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Everything posted by Jenny

  1. Grade Two Math California

    I answered your other post, but I would like to address the last point 2(c). Common Core is just a list of standards, i.e. topics to cover, along with some mathematical principles. It is not a curriculum. Whether a particular textbook is helpful to a student with dyslexia and how much visual or pictorial is included depends on the textbook writers, not Common Core. One of the tenets of Primary Mathematics is concrete to pictorial to abstract, and so there is quite a bit of pictorial. There is also minimal text (more pictures than words). So that could help too. One thing that has led to issues with Common Core is the interpretation of the Mathematical Principles by various writers/educators/whatever.. Explain your reasoning, does that mean write a paragraph in words to explain your answer or what? Well, it depends on the interpretation. Critiquing answers, does that mean finding why another student got an answer wrong, or... Some curricula writers might emphasize Explain your Reasoning over actually having them reason accurately even if they can't put it in words. Some kids who are very good at math do not yet have the verbal abilities, nor necessarily the words if their reasoning does not follow the accepted or taught methods. So explain your reasoning to them ends up meaning they have to solve it the way the teacher told them, because the teacher gave them a model writing sample for explaining....
  2. Grade Two Math California

    The books we label Common Core simply have added topics when needed, e.g. line plots of measurements to nearest fraction of an inch. All three editions are similar in basic content. And all three do not strictly follow Common Core requirements at grade level, fore example, there is multiplication in grade 2, which is not Common Core, (which is why Primary Mathis not on California's list of approved textbooks. This is based on Singapore's original Primary Mathematics, not on US math. I suppose whether it would be confusing for your child, which would be the case for any of our editions, is whether you plan to teach the content, or just hand him books and assume he knows how to do all the math from what he has learned or not learned in school. None of the editions will align exactly. Please see http://www.singaporemath.com/FAQ_Primary_Math_s/15.htm and http://www.singaporemath.com/v/PMSS_comparison.pdf I do recommend actually teaching/interacting/doing lessons with him, and I think that unless you are concerned about every topic being covered (those not covered in the other two editions are not essential to an understanding of math), the Standards edition has the best Home Instructor's Guide. The Standards edition follows what used to be California's standards (same caveat, it includes them all, but is still a bit advanced), but since California adopted Common Core, it no longer does.
  3. p. 47 #8: Draw the bars. Draw a bar for A. Draw a bar for B that is a bit shorter than A and mark the difference as 12. Draw a bar for C that is also a bit shorter and mark the difference as 8. Mark the total as 130 L. Now, make all 3 bars the same length. Since you are finding A, best to make them the length of A. To do so, you have to add 12 to B and 8 to C. That adds 12 and 8 to the total, so add 12 and 8 to to 130 L, which gives you 150 L. There are now 3 equal units, all equal to A, so divide by 3 to get the amount in A. p. 48 #10: Draw a part-whole bar showing Monica + Susan. Its total is 186. It has a monica unit and a susan unit. Then draw a bar showing Monica and Ruth under it. This has a monica unit same as the one above it, and then 4 susan units, since Ruth has 4 times as much as Susan. Its total is 372. Now you should be able to see that the difference between the two bars is 3 of those susan units. So find the difference between 372 and 186 and that is 3 susan units. Divide by 3 and you have 1 susan unit. p. 45 # 4. Draw a bar showing 5 units for spoons, and another under it showing 1 unit of same size for forks. That represents 5 times as many spoons as forks. Above that, draw a bar shorter than the 5 spoon units, and mark the difference as 20, since it lost 20 fewer knives than spoons, although it is not really necessary. The problem says 720 spoons and forks. There are 6 equal units for spoons and forks. So divide by 6 to get 1 unit, then multiply by 5 to get the 5 units of spoon, then subtract 20 to get how many knives were lost. p. 50 #8: Not a good problem yet, should have been after fractions. However, draw a bar for Cindy with 2 units, and one for Billy with 3 units. This shows Cindy as having 2/3 as much as Billy. There are 5 equal units in all. The total is 360. So divide by 5 to get the value of 1 unit. Multiply by 2 to get how much is for Cindy. That is how much money Cindy has. Divide that by 12 to get how many days.
  4. If the camera is 4 units, and the radio is 1 unit, the camera is 3 units more than the radio. So 1 unit is $120/3, not $120/4. Did you actually draw the bars? It helps.
  5. Extra practice Grade 6 US Edition

    You could reword the problem to make it the situation clearer. There are people at a track meet and they have to decide whether to participate in ....
  6. Extra practice Grade 6 US Edition

    You are not really supposed to write algebraic equations to solve yet. And this problem is tricky in its wording, as the total does not change. The 2 go from not participating to participating. So, draw the bar model showing 1 : 4 |______| |______|______|______|______| What happens is that 2 people leave the bottom bar (non participating) and join the top |______|2| |______|______|______|_____|X| And since that goes to a ratio of 1: 3, the after situation is: |________| |________|________|________| The total stays the same. Therefore 5 total units before becomes 4 total slightly longer units after. you want all units the same. Essentially, you want equivalent ratios where total is the same. Find lowest common multiple of 5 and 4, which is 20. So ratio before: 1 : 4 -> 4 : 16 and after 1 : 3 -> 5 : 16 And all the units are the same size, because there are 20 of them in both before and after, and the total does not change. Notice that now the top bar for the after situation is 1 unit more than the top bar for the before situation. And that is the number of people who switched. 1 unit = 2 20 units = 40 Which is the number of people at the track meet.
  7. Well, I don't think you did anything wrong. There is obviously something wrong with the diagram. Not possible. Can't drill a hole with diameter 8 cm in something only 5 cm wide. Nor 3 of them with length only 20 cm. Perhaps they meant the diameter is 4 cm. Except I still don't get the answer they got.
  8. Isn't 20 x 5 x 15 div 2 = 750, not 225?
  9. Draw bar models. 3 units for Sasha, 4 for Tess, then 4 and a bit more marked 30 cm for Wendy. If you subtract 30 from the total, i.e. from Wendy's height, you have 3 + 4 + 4 = 11 units. So 11 units = 470 - 30 = 440 Find the value of 1 unit, after which you can find Tess's and Wendy's height and then find the ratio.
  10. 6A workbook page 87 C

    Draw the bar models. Show 3 units for red and 2 units for blue. There is a total of 5 units. That 5 units is 3 L, or 3,000 ml So 1 unit = 3,000 รท 5 Once you find 1 unit, you can find 3 units and 2 units. Let me know if you need more detail.
  11. Unfortunately, this is apparently an issue with having moved content around. In the solution, they are using the Pythagorean Theorem to find the length of the sides of the triangle. But the Pythagorean Theorem is not covered until 8B. So best to ignore their solution and use a ruler to measure instead.
  12. 6B Textbook page 27, number 5

    You understand what part is a quarter circle, right? The yellow part is a quarter circle. So this picture is essentially two of them, one on top of the other, flipped. The radius is 10, not 5. On the image here, draw a line from upper left corner to lower right and that is half of the "eye" in the textbook. But you can get the pink area by subtracting the area of the quarter circle, with radius 10, from area of the square, with side 10. Once you get that, I think you can figure it out.
  13. 6B Textbook pg 27 #6

    That is just an interesting observation I made. Something that will be learned in a highschool geometry course. Maybe. Can look here http://www.gogeometry.com/pythagoras/lunes_hippocrates_area_01.htm for a start.
  14. 6B Textbook pg 27 #6

    And, if you put all the equations together, and use the distributive property, the answer is half of 12. Since the area of the larger semi circle is equal to the sum of the area of the two smaller ones, the area of the shaded part is the same as the area of the triangle - there is some geometric principles at work here. 1/2(pi(2^2 + 1.5^2 - 2.5^2) + (4)(3))
  15. 6B Textbook pg 27 #6

    If you actually use 3.14, the answer comes to 5.97. But if you use pi in the calculator, you get 6. Issue is rounding. Sorry for my poor checking of calculation. Method of solution is still the same. Area of the two semicircles, minus area of large one - triangle.
  16. 6B Textbook pg 27 #6

    Yes, sorry, I should proof better.
  17. 6B Textbook pg 27 #6

    To find the area of the shaded part, you need to find the area of the two semi-circles and subtract the part in the white semi-circle outside of the triangle. Area of large semi-circle (diameter of 5): 3.14 x 2.5^2/2 = 9.81 Area of smallest semi circle = 3.14 x 1.5 x 1.5 / 2 = 6.28 Similarly, area of middle one with radius 2 is 3.53 Area of triangle is 1/2 x 4 x 3 (base is 4, height is 3) = 6 So, area of shaded part = 6.28 + 3.53 - 6 = 6. The total length of the curved lines is simply the sum of the circumference of all three circles (one with diameter 5, one with diameter 4, one with diameter 3) 3.14 x (2.5 + 2 + 1.5) = 18.84
  18. 6B Workbook, page 22, number 4

    Think about cutting pieces and rearranging. Draw a line vertically up the middle. The shaded quadrilateral at the top becomes 2 triangles which fit in above the shaded part at the middle. nd then it the shaded part is actually a third of a circle.
  19. 6B Textbook page 24, numbers 6 and 7

    Each of the green sections in the thought bubble are quarter circles. Since there are 4 of them, altogether they make a circle, with a radius of 14. Trace the figure and cut it up if that helps. So you can find the area of the green part in the diagram by subtracting the area of a circle from that of the large square. Area of square is 28 squared, so area of green part is 784 = 616. Why would area of a circle be area of the green part? For 7, those are two quarter circles overlapping. So area of both minus one overlap, i.e. area of the square in the middle. You are adding in the area of the square in the middle. Plus, I don't know what you mean by larger square, those are quarter circles. Which together make a half circle. So could find area of half circle with radius 10 (314/2) and subtract area of square, 9. Which is what you are supposed to divine from the thought bubble.
  20. Yes, Sam's speed is 6 mi/h and Terry's is 15 mi/h.
  21. Textbook 6A page 149, Ex. 7

    Ben's total time is 250 min Paul's total distance is 375 km Paul's total time is 170 min Paul's average speed is 2.2 km/min = 132.4 km/h (Should actually round to 132, because of significant figures, but they don't know that yet.)
  22. Textbook 6A page 149, Ex. 7

    Paul's total distance of 375. That is correct. The Answer key is not correct.
  23. That is what it means, a calculator could be used. Only for those problems. I suggest not using a calculator at all for any, but that is what it means.
  24. help with math

    This forum is for customers of Singapore Math Inc. You need to go to a general math help forum for help like this. Or, do some research. Look here: http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueIneq.aspx
  25. NEM on Transcript

    What colleges are more concerned with is what level of math the student gets to while still in high school. And not what they did in grade 7. You can list it however you want, but I don't think NEM covers all of algebra 1. Nor geometry, at least not high school geometry. You could call it a credit of pre-algebra and 1 of introductory geometry, but I don't think it can honestly be listed as a full credit of high school algebra 1 and a full credit of high school geometry, because of the content. Regardless of how much time was put in to it. It partly depends on where she is going in the long run for later levels. Will you be doing NEM 3 and 4? NEM 1-4 maybe takes you through algebra 1, 2, and geometry, maybe half a credit of trigonometry. Note that in Singapore, NEM 1-4 is essentially Secondary 1-4, which is essentially grades 7-10 or 11 age wise, then they go to junior college, after taking Cambridge O level exams. Those kids wanting to sit for Cambridge A level exams take a second math course in Secondary 3 and 4, which would cover what we might consider grades 11 and 12 type of math if one wants some Calculus in grade 12.