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 Article posted June 2, 2012 at 12:33 PM GMT • comment • Reads 702 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 12:33 PM GMT • comment • Reads 702
 Article posted June 10, 2012 at 12:03 PM GMT • comment • Reads 378 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 12:03 PM GMT • comment • Reads 378
 Article posted June 11, 2012 at 12:40 AM GMT • comment • Reads 360 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 11, 2012 at 12:40 AM GMT • comment • Reads 360
 Article posted June 7, 2012 at 09:31 PM GMT • comment • Reads 319 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 09:31 PM GMT • comment • Reads 319
 Article posted June 7, 2012 at 09:49 PM GMT • comment • Reads 456 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 09:49 PM GMT • comment • Reads 456
 Article posted June 11, 2012 at 03:26 AM GMT • comment • Reads 270 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 11, 2012 at 03:26 AM GMT • comment • Reads 270
 Article posted June 11, 2012 at 03:34 AM GMT • comment • Reads 520 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 11, 2012 at 03:34 AM GMT • comment • Reads 520
 Article posted June 10, 2012 at 03:32 PM GMT • comment • Reads 259 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 03:32 PM GMT • comment • Reads 259
 Article posted June 10, 2012 at 04:24 PM GMT • comment • Reads 255 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 04:24 PM GMT • comment • Reads 255
 Article posted June 10, 2012 at 05:01 PM GMT • comment • Reads 268 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 05:01 PM GMT • comment • Reads 268
 Article posted June 11, 2012 at 07:06 PM GMT • comment • Reads 274 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 07:06 PM GMT • comment • Reads 274
 Article posted June 11, 2012 at 07:10 PM GMT • comment • Reads 466 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 07:10 PM GMT • comment • Reads 466
 Article posted June 12, 2012 at 02:04 AM GMT • comment • Reads 283 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 12, 2012 at 02:04 AM GMT • comment • Reads 283
 Article posted June 12, 2012 at 02:16 AM GMT • comment • Reads 274 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 12, 2012 at 02:16 AM GMT • comment • Reads 274
 Article posted June 11, 2012 at 02:00 AM GMT • comment • Reads 252 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 11, 2012 at 02:00 AM GMT • comment • Reads 252
 Article posted June 11, 2012 at 02:23 AM GMT • comment • Reads 375 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 11, 2012 at 02:23 AM GMT • comment • Reads 375
 Article posted June 12, 2012 at 10:29 PM GMT • comment • Reads 275 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 10:29 PM GMT • comment • Reads 275
 Article posted June 12, 2012 at 10:31 PM GMT • comment • Reads 379 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 10:31 PM GMT • comment • Reads 379
 Article posted June 9, 2012 at 03:48 PM GMT • comment • Reads 367 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H   Answer: X=13 H=8.1 Article posted June 9, 2012 at 03:48 PM GMT • comment • Reads 367
 Article posted June 4, 2012 at 11:43 AM GMT • comment • Reads 300 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 11:43 AM GMT • comment • Reads 300
 Article posted June 8, 2012 at 02:17 AM GMT • comment • Reads 251 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 8, 2012 at 02:17 AM GMT • comment • Reads 251
 Article posted June 11, 2012 at 03:45 AM GMT • comment • Reads 280 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 11, 2012 at 03:45 AM GMT • comment • Reads 280
 Article posted June 10, 2012 at 04:25 PM GMT • comment • Reads 247 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 04:25 PM GMT • comment • Reads 247
 Article posted June 9, 2012 at 07:17 PM GMT • comment • Reads 247 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 07:17 PM GMT • comment • Reads 247
 Article posted June 6, 2012 at 09:34 PM GMT • comment • Reads 273 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 09:34 PM GMT • comment • Reads 273
 Article posted June 11, 2012 at 01:47 AM GMT • comment • Reads 257 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 11, 2012 at 01:47 AM GMT • comment • Reads 257
 Article posted June 10, 2012 at 03:17 PM GMT • comment • Reads 230 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 03:17 PM GMT • comment • Reads 230
 Article posted June 14, 2012 at 05:40 PM GMT • comment • Reads 272 We finally made it!  Finals are around the corner, and so is summer!  I can’t wait to ditch my backpack and head to the beach.  But first, I have to survive finals.  To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below.     KL⎮⎮JM in isosceles trapezoid JKLM.  Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º.   m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3   m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11   I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all.  I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 05:40 PM GMT • comment • Reads 272
 Article posted June 14, 2012 at 05:40 PM GMT • comment • Reads 279 We finally made it!  Finals are around the corner, and so is summer!  I can’t wait to ditch my backpack and head to the beach.  But first, I have to survive finals.  To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below.     KL⎮⎮JM in isosceles trapezoid JKLM.  Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º.   m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3   m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11   I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all.  I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 05:40 PM GMT • comment • Reads 279
 Article posted June 6, 2012 at 09:33 PM GMT • comment • Reads 295 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 09:33 PM GMT • comment • Reads 295
 Article posted June 10, 2012 at 09:52 PM GMT • comment • Reads 252 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 09:52 PM GMT • comment • Reads 252
 Article posted June 10, 2012 at 09:11 PM GMT • comment • Reads 252 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 09:11 PM GMT • comment • Reads 252
 Article posted June 11, 2012 at 01:54 AM GMT • comment • Reads 250 Hello everyone! Hope your week has gone nicely!    Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review.    This problem was on a worksheet I did recently involving the Law of Sines:   sinA  =  sinB    a          b    A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite.    Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B.    Through substitution, I came out with this equation:    sin(35) = sin(B)    8            12   By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B:      sin-1(12sin(35)) = sin-1(sin(B))                 8    m   This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 11, 2012 at 01:54 AM GMT • comment • Reads 250
 Article posted June 11, 2012 at 02:02 AM GMT • comment • Reads 250 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now.    Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown:  a) 378 sq. ft     b) 189 sq. ft     c) 162 sq. ft     d) 27(square root of)85 sq. ft.  I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189  With that equation, I was able to find that the area was 189 sq. ft., or answer b.  I hope this has been a helpful review question! Have a great week! (: Article posted June 11, 2012 at 02:02 AM GMT • comment • Reads 250
 Article posted June 11, 2012 at 01:18 PM GMT • comment • Reads 317 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 01:18 PM GMT • comment • Reads 317
 Article posted June 11, 2012 at 12:04 AM GMT • comment • Reads 213 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 11, 2012 at 12:04 AM GMT • comment • Reads 213
 Article posted June 4, 2012 at 02:38 PM GMT • comment • Reads 296 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 02:38 PM GMT • comment • Reads 296
 Article posted June 6, 2012 at 12:53 AM GMT • comment • Reads 266 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house?   ***SPOILER ALERT***   Answer: Roughly 17 Feet   --Joe Article posted June 6, 2012 at 12:53 AM GMT • comment • Reads 266
 Article posted June 10, 2012 at 12:42 PM GMT • comment • Reads 229 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard.  Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum?   ANSWER: 4330 ft. squared   Have a great last week! --Joe Article posted June 10, 2012 at 12:42 PM GMT • comment • Reads 229