In Advanced Algebra, we discussed the Rational Zero Root Theorem. This theorem only finds rational roots, not irrational roots. The homework asks you to find the possible rational roots to a polynomial function. The homework is pg. 513 #16-21. We will spend another day on it. The Chapter 8 Test is next week. All with the homework of this week, Journal #2 is due on Friday as well.
In Geometry, we wrapped up our discussion on the Law of Sines. We might do another word problem on the Law of Sines. We will begin Law of Cosines. The Chapter 8 Test is next week. The homework for today is pg. 429 #25-32.
Be safe!
Monday, March 30, 2009
Friday, March 27, 2009
The End of the Card Marking!
Well everyone, it is the last day for the 3rd quarter. Another quarter to go before the close of the school year.
In Advanced Algebra, we worked on 8-4: Roots and Zeroes. We did some application problems. Question 7 of the worksheet is a journal problem due next Friday. I will put it up on Saturday. Next week we will begin 8-5: Rational Root Theorem.
In Geometry, we review 8-4: Angles of Depression and Elevation. We began 8-5: Laws of Sines. The homework is on pg. 429 #16-24. Next week we will talk about Laws of Cosines and take the test on Thrusday. It will be a busy week.
Be Safe and Enjoy the Weekend!
In Advanced Algebra, we worked on 8-4: Roots and Zeroes. We did some application problems. Question 7 of the worksheet is a journal problem due next Friday. I will put it up on Saturday. Next week we will begin 8-5: Rational Root Theorem.
In Geometry, we review 8-4: Angles of Depression and Elevation. We began 8-5: Laws of Sines. The homework is on pg. 429 #16-24. Next week we will talk about Laws of Cosines and take the test on Thrusday. It will be a busy week.
Be Safe and Enjoy the Weekend!
Thursday, March 26, 2009
It's Friday Eve, the day before books close!
Hi, Everyone.
In Advanced Algebra, we continued our discussion of 'Roots and Zeroes'. We mentioned the purpose of Decartes' Rule of Signs. Based on the number of sign changes with p(x) as x is positive and p(x) as x is negative, can determine the number of real and imaginary roots.
Read this:
Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial, assuming that you don't have the graph to look at. This topic isn't so useful if you have access to a graphing calculator because, rather than having to do guess-n-check to find the zeroes (using the Rational Root Test, Descartes' Rule of Signs, synthetic division, and other tools), you can just look at the picture on the screen. But if you need to use it, the Rule is actually quite simple.
Use Descartes' Rule of Signs to determine the number of real zeroes of:f (x) = x5 – x4 + 3x3 + 9x2 – x + 5.
Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect.
First, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case:
f (x) = x5 – x4 + 3x3 + 9x2 – x + 5
Ignoring the actual values of the coefficients, I then look at the signs on those coefficients:
f (x) = +x5 – x4 + 3x3 + 9x2 – x + 5
I draw little lines underneath to highlight where the signs change from positive to negative or from negative to positive from one term to the next:
Then I count the number of changes:
There are four sign changes in the "positive" case. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial f (x) = x5 – x4 + 3x3 + 9x2 – x + 5. However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex and thus not graphable as x-intercepts. Because of this possibility, I have to count down by two's to find the complete list of the possible number of zeroes. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).
Now I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case):
f (–x) = (–x)5 – (–x)4 + 3(–x)3 + 9(–x)2 – (–x) + 5
= –x5 – x4 – 3x3 + 9x2 + x + 5
I look at the signs:
f (–x) = –x5 – x4 – 3x3 + 9x2 + x + 5
...and I count the number of sign changes:
There is only one sign change in this "negative" case, so there is exactly one negative root. (In this case, I don't try to count down by two's, because the first subtraction would give me a negative number.)
There are 4, 2, or 0 positive roots, and exactly 1 negative root.
Some texts have you evaluate f (x) at x = 1 (for the positive roots) and at x = –1 (for the negative roots), so you would get the expressions "1 – 1 + 3 + 9 – 1 + 5" and "–1 – 1 – 3 + 9 + 1 + 5", respectively. But you would not simplify, and the numerical values would not be the point; you would analyze only the signs, as shown above.
Using Descartes' Rule of Signs, determine the number of real solutions to4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1 = 0.
I look first at the polynomial f (x) (this is the "positive" case):
f (x) = +4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1
There are two sign changes, so there are two or, counting down in pairs, zero positive solutions. Now I look at the polynomial f (–x) (this is the "negative" case):
f (–x) = 4(–x)7 + 3(–x)6 + (–x)5 + 2(–x)4 – (–x)3 + 9(–x)2 + (–x) + 1
= –4x7 + 3x6 – x5 + 2x4 + x3 + 9x2 – x + 1
There are five sign changes, so there are five or, counting down in pairs, three or one negative solutions. Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
There are two or zero positive solutions, and five, three, or one negative solutions.
In the above example, the maximum number of positive solutions (two) and the maximum number of negative solutions (five) added up to the leading degree (seven). It will always be true that the sum of the possible numbers of positive and negative solutions will be equal to the degree of the polynomial, or two less, or four less, or.... For instance, if I had come up with a maximum answer of "two" for the possible positive solutions in the above example but had come up with only, say, "four" for the possible negative solutions, then I would known that I had made a mistake somewhere, because 2 + 4 does not equal 7, or 5, or 3, or 1.
Use Descartes' Rule of Signs to find the number of real roots of f (x) = x5 + x4 + 4x3 + 3x2 + x + 1.
I look first at f (x):
f (x) = +x5 + x4 + 4x3 + 3x2 + x + 1
There are no sign changes, so there are no positive roots. Now I look at f (–x):
f (–x) = (–x)5 + (–x)4 + 4(–x)3 + 3(–x)2 + (–x) + 1
= –x5 + x4 – 4x3 + 3x2 – x + 1
There are five sign changes, so there are as many as five negative roots.
There are no positive roots, and there are five, three, or one negative roots.
Use Descartes' Rule of Signs to determine the possible number of solutions to the equation 2x4 – x3 + 4x2 – 5x + 3 = 0.
I look first at f (x):
f (x) = +2x4 – x3 + 4x2 – 5x + 3
There are four sign changes, so there are 4, 2, or 0 positive roots. Now I look at f (–x):
f (–x) = 2(–x)4 – (–x)3 + 4(–x)2 – 5(–x) + 3
= +2x4 + x3 + 4x2 + 5x + 3
There are no sign changes, so there are no negative roots.
There are four, two, or zero positive roots, and zero negative roots.
Descartes' Rule of Signs can be useful for helping you figure out (if you don't have a graphing calculator that can show you) where to look for the zeroes of a polynomial. For instance, if the Rational Roots Test gives you a long list of potential zeroes, and you've found one negative zero, and the Rule of Signs says that there is at most one negative root, then you know that you should start looking at positive roots, because there are no more negative roots, rational or otherwise.
Similarly, if you've found, say, two positive solutions, and the Rule of Signs says that you should have, say, five or three or one positive solutions, then you know that, since you've found two, there is at least one more (to take you up to three), and maybe three more (to take you up to five), so you should keep looking for a positive solution.
The class has a worksheet to try out for tomorrow's class discussion.
In Geometry, we continued our discussion of Angle of Depression and Angle of Elevation. The class was given a worksheet to practice at home. Remember to draw pictures to figure things out.
Be Safe!
In Advanced Algebra, we continued our discussion of 'Roots and Zeroes'. We mentioned the purpose of Decartes' Rule of Signs. Based on the number of sign changes with p(x) as x is positive and p(x) as x is negative, can determine the number of real and imaginary roots.
Read this:
Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial, assuming that you don't have the graph to look at. This topic isn't so useful if you have access to a graphing calculator because, rather than having to do guess-n-check to find the zeroes (using the Rational Root Test, Descartes' Rule of Signs, synthetic division, and other tools), you can just look at the picture on the screen. But if you need to use it, the Rule is actually quite simple.
Use Descartes' Rule of Signs to determine the number of real zeroes of:f (x) = x5 – x4 + 3x3 + 9x2 – x + 5.
Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect.
First, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case:
f (x) = x5 – x4 + 3x3 + 9x2 – x + 5
Ignoring the actual values of the coefficients, I then look at the signs on those coefficients:
f (x) = +x5 – x4 + 3x3 + 9x2 – x + 5
I draw little lines underneath to highlight where the signs change from positive to negative or from negative to positive from one term to the next:
Then I count the number of changes:
There are four sign changes in the "positive" case. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial f (x) = x5 – x4 + 3x3 + 9x2 – x + 5. However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex and thus not graphable as x-intercepts. Because of this possibility, I have to count down by two's to find the complete list of the possible number of zeroes. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).
Now I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case):
f (–x) = (–x)5 – (–x)4 + 3(–x)3 + 9(–x)2 – (–x) + 5
= –x5 – x4 – 3x3 + 9x2 + x + 5
I look at the signs:
f (–x) = –x5 – x4 – 3x3 + 9x2 + x + 5
...and I count the number of sign changes:
There is only one sign change in this "negative" case, so there is exactly one negative root. (In this case, I don't try to count down by two's, because the first subtraction would give me a negative number.)
There are 4, 2, or 0 positive roots, and exactly 1 negative root.
Some texts have you evaluate f (x) at x = 1 (for the positive roots) and at x = –1 (for the negative roots), so you would get the expressions "1 – 1 + 3 + 9 – 1 + 5" and "–1 – 1 – 3 + 9 + 1 + 5", respectively. But you would not simplify, and the numerical values would not be the point; you would analyze only the signs, as shown above.
Using Descartes' Rule of Signs, determine the number of real solutions to4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1 = 0.
I look first at the polynomial f (x) (this is the "positive" case):
f (x) = +4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1
There are two sign changes, so there are two or, counting down in pairs, zero positive solutions. Now I look at the polynomial f (–x) (this is the "negative" case):
f (–x) = 4(–x)7 + 3(–x)6 + (–x)5 + 2(–x)4 – (–x)3 + 9(–x)2 + (–x) + 1
= –4x7 + 3x6 – x5 + 2x4 + x3 + 9x2 – x + 1
There are five sign changes, so there are five or, counting down in pairs, three or one negative solutions. Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
There are two or zero positive solutions, and five, three, or one negative solutions.
In the above example, the maximum number of positive solutions (two) and the maximum number of negative solutions (five) added up to the leading degree (seven). It will always be true that the sum of the possible numbers of positive and negative solutions will be equal to the degree of the polynomial, or two less, or four less, or.... For instance, if I had come up with a maximum answer of "two" for the possible positive solutions in the above example but had come up with only, say, "four" for the possible negative solutions, then I would known that I had made a mistake somewhere, because 2 + 4 does not equal 7, or 5, or 3, or 1.
Use Descartes' Rule of Signs to find the number of real roots of f (x) = x5 + x4 + 4x3 + 3x2 + x + 1.
I look first at f (x):
f (x) = +x5 + x4 + 4x3 + 3x2 + x + 1
There are no sign changes, so there are no positive roots. Now I look at f (–x):
f (–x) = (–x)5 + (–x)4 + 4(–x)3 + 3(–x)2 + (–x) + 1
= –x5 + x4 – 4x3 + 3x2 – x + 1
There are five sign changes, so there are as many as five negative roots.
There are no positive roots, and there are five, three, or one negative roots.
Use Descartes' Rule of Signs to determine the possible number of solutions to the equation 2x4 – x3 + 4x2 – 5x + 3 = 0.
I look first at f (x):
f (x) = +2x4 – x3 + 4x2 – 5x + 3
There are four sign changes, so there are 4, 2, or 0 positive roots. Now I look at f (–x):
f (–x) = 2(–x)4 – (–x)3 + 4(–x)2 – 5(–x) + 3
= +2x4 + x3 + 4x2 + 5x + 3
There are no sign changes, so there are no negative roots.
There are four, two, or zero positive roots, and zero negative roots.
Descartes' Rule of Signs can be useful for helping you figure out (if you don't have a graphing calculator that can show you) where to look for the zeroes of a polynomial. For instance, if the Rational Roots Test gives you a long list of potential zeroes, and you've found one negative zero, and the Rule of Signs says that there is at most one negative root, then you know that you should start looking at positive roots, because there are no more negative roots, rational or otherwise.
Similarly, if you've found, say, two positive solutions, and the Rule of Signs says that you should have, say, five or three or one positive solutions, then you know that, since you've found two, there is at least one more (to take you up to three), and maybe three more (to take you up to five), so you should keep looking for a positive solution.
The class has a worksheet to try out for tomorrow's class discussion.
In Geometry, we continued our discussion of Angle of Depression and Angle of Elevation. The class was given a worksheet to practice at home. Remember to draw pictures to figure things out.
Be Safe!
Wednesday, March 25, 2009
X-Men Last Stand...It involves some mathematics?
Hi, everyone.
In Advanced Algebra, we discussed 8-4: Roots and Zeroes. We mentioned the fact that every polynomial function has at least one root: real or imaginary. The roots can take the form: a + bi. For instance, the roots to x^2 + 1 = f(x) are +i and -i. They appear in the form 0 + i and 0 - i. This means that (x - (0 - i))(x -(0 + i)) = x^2 + 1 = f(x). We also generated polynomial functions from roots. That is, if x = -2, 3 and 4i. Then the polynomial function (in factored from) looks like (x + 2)(x - 3)(x - (0-4i))(x - (0+4i)), a quartic function. The homework is on pg. 507 #32-40.
In Geometry, we discussed 8-4: Angles of Elevation and Depression. Using trigonometric ratios, we determined the values of angles or side measures in word problems.
The example used in class is the following....watch the clip
http://www.youtube.com/watch?v=eiL31ZBxoy8
Looking at the clip on youtube...Look at time 1:00. As Magneto is meters and seconds away from lowering the bridge on the guards, he thinks-"I wonder what the angle of depression is to the guard as I am about to finish them!" The image of the bridge and Alcatraz forms a special image. The distance from Magneto's eye line to the guard's eye line is 1,000 ft (the hypotenuse). Magneto is 100 ft above the guard (the altitude). Based on this information, you can use 'sine of x' to figure the problem out. So, sin (x) = (100/1,000). Once you take the inverse of each side, sin^-1(100/1,000) = x -which is the angle of depression. Make sure your calculator is in degrees and not radians. The homework is pg. 422 #22-25.
Be Safe. Please, BE SAFE!
In Advanced Algebra, we discussed 8-4: Roots and Zeroes. We mentioned the fact that every polynomial function has at least one root: real or imaginary. The roots can take the form: a + bi. For instance, the roots to x^2 + 1 = f(x) are +i and -i. They appear in the form 0 + i and 0 - i. This means that (x - (0 - i))(x -(0 + i)) = x^2 + 1 = f(x). We also generated polynomial functions from roots. That is, if x = -2, 3 and 4i. Then the polynomial function (in factored from) looks like (x + 2)(x - 3)(x - (0-4i))(x - (0+4i)), a quartic function. The homework is on pg. 507 #32-40.
In Geometry, we discussed 8-4: Angles of Elevation and Depression. Using trigonometric ratios, we determined the values of angles or side measures in word problems.
The example used in class is the following....watch the clip
http://www.youtube.com/watch?v=eiL31ZBxoy8
Looking at the clip on youtube...Look at time 1:00. As Magneto is meters and seconds away from lowering the bridge on the guards, he thinks-"I wonder what the angle of depression is to the guard as I am about to finish them!" The image of the bridge and Alcatraz forms a special image. The distance from Magneto's eye line to the guard's eye line is 1,000 ft (the hypotenuse). Magneto is 100 ft above the guard (the altitude). Based on this information, you can use 'sine of x' to figure the problem out. So, sin (x) = (100/1,000). Once you take the inverse of each side, sin^-1(100/1,000) = x -which is the angle of depression. Make sure your calculator is in degrees and not radians. The homework is pg. 422 #22-25.
Be Safe. Please, BE SAFE!
Tuesday, March 24, 2009
On This Day, Tommy Hilfiger and Peyton Manning were born.
Hi, Everyone.
Today was a busy and hectic day.
In Advanced Algebra, we took a group quiz on 8-1 thru 8-3. It involved the Remainder Theorem, and various functions with their tables of values. Tomorrow we will tackle 8-4: Roots of a Function.
In Geometry, we took a group quiz on 8-1 thru 8-3. It involved the geometric mean and the Pythagorean Theorem with some trigonometric functions mixed in. Tomorrow we will use trigonometric functions in word problems, 8-4.
Be Safe!
Today was a busy and hectic day.
In Advanced Algebra, we took a group quiz on 8-1 thru 8-3. It involved the Remainder Theorem, and various functions with their tables of values. Tomorrow we will tackle 8-4: Roots of a Function.
In Geometry, we took a group quiz on 8-1 thru 8-3. It involved the geometric mean and the Pythagorean Theorem with some trigonometric functions mixed in. Tomorrow we will use trigonometric functions in word problems, 8-4.
Be Safe!
Monday, March 23, 2009
It's Week 28! There is some stuff from WK 27 that I forgot to mention.
Hi, everyone.
Last week Advanced Algebra was assigned pg. 498 #29-32,37,38 on Friday.
Starting today, Advanced Algebra worked on 8-3 again. We talked about local minimums and local maximums, zeroes, and how to find them on the table of a function. The group quiz has been pushed back until tomorrow.
In Geometry, we reviewed Geometer Sketchpad Lab 3 which covered the two special triangles: 45-90-45 and 30-60-90. The CW/HW is pg. 416 #24-40. The group quiz has been pushed back until tomorrow.
Be Safe!
Last week Advanced Algebra was assigned pg. 498 #29-32,37,38 on Friday.
Starting today, Advanced Algebra worked on 8-3 again. We talked about local minimums and local maximums, zeroes, and how to find them on the table of a function. The group quiz has been pushed back until tomorrow.
In Geometry, we reviewed Geometer Sketchpad Lab 3 which covered the two special triangles: 45-90-45 and 30-60-90. The CW/HW is pg. 416 #24-40. The group quiz has been pushed back until tomorrow.
Be Safe!
Wednesday, March 18, 2009
I was away at a workshop. It's Week 27.
Hi, everyone.
It has been a busy week, thus far.
In Advanced Algebra, on Monday-we did pg. 489 #16-23, 30-37 in class. It is 8-2: The Remainder Theorem. On Tuesday-we did a worksheet in class. It covered 8-2. Today in class we reviewed the work from the last two days. The work from Monday is due tomorrow.
In Geometry, on Monday-we did a worksheet on Geometric Mean. On Tuesday-we did a worksheet on Pythagorean Theorem. Today we reviewed both sheets in class.
All of the sheets from today will be placed on the blog tomorrow.
A reminder was issued that all assignments are due by March 30th. Or it is a zero.
Do your work! Please study! Be safe!
It has been a busy week, thus far.
In Advanced Algebra, on Monday-we did pg. 489 #16-23, 30-37 in class. It is 8-2: The Remainder Theorem. On Tuesday-we did a worksheet in class. It covered 8-2. Today in class we reviewed the work from the last two days. The work from Monday is due tomorrow.
In Geometry, on Monday-we did a worksheet on Geometric Mean. On Tuesday-we did a worksheet on Pythagorean Theorem. Today we reviewed both sheets in class.
All of the sheets from today will be placed on the blog tomorrow.
A reminder was issued that all assignments are due by March 30th. Or it is a zero.
Do your work! Please study! Be safe!
Friday, March 13, 2009
I have a 4 day weekend! NOT!
Hi, everyone.
In Advanced Algebra, we began our discussion on 8-2: The Remainder/Factor Theorem. Remember that "Dividend = Quotient (Divisor) + Remainder" for a number or for a polynomial.
For example, 56 = 3 (18) + 2. And also, x^2 + 3x - 4 = (x + 8) (x - 5) + 36.
In Geometry, we finished our discussion on 8-1: Geometric Mean and Pythagorean Theorem. Throughout the school year, we have been discussing the Pythagorean Theorem.
There is no homework due on Monday. On Wednesday, I will expect all students to have turned in their comics, fractal or list project/sheet.
BE SAFE and DO NOT GIVE THE SUB A HARD TIME NEXT WEEK!
In Advanced Algebra, we began our discussion on 8-2: The Remainder/Factor Theorem. Remember that "Dividend = Quotient (Divisor) + Remainder" for a number or for a polynomial.
For example, 56 = 3 (18) + 2. And also, x^2 + 3x - 4 = (x + 8) (x - 5) + 36.
In Geometry, we finished our discussion on 8-1: Geometric Mean and Pythagorean Theorem. Throughout the school year, we have been discussing the Pythagorean Theorem.
There is no homework due on Monday. On Wednesday, I will expect all students to have turned in their comics, fractal or list project/sheet.
BE SAFE and DO NOT GIVE THE SUB A HARD TIME NEXT WEEK!
Thursday, March 12, 2009
It's Friday-Eve. The kids are coming back :)
Hi, everyone. It has been wierd without all of the students. Tomorrow will be a review day for the class on Chapter 8 Section 1. And we more on!
In Advanced Algebra, 8-1 is Graphing Polynomial Functions; we will briefly review it and start 8-2. Today we took a retake quiz on Conic Sections.
In Geometry, 8-1 is Geometric Mean and Pythagorean Thm. We did classwork on Geometric Mean-pg. 408 #15-29. Also, students turned in the Fractal Project and Primitive Pythagorean Triples List. Tomorrow, notebook check (binders) are due.
Be Safe!
In Advanced Algebra, 8-1 is Graphing Polynomial Functions; we will briefly review it and start 8-2. Today we took a retake quiz on Conic Sections.
In Geometry, 8-1 is Geometric Mean and Pythagorean Thm. We did classwork on Geometric Mean-pg. 408 #15-29. Also, students turned in the Fractal Project and Primitive Pythagorean Triples List. Tomorrow, notebook check (binders) are due.
Be Safe!
Wednesday, March 11, 2009
Second Day of Smaller Classes. I miss all my students!
In Advanced Algebra, we continued are discussion of 8-1: Graphing Polynomials. Tomorrow we will have a retake quiz on Conic Sections & complete are discussion on 8-1.
In Geometry, we continued are discussion of 8-1: Geometric Mean. We did an activity in finding the height of the walls using the geometric mean. The homework is to find 10 Primitive Pythagorean Triples and The Fractal Project.
BE SAFE!
In Geometry, we continued are discussion of 8-1: Geometric Mean. We did an activity in finding the height of the walls using the geometric mean. The homework is to find 10 Primitive Pythagorean Triples and The Fractal Project.
BE SAFE!
Tuesday, March 10, 2009
The First Day of ACT Testing! Smaller Classes :)
Hi, everyone!
Since there is ACT testing, all of my classes moved at a slower pace today.
In Advanced Algebra, we discussed conic sections (parabola, ellipse, hyperbola, and circle) and reviewed for the quiz retake on Thursday. Be sure to have yesterday's homework assignment completed for tomorrow's discussion.
In Geometry, (during 2nd, 4th, and 6th hours) we began 8-1: Geometric Mean/Pythagorean Theorem. We discussed the formula for geometric mean: (x/a)=(b/x) or x^2 = ab. We also did an activity to see the similarities between several triangles within a right triangle. The homework is to obtain a list of 10 Pythagorean Primitive Triples by Thursday. Use the internet to your advantage.
BE SAFE!!
Since there is ACT testing, all of my classes moved at a slower pace today.
In Advanced Algebra, we discussed conic sections (parabola, ellipse, hyperbola, and circle) and reviewed for the quiz retake on Thursday. Be sure to have yesterday's homework assignment completed for tomorrow's discussion.
In Geometry, (during 2nd, 4th, and 6th hours) we began 8-1: Geometric Mean/Pythagorean Theorem. We discussed the formula for geometric mean: (x/a)=(b/x) or x^2 = ab. We also did an activity to see the similarities between several triangles within a right triangle. The homework is to obtain a list of 10 Pythagorean Primitive Triples by Thursday. Use the internet to your advantage.
BE SAFE!!
Monday, March 9, 2009
It's Week 26. The week of Pi.
Hi. Today was an average day.
Advanced Algebra began learning about polynomial functions. We drew the parent graphs and some variations of the linear, quadratic, cubic, quartic, and quintic functions. The homework is pg. 483 #24-32.
Geometry took their test of Chapter 7, which they were unable to do on Friday. The Fractal Project is due on Thursday/Friday. The notebook check will not occur until Friday, due to my absence and ACT/MME testing.
Good luck, Juniors and remember to be safe!
Advanced Algebra began learning about polynomial functions. We drew the parent graphs and some variations of the linear, quadratic, cubic, quartic, and quintic functions. The homework is pg. 483 #24-32.
Geometry took their test of Chapter 7, which they were unable to do on Friday. The Fractal Project is due on Thursday/Friday. The notebook check will not occur until Friday, due to my absence and ACT/MME testing.
Good luck, Juniors and remember to be safe!
Thursday, March 5, 2009
It's Parent Teacher Conference Day1
Hi, Everyone.
In Advanced Algebra, we reviewed conic sections in preparation for the quiz tomorrow. Students also turned in their Comics Project.
In Geometry, we reviewed for the Chapter 7 Test - which is tomorrow. There will be two parts: one in class and one over the weekend. A new project was announced: create a fractal up to stage 4. Be creative.
Earlier I introduced a video on similar triangles. Here is part two.
www.youtube.com/watch?v=qO2cTx6DwCA
Copy it to the URL and watch it.
BE SAFE!
In Advanced Algebra, we reviewed conic sections in preparation for the quiz tomorrow. Students also turned in their Comics Project.
In Geometry, we reviewed for the Chapter 7 Test - which is tomorrow. There will be two parts: one in class and one over the weekend. A new project was announced: create a fractal up to stage 4. Be creative.
Earlier I introduced a video on similar triangles. Here is part two.
www.youtube.com/watch?v=qO2cTx6DwCA
Copy it to the URL and watch it.
BE SAFE!
Wednesday, March 4, 2009
Tomorrow is Parent Teacher Conference....Okay!
Well another TechnoDay has occured. In Geometry, we went to the computer lab and worked with Geometer Sketchpad and understanding more about iterations and similarity. The homework is to prepare for the test on Friday.
In Advanced Algebra, we discussed how to interpret the different conic sections by looking at the equations. For instance:
1. parabola equations - they always have at most one quadratic term with degree one for one variable and the other variable is degree one.
2. circle equations - they always have two quadratic terms with degree two and whose coefficients are positive one.
3. ellipse equations - they always have two quadratic terms with degree two and whose coefficients are primarily a number other than one.
4. hyperbola equations - they always have two quadratic terms with degree two and whose coefficients are primarily a number other than one; and either the coefficient is -1 for x^2 or for y^2.
The quiz is this Friday.
BE SAFE AND STUDY!
In Advanced Algebra, we discussed how to interpret the different conic sections by looking at the equations. For instance:
1. parabola equations - they always have at most one quadratic term with degree one for one variable and the other variable is degree one.
2. circle equations - they always have two quadratic terms with degree two and whose coefficients are positive one.
3. ellipse equations - they always have two quadratic terms with degree two and whose coefficients are primarily a number other than one.
4. hyperbola equations - they always have two quadratic terms with degree two and whose coefficients are primarily a number other than one; and either the coefficient is -1 for x^2 or for y^2.
The quiz is this Friday.
BE SAFE AND STUDY!
Tuesday, March 3, 2009
FRACTALS ARE FUN!
In Advanced Algebra, we discussed hyperbolas. There are vertical and horizontal sketches of these kinds of conic sections. Tomorrow we'll look at all the conic sections. The homework is pg. 446 #23-30.
In Geometry, we discussed Fractals again. We reviewed the process of iteration and found the dimensions of various fractals. We saw the relationship between Sierpenski's Triangle and Pascal's Triangle. The homework is to review for the test on Friday, which is on Chapter 7.
Here is a video on ratios; I think it might be helpful.
Tomorrow, I'll put up the other video that complements it!
Remember Be Safe!!
In Geometry, we discussed Fractals again. We reviewed the process of iteration and found the dimensions of various fractals. We saw the relationship between Sierpenski's Triangle and Pascal's Triangle. The homework is to review for the test on Friday, which is on Chapter 7.
Here is a video on ratios; I think it might be helpful.
Tomorrow, I'll put up the other video that complements it!
Remember Be Safe!!
Monday, March 2, 2009
It's Monday- Week 25!
Hello, Everyone!
I just saw Medea Goes to Jail"! I never laughed so loud!
Anyways, in Advanced Algebra, we discussed Ellipses. We talked about major and minor axises. The homework is pg. 437 #19-26.
In Geometry, we discussed FRACTALS! We did preliminary work on learning about fractals. We dealt with 2D objects and learned LENGTH ^2 = SIZE. We dealt with 3D objects and learned LENGTH ^2 = SIZE. What was the general equation that we developed?
The homework is 7-6 Practice sheet which will be placed on the blog tomorrow.
BE SAFE!
I just saw Medea Goes to Jail"! I never laughed so loud!
Anyways, in Advanced Algebra, we discussed Ellipses. We talked about major and minor axises. The homework is pg. 437 #19-26.
In Geometry, we discussed FRACTALS! We did preliminary work on learning about fractals. We dealt with 2D objects and learned LENGTH ^2 = SIZE. We dealt with 3D objects and learned LENGTH ^2 = SIZE. What was the general equation that we developed?
The homework is 7-6 Practice sheet which will be placed on the blog tomorrow.
BE SAFE!
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