My Site 9th Grade
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Cow problem
Process
The cow problem is using our basics skill to figure out how much space a cow can cover. The cow it tied to a rope that's 100ft attached to a 10 by 10 barn.
My first attempt at this cow problem was to break down the the 10 by 10 square barn by other simpler problems. The first square I tried to do was braking it down to use the pythagorean theorem, but then I realized the triangles add up to 180, which the two triangles make square that equals 360 that can be solved as well. For the second attempt I tried using the given circle, and using percentages, and within the 25% circles finding the degrees. The reason why my second circle didn't work was because I realized that we weren't working with a perfect circle. Next I broke my diagram into manageable pieces. One manageable piece that we had was 3/4 of a circle, so we calculated the radius(rope) with A=Pi times 100 squared and times that by 3/4. The final answer we got was A of 3/4 of a circle=23,550. Another manageable piece of the circle we looked at was the top right corner on the top right. Next, we broke that last fourth of the circle into into manageable triangles. This was useful to me because we can use Soh(sin)-Cha(coas)-Toa(tan) to calculate those triangles. Then, I realized I had two remaining sectors of the circle that needed to be calculated. The same way I found the area of 3/4 of the circle, is the same way I found the are of those two remaining sectors.
Self Evaluation
I For this tediuos Cow Problem I had to think of,"What can I do to break this problem down?" Starting off with basic knowledege, I automatically thought of pythagorean theorem. Pythagorean theorem, because that seemed easy to find angles when broken down to triangles. Then I realized I didn't have angles to work with. I got the most out of this problem by asking questions to refresh my memory on how to do certain equations. When it came to the group quiz it didn't go as well, as I thought it would. Cayla took over in guiding us us on how to solve the problem, while I asked questions in the process. If I were to grade myself on this unit I would give my self a 4/10 because I could've tried to solve the group quiz, and and make mistakes to learn from them as a group. When it came to the individual test I failed, but still got points for trying. When it came to coming in the mornings to get help for this DP, that I'm typing right now I think I deserve a 9/10, because I showed up in the mornings with Coach Marquez to get extra help understanding the problems. The picturs above is the work that we did to solve the Cow Problem.
Maximum rectangle problem statement
The Maximum problem statement is finding the area and perimeter of the rectangle as we graphed the parabolas. The parabolas give us something too start with, because we had to get the coordinates of the graph.
Process
.My first attempts at going at this problem was plugging in numbers for the given equation, which was Y=16-X^2. After creating the X and Y table we had all plugged in one number to get things done quicker onto the table. We had worked with plugging our points on the graph, which were (0,0),(0,4),(4,7),(2,12),(1,15), and (0,16) After doing this step I now had to find the area of the rectangle. The formula to the area was A=X*Y. So, the equation would look like this A=X(16-X^ 2). The final answer A=16X-X^3
Solution |
For the perimeter equation was P=2(X+Y), P=2(X+(16-X^2), and finally P=2X+32-2X^2.
Group/ indiv test
The group test was us graphing points after we got numbers from the equation y=4-x^2. I created a X, Y table and used 0,-1, 2, 3, and 4. The Y outcomes were 0, 3, 0, 5, and -11. Once we got these points we plugged in the equation y=4-x^2 getting a new outcome for X and Y because we realized we made a mistake. The new numbers we used for X were -2,-1, 0, 1, 2 and the outcome for Y were 0, 3, 4,3 ,0. From there we graphed the points from the X and Y table (-2, 0), (-1,3), (0,4),(1,3),(2,0). After the graphing, we now had to find the area using the formula A=X*Y. We found that the it was 6. On the graph you go up 3 and over 2, and times that to get 6.This looked like this A=(X*Y), 6=X*Y, 6=2^3, and to the last 6=6.
The individual test was fairly easy. Again, we had to plug numbers into the equation y=9-X^2. Again, I created a X and Y table using the numbers for X 0, 1, 2, 3, 4. The Y outcome numbers were 0,8,5,0,-1. The Y out come had to be checked by plugging it to the same equation, which was y=9-X . The new X and Y table points were (0,0),(8,8),(2,-1),(0,0), and (4,8). The area equation was A=X*Y, and X is given, which is 1.1. So, Y=9-!.1^2) equals Y=9-1.21, and the final answer is Y=7.79. Now we do 16=1.1* 7.79, with 7.79 rounded to the nearest tenth. Final answer for area was 16=8.67. For perimeter of the rectangle the equation is P=2x+2y, P=2(1.1)+2(7.79). Then, it went from P=2.2+15.58, and the final perimeter was P=17.78.
The individual test was fairly easy. Again, we had to plug numbers into the equation y=9-X^2. Again, I created a X and Y table using the numbers for X 0, 1, 2, 3, 4. The Y outcome numbers were 0,8,5,0,-1. The Y out come had to be checked by plugging it to the same equation, which was y=9-X . The new X and Y table points were (0,0),(8,8),(2,-1),(0,0), and (4,8). The area equation was A=X*Y, and X is given, which is 1.1. So, Y=9-!.1^2) equals Y=9-1.21, and the final answer is Y=7.79. Now we do 16=1.1* 7.79, with 7.79 rounded to the nearest tenth. Final answer for area was 16=8.67. For perimeter of the rectangle the equation is P=2x+2y, P=2(1.1)+2(7.79). Then, it went from P=2.2+15.58, and the final perimeter was P=17.78.
Self Evaluation
Even though this was hard at first, even if it was my group mates going too fast me or them just doing their own thing. I had to ask questions because they weren't asking if I needed help or them needing to slow down. I honestly don't like a group that only does their own thing when two people go fast with each other. I've become more confident in asking questions to better my understanding, no matter how dumb the question seems to be to me. I've pushed my thinking by getting out of my mindset that "I'm not good at this. I won't be able to understand it. I'll fail again." Eventually this got my to well on the individual part of the test. So my over grade I'd give myself is 97% because I've always asked questions ,and told my team to slow down when they went too fast.