3 non-collinear points determine how many distinct planes?
Well, let's look at Lesson 9-1 to find out:
Point-Line-Plane Postulate:
f. Through three noncollinear points, there is exactly one plane.
Therefore the correct answer is 1 -- and of course, today's date is the first.
Today I subbed in an seventh grade English class. Since I just did a "Day in the Life" yesterday, there's no need for me to do another one today. I will point out a few quick details. The students are reading "The Miracle Worker," a play about one of the world's most famous teachers -- Anne Sullivan, the tutor of Helen Keller. It's interesting that the students would read this story today, since this is the 50th anniversary of Keller's death. Yes indeed, the famous disability rights activist died back on June 1st, 1968.
Let's code some more music in Mocha, using our 18EDL scale.
The 18EDL scale:
Degree Ratio Note
18 1/1 white D
17 18/17 umber D#
16 9/8 white E
15 6/5 green F
14 9/7 red F#
13 18/13 ocher G
12 3/2 white A
11 18/11 amber B
10 9/5 green C
9 2/1 white D
http://www.haplessgenius.com/mocha/
10 CLS
20 N=8
30 FOR A=0 TO 6
40 B=4
50 X=A-INT(A/2)*2
60 IF X=0 THEN D=18 ELSE D=17
70 PRINT D;
80 L=RND(B)
90 SOUND 261-N*D,4*L
100 IF L>1 THEN FOR I=1 TO L-1:PRINT " ";:NEXT I
110 B=B-L
120 IF B>0 THEN D=19-RND(10):GOTO 70
130 PRINT
140 NEXT A
150 PRINT 18
160 SOUND 261-N*18,16
This randomizer is based on D (supermajor) and D# (diminished 7th) chords.
I've been thinking about the "Packet Song" this week after hearing that we might need to assign summer packets to supplement Edgenuity. Like most of the songs I played, I wrote the song in a major key, and so it fits the supermajor 18EDL scale. But if any song needs to be in a minor key to match the mood of the class -- the students groan that they have to do packets, and I groan as I have to create them when I shouldn't have to -- the "Packet Song" is the one. We know that 18EDL also supports a minor scale, but then we could just use 12EDL instead of 18EDL.
Here's a simple version of "The Packet Song" in Mocha. I decided to use the minor third for the first two verses and a supermajor third for the last verse (see line 70, which changes Degree 15 in the Verse 3 to Degree 14). So this demonstrates how 18EDL can be used to play both minor thirds and supermajor thirds in the same song. (Once again, I haven't decided yet whether I'll keep this tune, change it, or throw it out completely.)
NEW
10 N=12
20 FOR V=1 TO 3
30 FOR X=1 TO 5
40 FOR Y=1 TO 7
50 READ D
60 IF (V<3) OR (D<>15) THEN SOUND 261-N*D,4
70 IF (V=3) AND (D=15) THEN SOUND 261-N*14,4
80 NEXT Y
90 FOR I=1 TO 400
100 NEXT I,X
110 RESTORE
120 NEXT V
130 DATA 16,16,11,10,9,12,12
140 DATA 18,16,15,16,18,12,16
150 DATA 11,13,15,10,9,15,12
160 DATA 15,9,10,15,13,11,11
170 DATA 16,13,16,16,16,15,9
If we wanted to convert 18EDL to an EDO scale, here are the possibilities:
1,4, 6, 7, 11, 24, 34, 53, 72, 342, ...
We see that 24EDO makes the list. Last week, we found out how 24EDO supports the 13-limit, and then we can extend it to 17-limit by using an ordinary semitone for 17/16 and 18/17.
The next EDO on the list is 34EDO. This is one of the EDO's that was mentioned last week in that Dozens Online debate about 50EDO -- it was pointed out that even 34EDO is superior to 50EDO, and two more EDO's (53EDO and 72EDO) appear on both the debate thread and this EDO list.
Here's a link to the Xenharmonic site regarding 34EDO. (All links will be dead after July!)
http://xenharmonic.wikispaces.com/34edo
34edo divides the octave into 34 equal steps of approximately 35.29412 cents. 34edo contains two 17edo's and the half-octave tritone of 600 cents. It excels as a 5-limit system, with tuning even more accurate than 31edo, but with a sharp fifth rather than a flat one, and supports hanson, srutal, tetracot, würschmidt and vishnu temperaments. It does less well in the 7-limit, with two mappings possible for 7/4: a flat one from the patent val, and a sharp one from the 34d val. By way of the patent val 34 supports keemun temperament, and 34d is an excellent alternative to 22edo for 7-limit pajara temperament. In the 11-limit, 34de supports 11-limit pajaric, and in fact is quite close to the POTE tuning; it adds 4375/4374 to the commas of 11-limit pajaric. On the other hand, the 34d val supports pajara, vishnu and würschmidt, adding 4375/4374 to the commas of pajara. Among subgroup temperaments, the patent val supports semaphore on the 2.3.7 subgroup.
Unfortunately, Kite's color notation isn't mentioned at the above link.
Notice that 34EDO is not a meantone system. Therefore we can't just blindly assume that C-E is a major third. At the above link, we see the following:
Viewed in light of Western diatonic theory, the three extra steps (of 34-et compared to 31-et) in effect widen the intervals between C and D, F and G, and A and B [that is: 6 5 3 6 5 6 3], thus making a distinction between major tones, ratio 9/8 and minor tones, ratio 10/9.
This comes from an external link using Johnston notation. In Johnston notation, C-D-E-F-G-A-B-C is always a just major scale, so C-D is 9/8 while D-E is 10/9, so C-E is 5/4. But Xenharmonic prefers Pythagorean notation where F-C, C-G, G-D, and so on are all 3/2 perfect fifths. Thus C-D is indeed 9/8 but so is D-E, so that C-E is always 81/64 (which isn't 5/4 except in meantone).
Oh, and there's one more thing to explain about 34EDO. In our other non-meantone tuning (27EDO), it's theoretically possible to use sharps and flats (including double sharps and flats) to indicate all notes of the scale, though this might be misleading (such as Cx/C double sharp being enharmonic to Gbb/G double flat).
On the other hand, it's impossible to use sharps/flats to notate all notes of 34EDO. This is because the perfect fifth is 20 steps while the octave is 34 steps, and the GCF of 20 and 34 is 2, not 1. In this sense, 34EDO is more like last week's 24EDO -- the quarter-sharps and quarter-flats can never be reached starting with C and moving either fourthward or fifthward on the circle of fifths.
Thus at the Xenharmonic link above, the note a perfect third above D is F#v (F# down), rather than F# as it would be in meantone. The Xenharmonic site often begins its scales with D rather than C -- probably because starting on D is more symmetrical. (Recall that D Dorian is the only completely symmetrical mode on the white keys of a piano.)
Fortunately, starting on D works for us since this is the fundamental note of 18EDL as well. Here is the 18EDL scale converted to 34EDO:
The 18EDL scale (converted to 34EDO):
Degree Ratio Note
18 1/1 D
17 18/17 D#v (D# down)
16 9/8 E
15 6/5 F^ (F up)
14 9/7 F#^ (F# up)
13 18/13 Ab
12 3/2 A
11 18/11 A#
10 9/5 C^ (C up)
9 2/1 D
So "green" and "red" map to "up" while "umber" maps to "down." Meanwhile, "ocher" maps to "double-up" and "amber" maps to "double-down," but double-up and double-down notes are enharmonic in 34EDO to notes appearing on the circle of fifths (G^^=Ab, Bvv=A#). (Also, notice that 9/7 appears at the link as F#, not F#^, because that list isn't consistent on 7. So ratios involving 7 don't sound as good as those involving other primes.)
Here's a video in 34EDO:
So "green" and "red" map to "up" while "umber" maps to "down." Meanwhile, "ocher" maps to "double-up" and "amber" maps to "double-down," but double-up and double-down notes are enharmonic in 34EDO to notes appearing on the circle of fifths (G^^=Ab, Bvv=A#). (Also, notice that 9/7 appears at the link as F#, not F#^, because that list isn't consistent on 7. So ratios involving 7 don't sound as good as those involving other primes.)
Here's a video in 34EDO:
Question 27 of the SBAC Practice Exam is on quadrilaterals:
In the given figure, quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Ted claims that the two shaded triangles [ABC and DCE] are congruent. Is Ted's claim correct? Include all work and/or reasoning necessary to either prove the two triangles congruent or to disprove Ted's claim.
This is a Geometry question. Let's try to come up with a two-column proof of Ted's claim:
Given: Quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Prove: Triangle ABC = DCE.
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. ABCD is a pgram. 2. A rectangle is a parallelogram.
3. AB = DC, AC = DE 3. Opposite sides of a pgram are congruent.
4.
5. Angle BAC = ACD, 5. Alternate Interior Angles (parallel line consequences)
Angle ACD = CDE
6. Angle BAC = CDE 6. Transitive Property of Congruence
7. Triangle ABC = DCE 7. SAS Congruence Theorem [steps 3,6,3]
This proof is sound, but it's tricky to enter a two-column proof into the SBAC. Most likely, the SBAC expects students to enter a paragraph proof into the box. We might try the following:
Paragraph Proof:
Since ABCD is a rectangle, it's also a parallelogram. Since opposite sides of a parallelogram are congruent, we have AB = DC and AC = DE. Since opposite sides of a parallelogram are also parallel by definition,
Triangles ABC and DCE are congruent by SAS. QED
It's interesting to think of this problem in terms of transformations. Notice that if a diagonal divides a parallelogram into two triangles, then a 180-degree rotation about the midpoint of the diagonal maps one triangle to the other. Thus a 180-degree rotation about the midpoint of
Question 28 of the SBAC Practice Exam is on nonlinear functions:
Choose the domain for which each function is defined.
[The four functions are f (x) = (x + 4)/x, f (x) = x/(x + 4), f (x) = x(x + 4), 4/(x^2 + 8x + 16). The possible domains are all real numbers, x != 0, x != 4, x != -4. Here the "!=" means "does not equal" in ASCII and several computer languages.]
These functions are not linear. It will be hard-pressed to consider this an Algebra I question -- even though many texts (such as the Glencoe Algebra I text) contains a lesson on rational functions (Lesson 11-2), many teachers (including those in our district) skip Chapter 11 altogether. Thus even though students don't have to graph anything, this is mostly likely considered an Algebra II problem.
The first function has x in the denominator, so its domain is x != 0.
The second function has x + 4 in the denominator, so its domain is x != -4.
The third function has 1 in the denominator, so its domain is all real numbers.
The fourth function has (x + 4)^2 in the denominator, so its domain is x != -4.
Today is an activity day. The activity for Question 27 comes from Lesson 7-6 of the U of Chicago Geometry text, since this is the lesson on properties of a parallelogram and congruent triangles. The Exploration Question here is on congruent triangles in a regular pentagon. The activity for Question 28 comes from Lesson 13-4 of the U of Chicago Algebra I text, "Function Language," since this lesson discusses the domain and range of a function -- including rational functions like x/(x + 4). The Exploration Question here is about the range of the absolute value function. It's a great question for Common Core, since it involves translating the function up and down.
SBAC Practice Exam Question 27
Common Core Standard:
Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
SBAC Practice Exam Question 28
Common Core Standard:
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
Commentary: The properties of parallelograms and congruent triangles are easily found in Chapter 7 of the U of Chicago Geometry text. Even though rational functions are usually saved for Algebra II, the domain and range of functions -- including rational functions as one of the examples -- appear in many Algebra I texts, including the U of Chicago text.
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