complment and supplement?" how to solve it is completely beyond me i am terribly comfused. i can figure out x being the angle measure, the complement is 90-x, and the supplement is 180-x. your help is greatly appreiciated. thank you so much! Click here to see answer by MathTherapy(1423)
Question 625263: two angles are supplementary angles if the sum of their measures is 180. Find two supplementary angles such that the measure of the first angle is 54 less than three times the measure of the second Click here to see answer by rfer(12657)
Question 626532: A building in kokomo, indiana is 139 ft tall. On a particular day at noon it casta a 219 ft shadow. What is the sun's angle of elevation at that time?
a 32.4 degrees
b. 57.6 degrees
c 39.4 degrees
d. 50.6 degrees Click here to see answer by stanbon(57307)
Question 626388: The complement of an angle, A, in degrees, is equal to one-fifth of the supplement of the angle, also in degrees. Then A =
A) 15°
B) 20°
C) 30°
D) 40°
E) 45°
F) 50°
G) 60°
H) 72°
I) 80°
J) none of these
Question 626628: the measure of one of the angles in a certain linear pair is 9 more than double the measure of the other angle what is the mesure of each angle
please help ..i really need this tomorrow
i just need a equation thank you ..:D Click here to see answer by stanbon(57307)
Question 628308: Hi!!
I am trying to teach my boys over the summer and I have many worksheets with answers. However, these specific ones on Special Angles & Algebra don't have a formula or answer key, so I will give you an example and see if you might be able to help. Thank you so much!!
Find the value of x. Write and solve an equation!
1. x = __________
the angle created on top is 2x - 3
Question: If angle B has a measure of (2x + 8) degrees, what is its complement?
Answer: The complement of angle B is (90 - (2x + 8)) degrees.
Question: If the complement of an angle A is equal to one-fifth of its supplement, what is the measure of angle A?
Answer: Angle A is 45 degrees.
Question: What is the measure of the second angle in the triangle where the first angle is 54 degrees less than three times the second angle, and they are supplementary?
Answer: The second angle is 42 degrees.
Question: What is the sum of the measures of the two supplementary angles in the triangle where the first angle is 15 degrees more than twice the second angle, and the third angle exceeds the second by 25 degrees?
Answer: The sum of the measures of the two supplementary angles is 180 degrees. | 677.169 | 1 |
angles between them must be 120".
30 In three dimensions, Steiner minimizes the total distance
45 f(x, y) = In(1- xy)
+ + +
Ax, y, z) = dl d2 d3 d, from four points. Show that
grad dl is still a unit vector (in which direction?) At what
Find f,, fy, f,,, fxy,fyy at the basepoint. Write the quadratic
angles do four unit vectors add to zero?
approximation to f(x, y) - the Taylor series through second-
31 With four points in a plane, the Steiner problem allows order terms:
branches (Figure 13.18~). Find the shortest network connect-
ing the corners of a rectangle, if the side lengths are (a) 1 and
2 (b) 1 and 1 (two solutions for a square) (c) 1 and 0.1.
32 Show that a Steiner point (120" angles) can never be out- 50 The Taylor series around (x, y) is also written with steps
side the triangle. hand k:Jx + h, y + k)=f(x,y)+ h +k +
33 Write a program to minimize f(x, y) = dl + d2 + d3 by 3h2- +hk + --..Fill in those four blanks.
Newton's method in equation (5). Fix two corners at (0, O), 51 Find lines along which f(x, y) is constant (these functions
(3, O), vary the third from (1, 1) to (2, 1) to (3, 1) to (4, l), and have f,, fyy=fa or ac = b2):
compute Steiner points. +
(a)f = x2 - 4xy 4y2 (b)f = eXeY
34 Suppose one side of the triangle goes from (- 1,0) to (1,O). 52 For f(x, y, z) the first three terms after f(O, 0,0) in the Tay-
Above that side are points from which the lines to (- 1, 0) and lor series are . The next six terms are
(1, 0) meet at a 120" angle. Those points lie on a circular arc-
draw it and find its center and its radius. 53 (a) For the error f -f, in linear approximation, the Taylor
series at (0, 0) starts with the quadratic terms
35 Continuing Problem 34, there are circular arcs for all three (b)The graph off goes up from its tangent plane (and
sides of the triangle. On the arcs, every point sees one side of
the triangle at a 120" angle. Where is the Steiner point?
f > f d if- . Then f is concave upward.
Question: What is the quadratic approximation to f(x, y) in the text? Answer: The quadratic approximation to f(x, y) is not explicitly given in the text.
Question: What is the function given for f(x, y)? Answer: f(x, y) = ln(1 - xy) | 677.169 | 1 |
Cue poorly-hand-drawn and photographed image... It was easier for me than drawing it on the computer, and hopefully gets the point across:
Shape A is a convex shape with an arbitrary number of edges/verts (but, if it helps, I can make it a low maximum number, like 8). Shape B may or may not be concave. Depending on which makes the maths easiest, I can define either shape as just their outside points (clockwise or anti-clockwise), or as a "triangle soup" that makes up the object.
What I want to work out, is that when A intersects B, to get some kind of workable description of C (either as the outside edge points, or as a "triangle soup"), which is the shape that's left from subtracting A from B. How might I go about doing something like that? Scissoring something to a rectangle is relatively straightforward, but I don't know how to go about doing something like this.
I can't say this is the most efficient way to do it, but this is how I would do it. I would take each line segment of the intersecting object, and then use the line intercept formula and go over each line segment in the intercepted object, so line1 of interceptor would iterate over all line segments of the intercepted object then line2 and etc..(keep in mind some intercept points, naturally, won't be found). Then I would find all the points of the interceptor that are located withing the intercepted. That should give you all your points for your new
Question: How can Shape A and Shape B be defined? Answer: Either by their outside points (clockwise or anti-clockwise) or as a "triangle soup" that makes up the object.
Question: What is the maximum number of edges/vertices that Shape A can have, according to the text? Answer: 8 (but it can be an arbitrary number) | 677.169 | 1 |
Imagine an infinitely long, straight, two-lane highway and an infinitely long, straight power line propped up on utility poles. Further imagine that the power line and the highway center line are both infinitely thin, and that the power line doesn't sag between the poles. Suppose the power line passes over the highway somewhere. Then the center line of the highway and the power line define skew lines.
Points, Lines, Planes, and Space Practice Problems
PROBLEM 1
Find an example of a theoretical plane region with a finite, nonzero area but an infinite perimeter.
SOLUTION 1
Examine Fig. 7-6. Suppose the three lines PQ, RS , and TU (none of which are part of the plane region X , but are shown only for reference) are mutually parallel, and that the distances d1 , d2 , d3 , . . . are such that d2 = d1 /2, d3 = d2 /2, and in general, for any positive integer n , d( n +1) = d n /2. Also suppose that the length of line segment PV is greater than the length of line segment PT . Then plane region X has an infinite number of sides, each of which has a length greater than the length of line segment PT , so its perimeter is infinite. But the interior area of X is finite and nonzero, because it is obviously less than the interior area of quadrilateral PQSR but greater than the area of quadrilateral TUSR .
Fig. 7-6 . Illustration for Problem 1.
PROBLEM 2
How many planes can mutually intersect in a given line L?
SOLUTION 2
In theory, an infinite number of planes can all intersect in a common line. Think of the line as an "Euclidean hinge," and then imagine a plane that can swing around this hinge. Each position of the "swinging plane" represents a unique plane in space.
Question: What is the minimum number of planes that can mutually intersect in a given line L? Answer: An infinite number of planes can mutually intersect in a given line L.
Question: Can the perimeter of plane region X in Problem 1 be finite? Answer: No, the perimeter of plane region X is infinite. | 677.169 | 1 |
That would be 270*t^8 under the radical sign, and the 3, up in the checkmark on the outside of the radical sign. 1 solutions Answer 177399 by Fombitz(13828) on 2009-11-24 08:00:01 (Show Source):
Proportions/238383: The ratio of the areas of two squares is 3 : 4. What is the ratio of the
lengths of their corresponding diagonals? 1 solutions Answer 175256 by Fombitz(13828) on 2009-11-13 10:43:25 (Show Source):
You can put this solution on YOUR website! The area of a square of side s is
The diagonal of the square is
So area as a function of diagonal is
The area is proportional to the square of the diagonals.
You can put this solution on YOUR website! For every triangle,
where A,B,C are the angles.
An obtuse angle is an angle greater than 90 degrees.
Let A be an obtuse angle.
Then A=90+e, where e is a positive value, less than 90.
B and C cannot equal 0.
Then
which means they are both acute.
You can put this solution on YOUR website! Since it's parallel, the left hand side will be identical.
Only the right hand side will change.
4x+2y=G
2y=-4x+G
y=-4x+(G/2)
"y-intercept of 2"
G/2=2
G=4
4x+2y=4
Question: What is the relationship between the area of a square and the length of its diagonal? Answer: The area is proportional to the square of the diagonals. | 677.169 | 1 |
health who are resonsible if the doctors office suffers a power outage. the doctor, or the agency managing the computerized data for the doctor????
math Find a parametric description for the ellipse having the focus F=(0,-4), corresponding directrix y=3, and eccentricity e=3/4.
math What is the distance between the parallel planes: ax+by+cz=d1 and ax+by+cz=d2?
math Is it possible for the lengths of a right triangle to form a geometric sequence? If yes, determine the lengths and the measures of the acute angles. if it's not possible, explain why not.
math illuminated by the rays of the setting Sun, Kelly rides along on a merry-go-round casting a shadow on a wall. The merry-go-round is turning 36 deg per sec. Kelly is 25 ft from its center, and the Sun's rays are perpendicular to the wall. At what rate is the shadow moving w...
Math What is the distance between the parallel planes 2x+y-3z=13 and 2x+y-3z=35?
Question: What is the distance between the two parallel planes given by the equations 2x+y-3z=13 and 2x+y-3z=35? Answer: The distance is 22. | 677.169 | 1 |
Trig Ratio: Around the World With Soh-Cah-Toa(Recently Revised!) This is a game I use with my students to review trig ratios. The game is played like around the world... two students go against each other, the first to say (or write) the answer wins and moves on to challenge the next student. There are 50 slides, plenty to engage a whole class for 50 minutes. The revised version is bold, clear, and easy to read on a white background. There are also several types of right triangles in various orientations to keep things exciting. Enjoy! And I almost forgot... answers are included in the notes view on PowerPoint.
Presentation (Powerpoint) File
Be sure that you have an application to open this file type before downloading and/or purchasing.
3690.5 actually used these as separate problems for different teams. I have also used it in three groups - one for initial check for understanding - one for homework - and one for review. It is a very flexible product! Thanks.
Here's how I have played it in the past:
1. Students sit in their seats.
2. One student begins by standing next to one of the seated students.
3. Put up the question, and the first to say the correct answer wins. (I print out the slide show ahead of time so I don't have to calculate on the spot)
4. If the standing student wins, they continue to the next seated student
5. When the standing student loses, they take the seated person's place and the new winner continues around the room (hence the name, around the world)
6. My classes were 50 minutes, so after let's say 30 minutes of game play, whoever won against the most people wins!
My students really enjoyed this game and it was also a good way for them to review and have to quickly figure out the trig functions.
February 29, 2012
dmaslanko
This game was a lot fun to play in the class as teams...however, it would have been better if there were more different sizes/orientations of the triangles, and some of the fill colors for the triangles were too dark -- I needed to lighten them so the students could see the right angle sign in order to identify the hypotenuse.
Thanks for your question! The slides are currently set up so that students can find the ratios in simplest form. The answers that I provide are all based on ratios, not angle measures, but you could easily adapt any of the slides to ask about angle measures as well. I like the ratios because students don't need a calculator, but adding another round of angle measures could be fun! Hope this answers your question.
Question: What is the time duration for which the game can engage a whole class? Answer: 50 minutes
Question: What is the name of the game described in the text? Answer: Around the World With Soh-Cah-Toa | 677.169 | 1 |
Semiregular Tessellations 1: Adventurous Ideas for Floor Tiling
If you are planning to tile your your floor and want something different from the usual tiling which is usually made up of rectangles and squares, the following semiregular tessellations might be of good use to you.
In the first figure, the tessellation is made of squares and octagons (8-sided polygons).
The second figure is made of squares, equilateral triangles, and hexagons (6-sided polygons).
The third fiugre is made of equilateral triangles and dodecagons (12-sided polygons).
The tessellations above were made manually by yours truly using GeoGebra. :-)
Question: Is there a tessellation made only of squares and rectangles? Answer: No. | 677.169 | 1 |
Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?
Orthogonal Circle
Stage: 5 Challenge Level:
Why do this problem?
The problem naturally brings in the formula for a circle and it
could be used to introduce this formula.
The problem requires knowledge and understanding of the formula for
the distance between two points, the use of Pythagoras Theorem and
algebraic skills in solving simultaneous equations.
The problem introduces the concept of orthogonality which is very
important in higher mathematics.
Possible approach
Depending on their experience of problem solving and thinking for
themselves, if there is some lack of confidence then the class
could first discuss how they are going to set about solving the
problem. This should be in the context of the teacher making it
explicit that the learners should be developing problem solving
skills and growing out of dependence on the teacher to indicate
methods to use.
List what information is given, what they are tryng to find,
and then how to use the information given.
In trying to find the orthogonal circle the 3 unknowns are its
radius and the coordinates of its centre.
The diagram suggests the use of 3 right angled triangles and
the use of Pythagoras' Theorem to give 3 equations.
How can they solve the equations. How can they eliminate one of
the variables between pairs of equations.
The learners can then work independently having planned how to
work on the problem.
Key questions
What information is given in the question?
What are the unknowns? Have we assigned notation to refer to
the unknowns?
Can we use the given information, and our mathematical
knowledge, to write down some equations?
What methods have we used in the past to solve similar
equations? Are those methods useful here
Question: What are the three unknowns that need to be determined to find the orthogonal circle? Answer: The radius of the circle and the coordinates of its center.
Question: What mathematical concept is introduced in this problem that is important in higher mathematics? Answer: The concept of orthogonality. | 677.169 | 1 |
A affects the amplitude of the graph and makes the circle smaller or larger. Say A=4 will make the graph have its y-maximum at 4 and y-minimum at -4; the circle will have a radius of 4. Say A=0.5 will make the graph have its y-maximum at 0.5 and y-minimum at -0.5; the circle will have a radius of 0.5. This means that the point along the circle will have more or less height over a certain amount of time.
B affects the wavelength distance and the speed a point moves along a circle. Say B=4 will make the wavelength shorter because it will travel more distance over a certain amount of time. B=4 will make the point in a graph move faster. B=0.5 will make the graph have a longer wavelength. B=0.5 will make a point in a graph travel much slower. This means that the point along the circle will travel in a slower or faster pace the height in a certain amount of time.
Sine is the height of the point along the circle over time. *Signs of B do affect the shape of the sine graph because it is not symmetrical. This means that a point in the opposite side of another point, won't have the same height.
Student C - The next two questions, #3 and #4, are fairly straight forward. The kids need to explain how you could manipulate sine and cosine functions in order to make the circle bigger, smaller, or to turn the rotational direction around.
3. In order for you to create either a larger or smaller circle you have to change the number in front of the sine or cosine formula. If you want the circle to be bigger the number in front has to be greater than 1 and if you want the circle to be smaller the number in front has to be less than 1. For example for a circle to be smaller the number has to be less than 1 such as C=(0.25cos(t), 0.25sin(t)). This will make the circle be smaller and have a radius of 0.25. On the other hand if you want the circle to become larger the value has to be replaced by a number greater than 1 such as C=(2cos(t), 2sin(t)). This will make the circle larger by augmenting its radius to 2.
4. In order for the point of the circle to rotate in a clockwise rotation you have to change the formula by adding a (-) sign before the (t) value. An example to make the rotation clockwise is C=(cos(-t),sin(-t)).
Student D - Questions 5 and 6 ask the student to analyze a given sine or cosine equation to tell me about the circular motion that is (partially) described. I'm looking for them to describe the size of the circle, its speed, the direction of rotation along the circle, and the (x, y) point along the circle where the point starts rotating at 0 seconds.
Question: Which parameter, A or B, affects the amplitude of the graph? Answer: A affects the amplitude of the graph.
Question: What is the effect on the wavelength when B is increased to 4? Answer: The wavelength becomes shorter. | 677.169 | 1 |
Just the concept of triangles is being tested here. Also the fact that the rectangle has rt angles @ 4 corners
Hence if u get one angle u can derive all the other angles and sum of all angles of a triangle is 180 hence u can find the reminaing angles.
Darshan
bmwhype2 wrote:
What is the concept being tested? Can someone explain the answer step by step. thanks
Question: What is the main concept being tested in the text? Answer: The concept of triangles and their properties, specifically that the sum of angles in a triangle is 180 degrees. | 677.169 | 1 |
Trapezium, Central Median of Trapezium, Triangle
A quadrangle with only two opposite sides parallel is called a trapezium, or trapezoid.
The sides of the trapezium that are parallel, are called bases and those that are not parallel are called legs. If the legs are equal in length, then this is an isosceles trapezoid. The distance between the bases is called height of trapezoid.
Central Median of Trapezium
This is the line segment joining the middles of the two sides of a trapezium that are not parallel. The central median of a trapezium is parallel to its sides.
Theorem:
If a line crosses the middle of one of the legs of a trapezium and is parallel to its bases, then it crosses the middle of the other leg.
Theorem:
The central median of a trapezium is the mean of the lengths of the two parallel sides
MN || AB || DC
AM = MD; BN = NC
MN central median, AB and CD
are bases, AD and BC are legs
MN = (AB + DC)/2
Theorem:
The central median of a trapezium is the mean of the lengths of the parallel sides.
Basic assignment: Prove that the central median of a trapezium halves each segment the ends of which lie on the two bases.
Central Median of a Triangle
The line segment joining the middles of two of the sides of a triangle is called central median of a triangle. It is parallel to the third side and its length is half the length of the third side. Theorem: If a line segment crosses the middle of one side of a triangle and is parallel to another side of the same triangle, then this line segment halves the third side.
AM = MC and BN = NC =>
MN || AB
MN = AB/2
Application of the properties of the central medians in a trapezium and triangle
Division of the segment into a given number of equal parts
Assignment: Divide the segment AB given into 5 equal parts.
Solution:
Let p be an arbitrary ray with A being the beginning, which does not lie on the AB straight line. We draw consecutively five equal segments on p AA1 = A1A2 = A2A3 = A3A4 = A4A5
We connect A5 with B and draw lines through A4, A3, A2 and A1 that are parallel to A5B. They cross AB respectively in the points B4, B3, B2 and B1. These points divide the segment AB into five equal parts. Indeed, from the trapezium BB3A3A5 we see that BB4 = B4B3. In the same way, from the trapezium B4B2A2A4, we obtain B4B3 = B3B2
While from the trapezium B3B1A1A3, B3B2 = B2B1.
Then, from B2AA2, it follows that B2B1 = B1A. We finally obtain :
Question: What is the distance between the bases of a trapezium called? Answer: Height of trapezoid | 677.169 | 1 |
The first spiral was made by 3rd grader Chris Y .
He made 20° angles around the center, ending up with a 70° spiral. The
Nautilus is a spiral of about 79.5°. He found ratios of the radius vectors
(from the center O, to points on the curve (almost) that are 360° apart) OB1/OA1= 3.20 and OB/OA= 3.45. This is a measure of the growth of the
Nautilus. Frank Land in "The Language of Mathematics" gives this value
of 3.2 for the Nautilus shell. For 2 times around Chris found the growth to be
9.00.
Directions to make a spiral this way is shown
near the bottom of this page.
The second spiral was made by SarahP.
Fine job Chris and Sarah!
Directions to make the spirals:
This graph above was done in Mathematica. Notice that
when it goes from 1 on the x-axis, counterclockwise 360°
it hits the x-axis again at 3.2. This shows that its growth in 2D is 3.2 times
in 360°, and is always that.
Alex plotted the Nautilus on polar graph paper
below. It didn't quite get to 3.2 after 360° because we only used 2-digit
accuracy in the equation r = e^(.18*theta).
Question: What is the final angle of Chris Y's spiral? Answer: 70° | 677.169 | 1 |
Loci
Rethinking Pythagoras
by Daniel J. Heath (Pacific Lutheran University)
Abstract: The Pythagorean Theorem, perhaps the most widely known result in mathematics, has been proven in countless ways, and remains a basic building block of Euclidean geometry. Pythagorean Theorem analogs in non-Euclidean geometries can provide a gateway into understanding those geometries.
This article includes activities which run in Java WebStart -- you will need the Java Runtime Environment, version 1.5 or later, in order to run these activities.
1. Introduction
How can we use calculus to enhance geometric understanding across multiple geometries? Here we discuss the Pythagorean Theorem and its "evil twins" in alternate geometries, ending, in the appendix, with discovery activities appropriate either for college geometry students or for accelerated calculus students wanting to use their knowledge to explore alternate geometries.
Many of the proofs of the Pythagorean Theorem begin with Figure 1a, showing a right triangle with squares constructed on each side. These proofs usually proceed by matching pieces from the smaller squares to sub-regions of the larger square. However, since squares do not exist in non-Euclidean geometries, we use circles to relate lengths of sides to areas, as in Figure 1b. In this case, if we call the lengths of sides of the triangle \(a\), \(b\), and \(c\) in ascending order of length, then it is easy to see that the circles have area \(\pi a^2\), \(\pi b^2\), and \(\pi c^2\), at least in Euclidean geometry. We note that the switch from squares to circles comes at a cost: figure 1b is both harder to interpret than 1a, and does not constitute a rigorous proof in the Euclidean case as does 1a. However, the universality of circles will make up for the messiness of the picture, and will allow us to conduct numerical explorations that lead us to conjectures in Euclidean, spherical, and hyperbolic geometries. These conjectures can be followed up with rigorous proof, and much of this article describes the "rigorization" of some of those conjectures.
Figure 1a.
Figure 1b.
In Section 2, we discuss the Pythagorean Theorem in the Euclidean plane. We move to its twin in spherical geometry in Section 3, rigorously treating circumference and area of circles on the unit sphere, then use dynamic software, together with our formula for area, to conjecture the Spherical Pythagorean Theorem. We repeat the process in the Poincaré Disc Model of hyperbolic geometry in Section 4a, and in the hyperboloid model (also known as the Minkowski-Weierstrass model) in Section 4b (both of which may be considered advanced). In all cases, we make use of the tools of calculus and offer evidence in support of our conjectures (in the way of other known calculus formulas that are implied by our answers), though we do not prove all of the conjectures. We summarize in Section 5.
Question: What is the purpose of using dynamic software in Sections 3 and 4? Answer: To conjecture the Spherical Pythagorean Theorem and the Pythagorean Theorem in hyperbolic geometry, respectively.
Question: Is the Pythagorean Theorem only applicable in Euclidean geometry? Answer: No.
Question: What does the author aim to do with the conjectures made through numerical explorations? Answer: Follow them up with rigorous proof.
Question: What is the main tool used throughout the article to support the conjectures? Answer: The tools of calculus. | 677.169 | 1 |
In the figure above, the smaller circles each have
radius 3. They are tangent to the larger circle at
points A and C, and are tangent to each other at
point B, which is the center of the larger circle.
What is the perimeter of the shaded region?
The diameter of a smaller circle is the radius of the larger circle. So the larger circle has radius 6. The circumference of the large circle is then 12pi, and the circumference of a small circle is 6pi. Now if we travel along the perimeter of the shaded region we trace out half the circumference of the large circle, then half the circuference of a small circle, then half the circuference of a small circle once more. So the answer is 12pi/2 +6pi/2 + 6pi/2 = 12pi.
It's always good to test things out for SAT math - better safe than sorry, right?
20) Ok, so if j, k, and n are consecutive, so if the units' digit of jn is 9, j must end in 9 too... e.g. 9, 19, 29... etc, and n will be 1, 11, 21, etc. k, then, will be something like 10, 20, 30, etc. so its units digit is 0. SWAG. The answer is (A)!
<Any 2 points determine a line. If there are 6 points in a plane, no 3 of which lie on the same line, how many lines are determined by pairs of these 6 points?>
My answer was 30. This is because every point can form a line with one of the five other points, meaning the possibilities are 6 x 5 (assuming one point can be part of multiple lines).
I feel the difficulty lies in deciphering the prompt, is my interpretation of "how many lines are 'determined'" correct? What does the prompt mean, exactly?
A list consists of 1000 consecutive even integers. What is the difference between the greatest number in the list and the least number in the list?
Answer is................................................ .................................................. ..............1998.
I got 2000. Because I had 0 as my first even integer. Then 2000 as my largest. I got 2000 as my largest even integer because I noticed that 20 was the 10th consecutive integer and 30 was the 15th consecutive integer so to get the 1000th even integer I multiplied 1000 by 2. Where did I go wrong? Thanks in advance!
@tranman: You know where you went wrong? If you take 0 as your first and 2000 as your last value you create a range of 1001 integers not 1000.
Question: What is the relationship between the diameter of a smaller circle and the radius of the larger circle in the given figure?
Answer: The diameter of a smaller circle is equal to the radius of the larger circle.
Question: What is the perimeter of the shaded region?
Answer: 12π
Question: If there are 6 points in a plane, with no three of them lying on the same line, how many lines are determined by pairs of these points?
Answer: 15 (since each point can pair with 5 others, and there are 6 points, but we must divide by 2 to avoid double-counting: 6*5/2 = 15)
Question: If j, k, and n are consecutive integers and the units' digit of jn is 9, what must the units digit of j be?
Answer: 9 | 677.169 | 1 |
Fig. 5 "Semi-circle and circular segment" from [Heath, 392]
One first constructs the semicircle ABC centered at G and a second circular segment DEF such that the circumference of ABC is equal to that of DEF. Construct H as the center of the circle DEF and draw EHK and BG perpendicular to DF and AC, respectively. Finally, draw DH and the line LHM parallel to DF. As summarized in [Heath, 393], Pappus arrives at the following conclusions:
LH: AG = (arc LE): (arc AB)
= (arc LE) : (arc DE)
= (sector LHE) : (sector DHE)
and, because circles are as the squares on their radii,
LH2: AG2 = (sector LHE) : (sector AGB).
Therefore, it follows that
(sector LHE) : (sector DHE) = (sector DHE) : (sector AGB).
In a preliminary lemma, Pappus had shown that
(sector EDH): (half segment EDK) > (right angle) : (∠ DHE).
Note that this is equivalent to the modern statement that
2a
2a-sin2a
>
p/2
a
orto
q
q-sinq
>
p
q
,
where a = ÐDHE, q = 2a, and 0 < q < p. (You can convince yourself of the truth of the inequality by plotting both sides on a graphing calculator.)
Then
(sector EDH) : (half segment EDK) > (∠ LHE) : (∠ DHE)
> (sector LHE) : (sector DHE)
> (sector EDH) : (sector AGB).
"Therefore," concludes Heath, "the half segment EDK is less than the half semicircle AGB, whence the semicircle BC is greater than the segment DEF" and the semicircle is established as the figure of maximum area. Pappus's treatment of the isoperimetric problem was, therefore, an important step not only in developing ancient mathematical philosophy, but also in preserving and adding to the body of work already existing on Dido's now famous problem.
Question: What is the isoperimetric problem? Answer: The isoperimetric problem is a mathematical problem that seeks to find the shape with the maximum or minimum area for a given perimeter. | 677.169 | 1 |
taught, its value would scarcely have required
insisting on. But the didactic method hitherto
used in teaching it does not exhibit its powers
to advantage.
Any true geometrician who wjll teach practi-
cal geometry by definitions and questions there-
on, will find that he can thus create a far great-
er interest in the science than he can by the
usual course ; and, on adhering to the plan, he
will perceive that it brings into earlier activity
that highly-valuable but much-neglected power,
the power to invent. It is this fact that has in-
duced the author to choose as a suitable name
for it, the inventional method of teaching prac
tical geometry.
He has diligently watched its effects on both
*exes, and his experience enables him to say
g INTRODUCTION.
that its tendency is to lead the pupil to rely on
his own resources, to systematize his discoveries
in order that he may use them, and to gradually
induce such a degree of self-reliance as enables
him to prosecute his subsequent studies with sat-
isfaction: especially if they st'onld happen to
be such studies as Euclid's "Elen-ents," the nse
of the globes, or perspective.
A word or two as to using the definitions
and questions. Whether they relate to the
mensuration of solids, or surfaces, or of lines ;
\vhether they Belong to common square meas-
ure, or to duodecimals ; or whether they apper-
tain to the canon of trigonometry ; it is not the
author's intention that the definitions should be
learned by rote ; but he recommends that the
pupil should give an appropriate illustration oi
each as a proof that he understands it.
Again, instead of dictating to the pupil how
to construct a geometrical figure say a square
and letting him rest satisfied with being able
to construct one from that dictation, the author
has so organized these questions that by doing
justice to each in its turn, the pupil finds that,
INTRODUCTION. 9
when he comes to it, he can construct a square
without aid.
The greater part of the questions accompany-
ing the definitions require for their answers ge-
ometrical figures and diagrams, accurately con-
structed by means of a pair of compasses, a scale
of equal parts, and a protractor, while others
require a verbal answer merely. In order to
place the pupil as much as possible in the state
in which Nature places him, some questions
have been asked that involve an impossibility.
Whenever a departure from the scientific
order of the questions occurs, such departure
has been preferred for the sake of allowing
time for the pupil to solve some difficult prob-
lem ; inasmuch as it tends far more to the for-
mation of a self-reliant character, that the pupil
should be allowed time to solve such difficult
problem, than that he should be either hurried
or assisted.
Question: What tools are required for answering most of the questions accompanying the definitions? Answer: A pair of compasses, a scale of equal parts, and a protractor.
Question: What kind of questions are also included to place the student in a more natural state? Answer: Questions that involve an impossibility.
Question: What does the author prefer to allow time for, in order to foster a self-reliant character in the student? Answer: The author prefers to allow time for the student to solve difficult problems.
Question: Which famous work by Euclid is mentioned as a suitable subsequent study for the student? Answer: Euclid's "Elements". | 677.169 | 1 |
Assessment
- HW - Daily participation -Tests - Test corrections - Group Work
Outcomes
Students will: - Find the measure of an angle in either degrees or radians and to find coterminal angles; - Find the arc length and area of a sector of a circle and to solve problems involving apparent size; - Use the definitions of sine and cosine to find values of these functions and to solve simple trigonometric equations; - Use reference angles, calculators, or special angles to find values of the sine and cosine functions and to sketch the graphs of these functions; - Find values of the tangent, cotangent, secant, and cosecant functions and to sketch the functions' graphs; - Find values of the inverse trigonometric functions.
Question: How many trigonometric functions do students learn to find values of and sketch their graphs?
Answer: 8 (sine, cosine, tangent, cotangent, secant, cosecant, and their inverses) | 677.169 | 1 |
Let p be the perpendicular on C from a given origin 0, and let w be the inclination of p (we may put dw for d0), C will be a given function of p, w; and, integrating first for w constant, the whole number of cases for which w falls between given limits w', co" is 3 dwJC3dp; the integral fC 3 dp being taken for all positions of C between two tangents to the boundary parallel to PQ. The question is thus reduced to the evaluation of this double integral, which, of course, is generally difficult enough; we may, however, deduce from it a remarkable result; for, if the integral 3 f f C 3 dpdw be extended to all possible positions of C, it gives the whole number of pairs of positions of the points A, B which lie inside the area; but this number is S2 2; hence fJC3dpdw=3522, the integration extending to all possible positions of the chord C, - its length being a given function of its co-ordinates p, w.2 ' The line might be anywhere within the circle without altering the question.
This integral was given by Morgan Crofton in the Cornptes rendus (1869), p. 1469. An analytical proof was given by Serret, Annales scient. de l'ecole normale (1869), p. 177.
Hence if L, S2 be the perimeter and area of any closed convex contour, the mean value of the cube of a chord drawn across it at random is 3522/L.
86. Let there be any two convex boundaries (fig. 5) so related that a tangent at any point V to the inner cuts off a constant segment S from the outer (e.g. two concentric similar ellipses); let the annular area between them be called A; from a point X taken at random on this annulus draw tangents XA, XB to the inner. The mean value of the FIG. 5.
The following lemma will first be proved: If there be any convex arc AB (fig. 6), and if N 1 be (the measure of) the number of random lines which meet it once, N2 the number which meet it twice, A 6 2 arc AB = N1+2N2. FIG. 6. For draw the chord AB; the number of lines meeting the convex figure so formed is N i -1-N2 =arc +chord (the perimeter); but N1= number of lines meeting the chord = 2 chord; 2 arc + N i = 2N2 -12N2, .. 2 arc = Ni-(-2N2.
Question: What is the area of the closed convex contour in the text? Answer: S²
Question: What is the perimeter of the closed convex contour in the text? Answer: L | 677.169 | 1 |
90. The calculation of geometrical probability and expectation is much facilitated by the following general principle: If M be a mean value depending on the positions of n points falling on a space A; and if this space receive a small increment a, and M' be the same mean when the n points are taken on A+a, and M the same mean when one point falls on a and the remaining nI on A; then, the sum of all the cases being M'(A+ a) n , and this sum consisting of the cases (i) when all the points are on A, (2) when one is on a the others on A (as we may neglect all where two or more fall on a), we have M'(A+a) n = MAn+nMlan -1; .'. (M' -M)A=nA(M1 -M), as M' nearly =M. For example, suppose two points X, Y are taken in a line of length 1, to find the mean value M of (XY) n. If 1 receives an increment dl, ldM=2d1(M 1 -M). Now M 1 here= the mean nth power of the distance of a single point taken at random in 1 from one extremity of 1; and this is l n (n+I) 1 (as is shown by finding the chance of n other points falling on that distance); hence 1dM =2d1{l n (n+1) - 1 - M }; 1dM +2Mdl = 2 (n+ 1)-1lndl, or 1-1.d . M12 =2 (n +1)-1lndl; Ml 2 =z (n + 1) - lf l n+ l dl = 21n +2/(n +I) (n +2)+C; .'. N = J 21 n 1(n+ i) (n +2), C being evidently o.
Question: In the example given, what is the length of the line? Answer: 1
Question: What is the final formula for M in the example? Answer: M = 2l n + 2/(n +1) (n +2) + C, where C is a constant. | 677.169 | 1 |
91. The corresponding principle for probabilities may thus be stated: If p is the probability of a certain condition being satisfied by the n points within A in art. 90, p the same probability when they fall on the space A+a, and p the same when one point falls on a and the rest on A, then, since the numbers of favourable cases are respectively p'(A+a)", pA n, np l aA n - 1 , we find ( p '- p) A= na (p i - p). Hence if p' = p then p 1 = p. For example, if we have to find the chance of three points within a circle forming an acute-angled triangle, by adding an infinitesimal concentric ring to the circle, we have evidently p' = p; hence the required chance is unaltered by assuming one of the three points taken on the circumference. Again, in finding the chance that four points within shall form a convex quadri lateral, if we add to the triangle a small band between the base and a line parallel to it, the chance is clearly unaltered. Therefore we may take one of the points at random on the base (fig. 8), the others X, Y, Z within the triangle. Now the four lines from the vertex B to the four points are as likely to occur in any specified order as any other. Hence it is an even chance that X, Y, Z fall on one of the triangles ABW, CBW, or that two fall on one of these triangles and the remaining one on the other. Hence the probability of a re-entrant quadrilateral is Zpl+ 2P2, where p ' = prob. (Wxyz re-entrant), X, Y, Z in one triangle; P2 = do., X in one triangle, Y in the other, Z in either.
But 1,1=4. Now to find p 21 the chance of Z falling within the triangle WXY is the mean area of WXY divided by ABC. Now by par. 89, for any particular position of W, M(WXY) =WGG', where G, G' are the centres of gravity of ABW, CBW. It is easy to see that WGG' =;ABC =i, putting ABC = i. Now if Z falls in CBW, the chance of Wxyz re-entrant is 2M(IYW), for Y is as likely to fall in WXZ as Z to fall in WXY; also if Z falls in ABW the chance of Wxyz re-entrant is 2M(IXW). Thus the whole chance is p 2 =2M(IYW+IXW) = y. Hence the probability of a re-entrant quadrilateral is 1 .} 1 2 - 1 2 $+3 9 - 3 That of its being convex is 3.
Question: What is the probability of a re-entrant quadrilateral being convex? Answer: The probability is 3/9 - 3/4 = 1/12. | 677.169 | 1 |
A three-dimensional figure has plane symmetry if a plane can divide the figure into two congruent reflected halves.
Symmetry About an Axis
A three-dimensional figure has symmetry about an axis if there is a line about which the figure can be rotated (by an angle greater than 0 degrees and less than 360 degrees) so that the image coincides with the preimage.
Tessellation
A tessellation is a repeating pattern that completely covers a plane with no gaps or overlaps. The measures of the angles that meet at each vertex must add up to 360 degrees.
A regular tessellation is formed by congruent regular polygons. To form a regular tessellation, the angle measures of a regular polygon must be a divisor of 360 degrees.
A semiregular tessellation is formed by two or more different regular polygons, with the same number of each polygon occurring in the same order at every vertex. The angle measures around a vertex must add up to 360 degrees.
Dilations
A dilation is a transformation that changes the size of a figure but not the shape. The image and the preimage of a figure under dilation are similar.
In a dilation, the lines connecting points of the image with the corresponding points of the preimage all intersect at the center of dilation.
The scale factor of a dilation is the ratio of the linear measurement of the image to a corresponding measurement of the preimage. For a dilation with a scale factor of k, if k > 0, then the figure is not turned or flipped. If k < 0 (negative), then the figure is rotated by 180 degrees.
Question: In a semiregular tessellation, how many of each regular polygon occur at every vertex? Answer: The same number of each polygon | 677.169 | 1 |
Bisected hexagonal tiling
In geometry, the bisected hexagonal tiling is a tiling of the Euclidean plane. It is an equilateral hexagonal tiling with each hexagon divided into 12 triangles from the center point. (Alternately it can be seen as a bisected triangular tiling divided into 6 triangles.)
It is labeled V4.6.12 because each right triangle face has three types of vertices: one with 4 triangles, one with 6 triangles, and one with 12 triangles. It is the dual tessellation of the great rhombitrihexagonal tiling which has one square and one hexagon and one dodecagon at each vertex.
Related polyhedra and tilings
It is topologically related to a polyhedra sequence defined by the face configurationV4.6.2n. This group is special for having all even number of edges per vertex and form bisecting planes through the polyhedra and infinite lines in the plane, and continuing into the hyperbolic plane for any n.
With an even number of faces at every vertex, these polyhedra and tilings can be shown by alternating two colors so all adjacent faces have different colors.
Each face on these domains also corresponds to the fundamental domain of a symmetry group with order 2,3,n mirrors at each triangle face vertex.
Practical uses
The bisected hexagonal tiling is a useful starting point for making paper models of deltahedra, as each of the equilateral triangles can serve as faces, the edges of which adjoin isosceles triangles that can serve as tabs for gluing the model together.
Question: How many triangles are formed when each hexagon is divided? Answer: 12
Question: What is the tiling's Schläfli symbol? Answer: V4.6.12 | 677.169 | 1 |
Loci: Convergence
An Investigation of Historical Geometric Constructions
by Suzanne Harper and Shannon Driskell
Hippias' Attempt to Trisect an Angle
One of the most famous attempts to trisect an angle has been attributed to Hippias of Elis (born around 460 B.C.E.). He was a statesman and philosopher who traveled around being paid for his lectures on poetry, grammar, history, politics, archaeology, mathematics and astronomy. Plato described him as a vain man being both arrogant and boastful, having a wide but superficial knowledge (Burton, 2003 ). Hippias' contribution to mathematics was small, but significant. In his attempt to trisect an angle, he created a new transcendental curve which unfortunately could not be constructed with only a compass and unmarked ruler; but the curve can be used to divide an angle into not only three, but any number of congruent angles.
Question: What was the main achievement of Hippias in the field of mathematics? Answer: He attempted to trisect an angle | 677.169 | 1 |
Ozzie's Answer:
A rule of polygons is that the sum of the exterior angles always equals 360 degrees, but lets prove this for a regular octagon (8-sides).
First we must figure out what each of the interior angles equal. To do this we use the formula:
((n-2)*180)/n where n is the number of sides of the polygon. In our case n=8 for an octagon, so we get:
((8-2)*180)/8 => (6*180)/8 => 1080/8 = 135 degrees. This means that each interior angle of the regular octagon is equal to 135 degrees.
Each exterior angle is the supplementary angle to the interior angle at the vertex of the polygon, so in this case each exterior angle is equal to 45 degrees. (180 - 135 = 45). Remember that supplementary angles add up to 180 degrees.
And since there are 8 exterior angles, we multiply 45 degrees * 8 and we get 360 degrees.
This technique works for every polygon, as long as you are asked to take one exterior angle per vertex M
Question: What is the sum of the exterior angles of any polygon? Answer: 360 degrees | 677.169 | 1 |
I also created a PowerPoint to go along with the foldable. It's nothing fabulous, it's just a slide show of the example I will be doing in class, but you are welcome to it.
I was curious how the students would react to something like this. I find that students in this class are typically serious and not interested in stuff like this. I found that the students liked it. Not only could I tell by their reaction, but I came right out and asked them.
3 comments:
In General, Quadrilateral means four sides but in geometry we can say any four sided shape is quadrilateral with a condition sides have to be straight and it has to be 2 dimensional. example a square, rhombus, trapezium and rectangle are also parallelograms.
Question: What is a quadrilateral according to the text? Answer: A four-sided shape with straight sides and in 2 dimensions | 677.169 | 1 |
Examples of Different Rotational Symmetry Order
Is there Rotational Symmetry of Order 1 ?
Not really! If a shape only matches itself once as you go around (ie it matches itself after one full rotation) there is really no symmetry at all, because the word "Symmetry" comes from syn- togetherand metron measure, and there can't be "together" if there is just one thing.
Question: How many times does a shape with rotational symmetry of order 1 match itself when rotated? Answer: It matches itself only once. | 677.169 | 1 |
Even though you will rarely see the phrases "complementary angles" or "supplementary angles" in advanced geometry problems, the concepts themselves are crucial. Oftentimes, in advanced geometry problems, you might see something like "Prove that ABC + CBD = 90" instead of "Prove that ABC and CBD are complementary" because foreign students might not be familiar with the terminology. In either case, you will still need this concept.
Question: Which of the following is NOT a reason given in the text for why foreign students might not be familiar with the terminology? A) They don't understand English B) They might not be familiar with the terminology C) They are not good at geometry Answer: A) They don't understand English | 677.169 | 1 |
Calculating Radius at End of Taper
I think I'd like to mill it like this, with a radius at the end of the tapers:
...which raises the question of how I calculate the radius.
I found the following in Holbrook Horton's "Mathematics at Work", 4th ed., p. 13-2.:
Horton solves for the diameter bc, given radius de and the angle of taper A as known.
My question: I'm wondering if its possible to solve for the radius de given bc and the angle of taper A as known, or some other known values, such as the length from the beginning of the taper to b and c???? I want to be able to drop that radius value into a CNC G03 command.
The formula Horton gives to solve for bc is:
The derivation goes like this:
Thanks for any help with this. I'm just starting to learn trig, and get a little lost around cotangents, but I can plug known dimensions into formulas.
Thanks a million for that. Exactly what I was looking for. Really appreciate it.
I found your diagram super helpful. What did you create it with??? I'm a little limited in what I can work out with the software I have, and would love to get something inexpensive that would let me do exactly what you've done: draw angles and radii to scale.
I can't afford anything fancy, just want to draw angles and radii to scale, literally. Can you think of anything???
Question: What is the angle 'A' in the diagram? Answer: The angle 'A' is not explicitly stated in the text, but it is referred to as the angle of taper.
Question: Can the radius 'de' be calculated given the diameter 'bc' and the angle of taper 'A'? Answer: Yes, it is possible to solve for the radius 'de' given the diameter 'bc' and the angle of taper 'A'. | 677.169 | 1 |
Monday, September 14, 2009 at 10:24pm by MathMate Of the three possible solutions, which numbers can be used in the ...
Wednesday, October 21, 2009 at 12:59pm by Kelly
math Given that a sports arena will have a 1400 meter perimeter and will have semi- circles at the ends with a possible rectangular area between the semi-circles, determine the dimensions of the rectangle and semi-circles that will maximize the total area.
Wednesday, May 8, 2013 at 1:03am by sunny
Math kindergarten You can use circle..make 6 circles=6 8 circles=8
Tuesday, October 9, 2012 at 6:07pm by Anonymous
math Verify that the circles x2+y2 = 25 and (x−5)2+(y−10)2 = 50 intersect at A = (4, 3). Find the size of the acute angle formed at A by the intersecting circles. You will first have to decide what is meant by the phrase the angle formed by the intersecting ...
Sunday, April 25, 2010 at 7:40pm by Ella
GEOMETRY(CIRCLE) Three circles with different radii have their centers on a line. The two smaller circles are inside the largest circle, and each circle is tangent to the other two. The radius of the largest circle is 10 meters. Together the area of the two smaller circles is 68% of the area ...
Monday, March 25, 2013 at 2:11am by i need help from anyone!
Area of Circles Find the total area of three circles, each with a radious of 1 1/2 feet. Let ii stand for pie Te book gives the answer as 27ii/4 aprroimately 21.195 I know the equation and can calculate the 27. A=ii (1)^2 A=ii (1.5)^2=3 3 X 3 circles =27ii I am not clear where the 4 is coming...
Saturday, May 9, 2009 at 2:05pm by G
PLEASE HELP TANGENTS AND CIRCLES PROBLEM GEOMETRY PLEASE HELP I REALLY NEED HELP. Each of the three circles in the figure below is externally tangent to the other two, and each side of the triangle is tangent to two of the circles. If each circle has radius three, then find the perimeter of the triangle. The figure is ...
Saturday, February 2, 2013 at 3:44pm by Knights
Question: Who is the author of the first post? Answer: MathMate | 677.169 | 1 |
Midpoints of Segments Lesson Packet lesson develops the concept of the midpoint of a segment and then builds slowly upon the concept so that students will be able to set up algebraic equations to find lengths of segments. Students will be engaged in hands on and minds on learning.
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing.
606.01
Question: What is the file size of the lesson packet? Answer: 606.01 (likely in kilobytes or megabytes, but the unit is not specified) | 677.169 | 1 |
If a component points to the left or downwards, it is given a negative sign (-).
2
Add all the magnitudes of the horizontal components (or those along the x-axis) together. Separately, add all the magnitudes of the vertical components (or those along the y-axis). If a component has a negative sign (-), its magnitude is subtracted, rather than added.
3
Calculate the magnitude of the resultant using the Pythagorean Theorem. The theorem may be stated: c2=a2+b2, where c is the magnitude of the resultant vector, a is the magnitude of the sum of the components along the x-axis, and b is the magnitude of the sum of the components along the y-axis.
4
Calculate the angle that the resultant makes with the horizontal (or the x-axis). Use the formula θ=tan-1(b/a), where θ is the angle that the resultant makes with the x-axis or the horizontal.
5
Represent your resultant vector.
For example, if the vectors represented forces, then write "A force of x N at yo to the horizontal/x-axis/etc".
Method Three: Vector Subtraction
1
Subtract by adding a negative. Subtracting a vector from another can be seen as adding its "negative".
2
Find the negative of the vector.
Find the negative of the vector. This is the same in magnitude to the original vector, but opposite in direction. You can represent it by making the original vector, but drawing the arrow the other way round, so that the tail becomes the head and the head becomes the tail.
3
Follow either addition method above, using the negative. Use either of the two addition methods described above to add the "negative" of the vector to be subtracted and the vector it had to be subtracted from
Question: What is the direction of the negative of a vector? Answer: Opposite to the original vector. | 677.169 | 1 |
Even if a triangle center function is well-defined everywhere the same cannot always be said for its associated triangle center. For example let f(a, b, c) be 0 if a/b and a/c are both rational and 1 otherwise. Then for any triangle with integer sides the associated triangle center evaluates to 0:0:0 which is undefined.
Default domain
In some cases these functions are not defined on the whole of ℝ3. For example the trilinears of X365 are a1/2 : b1/2 : c1/2 so a, b, c cannot be negative. Furthermore in order to represent the sides of a triangle they must satisfy the triangle inequality. So, in practice, every function's domain is restricted to the region of ℝ3 where a ≤ b + c, b ≤ c + a, and c ≤ a + b. This region T is the domain of all triangles, and it is the default domain for all triangle-based functions.
Other useful domains
There are various instances where it may be desirable to restrict the analysis to a smaller domain than T. For example:
When differentiating between the Fermat point and X13 the domain of triangles with an angle exceeding 2π/3 is important,
in other words triangles for which a2 > b2 + bc + c2 or b2 > c2 + ca + a2 or c2 > a2 + ab + b2.
A domain of much practical value since it is dense in T yet excludes all trivial triangles (ie points) and degenerate triangles
(ie lines) is the set of all scalene triangles. It is obtained by removing the planes b = c, c = a, a = b from T.
Domain symmetry
Not every subset D ⊆ T is a viable domain. In order to support the bisymmetry test D must be symmetric about the planes b = c, c = a, a = b. To support cyclicity it must also be invariant under 2π/3 rotations about the line a = b = c. The simplest domain of all is the line (t,t,t) which corresponds to the set of all equilateral triangles.
so f is a triangle center function. Since the corresponding triangle center has the same trilinears as the circumcenter it follows that the circumcenter is a triangle center.
1st isogonic center
Let A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let AB'C and ABC' be similarly constructed equilateral triangles based on the other two sides of triangle ABC. Then the lines AA', BB' and CC' are concurrent and the point of concurrence is the 1st isogonic center. Its trilinear coordinates are
csc(A + π/3) : csc(B + π/3) : csc(C + π/3).
Expressing these coordinates in terms of a, b and c, one can verify that they indeed satisfy the defining properties of the coordinates of a triangle center. Hence the 1st isogonic center is also a triangle center.
Fermat point
1
Question: What is the defining property of the coordinates of a triangle center? Answer: The coordinates of a triangle center must satisfy the defining properties of the triangle center function.
Question: Is the triangle center function always well-defined for all triangles? Answer: No, it is not always well-defined.
Question: Which triangles are excluded from the set of all scalene triangles? Answer: The planes b = c, c = a, a = b are removed.
Question: Is the 1st isogonic center also a triangle center? Answer: Yes, it is, as its trilinear coordinates satisfy the defining properties of a triangle center. | 677.169 | 1 |
eIt
is Euler's number and is defined to be the limit of (1 +
1/n)n as n approaches
infinity.
Eccentricity
For a conic section, it is defined as the ratio c/a.
For the ellipse the eccentricity is between 0 and 1 (including
0, but not 1). The eccentricity is the amount of roundness.
If the eccentricity is 0, then the conic is a circle. For
the parabola it is 1 and for the hyperbola it is greater
than 1.
Edge A line or a
line segment that is the intersection of two plane faces
of a geometric figure, or that is in the boundary of a plane
figure.
Elementary
row operations There are four elementary row operations
for producing equivalent matrices:
(1) RowSwap Interchange any two rows.
(2) Row+ Row addition; add a row to any other row.
(3) *Row Scalar multiplication; multiply (or divide) all
the elements of a row by the same nonzero real number.
(4) *Row+Multiply all the entries of a row (pivot row) by
a nonzero real number and add each resulting product to
the corresponding entry of another specified row (target
row).
Ellipse The set
of all points in a plane such that, for each point on the
ellipse, the sum of its distances from two fixed points
(called the foci) is a constant.
Elliptic geometry
A non-Euclidean geometry in which a Saccheri quadrilateral
is constructed with summit angles obtuse.
Equal to Two
numbers are equal if they represent the same quantity or
are identical. In mathematics, a relationship that satisfies
the axioms of equality.
Equality,
axioms of For real numbers a, b, and c: Reflexive property: a
= a
Symmetric property: If
a = b, then b = a.
Transitive property: If
a = b and b = c, then a = c.
Substitution property: If a = b,
then a may be replaced throughout by b (or
b by a) in
any statement without changing the truth or falsity
of the statement.
Equation A statement
of equality. If always true, an equation is called an identity;
if always false it is called a contradiction. If it is sometimes
true and sometimes false, it is called a conditional equation.
Values that make an equation true are said to satisfy the
equation and are called solutions or roots of the equation.
Equations with the same solutions are called equivalent
equations.
Equation
of a graph Every point on the graph has coordinates
that satisfy the equation, and every ordered pair that satisfies
the equation has coordinates that lie on the graph.
Equation
properties There are four equation properties:
(1) Addition property: Adding the same number to both sides
of an equation results in an
equivalent equation.
(2) Subtraction property: Subtracting the same number from
Question: Are two numbers equal if they represent the same quantity? Answer: Yes.
Question: What are the four elementary row operations for producing equivalent matrices? Answer: The four elementary row operations are: (1) RowSwap, (2) Row+, (3) Row (scalar multiplication), and (4) Row+ (row multiplication and addition).
Question: What is Euler's number? Answer: Euler's number, e, is defined as the limit of (1 + 1/n)^n as n approaches infinity.
Question: What is an ellipse defined as? Answer: An ellipse is defined as the set of all points in a plane such that, for each point on the ellipse, the sum of its distances from two fixed points (called the foci) is a constant. | 677.169 | 1 |
Supplementary Topic
The Value of p ...
This section of the supplementary topic is purely for your own joy and interest
... it is not needed to understand any of the relevant topics that are discussed.
The value of p = 3.14159265 ... which appears in
defining the area of a circle and in measuring angles
in units of radians is really no mystery. This
value can best be thought of as the ratio of the circumference of a circle
to its diatmeter. One can simply try to "draw" a perfect circle
and measure the circumference and the diameter, and dividing the circumference
by the diameter will give the value of p. It turns
out that no matter how small or how big of a circle on chooses to draw, this
ratio comes out to be a constant value that is roughly equal to 3.14159265 ...
(within errors of measurement and drawing "perfect" figures and so
on ...).
The above geometrical idea is the best way at this level to understand where
the value of p comes from ... if you are inclined
to think practically you can always imagine doing an experiment like the above
and getting values for p and then trying to "perfect"
your experiment.
Another way to think about the value of p is perhaps
in terms of some of our trigonometric functions.
If you look back at these functions you will realize that we defined the sine,
cosine, and tangent functions with the geometry of a right triangle and we said
that these were functions that applied to angles. In other words, we realized
that once we have an angle we can take its sine, cosine or tangent and this
gave us the ratio of the sides of a right triangle (which sides depended on
which function we were looking at). Now consider a right triangle whose non-right
angles are 45º (this is the nicest right triangle you can picture). This
triangle will have both of its legs equal in length (since both of the non-right
angles are 45º, and are equal).
The tangent of an angle is equal to the ratio of opposite to adjacent side
in a right triangle. Thus, the tangent of 45º is equal to 1 since the opposite
and adjacent sides in a 45º right triangle are equal in length.
If you recall from units of angles, a 45º angle is equal to p/4
radians. Thus:
tan (p/4) = 1
We can try to find the value of p by taking the
inverse tangent (labeled Arctan) of 1 and multiplying it by 4. If you
do this on your calculator you must be in radians mode (since the angle
p/4 is being specified in radians):
p/4 = Arctan(1) p = 4 x Arctan(1) = 3.14159265 ...
Actually ... there is a way to calculate Arctan(1) without a calculator and
Question: What is the angle in radians that corresponds to 45 degrees? Answer: p/4 radians
Question: Can the value of p change based on the size of the circle? Answer: No, it remains a constant value regardless of the circle's size.
Question: What is the ratio that p represents? Answer: The ratio of the circumference of a circle to its diameter.
Question: Which trigonometric function can be used to find the value of p? Answer: Tangent | 677.169 | 1 |
If ABCD is a parallelogram what is the value of x?
In homework questions such as these, the value of x is usually the length of one of the sides, although it's not uncommon for it be the area or perimeter. Only rarely is x one of the angles, since θ is normally used for this purpose. Please include all the relevant information when asking for homework help in future.
Question: Which of the following could x represent? A) The number of sides B) The area C) The perimeter D) The number of angles Answer: B and C | 677.169 | 1 |
Thursday, February 9, 2012
Identify if a given graph is a scorpion or not.
A scorpion is an undirected graph with 3 special vertices: the sting, the tail, and the body.
The sting has degree one and is connected to the tail. The tail has
degree two and is connected to the sting and the body. The body has
degree n – 2 and is connected to all vertices except the sting. The
other vertices may be arbitrarily connected with each other. Identify if
a given graph is a scorpion or not.
Question: Which two vertices are connected to the tail? Answer: The sting and the body | 677.169 | 1 |
Question 241572: Two angles are complementary. The sum of the measure of the first angle and half the second angle is 72.5. Find the measure of the angles.
what is the measure of the smaller angle?
what is the measure of the other angle? Click here to see answer by stanbon(57307)
Each time that I tried to solve these two (seperate) problems....kept getting an answer that does not make any sense....I think that I am doing something wrong but not sure where I am screwing it up. Can someone show me what I am doing wrong?- thank you. Click here to see answer by stanbon(57307)
Question 242042: Please show me how to set it up Digicon prints digital phots for $.12 each plus $3.29 shipping and handling. Your weekly budget for the school year book is $22.00. How many prints can you have made if you have $22.00. I came up with 155 but I know I didn't set it up properly. Click here to see answer by Theo(3458)
Question: If the angles are complementary, what is the relationship between their measures? Answer: The sum of the measures of two complementary angles is 180 degrees. | 677.169 | 1 |
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