Monday, 19 June 2017

Friday, June 16

*Oops* had this in drafts all weekend. Here is an update on last week's classes.

In class: The last three days we have had about half the class gone on the Ashland field trip. On Wednesday we investigated the properties of s-curves and paper clips [see here]. On Thursday we played grudge ball and battled over questions like

What is the sum of the mean, median, and mode of the numbers 2, 3, 0 , 3, 1, 4, 0, 3? [answer: 7.5]

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ , $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs? [can you figure out the answer? it is less than 30…]

Friday, 16 June 2017

Friday, June 16

by nickhershman Add Your Comment

Hi All —

In class this week students have been working on their statistics project. The project asks them to analyze a question in statistics — either a question which I created from a data set of student height and gender in CPMS advisory classes or a question of their choosing with data they find or generate. Here is the project task sheets and rubric.

Statistics project is due on Tuesday.

Thursday, 8 June 2017

Thursday, June 8 + Facial Tissue Request 😤

by nickhershman Add Your Comment

It would be a great help if you can send any donations of facial tissue! Many students have runny noses this time of the year and our previous classroom stockpile of facial tissue has run out completely. If you can send a box to school with your student that would be a great help! Thank you!

Upcoming: Trigonometry Test on Tuesday, June 13th.

In class: Problem solving with trigonometry. In order to be Highly Proficient in trigonometry students need to be able to solve a variety of problems involving right and non-right triangles. They should be able to write and solve equations involving $sin, cos$ and $tan$ as well as their inverses, recognize how to construct right triangles and use them to find missing measurements in non-right triangles. Recognize and explain how S-S-A information about a triangle may be ambiguous and measure both cases. Also, HP students will understand and be able to prove the Law of Sines and Cosines for acute triangles.

Assignment: For Monday students should complete

Problem Set #1 ("June 7")

Problem Set #2 ("Trigonometry p-set #6")

Notes: I made a trig review worksheet (here)

Thursday, June 8

by nickhershman Add Your Comment

Assignment 9.4 Task + RSG. In class notes on standard deviation

Note: The formula for standard deviation involves $Sigma$ (Greek, "Sigma") or "sum" notation. $x_i$ refers to the ith element in the data set, $overline{x}$ is the mean x-value.

$SD = sqrt{dfrac{sum_{i=1}^{N}(x_i - overline{x})^2}{N}}$

Reading this formula means:

Find the difference between each value and the mean
$(x_i - overline{x})$
Square those differences
$(x_i - overline{x})^2$
Find the sum of all of the squares
$sum_{i=1}^{N}(x_i - overline{x})^2$
Divide that value by N (the number of data elements in the set)
$dfrac{sum_{i=1}^{N}(x_i - overline{x})^2}{N}$
The Standard Deviation is the square root of the result
$SD = sqrt{dfrac{sum_{i=1}^{N}(x_i - overline{x})^2}{N}}$

One way to understand the standard deviation is that it is kind of like the average distance from the mean in the data set.

Wednesday, 7 June 2017

Wednesday, June 7 + Facial Tissue request :)

by nickhershman Add Your Comment

In class we have started Module 9 which is about data and statistics. Today students are completing 9.3 Task and Ready Set Go. We have completed 9.1 and 9.2 as well as a "3 Chips" Problem write-up that was turned in today at the start of class.

The class has run out of facial tissue but there are still plenty of sniffles. So, if you would like to donate some facial tissue to the class that would be much appreciated and put to good use, keeping kids in class instead of heading to the restroom. Thank you thank you thank you!

Friday, 26 May 2017

Thursday, May 25

by nickhershman Add Your Comment

Assignment: 12.2 on p. 628: 1-16, 21

Thursday, 25 May 2017

Wednesday, May 24

by nickhershman Add Your Comment

In class we worked on converting functions into rational form.

Assignment: 9.8 on p. 555: 1-8, 11-12

Here is an example problem, similar to the one worked in class.

Let's analyze the function given by the equation below. We will try to work out its various characteristics and sketch a graph before resorting to computer graphing.

$y = dfrac{x-1}{(x+1)(x - 2)} + dfrac{2x - 1}{x^2 - 4}$

To obtain a common denominator we multiply the first term by $(x+2)/(x+2)$ and the second term by $(x + 1)/(x+1)$

$y = dfrac{(x - 1)(x + 2)}{(x + 1)(x - 2)(x + 2)} + dfrac{(2x - 1)(x + 1)}{(x + 1)(x - 2)(x + 2)}$

Now we can simplify the expressions in the numerators and re-write as a single fraction.

$y = dfrac{x^2 + x - 2 + 2x^2 + x - 1}{(x +1)(x - 2)(x + 2)}$

Which simplifies to

$y = dfrac{3x^2 + 2x - 3}{(x + 1)(x - 2)(x + 2)}$ or $y = dfrac{3x^2 + 2x - 3}{x^3 + x^2 - 4x - 4}$

We note that $3x^2 + 2x - 3$ , the numerator, cannot be factored. So, later on as we analyze the properties of this function we will need to solve $3x^2 + 2x - 3 = 0$ to determine the value of any x-intercepts.

Now that the function is in rational form, we can analyze its properties.

We note that there are no holes in the function. Because there are no factors common to both the numerator and denominator. i.e. $(x + 2)/(x + 2)$ . These common factors would work out to essentially multiplying the function by 1 for all values of $x$ except for when $x + 2 = 0$ at $x = -2$. For that single value $(x + 2)/(x + 2)$ becomes undefined and we get a hole in the graph. Anyway, since this is not the case, our function has no holes.

The function has vertical asymptotes at

$x = -1, x = 2$ and $x = -2$

Inline image 1

We can tell from the numerator that the function will have x-intercepts where the numerator evaluates to zero, so we solve $3x^2 + 2x - 3 = 0$ to get

$x = dfrac{-2 pm sqrt{2^2 - 4(3)(-3)}}{2(3)}$

$x = dfrac{-2 pm sqrt{40}}{6}$ = $dfrac{-2 pm 2sqrt{10}}{6} = dfrac{-1 pm sqrt{10}}{3}$ $approx -1.387,0.72$

So, we note the two x-intercepts.

Inline image 1

In addition, we note that this function has denominator of greater degree than the numerator, so as $x$ approaches $pm infty$ the denominator will overpower the numerator and the value of the function will approach $0$ , creating horizontal asymptotes at $y = 0$

Inline image 3

Through point testing and some more intuition we can identify the location of the 4 sections of this function and sketch a graph

Inline image 2

Wednesday, May 24

by nickhershman Add Your Comment

Students took the Similarity test (chapter 11) on Friday. This week we have learned the right-triangle definition of the trigonometric functions sine, cosine and tangent. As of now these functions take as input one of the non-right angles in a right triangle and their output is a specific ratio of sides in all such similar right triangles.

So, based on this definition, if $sin(35^circ) approx .57$ that means that in any right triangle with a $35^circ$ angle, the ratio of $dfrac{text{opposite side}}{text{adjacent side}}$ is approximately .57. In the figure below this means that $dfrac{AC}{BC} approx .57$ .