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905 MATH (2010)

Wednesday, May 4, 2011. Jomari's Post for May 4, 2011. Today we learned about perpendicular bisectors. These are lines that bisect straight lines at 90°. Here's how to make one:. We made these bisectors for several chords. In a circle, we don't need to make crossed arcs on one side of the chord because of the center. After wards, we answered a problem involving bisectors. It was similar to this:. This is what we have to work with:. WE COULD USE THE PYTHAGORAS THEOREM TO FIND B. 178; b²=c². 8730; b ².

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905 MATH (2010) | spmath90510.blogspot.com Reviews

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Wednesday, May 4, 2011. Jomari's Post for May 4, 2011. Today we learned about perpendicular bisectors. These are lines that bisect straight lines at 90°. Here's how to make one:. We made these bisectors for several chords. In a circle, we don't need to make crossed arcs on one side of the chord because of the center. After wards, we answered a problem involving bisectors. It was similar to this:. This is what we have to work with:. WE COULD USE THE PYTHAGORAS THEOREM TO FIND B. 178; b²=c². 8730; b ².

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905 MATH (2010) | spmath90510.blogspot.com Reviews

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Wednesday, May 4, 2011. Jomari's Post for May 4, 2011. Today we learned about perpendicular bisectors. These are lines that bisect straight lines at 90°. Here's how to make one:. We made these bisectors for several chords. In a circle, we don't need to make crossed arcs on one side of the chord because of the center. After wards, we answered a problem involving bisectors. It was similar to this:. This is what we have to work with:. WE COULD USE THE PYTHAGORAS THEOREM TO FIND B. 178; b²=c². 8730; b ².

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905 MATH (2010): January 2011

http://www.spmath90510.blogspot.com/2011_01_01_archive.html

Friday, January 28, 2011. Binesi's Blog for January 28th 2011. Today in Class we took some notes. I was half awake and on the verge of collapsing, so I didn't quite get all of the information. I borrowed Hannah's notes so I figure about half of this scribe is accredited to her. For homework we have to do a Power Point Presentation. Put an example, non example, picture, and a definition. We also have to do 6.3 CYU #2, either the odd or even for practice, Apply (any 3), and extend any 2. Cm 2m, 413m, 819m,...

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905 MATH (2010): October 2010

http://www.spmath90510.blogspot.com/2010_10_01_archive.html

Wednesday, October 27, 2010. MrBacke the email you sent me for the invitation expired so I couldn't really do anything. Sorry for not clicking it a while back so I asked Alvin too let me use his blogger account to do my post. So in class today we learned the following: add proper and improper fractions. 1/4 1 / 4 = 1 1/4. When your subtracting fractions. 1/2 - 1 /6 = 3/6 - 1/6. When adding mixed numbers, first add the whole numbers then add the fractions. 2 1/4 3 1/4 = 2 3 1/4 1/4. 3 ( 2/10 - 5/10 ).

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905 MATH (2010): Jomari's Post for May 4, 2011

http://www.spmath90510.blogspot.com/2011/05/jomaris-post-for-may-4-2011.html

Wednesday, May 4, 2011. Jomari's Post for May 4, 2011. Today we learned about perpendicular bisectors. These are lines that bisect straight lines at 90°. Here's how to make one:. We made these bisectors for several chords. In a circle, we don't need to make crossed arcs on one side of the chord because of the center. After wards, we answered a problem involving bisectors. It was similar to this:. This is what we have to work with:. WE COULD USE THE PYTHAGORAS THEOREM TO FIND B. 178; b²=c². 8730; b ².

4

905 MATH (2010): February 2011

http://www.spmath90510.blogspot.com/2011_02_01_archive.html

Monday, February 28, 2011. Noelle's Blog Post for Feb. 28th, 2011. What did we learn today? Turning a trinomial into a binomial:. Trinomial - - Binomial. How do you do it? To get from a trinomial to binomial you have to find the product of each term and seperate them into the two terms of a binomial. Here's the tricky part; The constants in your binomials must add up to the constant or coefficient in the middle term, but multiply to get the constant or coefficient in the last term of the trinomial. Homew...

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905 MATH (2010): Jem's Blog for March 22, 2011

http://www.spmath90510.blogspot.com/2011/03/jems-blog-for-march-22-2011.html

Wednesday, March 23, 2011. Jem's Blog for March 22, 2011. So yesterday for math class our test was. Due to lack of understanding questions and how to solve them. Anyways, to help us understand the "hard" questions Mr.Backe went over questions with us, and made sure that we. What the questions mean, and how to solve them. Lets begin :. Solve what you can and need to first ). From here you can now cross multiply ). 2 ( 2x 8 ) = 9x 18. Now you solve what you can and need to again ). 4x - 16 = 9x 18.

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This site belongs to the class of 9-05 at Sargent Park School any correspondence can be addressed to rbacke97@gmail.com. Scribe Post for May 27, 2010. Sunday, May 30, 2010. Hello 9-05. Sorry this is a late scribe . I honestly forgot . but this is for the day before the test, if you don't remember. Anyways, in class we went over Question #21 on Page 403 in the textbook. The radius is 3cm because the diameter is 6cm (double the radius)*. Have fun with what's left of the weekend! Scribe Post for May 20, 2010.

spmath90510.blogspot.com

905 MATH (2010)

Wednesday, May 4, 2011. Jomari's Post for May 4, 2011. Today we learned about perpendicular bisectors. These are lines that bisect straight lines at 90°. Here's how to make one:. We made these bisectors for several chords. In a circle, we don't need to make crossed arcs on one side of the chord because of the center. After wards, we answered a problem involving bisectors. It was similar to this:. This is what we have to work with:. WE COULD USE THE PYTHAGORAS THEOREM TO FIND B. 178; b²=c². 8730; b ².

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