Find the height of the water in this vat.
(The vat is shaped like an inverted cone. The slant height of the water is 13 from the top and 26 from the bottom. The radius of the whole cone is 16, and the height of the whole cone is 36.)
The basic assumption in this sort of problem is that the planes containing the top of the cone (the opening of the vat) and the water level are parallel. This is true provided that it's a right cone and its axis is vertical. Obviously, if we tilt the vat, its water level changes even without adding or removing any water.
This also means that if we take a vertical cross section, the vat and the water form two isosceles triangles that are similar. This is proved using the Corresponding Angles Consequence and AA~.
Our goal is to find the height of the smaller triangle. The height of the large triangle is 36, and the legs of the small and large triangles are 26 and 39 respectively. (Notice that 13 is not the length of a side of a triangle -- a common trap in triangle similarity problems.)
And so we set up a proportion:
x/36 = 26/39
39x = 936
x = 24
Therefore the height of the water in this vat is 24 -- and of course, today's date is the 24th. Notice that the radius of the vat was given but not needed. Even if the question had asked for the volume of the water in the vat, we could have calculated the radius using the Pythagorean Theorem.
Before we begin today's post, I wish to reflect on whether I even want to have a music break during my summer computer-based class. If there's no music break, then there's no music to play, and hence no need to create new scales in which to play music.
I've decided that yes, there will be room for music break. Yesterday, I learned that I'll be teaching two sections of Algebra I during blocks of just shy of two hours. This is a bit longer than the blocks we had at my old charter school last year. So there's definitely time for music break, but then again, I didn't have music break during IXL time, or Study Island time, or any other computer time.
But I think back to tricks I learned from other Algebra I teachers online, such as "Slope Dude" (used to help students distinguish between positive and negative slope). "Slope Dude" doesn't appear in any text, and I doubt that he'll show up in the Edgenuity videos. Yet I believe that it can be helpful to my summer students.
For example, a student might watch the slope video and ask for help, or insist that a line has negative slope when I know it's positive. It would be easy for me just to say "puff puff positive" and end it with that, but that only works if I'm actually teaching the lesson using "puff puff positive." Instead, a video that makes no mention of Slope Dude is the main source of instruction.
And so here's a solution -- have music break and sing a song about Slope Dude. Then if a student asks a slope question, I can just sing the part about "puff puff positive" (or whichever part is needed), and that should help the student quickly. So music break will still have a place in my summer class -- and so our study of EDL scales can proceed.
Let's look again at our most recent scale, 16EDL:
The 16EDL scale:
Degree Ratio Note
16 1/1 white E
15 16/15 green F
14 8/7 red F#
13 16/13 ocher G
12 4/3 white A
11 16/11 amber B
10 8/5 green C
9 16/9 white D
8 2/1 white E
Let's look for our main triad in this scale. For 12EDL, the main triad is 12:10:8 (or 6:5:4), a minor triad, and for 14EDL, it's 14:12:10 (or 7:6:5), a sort of subminor-diminished triad.
So continuing our pattern, the main triad for 16EDL should be 16:14:12 (or 8:7:6). But this doesn't really sound like a chord at all. The 8/7 ratio is a septimal whole tone, or supermajor 2nd. We don't expect to hear seconds in chords. Meanwhile, 8/6 = 4/3 is a perfect fourth. Again, no reasonable chord consists of the root, a second, and a fourth.
If we wish our triads to contain a third and a fifth, then 16:13:11 is the only possible triad. We recall from yesterday that 16/13 is a neutral third while 16/11 is a subfifth. Our main 16:13:11 consists of the notes white E, ocher G, amber B. We must keep our problems with ocher and amber in mind -- the interval G-B is spelled like a major third, but oG-aB sounds more like a minor third. But on the guitar, I just ignore all colors including amber and ocher, and so I just play an E minor triad.
Now we want a secondary triad, one that leads into the main triad. For 12EDL (and to some extent 14EDL), we simply took the next steps in the scale (descending leading tones). So just as 13:11:9 leads to 12:10:8 (that is, 6:5:4), we have 15:12:10 to lead to 16:13:11. Notice that 15:12:10 is a just major triad -- the simplest possible major triad playable in EDL. Our fundamental major triad has its root at green F. The trouble is that the F major triad sounds consonant on its own, rather than a chord that resolves to 16:13:11. In fact, I can't be sure what the otonal triad 16:13:11 even sounds like, since Mocha can only play notes in melody, not harmony.
Returning to my summer class, I still want the song I play on the first day to be the "Dren Song." (I repeat that even though I won't have Dren Quizzes this summer, I'll still use the concept of "dren" to drive home the fact that students need to learn basic math to be successful in life.)
But I composed the tune of "Dren Song" long before I'd ever heard of EDL scales (or likely even EDO scales other than 12EDO). There's no reason for me to make "Dren Song" fit any EDL scale, but then again since this is my song (rather than an already existing song), it's still interesting to see whether I can play it in Mocha, which requires that I convert it to EDL. Otherwise, I want to stay away from blinding converting existing 12EDO songs to EDL and introducing errors in translation.
In class last year, I played "Dren Song" in the key of E minor. This fits with the idea that minor scales match the mood of the students during math class. The middle of the song (sort of like a "bridge") is in G major (the relative major of E minor), and the lyrics during that part tells of the narrator dreaming that he really is good at math, until he wakes up from the dream and remembers that he's just a dren, in E minor.
Actually, I first wrote the song in C# minor, but converted it to E minor when I wanted to play it on the guitar (as E minor is easier to play than C# minor). Let me tell you what I was thinking of at the time I wrote the tune -- I was watching an episode of The Simpsons, the one in which Homer and three friends form a musical band called the "Be Sharps." The name is intended to be clever since B# is also the name of a musical note. But in real music, the note B# doesn't come up that often because it is enharmonic to C (in 12EDO). The only time musicians would ever use B# is when writing a song in C# harmonic minor, where the term "harmonic" indicates the use of B# as a leading tone to C#.
Notice that in scales other than 12EDO, B# can appear as a note distinct from C. This occurs in scales such as 19EDO and 31EDO. The interval 7/4, a harmonic seventh, is also called a barbershop 7th (which is relevant because the "Be Sharps" is supposed to be a barbershop quartet). In scales such as 19EDO and 31EDO, the note a harmonic/barbershop 7th above C is A#. (Technically this is true in 12EDO as well, except that A# is enharmonic to Bb, which is how it's usually written.)
Then the note a barbershop 7th above D is B#, which is not enharmonic to C. The songs that Homer's quartet sings in the episode are in other keys (C and Bb, if I recall correctly), but I think that in at least one of the songs, the quartet sings a D barbershop 7th chord, and so the note B# is sung. You might complain that I forced B# to appear by converting the song to 19EDO/31EDO, right after saying that I wouldn't convert songs to other scales. But I argue that 19EDO/31EDO are better suited for 7-limit barbershop harmony than 12EDO, and so it's actually forcing the songs to fit 12EDO that introduces errors in translation! (In Kite's color notation, the note a barbershop 7th above white D is still called "blue C," rather than any sort of B#.)
Anyway, the song I wrote in C# minor with the B# note has nothing to do with barbershop music -- since where would I find three other guys to sing it? And on the guitar, I converted it to E minor, which has nothing to do with the B# note at all.
The other idea I had when composing the song is that I wanted it to be symmetrical, in the same way that songs based on the Fibonacci sequence or digits of rational numbers (explained in previous posts, such as my April Fool's Day post) are symmetrical. And so I labeled the notes of the C# minor scale as follows: C#=1, D#=2, E=3, and so on. B#, the leading tone to C#, became 0 (since I wanted it to lead to C#=1).
B# C# D# E F# G# A
0 1 2 3 4 5 6
Then in the second half of the song, I subtracted the values in the first half from 6, so that the first five notes of my song (C#-D#-E-C#-B#) became (G#-F#-E-G#-A). The minor third C#-E in the first half of the song inverts to the major third G#-E, which explains how the first half of the song moves from minor to major while the second half moves from major to minor (that is, the middle of the song is major).
How could I code this song in Mocha? Of the EDL scales we've studied so far, 12EDL is the closest as it contains a minor triad on the root. But this reminds me of Bart Hopkin's scale from earlier this week -- a 12-based "mode" of 16EDL:
The 16EDL scale (mode starting on 12):
Degree Ratio Note
12 1/1 white A
11 12/11 amber B
10 6/5 green C
9 4/3 white D
8 (or 16) 3/2 white E
15 8/5 green F
14 12/7 red F#
13 24/13 ocher G
12 2/1 white A
This scale, just like true 12EDL, begins with a minor-like scale. But now it contains two sixths -- the supermajor sixth from 12EDL, plus the just minor sixth. The seventh is interesting here. Instead of the minor 7th (natural minor) or a major 7th (harmonic minor), we compromise and use a tridecimal neutral seventh here. The ocher 7th is a better use of 13 than the ocher 3rd of 16EDL, where our ear expects either a major or minor 3rd, not a neutral 3rd. In fact, the neutral intervals involving both 11 and 13 are used as leading tones (one descending, one ascending) to the tonic. This is probably why Hopkin created this scale -- it's like an improved minor scale.
I'd like to take advantage of the song's symmetry to make it easier to code in Mocha. We could have DATA lines for only one half of the song. Then we RESTORE the data and then have Mocha subtract the degrees from some constant in order to perform the inversions for the second half of the song.
Our steps are labeled 0-6, so step 1 could be Degree 12 while step 0 is Degree 13. Then steps 2-5 fit Degrees 11 down to 8 like clockwork. But step 6, the highest note, would have to be Degree 7 (or 14) in order for a simple inversion to work -- that is, Mocha would play a supermajor sixth even though I wrote the song using a minor sixth. I've decided that this change is worth it since I want the Mocha code to be as simple as possible. But on the guitar, I'll probably play it as a minor sixth as intended.
For this song, the fundamental note (8 or 16) is white E, which leads to 12 as white A, so the song plays as A minor. But on the guitar, I play the song in E minor, and Hopkin intended the fundamental note (8 or 16) to be B, which then leads to 12 as E just like on the guitar. Unfortunately, 8 or 16 as B and 12 as E isn't one of the playable Mocha scales:
Possible 16EDL root notes in Mocha:
Degree Note
16 white E
32 white E
48 white A
64 white E
80 green C
96 white A
112 red F#
128 white E
144 white D
160 green C
176 amber B
192 white A
208 ocher G
224 red F#
240 green F
256 white E
Since 16 is a power of two, this is easy -- the possible root notes of the 16EDL scale are the same as all the notes of the fundamental 16EDL scale.
We see that the only B that's available is amber B, but this is B quarter-flat and so would be out of tune with my guitar or Hopkin's tangular arc. The other close root note is green C. If it were possible to drop another octave, then red B would become available. It helps us that the lowest note we need is 16 rather than 13, but red B is the 21st subharmonic, and 13 * 21 = 273, which is out of range. Note that 13 * 20 = 260, and Degree 260 is the lowest playable note. Degree 260 is ocher Eb, which sounds between Eb and E. Then 8/16 becomes green C and 12 becomes green F, so the song sounds more like F minor than E minor. Between this and the use of the supermajor sixth, the Mocha version of the song won't sound exactly like the guitar version.
By the way, you wonder whether C# minor -- the original key of my song -- is available. If 12 were C#, then 8/16 is G#. We have ocher G available as a root, which is closer to G# than G, so we might try this scale (based on the 13th subharmonic). The lowest required note, 13, is Degree 169. This note isn't called B#, but deep ocher Bb. Despite the flat in its name, it's sharper than B, since "deep ocher" (double ocher) raises Bb by two third-tones. (If we had used Kite's alternative of defining the emerald 6th to be minor rather than major, then Degree 169 would indeed become "deep ocher B#".)
Question 17 of the SBAC Practice Exam is on inscribed angles:
Use the circle below to answer the question.
The circle is centered at point C. Line segmentPQ is parallel to SR. What is the measure of angle QPS?
(Other givens from the diagram: angle QRS = 68, quadrilateral PQRS is inscribed in the circle.)
A) 68
B) 112
C) 136
D) 158
For this Geometry question, This is a job for the Inscribed Angle Theorem. (Now you can see why I said that the Inscribed Angle Theorem is the only part of Chapter 15 that actually appears in any SBAC or PARCC questions.)
Angle QRS is 68, so arc QPS is 2 * 68. Since a full circle is 360 degrees, arc QRS is 360 - 2 * 68. So by using Inscribed Angle again, angle QPS is (360 - 2 * 68)/2 = 180 - 68. This is 112 degrees, and so the correct angle is B).
By the way, I intentionally avoided arithmetic until the last step, 180 - 68. If angle QRS had been x, then QPS would be 180 - x. The opposite angles of an inscribed quadrilateral are supplementary, and this is essentially the proof. The givens that two of its sides are parallel (i.e., that the quadrilateral is a trapezoid) and thatSR is a diameter are irrelevant.
Question 18 of the SBAC Practice Exam is on graphing equations:
Choose the ordered pair that is a solution to the equation represented by the graph.
(The graph shows the equation y = (2/3)x + 2, but only the graph is given, not this equation.)
A) (0, -3)
B) (2, 0)
C) (2, 2)
D) (-3, 0)
Since only the graph is shown, we must look at the graph to see which of these points appears to lie on the graph. The first three choices clearly miss the graph. Point D) is difficult to tell since the step size on the graph is 2 rather than 1, but since none of the others are correct, D) is the answer.
This reminds me of the awkward method that Lesson 3-2 of Glencoe and the first lesson in Edgenuity use to solve equations. We're given an equation like (2/3)x + 2 = 0, and we're asked to solve it by graphing the function f (x) = (2/3)x + 2 and seeing where it crosses the x-axis! I'd argue that it's easier just to solve the equation algebraically than to graph it. And suppose the answer wasn't -3, but something like -2.5 or -3.5. That would be hard to discern on the graph! At least here on SBAC, the equation is already graphed, as opposed to being given a (one-variable) equation and expected to solve it by graphing in the coordinate plane.
SBAC Practice Exam Question 18
16 white E
32 white E
48 white A
64 white E
80 green C
96 white A
112 red F#
128 white E
144 white D
160 green C
176 amber B
192 white A
208 ocher G
224 red F#
240 green F
256 white E
Since 16 is a power of two, this is easy -- the possible root notes of the 16EDL scale are the same as all the notes of the fundamental 16EDL scale.
We see that the only B that's available is amber B, but this is B quarter-flat and so would be out of tune with my guitar or Hopkin's tangular arc. The other close root note is green C. If it were possible to drop another octave, then red B would become available. It helps us that the lowest note we need is 16 rather than 13, but red B is the 21st subharmonic, and 13 * 21 = 273, which is out of range. Note that 13 * 20 = 260, and Degree 260 is the lowest playable note. Degree 260 is ocher Eb, which sounds between Eb and E. Then 8/16 becomes green C and 12 becomes green F, so the song sounds more like F minor than E minor. Between this and the use of the supermajor sixth, the Mocha version of the song won't sound exactly like the guitar version.
By the way, you wonder whether C# minor -- the original key of my song -- is available. If 12 were C#, then 8/16 is G#. We have ocher G available as a root, which is closer to G# than G, so we might try this scale (based on the 13th subharmonic). The lowest required note, 13, is Degree 169. This note isn't called B#, but deep ocher Bb. Despite the flat in its name, it's sharper than B, since "deep ocher" (double ocher) raises Bb by two third-tones. (If we had used Kite's alternative of defining the emerald 6th to be minor rather than major, then Degree 169 would indeed become "deep ocher B#".)
Question 17 of the SBAC Practice Exam is on inscribed angles:
Use the circle below to answer the question.
The circle is centered at point C. Line segment
(Other givens from the diagram: angle QRS = 68, quadrilateral PQRS is inscribed in the circle.)
A) 68
B) 112
C) 136
D) 158
For this Geometry question, This is a job for the Inscribed Angle Theorem. (Now you can see why I said that the Inscribed Angle Theorem is the only part of Chapter 15 that actually appears in any SBAC or PARCC questions.)
Angle QRS is 68, so arc QPS is 2 * 68. Since a full circle is 360 degrees, arc QRS is 360 - 2 * 68. So by using Inscribed Angle again, angle QPS is (360 - 2 * 68)/2 = 180 - 68. This is 112 degrees, and so the correct angle is B).
By the way, I intentionally avoided arithmetic until the last step, 180 - 68. If angle QRS had been x, then QPS would be 180 - x. The opposite angles of an inscribed quadrilateral are supplementary, and this is essentially the proof. The givens that two of its sides are parallel (i.e., that the quadrilateral is a trapezoid) and that
Question 18 of the SBAC Practice Exam is on graphing equations:
Choose the ordered pair that is a solution to the equation represented by the graph.
(The graph shows the equation y = (2/3)x + 2, but only the graph is given, not this equation.)
A) (0, -3)
B) (2, 0)
C) (2, 2)
D) (-3, 0)
Since only the graph is shown, we must look at the graph to see which of these points appears to lie on the graph. The first three choices clearly miss the graph. Point D) is difficult to tell since the step size on the graph is 2 rather than 1, but since none of the others are correct, D) is the answer.
This reminds me of the awkward method that Lesson 3-2 of Glencoe and the first lesson in Edgenuity use to solve equations. We're given an equation like (2/3)x + 2 = 0, and we're asked to solve it by graphing the function f (x) = (2/3)x + 2 and seeing where it crosses the x-axis! I'd argue that it's easier just to solve the equation algebraically than to graph it. And suppose the answer wasn't -3, but something like -2.5 or -3.5. That would be hard to discern on the graph! At least here on SBAC, the equation is already graphed, as opposed to being given a (one-variable) equation and expected to solve it by graphing in the coordinate plane.
SBAC Practice Exam Question 17
Common Core Standard:
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
SBAC Practice Exam Question 18
Common Core Standard:
Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).
Commentary: The Inscribed Angle Theorem is in Lesson 15-3 of the U of Chicago Geometry text, and it's the only Chapter 15 Theorem that appears on the SBAC. Meanwhile, Lesson 3-7 of the U of Chicago Algebra I text is arguably related to this standard. This standard might be confusing if students are required to learn about this before solving equations algebraically, which, unfortunately, is the way Edgenuity presents this material.
For future reference, here is the following Common Core Standard, which is also relevant here:
CCSS.MATH.CONTENT.HSA.REI.D.11
Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear....
Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear....
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