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GCSE: Hidden Faces and Cubes

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  1. GCSE Maths questions

    • Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
    • Level: GCSE
    • Questions: 75
  2. Border coursework

    This was done by drawing out the sequence.cofd fd" . "r se" . fd . "fd" . "w or". fd . " " . fd . "k infd fofd " . fd . "; Seq no 1 2 3 4 5 6 7 No of squares 1 5 13 25 41 61 85 I will use these numbers to try to create a type of formula to get any no of squares in any sequence.coba ba" . "r se" . ba . "ba" . "w or". ba . " " . ba . "k inba foba " . ba .

    • Word count: 1752
  3. shapes investigation coursework

    To save time, perimeter, dots enclosed, triangles etc. are written as their formulaic counterparts. My tables of recordings will include T, Q or H. This is because, whilst it will remain constant in any given table, I am quite sure that this value will need to be incorporated into any formulas. Triangles To find the P and D of shapes composed of different numbers of equilateral triangles, I drew them out on isometric dot paper. These tables are displayed numerically, starting with the lowest value of T.

    • Word count: 5002
  4. Painted Cube Investigation

    of Painted Faces 1 x 1 x 1 2 x 2 x 2 3 x 3 x 3 4 x 4 x 4 n x n x n 0 0 0 1 8 (n - 2)3 1 0 0 6 24 6 (n - 2)2 2 0 0 12 24 12 (n - 2) 3 0 8 8 8 8 unless 1 x 1 x 1 4 0 0 0 0 0 5 0 0 0 0 0 6 1 0 0 0 0 unless 1 x 1 x 1 The diagrams above justify the answers for 2 x 2 x 2, 3 x 3 x 3 and 4 x 4 x 4 cubes.

    • Word count: 1021
  5. Skeleton Tower Investigation

    4 since there were four sections (60 for a 6-high tower), and then added the cubes in the middle stack (66 for a 6-high tower). I repeated this procedure for a 12-high tower (66 cubes per section, 264 cubes on wings and 276 total cubes). Using the results I worked out an nth term for the Skeleton Tower: 2n� - n This rule can be proved as two 'arms' unite to form a rectangle with dimensions n by (n-1). There are four arms for the tower so this has to be multiplied by 2 and the center column is added and it has n blocks: The formula 2(n (n-1))

    • Word count: 1048
  6. volumes of open ended prisms

    I will 8cm 8cm use Pythagoras again and say that the height 2 = 82 - 42 h so, the height is 6.928cm, to find the area we must Multiply the height by the base: 8cm 6.928 x 8 = 55.426cm 2 = 27.713cm2 Now to find the volume: 27.713 x 32 = 886.816cm3 32cm H2 = 92 - 32 h= 8.49cm. Area = base (6) x height (8.49) 9cm 9cm 2 H = 25.455cm2 6cm Volume = area of cross section (25.455)

    • Word count: 2729
  7. painted sides of a cube

    There is a definite pattern for the cube and the sides painted. After looking at the first 4 cubes, the sides painted look as such: In a 2 x 2 x 2 cube there are: 0 blocks with 0 sides painted. 0 blocks with 1 side painted. 0 blocks with 2 sides painted. 8 blocks with 3 sides painted. In a 3 x 3 x 3 cube there are: 1 blocks with 0 sides painted.

    • Word count: 447
  8. Borders Investigation Maths Coursework

    and the total number of squares. Then I will test the formula on my pattern that I have drawn and the ones I haven't drawn. Finally I will extend my investigation to 3D and hopefully find a formula for it, test it and find total number of squares etc in pattern without drawing it. Patterns Pattern 1 = White 1 Black 0 Pattern 2 = White 1 Black 5 Pattern 3 = White 5 Black 8 Pattern 4 = White 13 Black 12 Patterns Pattern 5 = White 25 Black 16 Pattern 6 = White 41 Black 20 Pattern no No of white squares No of

    • Word count: 2991
  9. Maths Cw Borders Investigation Cubes

    61=25+36 =5*5+6*6 By tabulating all the above result I can derive the general result. Number/level (N) Mn(layer of cubes in the middle layer) 1 1=1=1*1=0*0+1*1 2 5=1+4=1*1+2*2 3 13=4+9=2*2+3*3 4 25=9+16=3*3+4*4 5 41=16+25=4*4+5*5 6 61=25+36=5*5+6*6 The first column values of the table were always 1 less than the corresponding cross shape where as the second column values are the cross shape numbers broken up. Therefore the general formula for the Nth cross shape must have (n-1)2+n2 cubes. Mn=(n-1)2+n2 cubes in it Mn=n2-2n+1+n2 Mn=2n2-2n+1, where n cross shape level Mn Number of cubes in the middle layer of nth cross-shape However the result obtained earlier for 2D cross shapes and proved alternatively using triangular numbers but now I have realized that the 3D cross shapes are built up based on 2D cross shapes.

    • Word count: 861
  10. Lines, regions and cross overs

    Then I will produce graphs to enable me to see if I can find any patterns and equations. I will the start working with the fewest number of lines and work up to the maximum and I shall make a table to show my results. Investigation 1 Line For 1 line the maximum you can have is 0 crossovers 0 closed spaces and 2 open Spaces. 2 Lines As you can see the maximum for 2 line is 1 crossovers 0 closed spaces and 4 open spaces. 3 Lines 4 Lines As you can see the maximum for 4 lines is 6 crossovers 3 closed regions and 8 open regions.

    • Word count: 905
  11. GCSE Maths - Cargo Project

    I will have a different colour to represent each label amount, e.g. blue for 3 labels, red for 2, etc. Then I will put my results into a table and then try to search for any patterns I can see. 3 labels 2 labels 1 label No labels 1st level - Birds Eye View 2nd level - Birds Eye View TOTAL 3 labels = 8 2 labels = 0 1 label = 0 No labels = 0 3 labels 2 labels 1 label No labels 1st level - Birds Eye View 2nd level - Birds Eye View 3rd level - Birds Eye View 3 labels 2 labels 1 label No labels 1st

    • Word count: 682
  12. Hidden Faces

    The sheet shows this as an example. It is a line of five cubes and it has 30 faces and 13 of the faces are hidden. I tried changing the amount of cubes, to see what results I Got. I used 7 cubes and noticed that I now had 42 faces and 19 of the faces are hidden. I used 6 cubes and noticed that I now had 36 faces and 16 of the faces are hidden. I used 1 cube and noticed that I now had 6 faces and 1 of the faces are hidden.

    • Word count: 643
  13. Persuasive essay

    4 4 4 0 0 0 0 19 15.8 D 2 Sutton, Ken 53 1 7 8 9 37 0 0 0 0 35 2.9 RW 14 *Dagenais *MNR* 9 3 2 5 1 6 1 0 1 0 20 15.0 D 6 *Commodore *MNR* 20 1 4 5 5 14 0 0 0 0 11 9.1 LW 19 McKenzie, Jim 53 2 2 4 0 119 0 0 0 0 32 6.3 D 3 Daneyko, Ken 77 0 4 4 8 87 0 0 0 0 50 0.0 LW 9 *Bicek *MNR* 5 1 0 1 0 4 0

    • Word count: 8892
  14. How far is it true to say that the work of solicitors and barristers has changed so much that it is no longer necessary for there to be two separate professions?

    Barristers in commercial fields might survive. This was a possibility as fewer solicitors specialised in those areas, but, in recent years more solicitors from big city firms have begun to specialise in this area too and therefore be a big threat to barristers as these firms offer large salaries and attract the graduates with the brighter futures therefore the Bar s reputation as the more prestigious with the best expertise would gradually fade. Again some thought that barristers with common law practices such as personaCHNKWKS P�����TEXTTEXTr=FDPPFDPP@FDPCFDPCBSTSHSTSHDhSTSHSTSHhE�SYIDSYID HSGP SGP -HINK INK "HBTEPPLC &HBTECPLC >HFONTFONTVHpSTRSPLC �H:PRNTWNPRITFRAMFRAMTM�TITLTITL�M2DOP DOP N"How far is it true to say that the work of solicitors and barristers has changed so much that it is no longer necessary for there to be two separate professions?

  15. Maths-hidden faces

    For each cube in the middle of the row, there are three faces showing and three faces hidden so the first part of my rule was 3n. At the end of each row there are only 2 hidden faces so that is why you have to -2 from the total number of hidden faces. 1.) Test and predict: Predict: Number of cubes=9 Using the rule, I predict that for 9 cubes there will be 25 hidden faces. 9x3=27-2=25 Test: I drew out the number of cubes showing the hidden faces, which gave the same answer as my prediction, which proves that my rule is correct.

    • Word count: 1980
  16. Alexander the Great's father was Philip. Philip's wife was Olympias. Their son Alexander was born in 356 B.C. Alexander had a younger sister, Cleopatra. When Alexander was sixteen his father

    What about me? Am I a bastard? Alexander shouted. Attalus threw his own goblet back and a general brawl ensued, during which Alexander and his father snarled at each other. After this incident Alexander and his mother left Macedon. Later they reconciled with Philip and returned home, but Alexander continued to mistrust his father. When Philip arranged for his retarded son Arridaeus to marry the daughter of a Persian governor, Alexander feared that this meant Philip intended to make Arridaeus his successor.

    • Word count: 1910
  17. A person is defined as being homeless when he/she lacks a fixed nightime

    This results in them not being able to find permanent accommodation of their own. Additionally One in seven Scottish households are overcrowded, so it is not unusual that friends and relatives would eventually want to enjoy the privacy of their own homes. It is natural that young people would eventually want to take the next step of leaving home, to begin lives of their own, they should not under any circumstance be forced out by overcrowding or family problems, resulting in them becoming homeless.

    • Word count: 1891
  18. Regular Polyhedra

    Regular Polyhedra is also known as Platonic Solids. There are only five platonic solids. Equilateral triangles -Tetrahedron 3 faces at each vertex -Octahedron 4 faces at each vertex -Icosahedron 5 faces at each vertex Squares -Cube (hexahedron) 3 faces at each vertex Pentagons - Dodecahedron 3 faces at each vertex Here are short explanations of some Polyhedra... Tetrahedron This is the simplest of all the polyhedra. It uses the least number of faces to enclose any three-dimensional space. The polygon that makes up these faces is also the simplest polygon. Tetrahedron has 4 vertices, 6 edges and 4 faces. Hexahedron Hexahedron is the most common polyhedron.

    • Word count: 578
  19. An experiment to find out if seeing the eyes of a well known persons face is a factor of face recognition

    The results indicate that the eyes of a person's face are an important factor of face recognition. Introduction / Background: Being able to recognise other people is of great significance in our lives. You can imagine how difficult life would be if you didn't recognise your mother, brother, best friend or partner. Human beings seem to have an amazing ability to recognise faces. Standing (1973) showed participants 10,000 faces over five days. When they were shown pairs of faces, one of which they had previously been shown, together with a new one, they were able to identify the face they had seen 98% of the time.

    • Word count: 2040
  20. Layers of cubes

    And the last column was the number of possible arrangements. So from that Table I found the formula for the arrangements of 2 layers to be GxL1, this is because the number of arrangements you could arrange 5 cubes on 6 square grid was 6 times, and the number of ways you could arrange 4 cubes on 5 cubes would be five. The Product of the grid size Number of cubes on layer 1 Number of cubes on layer 2 Total number of arrangements 2 1 2 3 2 1 6 4 3 2 12 5 4 3 20 6 5 4 30 7 6 5 42 8 7 6 56 9 8 7 72 10 9 8 90 (b)

    • Word count: 1254
  21. Shapes made up of other shapes

    As I progress, I will note down any obvious or less obvious things that I see, and any working formulas found will go on my �Formulas� page. To save time, perimeter, dots enclosed, triangles etc. are written as their formulaic counterparts. My tables of recordings will include T, Q or H. This is because, whilst it will remain constant in any given table, I am quite sure that this value will need to be incorporated into any formulas. Triangles To find the P and D of shapes composed of different numbers of equilateral triangles, I drew them out on isometric dot paper.

    • Word count: 8587
  22. Exposed cube sides

    When I have found a pattern I will try to work out a formula and test if it works. 4x4x4 3x3x3 Key 3 labels 2 labels 1 label No labels 6x6x6 5x5x5 Results 3x3x3 4x4x4 5x5x5 6x6x6 3 labels 8 8 8 8 2 labels 12 24 36 48 1 labels 6 24 54 96 No labels 1 8 27 64 Formula The pattern for 3 labels is 8, 8, 8... U3 = 8 = 0x3+8 U4 = 8 = 0x4+8 U5 = 8 = 0x5+8 Un = 0n+8 The pattern for 2 labels is 12, 24, 36, 48...

    • Word count: 582
  23. Hidden faces

    In fact both share the same pattern in that they both progress by 3 with each addition of a block. Finding the Nth term From this pattern and from the table I make the formula for the progression of visible faces: 3n + 2 e.g. Number of cubes = 2 Number of visible faces = 2 2 x 3 = 6 + 2 = 8 = 3n + 2 I have proved that this formula works by calculating the 9th and 10th terms by using it.

    • Word count: 450
  24. This report is about working out the formula to a hidden faces equation, I will find the nth term, put my results into a table and, figure out if the formula works.

    Method I am going to find out how many hidden faces (the faces that aren't visible from any angle) there on 1 cube 2 cubes 3 cubes 4 cubes 5 cubes 6 cubes 7 cubes 8 cubes 9 cubes I am doing this to try and find an easier way to find out how many hidden faces there are in....

    • Word count: 223
  25. Maths hidden faces

    Finally I will find formulas for the nth term for the number of hidden faces for each particular dimension. Hopefully, I will then try to find general formula for the total numbers of hidden faces in any given cube/cuboids. Maths Evaluation for hidden fa Evaluation I have used a systematic approach when drawing the cubes and cuboids. I started with smaller cubes and increased their size. I increased one dimension at a time while holding the other two exactly the same.

    • Word count: 1445
  26. "With reference to theories of visual object recognition outline the ways in which faces appear to be "special". How might such appearances be deceptive and in what ways does this bear on competing theories".

    Gregory applied this view to the explanation of visual illusions. He believed that when we experience a visual illusion, what we see may not be physically present in the stimulus. Therefore our attempt to make sense of the stimulus may be misplaced resulting in a visual illusion. This theory could be explained using the Muller-Lyer illusion: (Fig 1) Fig1 Gregory (1963 as cited in Levine, 2000) suggests that the way in which these arrows are placed provide cues from which certain things can be interpreted.

    • Word count: 3448

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