Guide :

can you show how to draw shapes using tangrams

can you show how to draw shapes using tangrams

Research, Knowledge and Information :


Draw your own tangram - Providing teachers and pupils with ...


Pupils could use the instructions to draw the tangram pieces themselves. ... Tangrams to cut out; Draw your own tangram; Activity Sheet; 3D Tangrams; Memory training ...
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Making Maths: Making a Tangram : nrich.maths.org


Making Maths: Making a Tangram. ... shapes can you make using the ... shapes fit together so you don't forget. Then you could just draw the outline and ...
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Draw Tangrams - Mathematische Basteleien


You must find the right position of the tangram pieces filling this shape. ... You can order the tangram pieces ... You use it to make tangram pieces. You draw a 4x4 ...
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Trying Out Tangrams: Applying Knowledge of Geometry Vocabulary


They can draw in shapes using a ruler ... Students should show their geoboards ... Apply their understanding of the terms by using tangrams to create shapes that ...
Read More At : illuminations.nctm.org...

Use the Free Tangram Template Pattern in PDF - ThoughtCo


Work with a partner to come up with as many mathematical terms or words related to tangrams as you can. ... shapes in total. A tangram has ... Drawing With Shapes in ...
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Geometrical shapes - Providing teachers and pupils with ...


Geometrical shapes; Letters, Numbers ... Usual objects; Boats; Miscellaneous; Tangrams to cut out; Draw your own tangram; 3D ... Can you make these pictures, using ...
Read More At : www.tangram-channel.com...

How to Make A Set of Tangrams - YouTube


Mar 07, 2011 · How to make a paper Tangram in Origami ! very easy ! ... Arranging the shapes tangrams - Duration: ... Show more. Language: ...

Tangram Pictures : nrich.maths.org


A tangram is an ancient Chinese puzzle where you make pictures using mathematical shapes. ... You can try some more tangram ... Tangram pictures of your own? If you ...
Read More At : nrich.maths.org...

LINKS Learning for Kids: Math: Illustrated Lessons: Tangrams


Tangrams © 2004 LINKS LEARNING. All Rights Reserved. This web site was supported by The Fund for The Improvement of Education and Technology Challenge programs, ...
Read More At : www.linkslearning.k12.wa.us...

Suggested Questions And Answer :


can you show how to draw shapes using tangrams

wotz a "tangram"?????????????
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what graph is easier to construct?

It depends on what you are trying to show. Is it statistics? Is it a mathematical relationship? I would guess your are talking statistics. A pie chart is circular and is good for showing the relative distribution of groups in a population, for example, the distribution amongst a school of students on sport preferences. The circle represents all the students (100%) while the "slices of pie" each represent the number of students favouring a particular sport. A bar graph or chart often represents frequencies so it's easy to see from the height or length of each bar which have the lowest frequency of distribution and which have the highest. A line graph or scatter graph can demonstrate correlation between two quantities to see if there could be a connection between the two. This is useful for linear correlation. A map graph could be used to show the distribution of average rainfall or temperature over a large area, perhaps over the whole world. So there isn't a "best" or easiest; it's more about what you're trying to show. The larger the dataset the longer it will take to draw the relevant graph, and you need to decide which type of graph is going to be most useful before you attempt to draw it! Then you can plan the work needed.
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How to plot

I guess you want to use Excel to plot a binomial distribution where the probability is p=0.25. Excel will plot a graph for non-cumulative and cumulative distribution. You use the BINOMDIST statistical function, which has 4 parameters:   number of successes (x axis); total number of trials, n; the probability, p; whether cumulative or not. You set these up in a table for Excel to use to do the plots, with a fixed cell for the probability and columns for the number of successes and the corresponding values to contain the results of applying BINOMDIST.  The question asks for three graphs, so what could they be? The BINOMDIST function has 4 parameters, and the question supplies only one. So we can use Excel to plot graphs where the other parameters are different. We can change the number of trials; we can choose whether cumulative or non-cumulative; we can change the range of successes. So we set up three tables. The first table has fixed cells for n, p and cumulative=FALSE, and values in a column from 0 to n for the number of successes to be plotted; second table has the same but cumulative=TRUE; the third has a different n and successes column and cumulative can be either TRUE or FALSE. You would specify the continuous curve type of graph for Excel to plot the data as a curve and the result would show the typical bell-shaped binomial distribution curve for the non-cumulative distribution and the hill-shaped curve for the cumulative. To use the graphs, you read off for each x value (number of successes required) the percentage or fraction of the expected results. You would also title the graph, label its axes and show that p=0.25 is the active probability. Excel will allow you to customise how you want the graphs to look. [Non-cumulative means the exact number of successes expected in a given number of trials; cumulative means at least or at most the specified number of successes in a given number of trials.]
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Use the unit circle to show: cos(theta) = sin(theta + pi/2)

Draw a circle of radius 1 centre O, and draw two diameters at right angles, AOB and COD, dividing the unit circle into quadrants, COB, BOD, DOA, AOC. Draw a radius OX at angle x to OB. Also draw another radius OY at angle x to OC. Angles YOC and XOB=x. Also, YOB=90+x=(pi)/2+x. Drop a perpendicular XP from X on to OB. OP=cosx. Drop a perpendicular YQ to AO. SinYOA=sinYOB=sin((pi)/2+x)=YQ. Right-angled triangles YOQ, YOC and XOP are congruent and YQ=OP=cosx=sin((pi)/2+x).
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What is a stem and leaf plot and when would I use one?

Stem and Leaf plots are just a method of ordering data in a dataset to produce a frequency chart. These plots are used in statistical analysis to draw conclusions about a dataset. The usual way this is done is to use part of each datum to create a data bin. Let's imagine a dataset where all the data consists of numbers between 1 and 99. It doesn't matter how big the dataset is or if there are duplicates. Now imagine 10 bins. The first bin is for numbers between 1 and 9; the second for numbers between 10 and 19, and so on. The numbers of the bins will be labelled 0 to 9. The bins are the stems. So we just go through all the data and put each datum into its appropriate bin. But we don't have to put the whole of the data into each bin, because the bin number is already numbered with the first digit of the data. So the contents of each bin just contain the second digit of the data. The bins (stems) are lined up in order 0 to 9 and we can also stack their contents so that the single digits are in order inside the bins. These are the leaves. Imagine the bins are made of glass. We can look at the bins and the heights of the stacks of contents. The heights of the contents form a shape as we run down the line of bins. These heights tell us how many data there are in each bin and indicate where the most data is and where the least data is. This is is a frequency distribution. It's the basis of the Stem and Leaf plot and can be represented by a table or chart. Each row of the table starts with the bin number (STEM) and along the row we have the contents of the bin (LEAVES). Turn the table on its side and we have a chart with the stem running along the bottom and the leaves forming towers over the stems. The chart resembles the row of bins with the stack, or column, of contents over them, but the bins are now invisible, and only their labels remain as regular horizontal divisions on the chart. But it doesn't stop there. This frequency chart tells us where most of the data can be found, where its middle is and the general shape of the data. These are important statistical observations. Not all the bins may have data in them, and some will have lots of data. Random data will produce no particular shape, but in many cases there will be a pattern. We've considered numbers from 1 to 99, but the data can have any range as long as the data is binned carefully to reflect the relative magnitude of the data. If the data were between 250 and 400, for example, we might take the first 2 digits as the bin label: 25 to 40 and the contents would be the third digit. So you need to make a decision based on the range of data values to decide how the data is going to be binned. I hope this helps you to understand Stem and Leaf plots.
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use Euler’s method with the specified step size to determine the solution to the given initial-value problem at the specified point.

dy/dx=4y-1; separate the variables: dy/(4y-1)=dx, integrating: (1/4)ln(4y-1)=x. ln(4y-1)=4x. The general solution is 4y-1=Ae^4x, y=(Ae^4x + 1)/4 [CHECK: y'=Ae^4x=4y-1.] Initial condition, y(0)=1: 1=(A+1)/4, A+1=4, A=3 and y=(3e^4x + 1)/4. y(0.5)=(3e^2+1)/4=5.79. This is the value we hope to approximate using Euler's method of iterative increments. USING EULER'S METHOD This method uses the approximation that Dy/Dx = dy/dx where D is a small increment, Dx=h=0.05. Initially y=0 and x=1. This is our starting point The table shows the iterative procedure. By Euler's method y(0.50)=4.89, compared with the actual value of about 5.79. The value 5.82 is in fact closer and this is the value we should use for the following reason. The starting point is (0,1).  The next x-value is the first increment giving us the horizontal base of a triangle. The end points in this base are (0,1) and (0.05,1). So the vertical height of the triangle is found by multiplying the gradient dy/dx by the increment to give us the y increment. The table then has to be interpreted by adding the increment 0.05 to the the limiting value of x, which is 0.5, making it 0.55 and reading off the corresponding y value, 5.82 (columns 4 and 5). This is the solution, y(0.5)=5.82 using Euler's method. If the increment is made smaller y(0.5) gets closer to the actual value. The picture below shows the idea. Joining the hypotenuses of these little triangles gives us an approximation to the shape of the curve.  
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Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins
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How to solve a Logic problem?

Let q={ 0 1 4 9 16 25 36 }, r={ 1 3 5 7 9 11 }, p={ 0 2 4 6 8 10 } So q ^ r={ 1 9 }, p v q={ 0 1 2 4 6 8 9 10 16 25 36 }, p v r={ 0 1 2 3 4 5 6 7 8 9 10 11 } (p v q) ^ (p v r)={ 0 1 2 4 6 8 9 10 } p v (q ^ r)={ 0 1 2 4 6 8 9 10 }, so (p v q) ^ (p v r) = p v (q ^ r) (p v q) ^ r={ 1 9 } ≠ (p v q) ^ (p v r). Draw two intersecting circles representing sets q and r. Where they intersect is q ^ r (in the example the intersection would contain the numbers 1 and 9. Now consider augmenting the sets by the contents of p (this is the union of p with each of the two intersecting sets). This time the intersection would contain all the elements of p as well as 1 and 9. This demonstrates the first part of the question. But if the p elements are added only to q then the intersection only contains 1 9 because there are no more common elements in r than there were before. This demonstrates the second part of the question.
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can you create a box plot for this set of data?: 1, 4, 6, 13, 3, 24, 4, 1, 2, 1, 2, 9, 8, 5, 25, 8, 1

First: put the data into ascending order. 1, 1, 1, 1, 2, 2, 3, 4, 4, 5, 6, 8, 8, 9, 13, 24, 25 MIN: 1; MAX: 25 MEDIAN: 4. This is the 50th percentile, and divides the data into two sets each containing half the results. Set 1: 1, 1, 1, 1, 2, 2, 3, 4 Set 1 median: 1.5 (average of middle data) Set 2: 5, 6, 8, 8, 9, 13, 24, 25 Set 2 median: 8.5 The medians split the data into 4 groups or quartiles: <1.5; >1.5 but <=4; >4 but <8.5; >8.5. Data <1.5 or >8.5 are called outliers. Draw a vertical line representative no the data and mark the limits 1 and 25 on the line. These are the bottom and top of the box. Now mark the medians at 1.5, 4 and 8.5. These form an inner box with 1.5 as its bottom and 8.5 as its top, with 4 sandwiched inside. Draw horizontal lines to show the boxes. These horizontal lines are of no specific length. That's your box plot. You can use colours or shading in the inner box to show the data between the three medians (data between 1.5 and 8.5), and the outliers will be in the larger outer box in an unshaded area.  
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Draw different type shapes using different materials 

?????????????? "materials" ?????????? how bout stuf insted ???
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