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# GCSE: Miscellaneous

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Meet our team of inspirational teachers Get help from 80+ teachers and hundreds of thousands of student written documents • Marked by Teachers essays 2
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.  ## Transformation Patterns. Our aim was to take different 3 digit number patterns and make a pattern that was instructed in the worksheet, and then find a correlation between the pattern of numbers and the line of symmetry and the order of rotation.

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Then we were to go y squares forward and turn 90 degrees clockwise and then go z square forward and turn 90 degrees. We were to repeat these instructions until we got the starting point. Methodology: We used the software MSWLOGO to create the shapes using the number patterns and formulae to repeat the pattern.

• Word count: 388
3.  ## Symmetry in Nature

Snowflakes can have either hexagonal or triangular symmetry although the hexagonal snowflake is most common. Beehive A beehive has translational symmetry meaning that it has a repeating pattern of hexagons. Individual cells of a honeycomb have rotational symmetry they can be rotated one sixth of a turn and still look like the same as before the rotation. A honey comb is built slanted so that honey doesn't fall over. The hexagonal shape of a cell gives strong construction and also uses less building material.

• Word count: 411
4. ## Guttering Investigation

I will be investigating guttering which will have a perimeter of "B" cm. In one case I will be using "18cm" which happens to be the length of the piece of paper I am using. Semi Circular Guttering Cross Section As this appears to be the most commonly used guttering and it is the type I have based my hypothesis on, I shall investigate its cross sectional area first so I can use it as a comparison for the other shapes I will use in my investigation. Perimeter of semi circle= pr ?B=pr So r=B/p Area of a semi circle = 1/2 p r2 = p B2/2p2 = B2/2p =0.1591549B2 This, as it

• Word count: 1634
5. make a graph using them and y = x2 to show the difference between the two equations: As we can see in the above graph, the vertex (turning point of a graph) is located every time on the y-axis. The parabola is the same size and shape though it is translated vertically, depending on the value of k. The reason why the parabola is translated rests with the fact that we either adding or subtracting a certain value from all points on the graph, thus pushing the graph up or down the y-axis.

• Word count: 1072
6. ## The Weight of Your School Bag

There was also another problem that students did not have a scale to measure the weight of the Bag in Kilograms (Kg) and/or Pounds (Lbs) and /or Stone (St,), so I had to decide what scale to use, and decide to Use Pounds (Lbs) as it was much user friendly, and also on Some Bathroom Scales. Copies of the Different Types of Questionnaires that I made are included. Plan: I planned to use 9 different Form groups from Year Ten at my School, and all nine were chosen at random.

• Word count: 1502
7. ## Number grids

2x2 12 13 22 23 (13x22) - (12x23) = 286-276 = 10 15 16 25 26 (16x25)-(15x26) = 400-390 = 10 77 78 87 88 (78x87)-(77x88) = 6786-6776 = 10 81 82 91 92 (82x91)-(81x92) = 7462-7452 = 10 33 34 43 44 (34x43)-(33x44) = 1462-1452 = 10 In all of the small 2x2 number grids above I have found out that the product of the top left and the bottom right numbers multiplied by the top right and the bottom left numbers, subtracted away from each other had a difference of 10.

• Word count: 2640
8. ## Equable shapes

This suggests that we can't have an equable rectangle with a width of 2. Now I will try to prove this algebraically. Area = 2L Perimeter = 2L+4 2L=2L+4 2L-2L=4 0L=4 L= Clearly, you can't have a rectangle that is infinitely long. I was right in thinking an equable rectangle with a fixed width of 2 doesn't exist. Now I will be looking at Rectangles with a fixed width of three. Rectangle Area Perimeter 3x1 3 8 3x2 6 10 3x3 9 12 3x4 12 14 3x5 15 16 I produced a graph to show these results.

• Word count: 1463
9. ## Box Coursework

My hypotheses are: 1. Boys are taller than girls 2. The taller a person is the heavier they are. To investigate these hypotheses I will need to focus on the height and weight of the students that I have collected in my random sample. The sample that I collected is presented on the next page. Now that I have my samples of data I will present then in the form of a cumulative frequency table. Boys Height (cm) Tally Frequency Cumulative Frequency 130?h<140 l 1 1 140?h<150 lllll l 6 7 150?h<160 lllll lllll lllll lllll lll 23 30 160?h<170 lllll lllll llll 14 44 170?h<180 lllll lll 18 62 180?h<190 l 1 63 190?h<200 0 63 200?h<210 l 1 64 Boys Weight (kg)

• Word count: 1285
10. ## number grid

I will first try with a 2 x 2 square. 12 13 22 23 a) 12 x 23 = 276 The difference of the two products are 10. 13 x 22 = 286 I worked the difference out by doing, 286-276 = 10 38 39 48 49 b) 38 x 49 = 1862 39 x 48 = 1872, I worked the difference out by 1872 - 1862 = 10 The product is again 10. My theory is that all 2 x 2 squares will have the product of 10, I will show one more 2 x 2 square to prove the theory.

• Word count: 2293
11. ## Maths Opposite corners

and the number ends in zero.To see if this changes I changed the first rectangle into a square by making it one square deeper. Both of the diagonal differences were 40. In a rectangle of 3 X 2 the diagonal difference was 20. This went up by 20 when I deepened the rectangle by one square. I then investigated whether the diagonal difference will go up by another 20 if I deepened the rectangle by another square. Both these were 60. So the diagonal difference has risen by 20 again. The diagonal differences are obviously going up in a sequence.

• Word count: 924
12. ## Maths -Borders

* In pattern 2, diagram 2 the difference between dark squares and blank squares is odd and the difference between the blank squares and total squares is also odd. The previous shape is the extended version of the original cross shape it is surrounded by 16 outer layer squares and contains a total of 41 squares. I will extend it one more time, which will look like this: The shape above is surrounded by 20 blank outer layer squares and 41 inner dark squares making a total number of 61 squares.

• Word count: 1032
13. ## T-Total Maths coursework

From this I predict that for T-Number 25 will have a T-Total of 62. I will now prove my theory. T-Number = 25 T-Total = 6+7+8+16+25 = 62 This shows that my prediction was correct, each time I move the T one place to the right that the T-Total increases by 5. This is because each time the T moves one place to the right each number increase by 1, there are 5 numbers in the T, so the T-Total will increase by 5.

• Word count: 2758
14. ## Layers investigation

So if you calculate 5*6 you get the answer 30. There is a theory behind this to find out the number of different arrangements without drawing the layers: "The number of squares filled in is always -1 of the number of arrangements and the number of possible empty squares." To test this theory out I did a made another grid of 3 by 2. Although this is the same as the 2 by 3 grid, I wanted to make the increase in grid size steady. I got the same results as the 2 by 3 grid.

• Word count: 2311
15. ## Tubes. I was given a piece of card measuring 24 cm by 32cm, and my task was to investigate and come up with as many open ended tubes by using this piece of card

Therefore, I have drawn this table to list all the combinations for the perimeter adding up to 24cm. As you can clearly see, the 6cm by 6cm cross-section gives me the largest volume. This tube is not a rectangle, as the sides are the same, so it must be a square. However to justify this, I will use figures around 6 * 6 to see if I can get a volume above 1152cm3. Length of rectangle Width of rectangle Area in cm2 Volume in cm3 5.5 6.5 35.75 1144 5.6 6.4 35.84 1146.88 5.7 6.3 35.91 1149.12 5.8 6.2 35.96

• Word count: 1753
16. ## Maths Investigative task on perimeter of a rectangle and volume of shapes

To do this, I used the area formula: As the area is kept constant at 1000 in the investigation, the formula can be made into, From this we can transfer the variable length to isolate width, thus getting a new equation. Substituting this in the original perimeter equation, To prove my formula right, I will make use of two examples. X=5 (when length is 5m) The perimeter will be: X=10 (when length is 10m) The perimeter will be: As seen above, my formula functions perfectly.

• Word count: 2973