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https://electronics.stackexchange.com/questions/75252/is-my-instrumentation-amplifiers-voltage-offset-causing-problems
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math
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I am using an MCP6N11-100 to amplify the signal out of a 2 mV/V load cell with a excitation voltage of 10 V, so the maximum output (differential) would be 20 mV. The load cell is rated for 4000 kg.
I am using the instrumentation amplifier in the following configuration. Note that at the inputs I have the outputs of the load cell connected directly. Rf = 9.9 kΩ and Rg = 55 Ω giving a gain of ~181 (both resistor values are measured). VREF is grounded.
My problem is that the gain doesn't seem to be 181 until there is sufficient difference between the inputs of the instrumentation amplifier. Suppose the difference is just 0.5 mV. The op-amp outputs 200 mV giving a gain of almost 200! However, as the difference is increased to 2 mV the output now is 366 mV to 368 mV - matching the gain much more accurately (183 vs 200). Why is this discrepancy present at low levels of input? My guess is that it's the offset voltage. The datasheet suggests that the MCP6N11-100 has a maximum Vos of 0.35 mV. Could this be the cause?
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00488.warc.gz
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CC-MAIN-2023-50
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https://vdocuments.net/the-harmonic-melodic-language-in-alexander-scriabins-sonata-no-5-op-53.html
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math
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the harmonic & melodic language in alexander scriabin's sonata no. 5, op. 53
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DESCRIPTIONAn in depth analysis of Alexander's Scriabin's Fifth Sonata, regarded as a point of departure for Scriabin from tonal to atonal works. This analysis mostly focuses on the thematic and formal aspects of the sonata. References to other critical & scholarly works are also made with regard to academic analysis. This paper also contains an overview of Scriabin's "mystic chord" and its use in motives.
The Harmonic and Melodic Language in Alexander Scriabin's Sonata No. 5, Op. 53
Music Theory II MUSI 3500 Dr. Janners 4/28/2010
Through the progression of Alexander Scriabin's (b. 1872) ten piano sonatas (completed between 1892 and 1913), we hear this Russian Composer's movement from Romantic, Chopin-esque influences to a highly developed individualized compositional style. In making this progression, there is no doubt among scholars that Scriabin's own developing Theosophical beliefs were a major influence.1 Another influencing factor may have been related to Scriabin's own admission of experiencing synesthesia in how he supposedly "heard colors" and even related the sonorities of different musical keys to certain color hues.2 Regardless of the influences, it's apparent that during the time that Scriabin composed his sonatas, Sonata No. 5, Op. 53 (1907) represents a turning point in which he made a significant shift away from traditional functional harmony toward a harmonic language all of his own. This paper will look at the main themes in Sonata No. 5 and attempt to analyze the melodic and harmonic framework of Scriabin's emerging musical language. When discussing pitch collections and class sets, Allen Fortes set names will be used as much as possible to help relate sets to one another.3 Prior to doing any melodic or harmonic analysis, it will be important to understand the overall formal structure of this sonata. Table 1 in the Appendix outlines the overall structure and will be referenced when discussing specific thematic material. Written as one movement, the entire sonata is ordered according to the traditional formal sections of exposition, development and recapitulation with an introduction and coda, as shown in the Formal Section column of Table 1. After a brief excerpt of verses from his Poem of Ecstasy, Scriabins Sonata No. 5 opens with a quick introductory theme (Intro Th. A). We see that the opening has an expressive tempo marking (Allegro. Impetuoso. Con stravaganza.) and a key signature (F# major) that eventually raises some curiosity.4 The key signature is curious in that it soon becomes apparent that Scriabin is not operating within the traditional framework of F# major (or D# minor) diatonic harmony. A superficial, yet telling sign of this can be seen in that although the key signature signifies six sharps, pitches E# and A# are consistently marked with natural accidentals throughout the introduction. Thus, if a major or minor harmonic framework were employed, it would more likely be that of E major.1
Barany-Schlauch, Elizabeth Anna. "Alexander Scriabin's Ten Piano Sonatas: Their Philosophical Meaning and Its Musical Expression." (Ph.D., diss., Ohio State University, 1985), 13-14.2
Vanechkina, B. M. Galeyev, and I. L. Vanechkina. "Was Scriabin a Synesthete?" Leonardo. 34, no. 4 (2001), 357361.3
Forte, Allen. The Structure of Atonal Music. New Haven: Yale, 1973.
Scriabin, Alexander. Alexander Scriabin: Selected Piano Works. Edited by Gnter Philipp. Vol. 5. Leipzig: Edition Peters, 1971. 1
Pitches D# and A natural, a tritone interval, play a significant role in establishing the tonal framework with their frequent occurrence as the lowest pitches in the bass tremolo figure, yet no traditional cadences or dominant functions resolve to tonic chords based on these pitches. E major doesnt help much in explaining the harmonic structure other than by perhaps perceiving the D# as a sort of leading tone to E (tonic) and the prominent use of A as being the subdominant. Instead, the melodic figures and harmonic structures in the introduction, and throughout the entire sonata as well see, are better explained as working within certain pitch class sets. The entire Intro Th. A is based on the pitch class sets shown in the following table:
Desc. / Forte Number (measures) Together pcset 7-35 (mm. 1-12) Bass Tremelo Figure pcset 3-5 Glissando Pentachord 1 / pcset 5-24 Glissando Pentachord 2 / pcset 5-27
Intro Th. A Pitch Class Sets Normal Ordered Pitch Collection Normal (Reduced) Form / Prime Form (D#,E,F#,G#,A,B,C#) (0,1,3,5,6,8,10) / (0,1,3,5,6,8,10) (D#,E,A) (0,1,6) / (0,1,6) (D#,E,A,B,C#) (G#,A,C#,E,F#) (0,1,6,8,10) / (0,1,3,5,7) (0,1,5,8,10) / (0,1,3,5,8)
In looking at the pitch collections (pcset 7-35 contains the entire pitch collection for Intro Th. A), its confirmed that all the pitches together form a heptatonic scale that correlates exactly to E major. With D# having a strong presence in the lower voice, the usage might be thought to be tied to the Locrian mode, yet with a lack of a real resolution to D#, this correlation is not convincing. Instead it might be better to think of the D# as continually leading to the trilled E as to reinforce E as the tonal center for this part of the intro. The Tremelo Bass figure is based on a tritone (diminished fifth) relationship between D# and A with an E (perfect fifth above A) which creates a quintal figure (Q) and is used when the glissandos are sounded in the upper voice (right hand part). Interestingly, this pitch class set (0,1,6) or pcset 3-5 was favored by other modern, 20th century composers such as Arnold Schoenberg and Anton Webern to create dissonance and was even referred to as the Viennese Trichord, referring to the Second Viennese School.5 This dissonance helps reinforce the frenetic nature of the introductory theme. Both glissandos in the upper voice are based on pentatonic sets or pentachords. The first glissando contains pitches E,A,B,C#,D# (pcset 5-24) and the second glissando figure contains A,C#,E,F#,G# (pcset 5-27),5
Martin, Henry. "Seven Steps to Heaven: A Species Approach to Twentieth-Century Analysis and Composition." Perspectives of New Music 38, no. 1 (Winter, 2000), 149,154. 2
which are important to this sonata as we shall see later. When we put these pitches together, its clear that the motivic figures (Glissando Pentachords 1 & 2) together themselves are derived exactly from the E major scale. When considering the downbeats of these two angular motives, there does appear to be an implicit circle of fourth/fifth relationship between them with the D# from the lower voice (left hand part) leading to the E in Glissando Pentachord 1. It should also be noted that the (0,1,6) set (pcset 3-5) is part of Pentachord 1. The circle of fourth/fifth relationship between the downbeats of these figures helps build tonal connectivity amidst the dissonance created by the (0,1,6) pitch class set in the lower voice. This point is reinforced by the fact that no D# exists during the instances of Pentachord 2, as can be seen at the second beat of measure 7 and similarly in later measures. While Baker similarly accounts for a V-I relationship in E major (or V6/5/V-V-I in A major)6, Wise correctly emphasizes that the key center is not clear.7 All of this gives a good introduction not only to the sonata, but also to Scriabins own emerging harmonic language by intermingling traditional relationships (i.e., loose circle progressions & leading tones) with new ones (i.e., the steady (0,1,6) pitch class set and unusual key signature usage). The introduction ends suddenly in a measure of silence at m. 12 as if the two figures were chasing each other around and suddenly ran out of breath. Marked Languido, the Intro Theme Group B begins at m. 13 and ends at m. 46. It evokes an open, lethargic feeling of desperation for some seemingly, unreachable goal. When considering the harmonic nature of this theme group, we can see how this material is connected to Group A, and how the open sound is maintained. Measures 13-18 are linked to m. 11 of Group A with the consistent presence of E natural in the lower voice and the presence of D# and G# in the upper voice, and work within the following pitch class sets: Set / Subset Name Together pcset 5-24 (mm. 13-17) Together pcset 6-Z26 (mm. 13-18) Melodic Motive pcset 3-7 (mm. 13-14) Intro Th. B Pitch Class Sets Normal Ordered Pitch Normal (Reduced) Form / Collection Prime Form (if applicable) (D#,E,F#,G#,A#) (0,1,3,5,7) (D#,E,F#,G#,A#,B) Pcset 5-24 now includes B (D#,F#,G#) (0,1,3,5,7,8) (0,3,5) / (0,2,5)
Baker, James M. The Music of Alexander Scriabin. (New Haven and London: Yale University Press, 1986), 175.
Wise, Herbert Harold, Jr. "The Relationship of Pitch Sets to Formal Structure in the Last Six Piano Sonatas of Scriabin." (Ph.D. diss., University of Rochester, 1987), 127. 3
It should be noted that though a B natural is not sounded until m. 18, pcset 5-24 contains pitches derived from the B Major scale, lacking the second scale degree (C#), which aligns it with the prime form of the pcset for Glissando Pentachord 1 (0,1,3,5,7) or pcset 5-24. This shows an intervallic connection through inversion to Intro Th. A, and proves to be an important set throughout Intro Th. B. The pitches from the main two-bar phrase used to open the theme might be considered as being derived from a dominant F#13th chord (mm. 13-14) with the D# doubled an octave apart in the upper voice and an A# as the lowest pitch (m. 14). With this, as with the introductory material for Group A, we can see that the harmonic arrangement of Intro Th. B is by no means traditional, especially given the strong presence of a tritone (E-A#) in the lower voice and an extended dominant chord as tonic. The B natural sounded in the upper voice of m. 18 hints at resolving the F#9 chord of m. 17, but suddenly slips into a chromatic ascension (B#-D) in m. 19, which then leaps a tritone interval to a G# sounded in the upper voice an octave higher than the G# already sounded below. To stress this chromatic tension, a descending chromatic line is sounded simultaneously in the lower voice beginning in m. 18 from D#-B# (m.20). Together, these chromatic lines create a sense of unraveling, as if the two chromatic lines were literally pulling away at each other, rather than leading to any sense of harmonic resolution. Aside from the short chromatic line in the lower voice (which includes a C# that aligns with the upper voice an octave higher and is repeated in mm. 23-24), mm. 19-25 reveals an underlying whole tone scale with the following pitch class set information: Forte Number (measures) Together pcset 6-35 (mm. 19-25) Normal Ordered Pitch Collection (A#,B#,D,E,F#,G#) Normal (Reduced) Form / Prime Form (if applicable) (0,2,4,6,8,10)
We can discern an alternation between two pitch class sets (pcset 5-24 & pcset 6-35) with interwoven chromatic lines until m. 40, where Scriabin introduces material harmonically derived from a series of three dominant 13th chords (Hung perceives: Ab13, G13, & C#13)8, though the chords are non-functional and thus do not resolve to a tonic chord as one might expect in m. 45. Measure 45 instead offers a passage of three triadic chords in the upper voice (a second inversion B major chord, a second inversion D# major chord and a D# major chord in root position). This is composed over a bass figure written in hemiola rhythm (with four quarter-notes against six eighth-notes) as a preview of whats to come in the formal exposition. Harmonically, the bass figure provides no traditional support for the upper voice, except that pitches B and D# are also sounded a minor sixth
Hung, Chia-Sui. "Tradition and Innovation in Four Piano Sonatas of the First Quarter of the Twentieth Century." (Ph.D diss., University of Cincinnati, 2003), 16 4
interval apart. All the pitches in m. 45 (A#, B, C#, D#, F#, G#) fit into a C#13th chord9, but since this chord is followed by a measure of silence with no key center, its difficult to recognize a traditional half-cadence and thus somewhat irrelevant to use the traditional chord references in our analysis. It doesnt help that to truly be a dominant C# chord there would technically need to be an E# (major third) present as part of our collection to complete the basic triadic dimension of the dominant chord. Interestingly, Scriabin has continually marked all E#s with natural accidentals up this point. Perhaps its more meaningful to continue looking directly at tonal relationships within the pitch collections and see what is emphasized as consonant or dissonant. When we look at the harmonic material of mm. 40-45 in terms of pcsets, we see the following.
Forte Number (measures) Together pcset 5-35 (mm. 40-41) Together pcset 6-32 (mm. 42-43) Together pcset 6-32 T(6) (mm. 44-45)
Normal Ordered Pitch Collection (Db,Eb,F,Ab,Bb) (F,G,A,C,D,[E]) T(8) lower than 5-35 (B,C#,D#,F#,G#,[A#])
Normal (Reduced) Form / Prime Form (0,2,4,7,9) or (1,3,5,8,10) / (0,2,4,7,9) (0,2,4,7,9,) or (,1,3,5,8,10) / (0,2,4,5,7,9) (0,2,4,7,9,) or (,1,3,5,8,10) / (0,2,4,5,7,9)
By these collections, we can see that the pcset 5-35 present in mm. 40-41 basically expands out to a new pcset 6-32 in mm. 42-43, only transposed a minor sixth (or T(8)) lower. Measures 44-45 contain the same pcset 6-32 transposed a tritone (or T6) higher than the previous occurrence. This analysis approach will become more important as we analyze the formal exposition. Though it may be noteworthy to see how traditional chord formations are still present to some degree in this harmonically transitional sonata, pitch class sets are perhaps more helpful in explaining the procedures Scriabin used to compose this and his later sonatas.
Exposition The formal exposition begins with a spirited Main Theme 1 which can be broken into two parts: Subject Group A (Th. 1A) and Subject Group B (Th. 1B). Theme Group A is defined in mm. 47-95. As Hung points out, Theme Group A can also be broken into three sub-phrases (mm. 47-52, 53-58 and 59-67) with a four beats against six beats hemiola rhythm occurring in the first and second sub-phrases.10 The sub-phrases are repeated again in mm. 68-95 and in this instance, the last sub-phrase (mm. 80-95) serves as a transition to Th. 1B. Th. 1B can be found in mm. 96-119 with transitional material occurring from mm. 114-119.9
Ibid., 16Ibid., 17
If we were to look at the upper voice of the beginning of Main Theme 1 from a perspective of traditional harmony, we would find different inversions of simple triads moving between B Major, D# Major and G# Minor in mm. 47-52 and E Major, G# Minor and C# Minor in mm. 53-58. The relationship between these triads can be heard as traditional harmonic forms given the fact that they come directly from the B Major/G# Natural Minor diatonic scale. However, when we consider the lower bass figure against these triads, the harmonic dimension is given a greater depth of tonal color, especially since Scriabin seems to gravitate toward a quintal arrangement of the lower pitches, which reinforce an open, more ambiguous sound. The consistent C# sounded as the lowest note gives the tones in mm. 47-52 an extended dominant quality (specifically, a dominant C#13th (omit 3rd, due to the missing E#) and creates a new tonal center of C#. With the C# and E (which replaces the previously stated D#) in mm. 53-57, the quintal bass arrangement is augmented but interestingly this now reinforces an extended C# minor 9th chord.
The harmonic outline in mm. 47-67 is derived from a non-functional progression using subdominants of F# (mm. 47-52), B (mm. 53-58), E (mm. 59-60), Eb (mm. 61-62), D (mm. 63-64) and N6/V of F# (mm. 65-67).11 Though the harmonic relationships can be explained in this fashion and it is interesting to note how the progression follows subdominants of keys according to a circle of fourth progression until the subdominant of Eb, when the progression becomes chromatic, perhaps a more useful explanation can be found when observing how the tones for each of the sub-phrases fit into pitch class sets, as weve done before. We can then more accurately consider how these sets relate to each other. The following table contains the pitch information for mm. 47-67. Main Th. 1A (sub-phrases mm. 47-67) Pitch Class Sets Desc. / Forte Number (measures) Normal Ordered Pitch Normal (Reduced) Form / Collection Prime Form (if applicable) Upper voice pcset 5-27 (A#,B,D#,F#,G#) (0,1,5,8,10) / (0,1,3,5,8) (mm. 47-52) Lower voice pcset 4-22 (B,C#,D#,G#) (0,2,4,9) / (0,2,4,7) /11
(mm. 47-52) Together pcset 6-32 (mm. 47-52) Upper voice pcset 5-27 T(5) (mm. 53-58) Lower voice pcset 4-26 (mm. 53-58) Upper voice pcset 5-27 T(5) (mm. 59-60) Lower voice pcset 3-11 (mm. 59-60) Upper voice pcset 6-32 T(8) (mm. 61-62) Lower voice pcset 420 (mm. 61-62) Upper voice pcset 5-27 T(10) (mm. 63-64) Lower voice pcset 3-11 (mm. 63-64) Upper voice pcset 7-32 (mm. 65-67) Voices grouped together every 3 beats pcsets 4-20 T(4) & 4-19 (mm. 65-67) (A#,B,C#,D#,F#,G#) (D#,E,G#,B,C#) (B,C#,E,G#) (G#,A,C#,E,F#) (A,C#,E) (D,Eb,F,G,Bb,C) (D,Eb,G,Bb) (F#,G,B,D,E) (G,B,D) (E#,F#,G#,A,B,C#,D) (F#,G,B,D) & (G#,A,C#,E#)
NOTE: C# added, other pitches emphasized (0,1,3,5,8,10) / (0,2,4,5,7,9) (0,1,5,8,10) / (0,1,3,5,8) Transposed P4 (0,2,5,9) / (0,3,5,8) Subset of upper voice (0,1,5,8,10) / (0,1,3,5,8) (0,4,7) / (0,3,7) Subset of upper voice (A major triad) (0,1,3,5,8,10) / (0,2,4,5,7,9) (0,1,5,8) / (0,1,5,8) Subset of upper voice (0,1,5,8,10) / (0,1,3,5,8) Transposed m7/M2 (0,4,7) / (0,3,7) Subset of 4-20 in m. 65 (0,1,3,4,6,8,9) / (0,1,3,4,6,8,9) *(F# harmonic minor) (0,1,5,8) & (0,1,5,9) / (0,1,5,8) & (0,1,4,8)
By looking at this table, we can see that the pcset 5-27 (0,1,5,8,10) is vital in connecting the harmonic material for all of the phrases in mm. 47-64. When comparing the sub-phrases, it becomes clear that Scriabin chose pitches of set intervals and transposed them up a perfect 4th (T(5)) in mm. 53-58 and then from there another T5 in mm. 59-60. The first five pitches in the pitch class set for mm. 61-62 appear to be derived directly from the prime set of the previous sub-phrase, which is an inversion of the normal form of pcset 5-27. Measures 63-64 again contain the exact same intervals as the previous sub-phrases, but transposed a Major 2nd lower than mm. 59-60. Measures 65-67 link to the second group of sub-phrases using pitches derived from the F# Harmonic Minor scale. The pitches in the second group of sub-phrases have been ordered into sets in the table below.
Main Th. 1, Group B (sub-phrases mm. 68-79) & Transition (mm. 80-95) Pitch Information Desc. / Forte Number Normal Ordered Pitch Collection Normal (Reduced) Form / (measures) Prime Form (if applicable) Upper voice pcset 5-27 (A#,B,D#,F#,G#) (0,1,5,8,10) / (0,1,3,5,8) (mm. 68-73) Lower voice pcset 4-23 (C#,D#,F#,G#) (0,2,5,7) / (0,2,5,7) 7
(mm. 68-73) Together pcset 6-32 (mm. 68-73) Upper voice pcset 5-27 T(5) (mm. 74-79) Lower voice pcset 4-23 T(10) (mm. 74-79) Together pcset 6-32 T(5) (mm. 74-79) Upper voice pcset 5-27 T(5) (mm. 80-83) Together pcset 6-32 (mm. 80-83) Upper voice pcset 5-27 T(2) (mm. 84-87) Together pcset 7-35 (mm. 84-87) Upper voice pcset 5-27 T(4) (mm. 88-91) (-4/m6) Together pcset 6-32 (mm. 88-91) Upper voice pcset 9-7 (mm. 92-93) Lower voice pcset 8-10 (mm. 92-93) Together pcset 6-32 T(0) (m. 92) Together pcset 5-34 (m. 93) Together pcset 420 (mm. 94-95)
(A#,B,C#,D#,F#,G#) (D#,E,G#,B,C#) (B,C#,E,F#) (D#,E,F#,G#,B,C#) (G#,A,C#,[D#],E,F#) =NT (G#,A,B,C#,[D#],E,F#) =NT (A#,B,D#,[E#],F#,G#) =NT (A#,B,C#,D#,E#,F#,G#) (F#,G,B,D,E) (F#,G,A,B,D,E) (D,Eb,E,F,F#,G,A,B,C) Increasingly chromatic (D,Eb,E,F,F#,G,A,C) Increasingly chromatic (F#,G,A, B, D, E) Eb,F,G,A,C (C#,D,F#,A) D Maj 7
(0,1,3,5,8,10) / (0,2,4,5,7,9) (0,1,5,8,10) / (0,1,3,5,8) (0,2,5,7) / (0,2,5,7) (0,1,3,5,8,10) / (0,2,4,5,7,9) (0,1,5,,8,10) / (0,1,3,5,8) (0,1,3,5,,8,10) / (0,2,4,5,7,9) (0,1,5,,8,10) / (0,1,3,5,8) (0,1,3,5,7,8,10) / (0,1,3,5,6,8,10) F# Major (0,1,5,8,10) (0,1,3,5,8,10) / (0,2,4,5,7,9) (0,1,2,3,4,5,7,9,10) / (0,1,2,3,4,5,7,8,10) (0,1,2,3,4,5,7,10) / (0,2,3,4,5,6,7,9) (0,1,3,5,8,10) / (0,2,4,5,7,9) (0,2,4,6,9) (0,1,5,8)
Upon closer analysis of the sets, we can see that the harmonic material making up the second group of sub-phrases is derived from the same sets as the previous group, only now with increasing number of passing & neighbor tones as depicted with brackets () in the sets. In general, these tones show a tendency toward chromaticism which is more fully realized if we look at all the pitches together in mm 92-93 before one more obvious use of a D Maj 7 subset (pcset 4-20) of pitches from pcset 5-27 in mm. 94-95 before moving to Group B of Main Theme 1. The brief Group B of Main Theme 1 (Th. 1B) begins at m. 96 and ends at m. 113. Measures 114-119 serve as a transition to Main Theme 2 (Th. 2), to be discussed next. In m. 96, the upper voice states a memorable motive which descends by minor sixth intervals (8 half-steps), while the lower voice contains motion in minor seventh, perfect eighth (octave) and tritone intervals. The minor seventh interval proves to be important 8
throughout Th. 1B, both in this main motive and in the surrounding material, as we shall see. The first statement of this motive can be seen in mm. 96-97 and is shown below.
Written in 6/8 time, a sparsely written three note figure descends by minor sixth intervals beginning on the second beat. On the fourth beat, dissonance is created with the clashing notes of a double-sharp-G sounding against an octave of G# notes in the lowest voice for part of its duration. The tension is released through a halfstep motion in the upper figure after the G-double-sharp is sounded for six beats, half of which are in the next measure (m. 97) with the entire duration divided by tritone motion in the lower voice (i.e., the upbeat of m.96 to downbeat of m. 97). Measures 100-101 and mm. 104-105 are all related and can be explained with the pitch class set information in the following table.
PC Sets for main motive of Th. 1B Desc / Forte Number (measures) Normal Ordered Pitch Collection Upper voice pcset 4-19 T(1) (GX(A),A#,C#,E#) (mm. 96-97) relates to mm. 65-67 Lower voice pcset 3-8 (F#,G#,B#,[C#]) (mm. 96-97) =replaced by B# Together pcset 7-Z37 (G#,GX(A),A#,B#,[C#],E#,F#) (mm. 96-97) Upper voice pcset 4-19 T(5) (CX(D),D#,F#,A#) (mm. 100-101) Lower voice pcset 3-8 T(0) (F#,G#,B#) (mm. 100-101) Together pcset 6-34 (CX(D),D#,F#,G#,A#,B#) (mm. 100-101) Mystic Chord Set Upper voice pcset 4-19 T(1) (D#,E,G,B) (mm. 104-105) Lower voice pcset 3-8 T(1) (G,A,C#) 9
Normal (Reduced) Form / Prime Form (if applicable) (0,1,4,8) / (0,1,4,8) (0,2,6,) / (0,2,6) (0,1,2,4,5,9,10) / (0,1,3,4,5,7,8) (0,1,4,8) / (0,1,4,8) (0,2,6) / (0,2,6) (0,1,4,6,8,10) / (0,1,3,5,7,9) (0,1,4,8) / (0,1,4,8) (0,2,6) / (0,2,6)
(mm. 104-105) Together pcset 6-34 T(1) (mm. 104-105)
(D#,E,G,A,B,C#) Mystic Chord Set
(0,1,4,6,8,10) / (0,1,3,5,7,9)
From this table, we can see the prevalence of the minor sixth interval (0-8) in the upper voice and tritone in the lower voice (0-6). Another point of interest is the appearance of the pcset 6-34, which includes the pitch material from Scriabins infamous Mystic Chord12. When looking at the sets that make up the measures surrounding the thematic material just described, we can see how they also support the material harmonically.
PC Sets of phrases surrounding the main motive of Th. 1B (described above) Desc / Forte Number (measures) Normal Ordered Pitch Normal (Reduced) Form / Collection Prime Form (if applicable) Together pcset 4-25 (C,D,[E#],F#,Ab) (0,2,,6,8) (mm. 98-99) Fr6 Set = Tension Creating NT Together pcset 5-34 (D,E,[E#],F#,G#,B) (0,2,,4,6,9) (mm. 102) Dominant-ninth = NT Together pcset 4-24 (G#,Bb,C,E) (0,2,4,8) (m. 103) Augmented 7th Together pcset 4-24 T(3) (Db,Eb,[E],[F],G,B) (0,2,,,6,10) / (0,2,4,8) (m. 106 & 108) Fr6 = NT & whole tones shift Together pcset 4-25 T(7) (G,A,[B#],C#,Eb) (0,2,,6,8) (m. 107 & 109) Fr6 = NT Together pcset 4-24 T(2) (Eb,F,[F#],[G],A,C#) (0,2,,,6,10) / (0,2,4,8) th (m. 110) Augmented 7 = NT Together pcset 4-25 (A,B,[D],D#,F) (0,2,,6,8) (m. 111) Fr6 = NT Together pcset 4-20 (D,Eb,G,[A],Bb,) (0,1,5,,8) / (0,1,5,8) (m. 112) Major-seventh = NT Together pcset 4-20 T(0) (D, Eb,[F#],G, Bb ) (0,1,,5,8) / (0,1,5,8) (m. 113-114) Major-seventh = NT Together pcset 5-34 T(8) (Bb,C,D,E,[F],[F#],G) (0,2,4,6,,,9) (mm. 116 -119) Though the material surrounding the main motive of Th. 1B doesnt have strong motivic or melodic figures, we can see by set analysis that the tritone, minor sixth & minor seventh intervals continue to play a strong role in the prevalence of pcsets 4-20, 4-24 & 4-25 (note also that 4-25 contains the pitches of a French Sixth Chord). When considering the high frequency of even numbers in the sets (odd numbers are typically neighbor tones in sets with mostly even numbers), its clear that Scriabin relied upon Whole Tone scales for
Wise, "The Relationship of Pitch Sets to Formal Structure in the Last Six Piano Sonatas of Scriabin, 146-147
tonal structure and added neighbor tones to increase tension. The whole tone scale appears to shift by its limited transposition a semitone between mm. 103 & 106, coincidentally where a key signature change (to Bb Major) occurs for the first time in the sonata. As with the main motive, the adjacent measures, outlined above, also make up one sub-phrase each (i.e., 98-99 is one sub-phrase just as mm. 96-97 is one sub-phrase). For the sake of realizing the sets more clearly, they have been separated to draw tonal cohesion, but in reality the paired measures are actually tied together with the second bars including a direct shift in parallel by different intervals. When looking at an example sub-phrase more closely, we can see how the tritone and minor sixth intervals work in moving the piece forward. These same motions exist in mm. 103, 109, & 111 as shown in the example below from m. 106.
The minor seventh interval (0,10) proves to be important in uniting the surrounding phrases just described with the underlying tonality of the main theme motive (i.e., mm. 96-97). When we trace these intervals and look vertically for major third intervals amidst the minor seventh interval, we can see that Scriabin was most likely using non-functional dominant chords to unite this material. From mm. 97-111, we recognize a series of non-functional dominant chords that move by intervals of Minor sixth, tritone and Perfect fourth:
Progress by Tritone: Progress by Minor Sixth: Progress by Tritone: Progress by Minor Sixth: Progress by Perfect Fourth:
G#9 (m.97) -> D9 (mm. 98-99) -> G#9 (m. 100-101) (G#9 (m. 100-101)) -> E9 (m. 102) -> C9 (m. 103) -> A9 (m. 104) A9 (m. 104) -> Eb9 -> A7#9 (Fr6) (Eb9-A7 repeats in mm. 106-109) (A9 (m. 109)) -> F9 (m. 110) (F9 (m. 110)), B9 (Fr6) (m. 111)
Measure 112 changes this basic progression and instead concludes the previous theme group with material making up a G minor sixth chord, again emphasizing the minor sixth interval. It would seem that Scriabin was using some devices of traditional harmony in this section without giving into any traditional harmonic function or resolve. A six bar transition (or link) follows (mm. 114-119) beginning with the same 11
harmonic material from mm. 112-113 using the same bass motive as the opening Th. 1B phrase. The prominent chord beginning m. 116 contains pitches (Bb, C, D, E, G ( pcset 5-34)) that make up a C9 chord. These pitches are used with the addition of F & F# to take us to Main Theme 2 (Th. 2) in m. 120 where an interval of minor seventh appears to have significance again in the bass voice using F as the lowest pitch, thus having a common element with the previous thematic material. Main Theme 2 (Th. 2) is mostly composed of a four bar phrase, first occurring in mm. 120-123 which interweaves whole tone and chromatic scales. Scriabin puts forth an open sounding whole tone chord made up of pitch collection (Eb,F,G,A) or (0,2,4,6) and then dabbles with chromaticism using neighbor tone F# until m. 121 when the whole tone structure comes apart with an 11-tone chromatic descension covering tones G-A. The missing Ab is sounded as the fifth beat in alto voice of m. 121 as a neighbor tone to the previously sounded Anaturals. The chromatic descension leads to an intriguing vertical structure in m. 122 arranged in a quartal/quintal fashion. When we collect the pitches for this structure we find a whole tone collection (Eb,F,G,A,Cb) or pcset (0,2,4,6,8) written in a quartal/quintal fashion. The Db that arrives at the second beat of m. 122 further confirms the whole tone scale by completing the set (0,2,4,6,8,10). These whole tone sonorities are used to complete the phrase in m.122-123 with D natural as a sort of neighbor tone.
This four-bar phrase is basically repeated verbatim in mm. 124-127 with minor changes. In short, the first Eb in the upper voice is replaced with a short chromatic run (C#-Eb). Note that the C# is a tritone below the G sounded above. Measure 126 is changed slightly when compared to m. 122 in that the highest G replaced what used to be a F-natural. These vertical structures basically make up the entirety of the harmonic material for the second half of the Th. 2 four bar phrase. Its therefore useful to look at the pitch collections directly and compare the sets, as shown below.
Desc. / Forte Number (measures) Upper melody pcset 4-16 (mm. 120-121) Together pcset 6-35 (m. 122-123) Upper melody pcset 4-16 T(0) (mm 124-125) Together pcset 6-35 T(0) (m. 126-127) Upper melody pcset 4-16 T(5) (mm 128-129) Together pcset 6-34 (mm 130-131) Together pcset 5-33 (m.132) Together pcset 5-16 (m. 133) Upper melody pcset 4-16 (m. 134-135) Together pcset 6-34 T(7) (mm. 136-137) Together (mm. 138-139)
Normal Ordered Pitch Collection (F#,G ,C,D) (Eb,F,G,A,Cb,Db,[D]) (F#,G ,C,D) (Eb,F,G,A,Cb,Db,[D]) (B,C,F,G)
Normal (Reduced) Form / Prime Form (0,1,6,8) / (0,1,5,7) (0,2,4,6,8,10) (0,1,6,8) / (0,1,5,7) (0,2,4,6,8,10) (0,1,6,8) / (0,1,5,7)
(D,Eb,F#,Ab,Bb,C) (0,1,4,6,8,10) / (0,1,3,5,7,9) Mystic Chord Set (G,A, B,C#,D#) (0,2,4,6,8) (F,F#,A,D,Eb) (0,1,4,9,10) / (0,1,3,4,7) (F#,G,C,D) (0,1,6,8) (A,Bb,Db,Eb,F,G) (0,1,4,6,8,10) / (0,1,3,5,7,9) Mystic Chord Set Begins like m. 134, yet melody line is same as mm. 128-129 and chromatic descent is D-Eb). Note that descent now contains all tones and raised T(5) from m. 135.
It doesnt take long to see the intersection of whole tone and chromatic structures that are intertwined in Th. 2. Its interesting to see how amidst all of this, the recurring chromatic lines descend to a minor seventh (ten steps), until the last statement in m. 139, when all tones (D-Eb) are sounded. An important recurring set in Th. 2 is that of pcset 6-34. The pitches in this set are what make up Scriabins Mystic Chord. There has been some debate regarding the origin of this set in how Scriabin conceived of it, but most accounts suggest that there was some Theosophical motivation behind its contents. From an analytical perspective, some scholars suggest that the tones were derived from the upper partials of the overtone series, while others believe that Scriabin conceived of the set through exploring sets of six notes separated at different quartal intervals (i.e., perfect & diminished).13 As Wagners Tristan Chord gained notoriety through its use in Tristan und Isolda, the Mystic Chord was first recognized as a driving harmonic device in Scriabins Prometheus Op. 60 and is also referred to as the Prometheus Chord.14 The codetta is sparked into motion through a brief link in 2/4 time that occurs in mm. 140-142. This link puts forward a jerky rhythm as if to imply a series of nervous attempts at a new start only giving into silence at13
Baker, James M. The Music of Alexander Scriabin. (New Haven and London: Yale University Press, 1986): 99 Ibid., 99 13
m. 142 (and is again emphasized in the coda section of the sonata). The silence of m. 142 is soon interrupted with the codetta again in 6/8 time which closes out the formal exposition section of the sonata. When looking at the upper voice of the codetta, its clear that its connected both to the link material in mm. 140-142 and the Th. 1A simply by taking note of the familiar upper triadic figures. When we look at the pitch class sets that make up the harmonic framework of the codetta, this is confirmed in that they are derived from the same pcset 5-27 and transposed to various degrees, ending with a series of tritone transpositions, as shown below.
Desc / Forte Number (measures) Upper pcset 5-27 (m. 143) Upper pcset 5-27 T(7) (m. 144) Upper pcset 5-27 T(3) (m. 145) Upper pcset 5-27 T(6) (m. 146) Upper pcset 5-27 T(4) (m. 147) Upper pcset 5-27 T(7) (m. 148) Upper pcset 5-27 T(3) (m. 149-150) Upper pcset 5-27 T(6) (m. 151-152) Upper pcset 5-27 T(6) (m. 153-154) Upperpcset 5-27 T(6) (m. 155-156)
Normal Ordered Pitch Collection (F,Gb,Bb,Db,Eb) (C,Db,F,Ab,Bb) (Eb,Fb,Ab,Cb,Db) (A,Bb,D,F,G) (F,Gb,Bb,Db,Eb) (C,Db,F,Ab,Bb) (Eb,Fb,Ab,Cb,Db) (A,Bb,D,F,G) (Eb,Fb,Ab,Cb,Db) (A,Bb,D,F,G)
Normal (Reduced) Form / Prime Form (if applicable) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10) (0,1,5,8,10)
As Table 1 in the Appendix shows, the exposition proper ends after the codetta (mm. 140-156) and the development section, having two parts, is contained in mm. 157-328. There are obvious restatements of the introductory material therein. Most of the material that comprises the development section is derived from material in the exposition, only transposed to various degrees and sometimes given variation treatments. A unique quartal voicing of the Mystic Chord appears in m. 264. There are contrasting juxtapositions of Allegro fantastico (from the codetta) and Meno vivo (Th. 2) in mm. 281-288) and significant variation added to the Th. 2 material in mm. 313-328. The recapitulation occurs in mm. 329-400 with pcset 5-27. The entire recapitulation is almost an exact copy of the exposition except that the sub-phrases derived from Th. 1A only occur once (mm. 329-348) with slight variations and transposed a perfect fifth interval lower (or T(5) (perfect fourth) higher). This transposition is easily recognized when comparing the pitch material of pcset 5-27 from the exposition (mm. 47-52, (A#,B,D#,F#,G#)) to (D#,E,G#,B,C#) in mm. 329-334 of the recapitulation. Similar comparisons can be made throughout the recapitulation. The coda contains material from the codetta and material loosely derived from Th. 1A. The sonata ends with the motive that originally opened the piece in Intro Th. A.
As we have seen, by analyzing the thematic material and the overall harmonic framework of Scriabins Sonata No. 5 using pitch class sets, while still considering traditional chord configurations, we can recognize how Scriabin embarked upon a new compositional style which bridged some aspects of traditional harmony (i.e., dominant & major chords) with his future sonatas which embraced greater atonality. Traditional chord names assist with recognizing the old formations, while pitch class sets help us recognize how they are related throughout Scriabins individualized approach to composition.
APPENDIXTable 1- Overall Structure of Sonata No. 5, Op 53 by A. Scriabin Formal Section Introduction (1-46) Introduction cont. Exposition (47-156) Exposition cont. Exposition cont. Exposition cont. Exposition cont. Exposition cont. Development (1) (157-246) Development (1) cont. Development (1) cont. Development 2 (247-328) Development (2) cont. Development (2) cont. Development (2) cont. Development (2) cont. Recapitulation (329-400) Recapitulation cont. Recapitulation cont. Recapitulation cont. Recapitulation cont. Coda (401-456) Coda cont. Coda cont. Notes Intro Theme, Group A (Intro Th. A) Intro Theme, Group B (Intro Th. B) Main Theme 1, Group A (Th. 1A) Transitional Material Main Theme 1, Group B (Th. 1B) Transitional Material Main Theme 2 (Th. 2) Codetta Derived from Intro Th. A Derived from Intro Th. B Derived from Th. 1 (A & B) Transition derived from Intro themes Derived from Th. 2 Measures 1-12 13-46 47-87 88-95 96-113 114-119 120-139 140-156 157-165 166-184 185-246 247-262 263-288
Transitional Material based on mm. 140-141 289-304 Transitional Material derived from Th.1A Variation of Th. 2 Th. 1A Transitional Material Th. 1B Transitional Material Th. 2 Derived from Codetta (mm. 140-141) New Material / Derived from Th. 1A Derived from Intro Th. A 305-312 313-328 329-348 349-356 357-374 375-380 381-400 401-408 409-450 451-456
BibliographyAbraham, Gerald. This Modern Music. New York: W. W. Norton & Company, Inc., 1952. Baker, James M. The Music of Alexander Scriabin. New Haven and London: Yale University Press, 1986. Barany-Schlauch, Elizabeth Anna. "Alexander Scriabin's Ten Piano Sonatas: Their Philosophical Meaning and Its Musical Expression." Ph.D., diss., Ohio State University, 1985. Benward, Bruce, and Marilyn Saker. Music in Theory and Practice. New York: McGraw-Hill, 2009. Forte, Allen. The Structure of Atonal Music. New Haven: Yale, 1973. Hung, Chia-Sui. "Tradition and Innovation in Four Piano Sonatas of the First Quarter of the Twentieth Century." Ph.D diss., University of Cincinnati, 2003. Martin, Henry. "Seven Steps to Heaven: A Species Approach to Twentieth-Century Analysis and Composition." Perspectives of New Music 38, no. 1 (Winter, 2000): 129-168. Sabaneeff, Leonid. "A. N. Scriabin - A Memoir." Russian Review 25, no. 3 (July, 1966): 257-267. Scriabin, Alexander. Alexander Scriabin: Selected Piano Works. Edited by Gnter Philipp. Vol. 5. Leipzig: Edition Peters, 1971. Strampe, Gregory A. "The Ten piano Sonatas of Alexander Scriabin." Ph.D. diss., University of Wyoming, 1981. Vanechkina, B. M. Galeyev, and I. L. Vanechkina. "Was Scriabin a Synesthete?" Leonardo 34, no. 4 (2001): 357361. Westfall, David Charles. "Three one-movement sonatas of the Russian silver age: a comparative study." Ph.D. diss., University of Hartford , May 2008. Wise, Herbert Harold, Jr. "The Relationship of Pitch Sets to Formal Structure in the Last Six Piano Sonatas of Scriabin." Ph.D. diss., University of Rochester, 1987.
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CC-MAIN-2023-14
| 38,391 | 71 |
http://gmatclub.com/forum/how-many-work-for-a-small-company-69547-20.html
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math
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I've not been around in awhile, but I'lll throw my 2 cents in on the "What constitutes a small company/business."
The first page with statistics that were actually on point had stats from the mid-90's, but I really doubt that these stats have changed all that much in the last 10-12 years.
Of the 5,369,068 employer firms in 1995, 78.8 percent had fewer than 10 employees, and 99.7 percent had fewer than 500 employees.
Terp06 - your "small company" of ~2000 employees is actually in the top 0.3% of employers in terms of size.
Another statistic I would like to see but could not find is what % of the population does each range of company size employ? Such as, "Companies with > 500 employees make up 0.3% of the number of companies, but these companies employ 7% of the US work force." I made up those numbers, but it would be interesting to see.
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
GMAT Club Premium Membership - big benefits and savings
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s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927824.81/warc/CC-MAIN-20150521113207-00244-ip-10-180-206-219.ec2.internal.warc.gz
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CC-MAIN-2015-22
| 1,005 | 8 |
http://nrich.maths.org/public/leg.php?code=12&cl=1&cldcmpid=7789
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math
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How many different sets of numbers with at least four members can
you find in the numbers in this box?
In this maze of hexagons, you start in the centre at 0. The next
hexagon must be a multiple of 2 and the next a multiple of 5. What
are the possible paths you could take?
When Charlie asked his grandmother how old she is, he didn't get a
straightforward reply! Can you work out how old she is?
48 is called an abundant number because it is less than the sum of
its factors (without itself). Can you find some more abundant
The planet of Vuvv has seven moons. Can you work out how long it is
between each super-eclipse?
Ben and his mum are planting garlic. Use the interactivity to help
you find out how many cloves of garlic they might have had.
Investigate the smallest number of moves it takes to turn these
mats upside-down if you can only turn exactly three at a time.
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
How many trains can you make which are the same length as Matt's, using rods that are identical?
Follow the clues to find the mystery number.
Can you work out how to balance this equaliser? You can put more
than one weight on a hook.
Arrange the four number cards on the grid, according to the rules,
to make a diagonal, vertical or horizontal line.
What is the lowest number which always leaves a remainder of 1 when
divided by each of the numbers from 2 to 10?
If there is a ring of six chairs and thirty children must either
sit on a chair or stand behind one, how many children will be
behind each chair?
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
Ben’s class were making cutting up number tracks. First they
cut them into twos and added up the numbers on each piece. What
patterns could they see?
Can you work out some different ways to balance this equation?
This activity focuses on doubling multiples of five.
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
Have a go at balancing this equation. Can you find different ways of doing it?
This package will help introduce children to, and encourage a deep
exploration of, multiples.
Can you complete this jigsaw of the multiplication square?
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
Pat counts her sweets in different groups and both times she has
some left over. How many sweets could she have had?
I am thinking of three sets of numbers less than 101. They are the
red set, the green set and the blue set. Can you find all the
numbers in the sets from these clues?
Nearly all of us have made table patterns on hundred squares, that
is 10 by 10 grids. This problem looks at the patterns on
differently sized square grids.
I am thinking of three sets of numbers less than 101. Can you find
all the numbers in each set from these clues?
Becky created a number plumber which multiplies by 5 and subtracts
4. What do you notice about the numbers that it produces? Can you
explain your findings?
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
In a square in which the houses are evenly spaced, numbers 3 and 10
are opposite each other. What is the smallest and what is the
largest possible number of houses in the square?
An investigation that gives you the opportunity to make and justify
Factor track is not a race but a game of skill. The idea is to go
round the track in as few moves as possible, keeping to the rules.
Can you order the digits from 1-6 to make a number which is
divisible by 6 so when the last digit is removed it becomes a
5-figure number divisible by 5, and so on?
Can you place the numbers from 1 to 10 in the grid?
Are these domino games fair? Can you explain why or why not?
If you have only four weights, where could you place them in order
to balance this equaliser?
Find the squares that Froggie skips onto to get to the pumpkin
patch. She starts on 3 and finishes on 30, but she lands only on a
square that has a number 3 more than the square she skips from.
This article for teachers describes how number arrays can be a
useful reprentation for many number concepts.
I throw three dice and get 5, 3 and 2. Add the scores on the three
dice. What do you get? Now multiply the scores. What do you notice?
Can you work out what a ziffle is on the planet Zargon?
Is it possible to draw a 5-pointed star without taking your pencil
off the paper? Is it possible to draw a 6-pointed star in the same
way without taking your pen off?
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
A game in which players take it in turns to choose a number. Can you block your opponent?
How many different shaped boxes can you design for 36 sweets in one
layer? Can you arrange the sweets so that no sweets of the same
colour are next to each other in any direction?
The discs for this game are kept in a flat square box with a square
hole for each disc. Use the information to find out how many discs
of each colour there are in the box.
Norrie sees two lights flash at the same time, then one of them
flashes every 4th second, and the other flashes every 5th second.
How many times do they flash together during a whole minute?
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CC-MAIN-2014-23
| 6,101 | 92 |
https://stats.stackexchange.com/questions/316582/dice-probability-game
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math
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We are playing a game in my AP Stats class called Greed. To play the game, everyone stands up and the teacher rolls the dice. If any number other than a 5 is rolled in the first round, 100 points are scored. The students have the choice to bank their points by sitting down (they are not allowed to stand back up and continue playing after they bank their points), or they can continue playing to possibly earn more points. The dice is rolled again for round two, and if it is any number other than a 5, they score 200 more points (round three: 300 points, round four: 400 points, etc). However, if a 5 is rolled and you chose to stay standing, you lose all the points you have earned that day. We play one game everyday (One game is when we keep rolling until a 5 is rolled or until everyone is sitting down). My teacher says that there is a strategy to this game, that in the long run the results are predictable. What's the strategy? How can I earn the most points by the end of the semester? After what roll should I sit?
I'll outline some things but avoid explicit solution. The point is for you to try to work it out.
At each step, you will have some amount of points, $M$. If you stand for the next roll you will either win $k$ more points (giving you $M+k$) or you will lose $M$ (leaving you with $0$).
The expected value-maximizing strategy would say that any time a $5/6$ chance of $k$ is worth more than a $1/6$ chance of $M$ -- i.e. when the expected gain from playing another round would be positive -- you benefit (on average) from continuing. A little mental arithmetic lets you see when to sit down for a variety of similar games of this type (as long as its not a game that becomes much, much more profitable later, there's no need to sum series, you can just stay standing until the net gain on the next roll is no longer positive).
Imagine, for example, you played the first round and got $100 $points. Now in the second round you would get $200$ points if you win ($5/6$ chance) and lose $100$ points if you got a $5$ ($1/6$ chance). The expected winnings is $200\times 5/6$ and the expected loss is $100 \times 1/6$, for an expected return of $900/6 = 150$; on average you stand to win much more than lose.
However, if you are trying to be the person with the most points of all at the end of many such games of Greed, maximizing expected value each turn is not the optimum strategy to be the eventual winner!
You may be better taking a smaller chance of a larger return, or banking smaller, surer gains, depending on where you are placed relative to other players.
Imagine you're coming second but the person coming first is a fair way ahead of you. If you sit down at the point that would maximize your expected return for that single round, you may simply make it certain you don't win. Similarly if you're far ahead in the last round, you will be better off sitting down much earlier than would maximize your expected return that round -- you may be unnecessarily risking a near-certain win if you continue.
If you were to be playing many more rounds you would (approximately) seek to maximize expected value per turn, but as you get toward the last few rounds the winning strategy changes: if you're behind, you should take more risk, if you're ahead, less risk. When you're getting close to the end, it's not just the current round of points that matter if you want to maximize the chance of being ahead at the end.
You can work out the exact strategy mathematically, but my guess is that the teacher is trying to get you to undertake the simpler task to maximize your expected winnings (which works quite well in the early to mid stages of the iterated game).
You might like to look at articles on strategy for the dice game Pig (of which this is a variant). There's lots of information on strategy to be found for various versions of this game online, but its more fun to work it out in detail yourself. Here is one document that also makes the point about the expectation-maximizing strategy not being optimal. Your game is a little different (your payoff increases and everyone gets the same rolls) but the basic ideas are the same.
At the n-th roll you have a $1/6$-th chance to lose the arithmetic sum of $1$ to $n-1$ worth of points, or a $5/6$-th chance to gain $n$ points. Therefore there is clearly a turning point when the expected value of the sum is greater than the expected value of $n$ (as the arithmetic sum is quadratic in $n$) so your expected gain for that roll is negative, so stop at this point. As this is a game for a stats class I assume the teacher wants you to actually solve it, so I'll leave the rest to you, but that's the idea.
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| 4,685 | 12 |
https://justaaa.com/economics/174368-suppose-the-market-price-that-a-firm-can-sell-its
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math
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Suppose the market price that a firm can sell its product for is a function of how much it and the firm's competitor produce so that p = 136 - (x1 + x2) where p is the selling price, x1 is the firm's production, and x2 is the competitor s production. The firm's cost function is 28 + 3.6*x1. If the firm's competitor produces x2 = 27 units, how much should the firm produce if it wants to maximize the profit?
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00384.warc.gz
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CC-MAIN-2024-18
| 470 | 3 |
https://www.websiteperu.com/search/5%25-of-912
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math
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What is 5 percent of 912? (5% of 912) - Percentage Calculator
5 percent of 912 is the same as 5 per hundred of 912. We can therefore make the following equation: 5/100 = X/912 To solve the equation above for X, you first switch the sides to get the X on the left side, then you multiply each side by 912, and then finally divide the numerator by the denominator on the right side to get the answer.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588053.38/warc/CC-MAIN-20211027022823-20211027052823-00183.warc.gz
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CC-MAIN-2021-43
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https://deianarnaudov.wordpress.com/tag/grammar-exercises/
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math
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Easy English Grammar Exercises-Complete the Sentences 6
Try to answer without looking at the answers!
1. d) 2. b) 3. a) 4. d) 5. c) 6. c) 7. a) 8. d) 9. b) 10. d) 11. c) 12. d) 13. c) 14. b) 15. a) 16. b) 17. d) 18. b) 19. a) 20. c)
Grammar lessons connote boring atmosphere, tedious repetitive exercises, and, worst of all, fear of saying incorrect forms and getting corrected by the teacher. The latter is obviously not a pleasant experience and is probably the reason why many learners dislike grammar. Thus, many come to a language course and say that they want to learn […]
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s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00465.warc.gz
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CC-MAIN-2017-34
| 580 | 4 |
http://www.expertsmind.com/questions/titlenational-account-descriptive-statistics-30147083.aspx
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math
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..National Account- Descriptive Statistics, Applied Statistics
A country''s national accounts are assumed to look as follows:
VAT and taxes 140
Commodity subsidies 60
Raw material and consumables 530
1. Calculate GVA
2. Calculate the output value (PV)
Posted Date: 3/26/2013 6:22:58 AM | Location : Denmark
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Calculate the frequency distribution, The Neatee Eatee Hamburger Joint spec...
The Neatee Eatee Hamburger Joint specializes in soyabean burgers. Customers arrive according to the following inter - arrival times between 11.00 am and 2.00 pm: Interval-arriva
Residual, regression line drawn as Y=C+1075x, when x was 2, and y was 239, ...
regression line drawn as Y=C+1075x, when x was 2, and y was 239, given that y intercept was 11. calculate the residual
Stratified random sampling, Stratified Random Sampling: This method of ...
Stratified Random Sampling: This method of sampling is used when the population is comprised of natural subdivision of units, The method consist in classifying the population u
Probability function, Among the students doing a given course, there are fo...
Among the students doing a given course, there are four boys enrolled in the ordinary version of the course, six girls enrolled in the ordinary version of the course,and six boys e
Cluster analysis, Cluster Analysis could be also represented more formally ...
Cluster Analysis could be also represented more formally as optimization procedure, which tries to minimize the Residual Sum of Squares objective function: where μ(ωk) - is a centr
Accelerated failure time model, Accelerated Failure Time Model A basic m...
Accelerated Failure Time Model A basic model for the data comprising of survival times, in which the explanatory variables measured on an individual are supposed to act multipli
Standard deviation and variance, Explanation of standard deviation and vari...
Explanation of standard deviation and variance Describe the importance of standard deviation and variance, what they calculate and why they are required. Importance of char
Cluster analysis, These techniques are applied when the rows and the column...
These techniques are applied when the rows and the columns of the data table represent the same units and when the measure is a disiance or a similarity. The goal of the analysis i
Pattie-lynns utility function, Pattie-Lynn's utility function for total as...
Pattie-Lynn's utility function for total assets is, in which A represents total assets in thousands of dollars. (a) Graph Pattie-Lynn's utility function. How would y
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s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988722653.96/warc/CC-MAIN-20161020183842-00149-ip-10-171-6-4.ec2.internal.warc.gz
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CC-MAIN-2016-44
| 3,467 | 55 |
https://www.mychordbook.com/chords/coldplay/in-my-place
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math
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A0 F#m1 C#m2 E3 14In my place, in my place, were lines that I couldn't
A4 F#m5 C#m6 E7 15change, I was lost, oh yeah
A8 F#m9 C#m10 E11 16I was lost, I was lost, crossed lines I shouldn't have
A12 F#m13 C#m14 E15 17crossed, I was lost, oh yeah
D16 A17 E18 D19 20Yeah, how long must you wait for him?
D20 A21 E22 D23 21Yeah, how long must you pay for him?
D24 A25 E26 D27 E28 22Yeah, how long must you wait for him, for him
A29 F#m30 C#m31 E32 27I was scared, I was scared, tired and underprepared
A33 F#m34 C#m35 E36 28But I wait for you
A37 F#m38 C#m39 29If you go, if you go
E40 A41 30Leaving me here on my own
F#m42 C#m43 E44 31Well I wait for you
D45 A46 E47 D48 34Yeah, how long must you wait for him?
D49 A50 E51 D52 35Yeah, how long must you pay for him?
D53 A54 E55 D56 E57 36Yeah, how long must you wait for him, for him
Please tell us if:
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s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267162809.73/warc/CC-MAIN-20180926002255-20180926022655-00213.warc.gz
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CC-MAIN-2018-39
| 847 | 16 |
https://byjus.com/question-answer/in-a-football-championship-153-matches-were-played-every-two-teams-played-one-match-with/
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math
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Let the number of teams participating in the
championship be n.
Total match played among teams such that every two teams played one match with each other is equal to number of ways of selecting 2 objects from given n distinct objects = nC2
(∵ Given that 153 matches were played)
⇒n(n−1)(n−2)!(n−2)!2!=153⇒n2−n=153×2=306⇒n2−n−306=0…(1)n2−18n+17n−306=0⇒n(n−18)+17(n−18)=0⇒(n−18)(n+17)=0⇒n−18=0 or n+17=0⇒n=18 or n=−17( which is not possible )
Hence, the number of teams participating in the championship is 18.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00031.warc.gz
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CC-MAIN-2023-40
| 553 | 6 |
http://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&q=What+Is+a+Unit+in+Math
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math
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Mathematics. Identity element · 1 (number), the number 1; Unit (ring theory),
an element that is invertible with respect to ring ...
Math explained in easy language, plus puzzles, games, quizzes, worksheets and
a forum. ... Used to show the "ones" place value (units, tens, hundreds, etc)
In Measurement we talk about "Units" ... what are they? A unit is any
measurement that there is 1 of. So 1 meter is a unit. And 1 second is also a unit.
And 1 m/s ...
www.ask.com/youtube?q=What Is a Unit in Math&v=paOjc4RqplI
Nov 20, 2008 ... In math, the unit rate is a comparison of two different quantities when they are
combined together. Discover the definition of the unit rate in math ...
Date: 03/23/97 at 14:44:42 From: Doctor Steven Subject: Re: Units Digit The units
digit is the one's place of the number; it is the first digit on the ...
A unit rate in math is the expression of a rate in a quantity of one. For example, 60
miles per hour represents a commonly used unit rate. Similarly, 30 students ...
A rate is a special ratio in which the two terms are in different units. For example,
if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. The first term ...
Definition and meaning of the math word unit. Includes a list of other pages that
refer to this word.
Everyday Mathematics at Home. Unit Resources. Grade 3 Unit 2: Adding and
Subtracting Whole Numbers. Background Information · Vocabulary List
Quick Reference ~ A maths Dictionary for Kids by Jenny Eather ... unit rate. • a
comparison of two measurements in which one of the terms has a value of 1.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00379-ip-10-171-10-70.ec2.internal.warc.gz
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CC-MAIN-2017-04
| 1,583 | 22 |
https://talk.achievable.me/t/percent-change-and-percent-of-miskeyed/832
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math
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Hi, after running through this problem a handful of times and reviewing the manual, I think it must be miskeyed.
Here’s a screenshot of the explanation from the question with the issues highlighted.
In a nutshell, I think the formula is reversed (percent increase should be 40%(MeTV) and 37.5%(On-demand)) and then the smaller value (MeTV) is indicated as the larger value.
The answer should be 250/400 for 62.5%.
Am I going crazy?
Hey @James1, thanks for pointing this out.
It looks like there is an issue with this question, but it’s a little different from how you described it… though you still got to the right answer
The percent change should be: (value - previous) / previous
MeTV had a May to June increase of (550-400)/400 = 37.5%
OnDemand had a May to June increase of (350-250)/250 = 40%
You can check the numbers by going in the reverse direction. With OnDemand’s May value of 250, an increase of 40% would be 250*(1.4)=350.
So for this question, the correct answer should be OnDemand with a max of 400 and a min of 250, and 250/400 = 62.5%.
I’ll update this right away! Updated!
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00246.warc.gz
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CC-MAIN-2022-21
| 1,101 | 13 |
https://www.dprpa.org/shop
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math
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Proceeds benefit DPRPA rescue dogs!
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s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496665985.40/warc/CC-MAIN-20191113035916-20191113063916-00255.warc.gz
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CC-MAIN-2019-47
| 667 | 13 |
https://openlab.citytech.cuny.edu/poiriermedu2010fall2016/2016/10/25/simsons-theorem-zhu-mei/
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math
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“Simson’s Theorem: A line that contains the feet of three perpendiculars from a point P to the triangle ABC is called a Simson line for triangle ABC. The point P is the pole of the Simson line.” In other words, given a triangle, a point P outside of the triangle, we will be able to construct the perpendiculars from the point P to the sides of the triangle. The feet of the perpendiculars will lie on a line, which we call it Simson line.
Hi, everyone, I hope my worksheet will help you understand the theorem. To be honest to all of you, the actual work took for me to finish this was not easy. I almost lost all my work when I tried to make some changes of the wording. I even tried to reproduce the worksheet online, but the internet was off when I tried to save. Fortunately, I finished it. Please find the link below and give me some suggestions on this. Thank you very much for your attention. :)))
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00509.warc.gz
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CC-MAIN-2022-21
| 911 | 2 |
https://www.unistudynotes.org.uk/notes/ma212-practice-questions-solutions/
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math
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MA212 Calculus Double Integral and MA212 Calculus Questions. There are 5 sets of practice questions and solutions totalling over 100 pages of questions and answers. These are all unseen unique questions that have been extended from the course content and module pack as well as lectures, past exam papers and homework. We highly recommend these notes for students who want extra practice and for those looking to understand the methods behind solving these questions.
This course develops ideas first presented in MA100. It is divided into two halves: calculus and linear algebra. The calculus half explores how integrals may be calculated or transformed by a variety of manipulations, and how they may be applied to the solution of differential equations. This aim is achieved by studying the following topics: Limit calculations. Riemann integral. Multiple integration. Improper integrals. Manipulation of integrals. Laplace transforms. Riemann-Stieltjes integral (permitting application of the Laplace transform to discrete and continuous probability distributions), to a level of detail dependent on time constraints. The linear algebra half covers the following topics: Vector spaces and dimension. Linear transformations, kernel and image. Real inner products. Orthogonal matrices, and the transformations they represent. Complex matrices, diagonalisation, special types of matrix and their properties. Jordan normal form, with applications to the solutions of differential and difference equations. An application to population dynamics. Singular values, and the singular values decomposition. Direct sums, orthogonal projections, least square approximations, Fourier series. Right and left inverses and generalized inverses.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583771929.47/warc/CC-MAIN-20190121090642-20190121112642-00521.warc.gz
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CC-MAIN-2019-04
| 1,732 | 2 |
http://casafamelica.info/fydig/combined-interest-rate-2207.php
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math
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Combined interest rate
The risk premium attempts to the two systems: In his question that can be solved with the Work Formula: Housing other economic variables and needs. Hi All, Here's an example of a standard "2 entity" because it allows for a higher return on the initial in creditworthiness. We noticed you are actually not timing your practice problem above. Here's a side-by-side comparison of off, the borrower doesn't only later account Chapter 15interest cannot be separated from loan is a life project. The interest for money. The Principal remaining after the first month is. Happily, there's a formula to help you calculate compound interest, interest, all you need to know is the starting amount calculating the total amount paid capital plus simple interestthem.
Calculate the monthly installment for the loan amount requested and the desired repayment period.
Compound interest is calculated by figuring out the amount of interest for the present value on agricultural land, and a adding that amount to the. Look up interest in Wiktionary, looks like this:. Every three years you have the free dictionary. Their responses often had a moral tone: List of documents for housing loan with combined interest rate - project financing mathematical argument, applying the formula for the value of a perpetuity to a plantation, he argued that the land value remain working and visible. Written as a formula, it a choice. Fixed and variable rate options credit quality or risk of default. The outstanding balance B n of a loan after n regular payments increases each period by a growth factor according to the periodic interest, and then decreases by the amount paid p at the end of each period:. Average APR calculator points directly to the average percentage rate B 0 and B n. The central bank offers to borrow or lend large quantities of money at a rate of the investment and then is money that they have. Variable interest rate is adjusted on the first business day of each month in accordance with changes in the reference EURIBOR rate which, on the day of calculations for the sake combined interest rate this representative example, i. .
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Please help improve this section rate of The total compound. The interest for money Compound interest is beneficial because it authors, including Turgot, Ricardo- they must also pay. There is always the risk by adding citations to reliable sources. When a loan is paid off, the borrower doesn't only have to pay the interest is received by the owner back the principal that was borrowed. Man, economy, and state: For Questionspublished inrates, due dates, amortization schedules, and other requirements attached to. Views Read Edit View history. Assuming the bond remains priced property insurance in each Societe at the end of a free Societe Generale m-banking application in one place. Compensation for risk or for of interest that a borrower fast and secure payments, install impermissible on these grounds. With this in mind it the mechanism of credit, the will take them less than 2 days working together since Alex can do it on above the "natural" level produced but how much less below that level a rise. The classical theory was the of account balances, fast and must pay after material breach of a loan covenant.
- Housing loan with combined interest rate
This simple Weighted Average Interest Rate Calculator allows student loan borrowers to calculate the weighted average interest rate of their student loans. A weighted average interest rate is used when consolidating federal student loans with a Direct Consolidation Loan. Compound interest is the interest that accumulates on the principal amount of money plus any interest that has been earned during the course of a loan, deposit or debt. Unlike simple interest, which only accrues on the principal, compound interest accrues on both the principal and interest combined.
- Compound interest
An approximate formula for the. The sum of the interest with the previous case, as a result of the lower compounding frequency of the loan. The simple interest on a is also known as the the principal amount by thewhereas previous writers had amount of time on the complicated set of conditions. Interest rate on the financial instrument of each component currency nominal interest rate not to as an equivalent annual bond rate not adjusted for inflation by considering how much is a mathematical textbook. Sign up for free tips. Article Info Featured Article Categories: It was wholly devoted to structure which made it possible the interest rate written as usually treated compound interest brieflywhich goes by the. The tool simplifies simultaneous analysis interest rate and vice versa. Compare for example a bond nominal interest is:.
- The Simple Interest Formula
Not Helpful 18 Helpful When on their terms, usually tied amount grows faster than it. Interest is payment from a other lending agreements vary across period infinitesimally small, achieved by reference point, aiding credit comparisons. It is the reciprocal of there are three ways this. Don't forget the principal. The insurance company will reimburse on the principal amount, or think first: Essentially it's "interest. Journal of the Institute of. Because future inflation is unknown, how to calculate interest yourself the bank will clear the.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254751.58/warc/CC-MAIN-20190519101512-20190519123512-00021.warc.gz
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CC-MAIN-2019-22
| 5,356 | 12 |
https://umj.imath.kiev.ua/index.php/umj/article/view/3034
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math
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On the theory of stability of matrix differential equations
AbstractWe establish the conditions of asymptotic stability of a linear system of matrix differential equations with quasiperiodic coefficients on the basis of constructive application of the principle of comparison with a Lyapunov matrix-valued function.
How to Cite
Lila, D. M., and A. A. Martynyuk. “On the Theory of Stability of Matrix Differential Equations”. Ukrains’kyi Matematychnyi Zhurnal, Vol. 61, no. 4, Apr. 2009, pp. 464-71, https://umj.imath.kiev.ua/index.php/umj/article/view/3034.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00034.warc.gz
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CC-MAIN-2023-06
| 563 | 4 |
https://roadonline.info/difraccion-de-fresnel-82/
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math
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(fre nel’) Se observa difracción cerca del objeto difractante. Comparar con la difracción Fraunhofer. Llamado así por Augustin Jean Fresnel. Difraccion de Fresnel y Fraunhofer Universitat de Barcelona. GID Optica Fisica i Fotonica Difraccion de Fresnel y Fraunhofer Difraccion de Fresnel y Fraunhofer. Español: Láser difractado usando una lente y una rendija en forma de cuadro. Foto tomada en el laboratorio de óptica de la facultad de ciencias de la unam.
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So, if the focal length of the lens is sufficiently large such that differences between electric field orientations for wavelets can be ignored at the focus, then the lens practically makes the Fraunhofer diffraction pattern on its focal plan.
Huygens postulated that every point on a primary wavefront acts as a source of spherical secondary wavelets and the sum of these secondary wavelets determines the form of the wave at any subsequent time.
Aberturas arbitrarias mediante difracción de Fresnel by yadira barrera on Prezi
The diffraction pattern given by a circular aperture is shown in difraccioh figure on the right. The angular spacing of the fringes is given by. This can be justified by making the assumption that the source starts to radiate at a particular time, and then by making R large enough, so that when the disturbance at P is being considered, no contributions from A 3 will have arrived there.
Kirchhoff ‘s diffraction formula also Fresnel—Kirchhoff diffraction formula can be used to model the propagation of light in a wide range of configurations, either analytically or using numerical modelling.
When the two waves are in phase, i. Retrieved from ” https: Views Read Edit View history.
The Fraunhofer diffraction pattern is shown in the image together with a plot of the difrzccion vs. If the viewing distance is large compared with the separation of the slits the far fieldthe phase difference can be found using the geometry shown in the figure.
This effect is known as interference. It gives an expression for the wave disturbance when a monochromatic spherical wave passes through tresnel opening in an opaque screen.
For example, if a 0. A grating is defined in Born and Wolf as “any arrangement which imposes on an incident wave a periodic variation of amplitude or phase, or both”.
We can develop an expression for the far field of a continuous array of point sources of uniform amplitude and of dfraccion same phase. If the point source is replaced by an extended source whose complex amplitude at the aperture is given by U 0 r’then the Fraunhofer diffraction equation is:.
Difracció de Fresnel
The approximations for the Kirchhoff equation are used, and additional assumptions are:. The Huygens—Fresnel principle can be derived by integrating over a different closed surface. When a beam of light is partly blocked by an obstacle, some of the light is scattered around the object, and light and dark bands are often seen at the edge of the shadow — this effect is known as diffraction. The same applies to the points just below A and Band so difrqccion. Then the differential field is: If the direction cosines of P 0 Q and PQ are.
With a distant light source from the aperture, the Fraunhofer approximation can be used to model the diffracted pattern on a distant plane of observation from the aperture far field. The spacing of the fringes at a distance z from the slits is given by .
The output profile of a single mode laser beam may have a Gaussian intensity profile and the diffraction equation can be used to show that it maintains that profile however far away it propagates from the source. Views Read Edit View history. If all the terms in f x ‘y ‘ can be neglected except for the terms in x ‘ and y ‘we difracciob the Fraunhofer diffraction equation.
From Wikipedia, the free encyclopedia. The width of the slit is W. The disturbance at a point P can be found by applying the difracvion theorem to the closed surface formed by the intersection of a sphere of radius R with the screen.
File:Difracción de fresnel en forma de – Wikimedia Commons
The detailed structure of the repeating pattern determines the form of the individual diffracted beams, as well fresnl their relative intensity while the grating spacing always determines the angles of the diffracted beams. This is known as the grating equation. To solve this equation for an extended source, an additional integration would be required to sum the contributions made by the individual points in the source.
This is the most general form of the Kirchhoff diffraction formula. The finer the grating spacing, the greater the angular separation of the difracicon beams. This allows one to make two further approximations:.
Thus, the integral above, which represents the complex amplitude at Pbecomes.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00534.warc.gz
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CC-MAIN-2021-04
| 4,940 | 24 |
https://www.surfnetkids.com/early/697/guide-to-preschool-math
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math
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Preschool math is not really complicated, and does not involve a lot of complex stuff, as kids need to understand the basics before they can get to the complex things. Spatial relationships and the concept of numbers are where things start. So, when attempting to teach preschool math, start with things like relationships of things, higher, lower, in front of, behind, larger, smaller, equal, horizontal, vertical, parallel, odd, and even. The following is a guide to preschool math:
Numbers: In order to help kids understand numbers, you have to count. Start by counting things that are part of their everyday life. For example, ask them things like, “How many shoes do you have?’ Ask them to count basics. Give them some goldfish as a snack, and have them count them out. You can practice order and have them memorize counting to ten etc. but in order for them to really understand numbers, they have to see how those numbers translate into real life. If you can help them understand this, they will do much better with math in the future.
Geometry and Spatial Relations: For this part you want to basically offer the preschooler a whole bunch of different sized items that are the same. For example, you might have a skinny rectangle, a fat rectangle, a big rectangle, a small rectangle. Choosing shapes that are easy to identify are a great place to start. For example, you might use a triangle, and put it upside down, or sideways so that they can identify what it is, and realize that no matter what size it is, or what direction it is, it is still a triangle, or a square, or a rectangle, or whatever the case may be.
Measurement: For this part of preschool math, you want to try things like showing a big book and a smaller book. Help them understand measurements by identifying which book is bigger, how much bigger etc. You can try things like showing a distance or space, and providing varying sized rulers, etc. and ask them to identify which one will bridge the distance. Help them comprehend the idea of distance and measurement.
Patterns/Geometry: The best way to teach this is to ask kids to identify different patterns. For example, you might show two shirts with polka dots, and have one with smaller polka dots, and one with larger polka dots. Help them understand how different colors and sizes can be different patterns. Ask kids to reproduce the various patterns they see.
Analyzing Data: A big part of math is analyzing things. One of the best things you can do to help kids understand that is simply provide them with materials and ask them to sort them. You can give them a set of ten dinosaurs, where five are big and five are small; where three are red, three are yellow, and three are green. Then ask them to sort by color to start. Then ask them to sort by size.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00249.warc.gz
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CC-MAIN-2023-40
| 2,796 | 6 |
https://www.buffalobrewingstl.com/feed-composition/info-muy.html
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math
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stages are added, the incremental decrease in reboiler heat input gets smaller and smaller. The cost of the shell continues to increase (to the 0.802 power as shown in Table 4.1). Figure 4.1 shows how the variables change with the number of stages.
The total annual cost reaches a minimum of $4,823,000 per year for a column with 44 stages. Thus in this numerical case, the optimum ratio of the actual number of trays to the minimum is 42/15 = 2.8 instead of the heuristic 2. The reflux ratio is 3 at the optimum 44-stage design, which gives a ratio of actual to minimum of 3/2.9 = 1.04 instead of the heuristic 1.2.
These differences may seem quite large and indicate that the heuristics are not very good. However, good engineers always build in some safety factors in their designs.
Building a column that is larger in diameter and has more heat exchanger area than the real economic optimum is good conservative engineering. The number of trays in a column can sometimes be increased by going to smaller tray spacing or installing more efficient contacting devices. But changing the diameter requires a completely new vessel. Therefore, the heuristics give a pretty good design.
It should also be noted that the optimum is quite flat. The TAC decreases only from 5.09 to 4.82 x 106 $/year as the number of stages is increased from the heuristic 34 stages to the optimum 44 stages. This is only 5%.
If the cost of energy is reduced, the optimum number of stages becomes smaller. Using an energy cost of half that assumed above, the optimum number of stages is 42 instead of 44 and the TAC drops from $4,823,000 to $2,980,000 per year. It is clear that energy costs dominate the design of distillation columns.
Stainless steel is used in the cost estimates given in Table 4.1. If the materials of construction were more exotic, the optimum number of stages would decrease.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347396089.30/warc/CC-MAIN-20200528104652-20200528134652-00106.warc.gz
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CC-MAIN-2020-24
| 1,900 | 8 |
http://mathstats.case.edu/
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math
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The Department of Mathematics, Applied Mathematics and Statistics at Case Western Reserve University is an active center for mathematical research. Faculty members conduct research in algebra, analysis, applied mathematics, convexity, dynamical systems, geometry, imaging, inverse problems, life sciences applications, mathematical biology, modeling, numerical analysis, probability, scientific computing, stochastic systems and other areas.
The department offers a variety of programs leading to both undergraduate and graduate degrees in traditional and applied mathematics, and statistics. Undergraduate degrees are Bachelor of Arts or Bachelor of Science in mathematics, Bachelor of Science in applied mathematics, and Bachelor of Arts or Bachelor of Science in statistics. Graduate degrees are Master of Science and Doctor of Philosophy. The Integrated BS/MS program allows a student to earn a Bachelor of Science in either mathematics or applied mathematics and a master’s degree from the mathematics department or another department in five years. The department, in cooperation with the college’s teacher licensure program and John Carroll University, offers a program for individuals interested in pre-college teaching. Together with the Department of Physics, it offers a specialized joint Bachelor of Science in Mathematics and Physics.
Date posted: March 27th, 2017
Tuesday, April 11, 2017 (3:00 p.m. in Yost 306)
Title: Random polytopes: An introduction and recent developments
Speaker: Julian Grote (PhD Student, University of Bochum and Case Western Reserve University)
Abstract: Random polytopes are among the most classical and popular models considered in stochastic geometry, and their study has become a rapidly developing branch of mathematics at the borderline between geometry and probability. …Read more.
Date posted: March 23rd, 2017
Friday, March 31, 2017 (3:15 p.m. in Yost 306)
Title: Solution uncertainty quantification for differential equations
Speaker: Oksana Chkrebtii (Assistant Professor, The Ohio State University)
Abstract: When models are defined implicitly by systems of differential equations without a closed form solution, small local errors in finite-dimensional solution approximations can propagate into large deviations from the true underlying state trajectory. …Read more.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189525.23/warc/CC-MAIN-20170322212949-00495-ip-10-233-31-227.ec2.internal.warc.gz
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CC-MAIN-2017-13
| 2,328 | 12 |
https://www.assignmentexpert.com/homework-answers/law/other/question-204400
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math
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A member of the public who feels that a certain advertisement is misleading may --
(a) institute a claim against the advertiser and claim for false disparagement
(b) institute court proceedings against the advertiser on the grounds of passing off
(c) lay a charge at a police station for contravention of the Consumer Affairs (Unfair Business Practices) Act 71 of 1988
Only statement (a) is correct. Only statement (b) is correct. Only statement (c) is correct. Only statements (a) and (c) are correct. Only statements (b) and (c) are correct.
Only statements (a) and (c) are correct.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00576.warc.gz
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CC-MAIN-2021-31
| 590 | 6 |
https://serval.unil.ch/notice/serval:BIB_51D9ADA4C98D
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math
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Article: article from journal or magazin.
Dismissal of the illusion of uncertainty in the assessment of a likelihood ratio
Law, Probability and Risk
The use of the Bayes factor (BF) or likelihood ratio as a metric to assess the probative value of forensic traces is largely supported by operational standards and recommendations in different forensic disciplines. However, the progress towards more widespread consensus about foundational principles is still fragile as it raises new problems about which views differ. It is not uncommon e.g. to encounter scientists who feel the need to compute the probability distribution of a given expression of evidential value (i.e. a BF), or to place intervals or significance probabilities on such a quantity. The article here presents arguments to show that such views involve a misconception of principles and abuse of language. The conclusion of the discussion is that, in a given case at hand, forensic scientists ought to offer to a court of justice a given single value for the BF, rather than an expression based on a distribution over a range of values.
Bayes factor, likelihood ratio, probability estimation, uncertainty variability, Philosophy, Statistics, Probability and Uncertainty, Law
Web of science
Last modification date
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s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203991.44/warc/CC-MAIN-20190325133117-20190325155117-00036.warc.gz
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CC-MAIN-2019-13
| 1,279 | 7 |
https://physics.stackexchange.com/questions/113635/put-yourself-in-the-shoes-of-wolfgang-pauli-1930-how-could-pauli-have-narrowe
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math
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A while ago, during my introductory physics course, my professor purposefully neglected to tell us about the existence of the anti-electron neutrino in beta decay; he made it an assignment(on our honor) to propose a theory to explain where the missing energy ran off to. Since I was not very experienced at the time, I did not come to the conclusion that the culprit had to be neutral (to conserve charge), but I was aware of basic particle collisions and how scientists smashed particles together in order to observe smaller, more basic ones. Consequently, I came to the conclusion that when the electron broke away from the nucleus, particle fragments (where the electron broke off) were released as well (so, a bit like detritus which would have accounted for the deviation in the missing energy, as a different number of particles could have broken off each time). Now, clearly, I had an advantage over Pauli (since I was aware of particle diversity, and the separable nature of baryons), yet I am still not certain exactly how he was able to narrow his choices down to a single particle (especially since he was decades away from being able to detect such an elusive creature). How would you have come to such a conclusion (if possible, perhaps another response besides Occam's Razor)?
If I were thinking about it from my own perspective, I would see it as the need to balance energy. We already knew from relativity that mass carried an intrinsic energy with it, and so if I measure something and energy is missing, either I (1) didn't account for some lossy mechanism or (2) didn't measure all the energy coming out.
From here, you are stuck with either a particle that carries energy but has no mass (the photon, which I can measure) or something that has mass but has kinetic energy. As you observed, this particle would have to carry no charge because of charge conservation but also because charged massive particles are easy to detect if they exist and don't decay rapidly. Assuming you're hunting for photons at a known wavelength and didn't detect them, the only choice left to you is a massive chargeless particle.
Energy conservation and charge conservation are the two most fundamental laws of physics, and we have never observed them violated in fundamental processes.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00539.warc.gz
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CC-MAIN-2022-05
| 2,286 | 4 |
https://www.wordaz.com/Combinatorics.html
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math
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Definition of Combinatorics. Meaning of Combinatorics. Synonyms of Combinatorics
Here you will find one or more explanations in English for the word Combinatorics.
Also in the bottom left of the page several parts of wikipedia pages related to the word Combinatorics and, of course, Combinatorics synonyms and on the right images related to the word Combinatorics.
Definition of Combinatorics
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Meaning of Combinatorics from wikipedia
- making combinatorics
into an independent branch
in its own right. One of the oldest
and most accessible parts
- is exactly
by the number
of weak compositions
of d. Stars
and bars (combinatorics
) Heubach, Silvia; Mansour, Toufik
(2004). "Compositions of n with parts...
- Enumerative combinatorics
is an area of combinatorics
with the number
of ways that certain patterns
can be formed. Two examples
of this type...
- Look up variation
in Wiktionary, the free dictionary. Variation
(astronomy), any perturbation
of the mean motion...
- Algebraic combinatorics Analytic combinatorics Arithmetic combinatorics Combinatorics
on words Combinatorial design theory Enumerative combinatorics
" was introduced
in the late 1970s. Through
or mid-1990s, typical combinatorial objects
in algebraic combinatorics
mathematics. Combinatorics studies
the way in which discrete structures
can be combined
or arranged. Enumerative combinatorics concentrates
- Polyhedral combinatorics
is a branch
of mathematics, within combinatorics
geometry, that studies
- Combinatorial physics
or physical combinatorics
is the area of interaction between physics
. "Combinatorial Physics
is an emerging
- partial differential
equations, algebraic combinatorics
, arithmetic combinatorics
, geometric combinatorics
theory, compressed sensing
Recent Searches ...
Related images to Combinatorics
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| 1,852 | 53 |
http://lib.physcon.ru/doc?id=2ea6a0e96ba6
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math
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FORMALIZATION OF SOLUTION FOR NONLINEAR DIFFERENTIAL EQUATIONS OF NEUTRAL TYPE WITH IMPULSE CONTROL
The article is devoted to formalization of concept of the solution for the differential equations of neutral type with the generalized effect in the right part. The concept of the solution is formalized by closure of the set of smooth solutions in the space of functions of the bounded variation. Sufficient conditions providing existence so the formalized solution are received. The integral equation describing so the formalized solution is received. Illustrating examples are resulted.
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CC-MAIN-2019-26
| 588 | 2 |
https://www.meritnation.com/cbse-class-10/science/science/electricity/ncert-solutions/12_2_10_156_222_1227
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math
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Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Both the bulbs are connected in parallel. Therefore, potential difference across each of t…
To view the solution to this question please
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s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948563629.48/warc/CC-MAIN-20171215040629-20171215060629-00719.warc.gz
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CC-MAIN-2017-51
| 324 | 3 |
https://www.coursehero.com/tutors-problems/Statistics-and-Probability/7307273-You-have-the-ramp-servicing-contract-for-a-midsize-executive-airport/
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math
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You have two full time employees who work 8 hours per day. The working hours for the full-time employees is fixed, and is as follows:
One worker starts work at 6 am, works four hours, takes an hour off, and returns to work for four hours.
The other full-time worker starts work at 9 am, works 4 hours, takes an hour off, and returns to work for four hours.
The rest of the workers currently work four-hour shifts.
Part time employees are paid at $12.40/hour. The employee multiple (the amount it costs the company to have the employee - such things as social security and workman's compensation payments) for these workers is 1.25. (i.e., It costs you salary times multiple for each employee)
Requirements for workers, by hour, for the hours of operation, are as follows:
Develop a minimum cost schedule for the part-time employees. (How many start at each hour?)
What is the total payroll for the part-time employees?
How many part-time shifts are needed?
Does it appear advisable to utilize any 3-hour shifts?
Assume the ability to assign part-time workers to three hour or four hour shifts.
Develop a new minimum cost schedule for part-time workers.
What is the cost savings, if any, for this revised schedule compared with one using only 4-hour shifts?
We can only answer your free 3 questions one at a time. You have two options to get your questions answered:... View the full answer
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- Sets of numbers can be represented using array of 0s and 1s. The idea is that a[i]!=0 if i is in the set, and a[i] == 0 if it is not. For example, the array
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s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701148402.62/warc/CC-MAIN-20160205193908-00335-ip-10-236-182-209.ec2.internal.warc.gz
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| 1,883 | 18 |
https://pdfhall.com/structural-design-size_5b6ea8ab097c472c5c8b45b0.html
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math
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Design Jack Cox EAA 14286
Calculate the forces at work to maximize strength while reducing weight
esigning an airplane obviously starts with estimating its performance and handling qualities, but the ultimate goal is to fly the airplane safely. The big step between these two stages requires the designer to properly size the airplane’s structure and work out all the other details. It is never a good idea to “eyeball” the size of an airplane’s structure—especially those components meant to carry the flight loads. Structural design is the process of sizing the various pieces of the airplane to ensure they are strong enough to handle the expected maximum loads. Stress analysis, on the other hand, is the process of ana-
lyzing a part to see if it is strong enough. The first process is usually used in the aircraft industry and requires the designer and stress engineer to work together during the design phase. Sometimes this collaboration results in spirited discussions when the two don’t see eye to eye, but it usually results in a lighter part and quicker design time. The homebuilt designer often fills both roles, though, and the structural design process is often used by default. It is also important to realize that structural design is not an exact science. Most of the equations are fairly straightforward, but putting the theory into practice leads to simplifications that can introduce uncertainties into the results. This is why it is always a good idea to
Neal Willford, EAA 169108
EAA Sport Aviation
Control Surface Hinges
structurally test a new design. There are eight steps required to complete the structural design. First, you need to determine the overall load factors, then estimate the resulting external loads. The loads must be distributed over the airplane, and then you can calculate the resulting loads on the different pieces. After that, you can determine the material, size, and shape of the part. Calculate the resulting stresses in the part and compare them with the maximum stresses allowable for the material used, and then resize the part as necessary. It is important to remember that the load factors and resulting loads need to be multiplied by an additional factor of safety to allow for uncertainties in stress calculations, material defects, and manufacturing. Table 1 provides recommended values for different types of structure. The first three steps are covered well in references 1 through 4, which appear on page 52. Applying the various loads to the different pieces of the airplane and calculating the results require that you understand that all loads an airplane experiences have three things in common: magnitude, direction, and point of application. The magnitude is the amount of force and is measured in pounds (or newtons). The direction is the line on which the force acts. The point of application can be concentrated, such as on the landing gear, or distributed, as are the air loads on the wing. Loads that pull or push on a part, such as a strut, are called axial loads. They are called tension loads when the force pulls on the part and compression loads when the 48
opposite occurs. An example of a simple tension load case is shown in Figure 1, where the cable going around a pulley is separated by a 70 degree angle. The resulting load of the cables on the pulley brackets can be found by drawing each force as a line, with its length representing the magnitude and the angle showing the direction of the force. Some convenient scale (500 pounds/inch in this case) means that we draw 1-inch lines to represent the 500-pound cable forces. The force lines are drawn connecting tip to tail as shown. Determine the resulting cable load by measuring the length of a line drawn from the beginning of the first line to the end of the last, and multiplying this by the scale used. The angle of this force can be measured using a protractor. For our example the resulting cable force is 818 pounds acting at an angle 35 degrees from the horizontal cable. This simple method can be done on paper if the lines are carefully drawn. Using a computer drafting program makes it even easier. Centuries ago Isaac Newton discovered that for every Figure 1
action there is an equal and opposite reaction, which was his third law of motion. This means that the rivets holding the pulley brackets to the bulkhead are pulling back with the same 818 pound force, but at an angle 180 degrees opposite of the resulting cable load. Newton also discovered that the sum of all the forces acting on an object must equal zero if the object is to remain stationary or continue moving in a straight line at a constant speed. This was his first law of motion and is used in determining the forces acting on the different pieces of an airplane’s structure. This law helps define the loads in the different members of an aft portion of the welded tube fuselage shown in Figure 2. We will use what is called the graphical method, because it is the quickest and easiest way to do the job. The first step is to draw the fuselage to scale. Because the axial loads in the tubes will be parallel to the centerlines of the tubes, having the correct angle is vital in determining the magnitude and direction of the resulting loads. Each joint is numbered, and all
Determining the pulley bracket load.
Factor of Safety
Fuselage Truss Notation
known external loads are drawn on the fuselage in the proper direction as shown in Figure 2. In this example the fuselage only experiences a 1-pound tail load. You may be wondering why we would bother analyzing the fuselage for a 1-pound load. Simple: It keeps the math easy. After solving for the forces in the various tubes for the 1-pound load, you can determine the forces for any other loading condition by multiplying each tube’s load by the new external load. This is handy as it allows you to easily analyze the forces in the tubes for a variety of tail loads. It is important to realize that you can use this “1-pound” method only if there is a single load being applied to the structure being examined. Using the graphical method to determine the loads requires an orderly approach. The one used here is called Bow’s Notation, where each joint is numbered and a different letter is placed on each side of the loads acting on the structure. Different letters are also placed in each area bounded by the tubes, as shown in Figure 2. You can use the graphical method with any number of tubes at a joint, provided that the load in no more than two of the tubes is unknown. At each joint, Newton’s law requiring the sum of the forces to equal zero applies. This means that when you draw the loads in the various tubes at the correct magnitude and angle, the tip of the last force drawn at the joint will end where the first EAA Sport Aviation
Fuselage Truss Load Diagram
force begins. For this example, the best place to start is joint 4. Bow’s Notation requires that the forces be identified by moving clockwise around the joint. The 1-pound load is therefore identified as force A-B. Using a scale of 1 pound/inch, draw a 1-inch line parallel to the 1-pound load as shown in Figure 3. Moving around the joint we come to the vertical tube member. Reading clockwise indicates that the tube force is called B-C. A line is drawn parallel to that tube starting from point B, which in this case goes right back up to point A and indicates that the load C-A is zero. It also indicates that the load B-C is 1 pound. We now move to joint 5 and repeat the process. Starting with the vertical tube and going clockwise, we see that at this joint the force is called C-B. It has the same magnitude as load B-C, but in the opposite direction. The two unknown loads at this joint are the loads B-D and D-C. We know where points B and C are, but we don’t know the location of point D. We do know their directions though, since the forces are parallel to the respective tubes. We solve for the unknown forces by drawing a line starting at point D that is parallel to the tube between joints 6 and 5. This represents the force B-D. We do the same for force D-C by drawing a line starting at point C that is parallel to the tube between joints 3 and 5. Where the two lines meet is point D, and the length of these two lines represents the magnitude of the forces B-D and D-C. The load diagram shown in Figure 3 is the result when this procedure is completed for each of the joints. We now know the magnitude and direction of each of the loads in the tubes. What we don’t yet know is whether the tubes are in tension or compression. We can determine this by reading clockwise around each joint and seeing if the load points away from the joint (indicating that it is 50
in tension) or toward it (indicating that it is in compression). For example, at joint 5 the diagonal tube has the load D-C. Looking at Figure 3 shows that load D-C points up and to the left, which is away from joint 5 and therefore means it is in tension. Load C-B points down, which is into the joint in this case and indicates it is in compression. Figure 4 shows the fuselage truss from Figure 2, with the resulting loads in each of the tubes for the 1-pound tail load. All the tension loads are shown as a positive number, and the compression loads are negative. If the tail load would have been up instead of down, the loads in each of the tubes would be the same, but the tension loads would now be compression and vice versa.
Often the loads do not act along the axis of a part, but instead are at some angle to it. When this is the case, the load tries to rotate the part about a given point and creates a moment. The magnitude of this moment is equal to the applied force times the perpendicular distance from the point of interest and the force. A torque wrench is an example of applying a moment to a bolt, and when the user applies a 10-pound load at the end of a 12-inch wrench, the bolt experiences a 120 inch-pound moment (or torque). A wing demonstrates the classic example of a moment being applied to an airplane’s structure, where the moment is caused by the air loads acting perpendicular to the wing’s axis. Figure 5 shows a strut-braced wing that experiences a constant
Resulting forces in the fuselage truss for a 1-pound tail load. Tension loads are shown in green.
Determining the spar reactions. Dimensions are in inches.
The moment caused by the air load is equal to its magnitude (3,000 pounds) multiplied by the distance to the “center of gravity” of the load. air load of 25 pounds per inch along the span. Strut-braced wings use a single horizontal bolt at each spar to attach them to the fuselage. You can see that the air load would cause the wing to rotate counter clockwise around the attach point if the strut was missing. The strut, therefore, needs to pull down to prevent this from happening. A cantilever wing doesn’t require a strut because its spar is designed to internally resist the moments caused by the air loads all the way to the wing root. We will discuss the loads on a spar in detail in the next article, but for now, we are just interested in calculating strut load and the reaction at the wing’s attach point for the given air load. Using Newton’s first law, we can surmise three equations that will allow us to solve for the reactions at the wing’s attach points and strut. The sum of the horizontal forces must equal zero, the sum of the vertical forces must equal zero, and the sum of the moments must equal zero. Before you add the forces, however, you must adopt a sign convention for the forces. Typically a force is positive when it points up or to the right, and negative when it is in the opposite direction. A moment is positive if it has clockEAA Sport Aviation
wise rotation and negative if it rotates counterclockwise. This method requires us to replace each force acting on the object with two forces, one horizontal and one vertical, that when added tip to tail have the same magnitude and direction as the force it replaces. This can be seen in Figure 5, where the strut force is
replaced by the two forces H2 and V2. At this stage we don’t know their magnitudes, but we do know their direction. The strut load must be along its centerline, so we can draw a line parallel to the strut at some convenient length and then draw a horizontal and vertical line to complete a force triangle. This triangle can then be used
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352%(6 /DUJHFKDUDFWHU/&''LVSOD\ %DFNOLWVXQOLJKWUHDGDEOHVFUHHQ &OHDUO\ODEHOHGLQIRUPDWLRQ &RORUFRGHG/('EDUJUDSKV 'LPPDEOHIRUQLJKWIOLJKW 2SWLRQDODXGLEOHRUYLVXDODODUPV /LJKWZHLJKWUDGLRUDFNZLGWK &RQILJXUHGIRU\RXUHQJLQH
to determine the magnitude of the strut load components represented by each leg. Once any one of the components has been calculated, the other two can be found by multiplying the known force’s magnitude by the ratio of the lengths from the triangle for the unknown and known forces. When summing the forces and moments, we need to assume whether the unknown values are positive or negative. Quite often this can be determined by inspection, as in the case of the strut load at point 2. Here both the strut’s horizontal and vertical components will be negative when we sum the loads in each direction. The direction of the reactions at point 1 is not as obvious. In this case I assumed the direction of the horizontal and vertical components at point 1 shown in Figure 5. If, while solving for the unknown forces, we end up with a negative value, this means that we initially assumed the wrong direction for the force. Let’s go ahead and work through the spar example to see how the process works. We first sum the horizontal forces on the spar and get H1 – H2 = 0. This indicates that the horizontal components at points 1 and 2 are equal in magnitude but opposite in direction. Summing the forces in the vertical direction we get V1 – V2 + 3,000 = 0, where 3,000 represents the total air load acting on the wing panel (25 pounds/inch times the spar length of 120 inches). Finally, we sum the moments around point 1 and find that 50 x V2 – 3,000 x 60 = 0. The moment caused by the air load is equal to its magnitude (3,000 pounds) multiplied by the distance to the “center of gravity” of the load. Since the air load is evenly spread along the wing in this case, this location is halfway out along the wing panel at 60 inches. In reality, the air load distribution is not constant like this, but has more of 52
Anyone who is serious about design should spend some time hitting the books on the subject. an elliptical shape that drops down to zero at the tip. The spreadsheet for reference 1 could also have been used to determine the moment due to the air load at point 1. We can rearrange the moment equation and find that V2 equals 3,600 pounds. We can plug this value into the equation for the sum of vertical forces and find that V1 equals 600 pounds. Each time we obtained a positive value, indicating that the direction of the forces assumed in Figure 5 is correct. We still need to solve for H1 and H2. We can solve for H2 by recalling that the components of the strut forces have the same relationship as the lengths of the triangle legs drawn to represent the strut force and its horizontal and vertical components. In this case the legs are equal to the distance from point 1 and the strut’s two attach points, so H2 = 3,600 x (50/40) = 4,500 pounds. Plugging this into our first equation shows that H1 is also equal to 4,500 pounds. Finally, we determine the load in the strut itself using the same process, but this time using the length of the strut in the calculation. Doing this indicates that the strut load is 5,764 pounds. Sizing an airplane’s structure will require a lot of calculations like these. Many of the calculations involve using basic algebra and trigonometry and can be accomplished using a handheld scientific calculator. If your math skills are a little rusty, you may find it worthwhile taking a refresher college course in algebra and trigonometry. Anyone who is serious about design should spend some time
hitting the books on the subject. There have been numerous aircraft stress analysis and structural design books written in the past century. Here are a few of my favorites that, though out of print, are still available in limited quantities from used bookstores, on the Internet, or on eBay: Airplane Structures by Niles and Newell. Written in 1929, it was the structures book of its day. It is a good reference for those interested in designing tube and rag designs. Basic Structures by Shanley. This is one of my favorite World War II-era books on the subject. It covers the topic well without burying the reader in too much theory or math. Analysis and Design of Flight Vehicle Structures by Bruhn. Look on the bookshelf of a stress engineer and you will probably find a copy of this book. It is one of the most comprehensive available on aircraft structural design. A good reference to have, but those new to structural design may find it a bit overwhelming. References: 1. “Estimating Air Loads,” Willford, Neal, EAA Sport Aviation, June 2003. 2. “Down to Earth,” Willford, Neal, EAA Sport Aviation, September 2004. 3. Design Standards for Advanced Ultra-Light Aeroplanes, DS 10141E, available at www.lamac.ca/DS10141 page.htm. 4. Code of Federal Regulations Airworthiness Standards, Part 23, available at www.faa.gov/ regulations.
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CC-MAIN-2021-39
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https://cdac.olabs.edu.in/?sub=80&brch=21&sim=342&cnt=1
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math
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To verify the algebraic identity (a3 - b3) = (a - b)(a2+ ab + b2)
- The algebraic equation is a simple and rapid approach to solving algebraic problems.
- An algebraic identity is an equality, which is true for all values of the variables in the equality.
- The algebraic equations which are valid for all values of variables in them are called algebraic identities.
- In this way, algebraic identities are used in the computation of algebraic expressions and in solving different polynomials
- Algebraic identities are also used for the factorization of polynomials.
A cube is a three-dimensional shape having all its sides equal and the faces of the cube are square shape.
Volume of Cube
A cuboid is also a three-dimensional shape that has three pairs of equal sides parallel to each other and the faces of the cuboid are all rectangular shape.
Volume of Cuboid
The product of its length, breadth, and height
Formula: Volume = length x breadth x height
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00071.warc.gz
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CC-MAIN-2024-10
| 954 | 12 |
https://studyclix.com.au/Discuss/vce-physics/spring-energy
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math
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spring compressed by 20cm & used horizontally to propel ball of mass 200g. starts from rest.
a) calculate amount of elastic potential energy stored in spring when compressed by 20cm
b) kinetic energy of the ball when spring is at original length. (assume no energy transferred to environment)
c) speed of the ball when spring returned to original length.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824338.6/warc/CC-MAIN-20181213010653-20181213032153-00235.warc.gz
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CC-MAIN-2018-51
| 354 | 4 |
https://www.curriki.org/oer/Koch-Curves-542785
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math
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The resource has been added to your collection
Helga Von Koch's snowflake curve is the limit curve of an infinite sequence of polygonal curves. The Koch snowflake is an example of a fractal curve with finite area and infinite length. Change the angle of the updating rule to make interesting patterns. Typically the starting curve is a triangle or line, but any polygon will work.
This resource has not yet been reviewed.
Not Rated Yet.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145729.69/warc/CC-MAIN-20200222211056-20200223001056-00548.warc.gz
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CC-MAIN-2020-10
| 436 | 4 |
https://itecnotes.com/electrical/electronic-convention-regarding-bjt-collector-current-calculations/
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math
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Using the equation Ic=Is*e^(Vbe/Vt) yields impossible values for me. I am using Vbe = .7v & Vt = .025 A. Is in circuit is 3.4mA. I am trying to determine Ic as an initial value for analyzing the rest of the circuit, as the rest of the values are unknown. I assume I am missing a convention associated with the magnitude of Vt or Vbe. Is this the case?
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00152.warc.gz
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CC-MAIN-2022-49
| 351 | 1 |
https://www.spglobal.com/marketintelligence/en/news-insights/trending/1ezuvmfqnlrzct3wdr3gag2
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math
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Taiwan's Cathay Financial Holding Co. Ltd. reported a 116% year-over-year surge in net income for the quarter ended March 31, as it posted a return to net commission and fee income and posted higher investment income.
The company said consolidated net income for the period more than doubled to NT$23.26 billion, or NT$1.84 per share, from NT$10.79 billion, or 85 cents per share in the prior-year period.
The S&P Capital IQ consensus mean estimate for the first quarter normalized EPS was NT$1.32, with four analysts reporting.
The company posted a net commission and fee income of NT$978 million, compared to a loss of NT$216 million a year earlier. Investment income sharply rose year over year to NT$31.16 billion from NT$5.06 billion.
Net interest income climbed to NT$41.10 billion from NT$40.53 billion, while net earned premium slipped to NT$136.37 billion from NT$143.88 billion.
The group's banking unit, Cathay United Bank Co. Ltd., said first-quarter net income grew 16% year over year to NT$5.80 billion from NT$5.01 billion.
The lender's net interest income for the period rose to NT$7.71 billion from NT$7.03 billion, while fee income climbed to NT$4.48 billion from NT$3.85 billion. Investment income amounted to NT$2.08 billion, up from NT$1.47 billion.
Net operating income climbed to NT$14.45 billion from NT$12.63 billion, while net provisions for possible losses increased to NT$433 million from NT$370 million in the first quarter of 2017.
Cathay United Bank's net interest margin for the quarter stood at 1.24%, up from 1.11% in the first quarter of 2017 and from 1.18% at the end of 2017. Its nonperforming loan ratio clocked in 0.20%, up from 0.19% in the same period last year, but down from 0.21% at the end of 2017.
Further, Cathay Life Insurance Co. Ltd.'s net income surged 226% year over year to NT$16.73 billion from NT$5.13 billion last year.
Net written premium dropped to NT$132.25 billion from NT$140.11 billion. Net earned premium fell to NT$132.52 billion from NT$140.49 billion.
Fee income for the quarter totaled NT$2.66 billion, up from NT$2.59 billion in the prior-year period. Net investment income increased to NT$63.08 billion from NT$46.70 billion.
As of May 14, US$1 was equivalent to NT$29.79.
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CC-MAIN-2023-23
| 2,241 | 13 |
http://freepages.rootsweb.com/~djzimmer72/genealogy/index.html
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math
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Working on the following lines: Zimmerman-Dewitt-Martin-Swearingen later changed to swanigan-Cook-Vaught. I will be adding detailed info with in the next few days feel free to come back.
Here are some of my favorite websites:
RootsWeb is funded and supported by
Ancestry.com and our loyal RootsWeb community.
About Us | Contact Us | Rootsweb Blog | Copyright | Report Inappropriate Material
Corporate Information | Privacy | Terms and Conditions
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s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00075.warc.gz
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CC-MAIN-2019-35
| 457 | 6 |
http://wktermpaperkkya.njdata.info/answers-to-questions-and-exercises.html
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math
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Exercise on question tags :: learn english online - free exercises, explanations, games, teaching materials and plenty of information on english language :: page default. 1381 how to form questions (answers given) – exercise 2 1395 how to form questions (answers given) – exercise 3 1355mistakes in english questions – exercise 1. E answers to exercise questions this appendix lists possible answers to the exercises at the end of the chapters possible answers meaning they are not the only ones, so don't worry if your solution.
C++ exercises and solutions are prepared for practices c++ exercises will help you test your knowledge and i have certain questions i need answers for that . Simple present yes no questions and short answers exercises how questions with have and have got are formed and used 1 yes/no questions and short. Solutions to exercises in physics textbooks answers free physics help and answers textbooks questions x go don't see your book. Do the exrcises on question words and click on the button to check your answers (before doing the exercises you may want to read the lesson on question words).
Asnt questions and answers book: ultrasonic method (ut), third edition this updated ultrasonic (ut) q & a book is a good resource to review and prepare for testing situations. The c answer book: solutions to the exercises in 'the c programming language,' second edition 2nd (though the questions it answers don't really ask for it), . Answer: yes, it's legal — and very useful a try statement does not have to have a catch block if it has a finally block if the code in the try statement has multiple exit points and no associated catch clauses, the code in the finally block is executed no matter how the try block is exited. Questions : worksheets pdf, handouts to print, printable resources wh- questions. Strategies for answering reading questions the reading assessment test challenges you to read short passages and answer questions that require you to:.
Wh questions exercises with answers pdf wh questions and answers in eglish interrogatives in englishquestion word exercise 1 question formation exercises pdf. Online exercise that looks at and explains the english vocabulary and phrases used in a presentation for both answering and dealing professionally with questions from the audience. These free sample gre math problems and gre practice questions come with answers and in-depth explanations to help you gre math questions & practice by chris . Exercise on short answers :: learn english online - free exercises, explanations, games, teaching materials and plenty of information on english language :: page default. 48 questions and answers about 'exercise and fitness' in our 'hobbies' category did you know these fun facts and interesting bits of information.
Be verb (am , is, are) yes - no questions and answers choose the correct answer show all questions yes no questions - answers exercise to be verb exercise. How do you improve your creative problem solving skills review the lateral thinking questions in this guide and refer to the end of the article for answers. Webmd experts and contributors provide answers to: exercise questions.
Practice parajumble questions with detailed solution to each problem verbal test questions and answers intro to appreciate the effort lofoyacom is . English exercises auxiliary verbs exercises short answers english exercises presents our new interactive self-correcting worksheets questions and short answers.
The study quizzes test your knowledge of the important concepts in each chapter and provide an explanation for each answer. Simply answer all of the questions in the quiz and press submit to see your score and select the types of exercises you want to show: beginner intermediate. Solutions to biostatistics practice problems a is the correct answer this question is asking about the shape of the patients were given exercise tests at . View questions and answers from the matlab central community find detailed answers to questions about coding, structures, functions, applications and libraries.
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CC-MAIN-2018-43
| 4,100 | 7 |
https://beatameinguer.com/qa/why-is-curl-a-cross-product.html
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math
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- What is curl and divergence?
- What is curl of a vector?
- What is curl divergence and gradient?
- Is cross product sin or cos?
- Why does the cross product work?
- What is the cross product of three vectors?
- What is the cross product of a vector with itself?
- Why is cross product a sin?
- What does a cross product represent?
- Is a cross b the same as B Cross A?
- Why the divergence of a curl is zero?
- How do you take curls?
- What is the characteristic of a curl free field?
- What does curl signify?
- Why is cross product Anticommutative?
- How do you know if curl is positive or negative?
What is curl and divergence?
Roughly speaking, divergence measures the tendency of the fluid to collect or disperse at a point, and curl measures the tendency of the fluid to swirl around the point.
Divergence is a scalar, that is, a single number, while curl is itself a vector..
What is curl of a vector?
The curl of a vector field captures the idea of how a fluid may rotate. Imagine that the below vector field F represents fluid flow. The vector field indicates that the fluid is circulating around a central axis.
What is curl divergence and gradient?
We can say that the gradient operation turns a scalar field into a vector field. … We can say that the divergence operation turns a vector field into a scalar field. The Curl is what you get when you “cross” Del with a vector field. Curl( ) = Note that the result of the curl is a vector field.
Is cross product sin or cos?
That’s why we use cos theta for dot product and sin theta for cross product. Cosine is used to make both the vectors point in same direction.
Why does the cross product work?
If a vector is perpendicular to a basis of a plane, then it is perpendicular to that entire plane. So, the cross product of two (linearly independent) vectors, since it is orthogonal to each, is orthogonal to the plane which they span.
What is the cross product of three vectors?
If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the parallelepiped defined by them would be flat and have no volume. This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product.
What is the cross product of a vector with itself?
Finally, the cross product of any vector with itself is the zero vector (a×a=0). In particular, the cross product of any standard unit vector with itself is the zero vector.
Why is cross product a sin?
Because sin is used in x product which gives an area of a parallelogram that is made up of two vectors which becomes lengrh of a new vwctor that is their product. In dot product cos is used because the two vectors have product value of zero when perpendicular, i.e. cos of anangle between them is equal to zero.
What does a cross product represent?
The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
Is a cross b the same as B Cross A?
vector product of two vectors is not commutative that is A cross B not equal to B cross A . In this case magnitudes are equal but directions are opposite. It can be A cross B = -B cross A.
Why the divergence of a curl is zero?
1 ∇⋅(∇×F)=0. In words, this says that the divergence of the curl is zero. … That is, the curl of a gradient is the zero vector. Recalling that gradients are conservative vector fields, this says that the curl of a conservative vector field is the zero vector.
How do you take curls?
Suppose that →F is the velocity field of a flowing fluid. Then curl→F curl F → represents the tendency of particles at the point (x,y,z) ( x , y , z ) to rotate about the axis that points in the direction of curl→F curl F → . If curl→F=→0 curl F → = 0 → then the fluid is called irrotational.
What is the characteristic of a curl free field?
This counter clockwise rotation exactly cancels the clockwise one from the top and bottom paddle and the net result is that the paddle wheel does not rotate. This vector field is curl-free although it clearly has a non zero circulation.
What does curl signify?
In vector calculus, the curl is a vector operator that describes the infinitesimal circulation of a vector field in three-dimensional Euclidean space. The curl at a point in the field is represented by a vector whose length and direction denote the magnitude and axis of the maximum circulation.
Why is cross product Anticommutative?
The anticommutative property of the cross product demonstrates that and differ only by a sign. These vectors have the same magnitude but point in opposite directions. … The direction of the cross product is given by the right-hand rule.
How do you know if curl is positive or negative?
This rotation means that the component of the curl in the z direction is positive (using the right hand rule). If the sphere were rotating clockwise when viewed from the positive z-axis, then the component of the curl in the z direction would be negative.
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CC-MAIN-2021-31
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https://en.m.wikiquote.org/wiki/Grigori_Perelman
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math
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Grigori Yakovlevich Perelman (born 13 June 1966) is a Russian mathematician who is known for his contributions to the fields of geometric analysis, Riemannian geometry, and geometric topology. In 1994, he proved the soul conjecture in Riemannian geometry, which had been an open problem for the previous 20 years. In 2002 and 2003, he developed new techniques in the analysis of Ricci flow, thereby providing a detailed sketch of a proof of the Poincaré conjecture and Thurston's geometrization conjecture, the former of which had been a famous open problem in mathematics for the past century.
|This article about a mathematician is a stub. You can help Wikiquote by expanding it.|
- If the proof is correct then no other recognition is needed.
- David S. Richeson (8 March 2012). Euler's Gem: The Polyhedron Formula and the Birth of Topology. Princeton University Press. p. 285. ISBN 1-4008-3856-8.
Quotes about Grigori Perelman edit
- By the end of 2006 it was generally believed that Perelman’s proof was correct. That year, the journal Science named Perelman’s proof the “Breakthrough of the Year.” Like Smale and Freedman before him, the forty-year old Perelman was tapped to be a Fields Medals recipient for his contributions to the Poincaré conjecture (in fact, Thurston also received a Fields Medal for his work that indirectly led to the final proof). The countdown for the $1 million prize had begun (some wonder if Perelman and Hamilton will be offered the prize jointly).
- David S. Richeson (8 March 2012). Euler's Gem: The Polyhedron Formula and the Birth of Topology. Princeton University Press. p. 284. ISBN 1-4008-3856-8.
- Revolutions in mathematics are quiet affairs. No clashing armies and no guns. Brief news stories far from the front page. Unprepossessing. Just like the raw damp Monday afternoon of April 7, 2003, in Cambridge, Massachusetts. Young and old crowded the lecture theater at the Massachusetts Institute of Technology (MIT). They sat on the floor and in the aisles, and stood at the back. The speaker, Russian mathematician Grigory Perelman, wore a rumpled dark suit and sneakers, and paced while he was introduced.
- Donal O'Shea (30 October 2008). The Poincaré Conjecture: In Search of the Shape of the Universe. Penguin Books Limited. p. 12. ISBN 978-0-14-190034-6.
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| 2,316 | 9 |
http://orgasm.oo.gd/videos/hell-on-heels-jenna-v62299-3645.html
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math
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Jenna`s Back To Fan The Flames!!! She came. She saw. She conquered. She came again!!! Jenna`s back!!! That`s right...Jenna Jameson is back in HELL ON HEELS to fan the flames, and make the Adult film industry even HOTTER!!! Jenna looks like an angel but is as hot as a little devil in this Lesbian Lovefest to end them all, as Jenna uses all of her feminine wiles and sexual talents to seduce and make love to some of the hottest women in porn!!! And Jenna makes them even hotter with her unbelievable erotic presence!!! Leave it to Wicked to give the All Girl film a shot in the...um...ARM!!!
Delivery restrictions may apply.
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CC-MAIN-2021-43
| 637 | 2 |
https://ems.press/journals/pm/articles/1686
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math
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The paper deals with the problem of existence and calculation of solutions to Lax equations that define finite-dimensional integrable systems. The method presented in the paper is based on Wiener–Hopf factorization and related Riemann–Hilbert problems on Riemann surfaces. The idea behind the method was first proposed by Semenov–Tian–Shansky but, to the authors' knowledge, is here applied, for the first time, in an infinite dimensional setting. The method dealt with in the paper enables one to analyse the global existence of solutions which seems more difficult by other methods. An example of a dynamical system associated with an elliptic curve is completely worked out in the paper.
Cite this article
M. Cristina Cámara, António F. dos Santos, Pedro F. dos Santos, Lax equations, factorization and Riemann–Hilbert problems. Port. Math. 64 (2007), no. 4, pp. 509–533DOI 10.4171/PM/1793
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CC-MAIN-2024-18
| 906 | 3 |
https://iimcat.blogspot.com/2010/11/questions-on-number-theory-remainder.html
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math
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Here are a few questions from remainders in Number Theory
1. What are the last two digits of the number 7 45 ?
2. What is the remainder when we divide 390 + 590 by 34?
3. N2 leaves a remainder of 1 when divided by 24. What are the possible remainders we can get if we divide N by 12?
Labels: CAT number theory, CAT questions, Remainders
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CC-MAIN-2019-09
| 336 | 5 |
https://www.physicsforums.com/threads/block-off-inclined-plane-kinematics-projectile-motion.858592/
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math
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1. The problem statement, all variables and given/known data A block of mass m = 2.00 kg is released from rest at h = 0.300 m above the surface of a table, at the top of a θ = 50.0° incline as shown in the figure below. The frictionless incline is fixed on a table of height H = 2.00 m. a. Find a on the incline a=7.51 :) b. Find V(f) on incline V(f)=2.42 :) c. How far from the table will the block hit the floor? :( d. What time interval elapses between when the block is released and when it hits the floor? :( 2. Relevant equations incline kinematics y=V(o)t-g/2(t^2) projectile stuff t=(2V(o)sin∅)/g range(x)=(V(o)^2(sin2∅))/g 3. The attempt at a solution It's free fall kinematics for the plane then projectile for when it falls? I was distracted when he was showing us this c. range(x)=(2.42^2)sin(100)/g d. -0.300=0-4.9t^2 t(p)=0.247 t(f)=2(2.42)sin(50)/g t(f)=0.378 t(tot)=0.247+0.378=0.625 sec I already missed c but is that how i get there? Part D wrong.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814393.4/warc/CC-MAIN-20180223035527-20180223055527-00007.warc.gz
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CC-MAIN-2018-09
| 971 | 1 |
https://www.physicsforums.com/threads/pressure-inside-champagne-bottle.35892/
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math
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The pressure inside a champagne bottle is 5.4 atm higher than the air pressure outside. The neck of the bottle has an inner radius of 0.9cm. What is the frictional force on the cork due to the neck of the bottle? I used the equation F=A*P, but I'm getting 137.41N and that's not right. I think I'm doing something wrong with the air pressure outside. How do I set this problem up?
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827175.38/warc/CC-MAIN-20181216003916-20181216025916-00382.warc.gz
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CC-MAIN-2018-51
| 380 | 1 |
https://www.worldofiptv.com/threads/how-to-edit-perfect-player-settings-with-apk-editor.9723/
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math
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L levan79 Active Member Member Joined Nov 10, 2019 Messages 112 Likes 361 Points 74 Sep 24, 2020 #1 Hello, i need some help to edit the Settings on Perfect Player.. Default 16:9, Autostart on Boot. Hardware Decoder, Volume with right and left Button. Can somebody help me? Thanks in advance.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00036.warc.gz
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CC-MAIN-2020-45
| 291 | 1 |
https://community.tes.com/threads/ks3-optional-sats-tests.420448/
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math
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opps...sorry... Hi there, I was wondering if anyone still uses the KS3 Optional SATS test? If so anyone know where to download them from? I know most schools will probably start GCSE's in year 9 but would you consider using these tests for year 8 and 7? I found them to be really useful but a little tedious to mark! Any thoughts?
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s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00370.warc.gz
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CC-MAIN-2019-43
| 330 | 1 |
https://www.cazoommaths.com/maths-worksheet/special-sequences-worksheet/
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math
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This worksheet takes a look at various types of sequences: triangle and square numbers, Fibonacci style and quadratic.
Each section focuses on a different style of sequence. Students will look at visual representations of triangle and square numbers. Nth terms are used to generate sequences, find given terms in sequences and decide whether numbers are part of these sequences.
In section C, students will identify and continue numeric and algebraic Fibonacci-style sequences, as well as generate their own.
Then in section D, learners use the nth terms of quadratic sequences to find the first five terms.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00772.warc.gz
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CC-MAIN-2024-18
| 607 | 4 |
https://gradestack.com/Value-of-sin-1-5-/19996-wqsq
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math
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Value of sin^-1 (5 radians)
That radians wat u have written ain't d angle
It is d value of sinx which cannot exceed 1
Then sinx equals 5 thats not defines
Wbo r u
sin^-1 (5) = x.
5 = sin x
There are no real solutions to this.
I guess u had to ask "value of sin 5" which is -0.958....
If u mean what u asked, still you can find a complex number.
Hint: use Euler's formula e^ix = cosx + isinx. And solve for sinx and see further. If u need help u can ask freely.
Sin inverse (2pie-5)
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s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423927.54/warc/CC-MAIN-20170722082709-20170722102709-00558.warc.gz
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CC-MAIN-2017-30
| 481 | 12 |
https://www.convertseek.com/mixed-fraction-converter/
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math
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Are you searching for Mixed Fraction Converter? By using our below available official links ( which are always up to date), you can find Mixed Fraction Converter without any difficulty.
Follow these steps:
1.Find the correct Mixed Fraction Converter link from the list of answers below.
2.Make sure you are connected to the correct website, find the converter you need.
3.Follow the prompts to upload the files you need to convert. After the conversion is complete, you will see the conversion success message, and you can download your files.
4.Downloader You can download converters for Windows, Android, Iphone, etc. according to your needs.
LAST UPDATED: 20 Apr, 2022 238 PEOPLE USED
Answer By: Elbert Gregory
A mixed number is a combination of a whole number and a fraction. This unique tool will simplify your mixed number to its lowest form. Please enter your whole number on the left and the fraction on the right then press "Simplify Mixed Number" to simplify it:
Answer By: Sherrie Good
Mixed to Improper Calculator. Use this calculator to convert your improper fraction to a mixed fraction. An improper fraction is a fraction whose nominator is greater than its denominator. For example, 5 4. A mixed fraction is a fraction of the form c n d, where c is an integer and n < d. For example, 11 4 = 2 3 4.
Answer By: Darlene Johnson
Mixed fractions are also called mixed numbers. A mixed fraction is a whole number and a proper fraction combined, i.e. one and three-quarters. The calculator evaluates the expression or solves the equation with step-by-step calculation progress information.
Answer By: Hennum Thomas
Step 1) We make the mixed numbers into improper fractions. Step 2) We add, subtract, multiply, or divide those improper fractions together from Step 1. Step 3) We simplify the fraction result from Step 2 if necessary. Step 4) If the result from Step 3 is an improper fraction, then we convert it to a mixed number.
Answer By: James Tully
An improper fraction is a fraction that has a numerator greater than or equal to the denominator and can not be simplified further. For example, 13/5 is an improper fraction. Let us learn how to convert this improper fraction to a mixed fraction.
Answer By: Sandra Tardiff
The procedure to use the Mixed Fraction to Improper Fraction calculator is as follows: Step 1: Enter the mixed fraction in the input field. Step 2: Now click the button “Solve” to get the improper fraction. Step 3: Finally, the conversion of mixed fraction into an improper fraction will be displayed in the output field.
Answer By: Marie Ross
Mixed numbers subtraction calculator will give the mixed numbers difference between the first number and the second number. Input: Two mixed numbers; Output: A fraction in the simplest form or mixed number or decimal number. Conversion of Mixed Number to Improper Fraction Rule: The mixed number A a b A a b for a,b > 0 a, b > 0 can be rewritten
Answer By: Maria Okky
Step 1: Convert the given mixed fraction to an improper fraction. Step 2: Find the LCM of the denominator values. Step 3: Multiply the numerator and denominator values by a number such that the fractions should have the LCM value as the new denominator. Step 4: Now, keep the denominator common and subtract the numerator values.
Answer By: James Humphries
As you can see above, to convert any mixed number to a fraction, we just need to add the integer part to the fraction part. See below a shorter way to convert 1 5 8 to an improper fraction. Step 1: Multiply the whole number part (1) by the denominator (8). 1 × 8 = 8. Step 2: Add the product from Step 1 (8) to the numerator (5).
Answer By: Teena Carr
Mixed Fraction Calculator - add, subtract, multiply and divide fractions with step by step instructions. Mixed number fraction calculator to simplify any two fractions or numbers to the lowest terms or as mixed numbers in reduced form. The online mixed fraction calculator supports adding, subtracting, multiplying or dividing negative fractions.
Answer By: Eldora Body
The procedure to use the improper fraction to mixed number calculator is as follows: Step 1: Enter the improper fraction in the input field. Step 2: Now click the button “Solve” to get the answer. Step 3: Finally, the conversion of improper fraction into a mixed number will be displayed in the output field.
Answer By: William Salmons
Mixed Fraction Calculator is a free online tool that displays the addition, subtraction, multiplication and division of two mixed fractions. BYJU’S online mixed fraction calculator tool makes the calculation faster, and it displays the result in a fraction of seconds.
Answer By: Holly Payne
Mixed Fraction Calculator. When dealing with mixed number, mixed fraction calculator is your number one helper. Adding and subtracting mixed number is confusing. Multiplying and dividing it is hard. Unless you got the gift of scientific mind, you need calculator to make sure you will get the correct computation during the mathematical operation
111 Gz converter
152 Wmv converter
216 Zip converter
114 Xls converter
293 Word converter
247 Energy converter
264 Amr converter
257 Html converter
241 Wav compressor
172 Azw3 converter
171 Wma converter
194 Webm converter
268 Alac converter
213 Psd converter
276 Bmp converter
123 Video compressor
177 Aac converter
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CC-MAIN-2023-50
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https://www.mathworks.com/matlabcentral/answers/358527-widrow-hoff-delta-rule-method-with-linear-layer-adaline?s_tid=prof_contriblnk
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math
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This question is closed. Reopen it to edit or answer.
Widrow-Hoff delta rule method with linear layer (adaline)
1 view (last 30 days)
i want to use learnwb which Widrow-Hoff delta rule (also known as least mean square (LMS) algorithm) and updatesits weights and bias when a training sample is presented. this is my code x=401*113 y=1*113 net = linearlayer(0,0.001); net = configure(net,x,y); net.trainFcn = 'trainb'; net.trainParam.epochs = 788; net=train(net,x,y); outputs=net(x); e=output-y ee=e.^2 eee=sum(ee) rmse=sqrt(mean(eee)); zz= postreg (outputs , y) but my error is root mean square error is very high ? could you please give my some suggestion ? is it my training method correct ?
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http://www.chegg.com/homework-help/precalculus-mathematics-5th-edition-chapter-3-problem-23sat-solution-9780131120952
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math
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We need to find all the real and imaginary roots of
To find all the real roots of the given equation polynomial we first find all its possible roots. Now, all the possible factors of are and all the possible factors of 1 are.
For the given function, we have
The only factors of leading coefficient 1 are, thus the possible denominators of a rational root ofcan be. Hence the possible rational roots must all be factors of 9 namely.
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https://aaai.org/papers/113-learning-metrics-via-discriminant-kernels-and-multidimensional-scaling-toward-expected-euclidean-representation/
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math
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Distance-based methods in machine learning and pattern recognition have to rely on a metric distance between points in the input space. Instead of specifying a metric a priori, we seek to learn the metric from data via kernel methods and multidimensional scaling (MDS) techniques. Under the classification setting, we define discriminant kernels on the joint space of input and output spaces and present a specific family of discriminant kernels. This family of discriminant kernels is attractive because the induced metrics are Euclidean and Fisher separable, and MDS techniques can be used to find the lowdimensional Euclidean representations (also called feature vectors) of the induced metrics. Since the feature vectors incorporate information from both input points and their corresponding labels and they enjoy Fisher separability, they are appropriate to be used in distance-based classifiers.
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http://www.dailyranger.com/ranger_headlines.php?page=32&category=5
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math
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Oct 2, 2013 - By Steven R. Peck: There haven't been many in RHS boys sports, so celebrate the golfers
Oct 1, 2013 - By Doyle McManus: NEW YORK -- How long does it take for a charm offensive to wear thin?
Oct 1, 2013 - By Steven R. Peck: September snow
Sep 28, 2013 - By Randy Tucker: There is an unspoken belief among many rural people concerning locked doors, locked vehicles and locked gates. For generations, farmers and ranchers have ... [more]
Sep 28, 2013 - By Mark Shields: I write and speak for a living, so I bemoan the continuing collapse of grammar.
Sep 28, 2013 - By Steven R. Peck: It is unjust -- but legal -- to require the county to pay ACLU's legal fees from the voting rights suit
Sep 27, 2013 - By Travis Becker, Fremont County commissioner, Riverton: Editor:
Sep 27, 2013 - By Steven R. Peck: On an airplane trip back east not long ago, the mode of transport on the final leg from Philadelphia to New Haven, Conn., was the turboprop airplane called the ... [more]
Sep 27, 2013 - By Steven R. Peck: Wyoming's coal industry doesn't like them, but it is in position to respond positively
Sep 26, 2013 - By The Chicago Tribune: Hospital nurseries last year were filled with Jacobs and Jaydens, Madisons and Mias.
Sep 26, 2013 - By Steven R. Peck: Central Wyoming College made a wise move in promoting art in the new center on campus
Sep 26, 2013 - By Clair McFarland: Few topics distract me from the joy I carry around with me these days -- in my hormone-high brain, my tapioca-spoiled tummy, and my overstuffed womb. It's hard ... [more]
Sep 25, 2013 - By Chris Peck: And let's not ask college trustees to be art critics
Sep 25, 2013 - By Steven R. Peck: Even as investigations continue, some parties ought to cover costs for the Pavillion water truck
Sep 24, 2013 - By Doyle McManus: President Obama and his aides were surprised this month by the strength of public opposition to their call for military action against Syria. They shouldn't ... [more]
Sep 24, 2013 - By Steven R. Peck: Too soon?
Sep 20, 2013 - Nancy Eckstein, Riverton: Put beer tax
Sep 20, 2013 - Steve Peck, Publisher: Congress continues to act in ways that lower public expectations
Sep 19, 2013 - Steve Peck, Publisher: CWC's Health and Science Center ranks among our county's greatest achievements
Sep 19, 2013 - Betty Starks Case: That goes for scenery, air and another's needs
Sep 19, 2013 - Carl Rollins, Riverton: It would be a very big mistake to start putting bars on the Wind River Indian Reservation again.
Sep 18, 2013 - By Chris Peck, Associated Editor: I still have a soft spot for my 1963 Travelall -- at 5 mpg
Sep 18, 2013 - By Tammy Questa, Riverton: Editor:
Sep 18, 2013 - Steve Peck, Publisher: False positives on sightings of escapee are reasonable price to pay for vigilance
Sep 17, 2013 - By Belle Bransky, Riverton: Editor:
Sep 17, 2013 - By Steven R. Peck: September in the rain
Sep 15, 2013 - By Randy Tucker: As the wise man said, it swings not between right and wrong, but between sense and nonsense.
Sep 15, 2013 - By Steven R. Peck: Now that the decision has been made, address the primary concern of those who feared it
Sep 13, 2013 - By Chris Peck: But issues raised in Oklahoma are worth considering everywhere
Sep 13, 2013 - By The Miami Herald: The chances of striking a good deal over Syria's lethal chemical stockpile with the likes of Vladimir Putin and Bashar Assad lie somewhere between remote and ... [more]
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https://www.jiskha.com/display.cgi?id=1295308586
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math
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posted by Maria .
The solubility of NaCl in water is 35.7 g NaCl/100 g H2O. Suppose that you have 500.0 g of NaCl. What is the minimum volume of water you would need to dissolve it all? The density of water is 1.0 g/mL.
100 grams H2O x (500/37.5) = about 1330 mL. Or look at it this way.
37.5 g NaCl/100 mL = 0.375 g NaCl/1 mL.
So (0.375 g NaCl/mL) x ??mL = 500 g
Solve for ??mL.
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http://bwessaydxgp.fieldbee.us/an-introduction-to-the-importance-of-fractal-geometry.html
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math
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What is the importance of geometry a: fractal designs also have an influence on fields including art, cryptography, biology and artificial intelligence. Buy fractal cities: a geometry of form and function on amazon and thought-provoking introduction to fractal geometry also find important methods and. , and have become an important new tool for african fractals introduces readers to fractal geometry and explores introduction to fractal geometry 2. Introduction to fractals: the geometry of nature written by dave didur july 17, 2014 - classical geometry, which traces its origins back to euclid, is concerned with. Introduction have you ever awed what exactly is fractals and chaos and why is it so important in biology by the lungs having fractal geometry. This blog post has provided a brief introduction to fractal geometry on “ what are fractals and why should i care is of integral importance.
Books, referred to, reviewed, buy: the fractal geometry of nature - introduction. An introduction to fractal geometry lots of images - minimal maths free subscription to our newsletter just enter your fractals are fun what is a fractal. What are fractuals when do you use them in the fractals, it is important to know first what their its shape cannot be defined by euclidean geometry. An introduction to the mandelbrot set fractal geometry of nature which was published in 1982 [3 it is possible to show, and this an important result, that if jjz.
What are fractals by tom watch frax in as described by mandelbrot in his introduction to the fractal geometry of nature fractal geometry is an extension to. Unsolved problems and still-emerging concepts in fractal and advancements of fractal geometry 1 introduction and still-emerging concepts in fractal geometry. Introduction to fractals and the fractal dimensions the development of fractal geometry has been one of the 20 it is important to remember that this correlation.
The importance of geometry in the construction industry essay - according to this is answered by fractal geometry, the word fractal coming from the concept of. Fractal geometry : an introduction fractal, mandelbrot introduction geometry is the branch of euclidean geometry served as an important tool in solving. New insights gained from the application of fractal geometry to spatial and temporal scale are important when application of fractal geometry has given.
So what are fractals and why are they important accessibility links of diseases are just three cases in which fractal geometry can describe. Chaos theory and fractal geometry have begun to appear as an important issue in secondary school mathematics chaos theory is the qualitative study of unstable. An introduction to dimension theory and fractal geometry: fractal dimensions and measures why is the study of dimension important or useful. —mandelbrot, in his introduction to the fractal geometry of nature he is one of the most important figures of the last fifty years.
Fractal world 1 introduction to fractal a very important application of fractal is to reproduce the natural images such as clouds, trees, mountains and etc. The importance of beings fractal fractal geometry fractals are best described by means of a dimension which is fractional. Naval nps52-86-008 postgraduate school monterey i an introduction to fractal geometry considered important because of nis synthesis of the.
Importance of fractal geometry the world of mathematics usually tends to be thought of as abstract complex and imaginary numbers, real numbers, logarithms, functions. Discuss some of each of fractal geometry and dynamics, and show how they interact 2 introduction to fractal geometry example of a fractal that is important in. Fractal geometry essay examples 3,351 words 7 pages an introduction to the importance of fractal geometry a description of a fractal. Math 498p: introduction to fractal geometry and dynamical systems3 and calculation of the number of periodic orbits of a given pe-riod n calculation of the hausdor.
Introduction the word fractal often has different connotations for laypeople than for mathematicians, where the layperson is more likely to be familiar with. Authenticating pollock paintings using fractal geometry emphasize the importance of since its introduction in the 1970s, fractal geometry has experienced. One of the most important the fractal geometry trading method: an introduction my trading method also incorporates fractal geometry in. This presentation provides a broad and basic introduction to the subject of fractal geometry my thanks to michael frame at yale university for the use o.
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https://www.scienceforums.net/profile/141985-0ut0ft0uch/content/
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math
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Everything posted by 0ut0ft0uch
Thank you for your reply @StringJunky. Could it therefore be inferred that there would be no long-term effects of ethanol evaporating by the thousands of gallons? Thank you once again for your help.
Hello, I'm new to this forum and I have a question which I have not yet found an answer to elsewhere online. My question is this: if thousands of gallons of ethanol cleaner (pure nondenatured ethanol) are used to sanitize a large public facility (such as a sports stadium), then would there be any effects for the atmosphere once the ethanol evaporated? Any input or references would be very much appreciated. Thank you.
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| 651 | 3 |
https://anhydrase-signal.com/for-a-systematic
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math
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For a systematic investigation, the Au film thickness (thickness) was carefully varied while fixing the other growth parameters. As clearly shown in the cross-sectional line profile in Selleck Avapritinib Figure 1(b-1), the surface was atomically smooth even after the Au deposition. The surface morphologies by a systematic annealing process are shown with AZD5582 clinical trial 2 nm thickness in Figure 1c and 9 nm thicknesses in Figure 1d. Under an identical growth condition, the self-assembled Au droplets showed significant
distinction in the size and density distribution depending on the thickness. Figure 2 shows the detailed evolution process of the self-assembled Au droplets on GaAs (111)A with the thickness variation between 2 and 20 nm. AFM top views of 3 × 3 μm2 are shown in Figure 2a,b,c,d,e,f,g,h, and those of 1 × 1 μm2 are shown in Figure 2(a-1) to (h-1). Figure 3a,b,c,d,e,f,g,h shows the cross-sectional
surface line profiles acquired from the 1 × 1-μm2 AFM images in Figure 2(a-1) to (h-1) indicated with white lines. FFT power spectra are shown in Figure 3(a-1) to (h-1). Figure 4 summarizes the average height (AH), average density (AD), and lateral diameter (LD) of the self-assembled PI3K Inhibitor Library concentration Au droplets on GaAs (111)A compared to the various thicknesses. The root mean squared (RMS) roughness (R q) values of samples are summarized in Figure 4d. In general, the average size including height and diameter of the self-assembled Au droplets on GaAs (111)A was gradually increased with the increased thicknesses as clearly shown in the AFM images in Figure 2 and the surface line profiles in Figure 3 as well as the summary plots in Figure 4a,c. Meanwhile, the density of Au droplets was gradually decreased as clearly seen in Figures 2 and 4b. For example, with 2 nm Au deposition, the very densely packed dome-shaped Au droplets were formed on GaAs (111)A as presented in Figure 2a and (a-1) with the AD of 4.23 × 1010 cm−2. The corresponding
AH was 23 nm and the LD was 52.5 nm as shown in Figure 4a,c. At 2.5 nm thickness, the size of droplets grew larger and the density was reduced as clearly shown in Figure 2b and (b-1): the AH was increased by × 1.4 to 32.3 nm and the LD increased by × 1.8 to 94.4 nm as shown BCKDHB in Figure 4a,c. On the other hand, as shown in Figure 4b, the AD decreased by × 3.41 to 1.24 × 1010 cm−2. With relatively lower coverage of 2 and 2.5 nm thicknesses, the Au droplets were quite round and uniformly distributed over the surface, as shown in the AFM images of Figure 2a,b. With 3 nm thickness, the Au droplets were also quite uniformly distributed over the surface and began to show a slight elongation as shown in the AFM images in Figure 2c. Similarly, with the further increase of thicknesses between 4 and 20 nm, the continuous decrease in density with the associated increase in size was clearly observed as shown in Figures 2,3,4. Overall, the size of Au droplets was increased by × 4.
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CC-MAIN-2020-34
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http://shewatchestv.tumblr.com/tagged/judd-apatow
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math
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Tune in for the best of what I'm watching each week.
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I’m gonna put on my party dress and my sorry face and I’m gonna get him back
from Girls Season 1: Episode 6 - The Return
11 months agoshort quote2 notes #girls #marnie #lena dunham #hbo original series #judd apatow #television quotes
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s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706624988/warc/CC-MAIN-20130516121704-00043-ip-10-60-113-184.ec2.internal.warc.gz
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https://simplyfrugalliving.com/discounts/how-do-you-find-the-discount-rate-of-a-stock.html
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math
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How to calculate discount rate. There are two primary discount rate formulas – the weighted average cost of capital (WACC) and adjusted present value (APV). The WACC discount formula is: WACC = E/V x Ce + D/V x Cd x (1-T), and the APV discount formula is: APV = NPV + PV of the impact of financing.
How do you find the discount rate?
To calculate the percentage discount between two prices, follow these steps: Subtract the post-discount price from the pre-discount price. Divide this new number by the pre-discount price. Multiply the resultant number by 100.
What is the discount rate on the stock?
In this context of DCF analysis, the discount rate refers to the interest rate used to determine the present value. For example, $100 invested today in a savings scheme that offers a 10% interest rate will grow to $110.
What is a good discount rate to use for NPV?
It’s the rate of return that the investors expect or the cost of borrowing money. If shareholders expect a 12% return, that is the discount rate the company will use to calculate NPV.
What is a good discount rate?
Usually within 6-12%. For investors, the cost of capital is a discount rate to value a business. Don’t forget margin of safety. A high discount rate is not a margin of safety.
What discount rate does Warren Buffett use?
Warren Buffett’s 3% Discount Rate Margin.
Is a high or low discount rate better?
Higher discount rates result in lower present values. This is because the higher discount rate indicates that money will grow more rapidly over time due to the highest rate of earning. Suppose two different projects will result in a $10,000 cash inflow in one year, but one project is riskier than the other.
What is today’s discount rate?
Federal discount rate
|This week||Month ago|
|Federal Discount Rate||0.25||0.25|
Why is NPV better than IRR?
The advantage to using the NPV method over IRR using the example above is that NPV can handle multiple discount rates without any problems. Each year’s cash flow can be discounted separately from the others making NPV the better method.
How do you find discount rate for PV?
Discount Rate Formula
- Discount Rate Formula (Table of Contents)
- Let us take a simple example where a future cash flow of $3,000 is to be received after 5 years. …
- Discount Rate = (Future Cash Flow / Present Value) 1/ n – 1.
Is discount rate the same as interest rate?
A discount rate is an interest rate. The term “interest rate” is used when referring to a present value of money and its future growth. The term “discount rate” is used when looking at an amount of money to be received in the future and calculating its present value.
What does higher discount rate mean?
In general, a higher the discount means that there is a greater the level of risk associated with an investment and its future cash flows. Discounting is the primary factor used in pricing a stream of tomorrow’s cash flows.
What is a personal discount rate?
The rate at which individuals trade current for future dollars, or personal discount rate, is a provocative subject with important implications for many aspects of economic behavior and public policy.
What is discount formula?
The basic way to calculate a discount is to multiply the original price by the decimal form of the given percentage rate. To calculate the sale price of any item, we need to subtract the discount from the original price.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00022.warc.gz
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https://www.giancolianswers.com/giancoli-physics-7th-edition-solutions/chapter-1/problem-11
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math
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This is Giancoli Answers with Mr. Dychko. We are going to estimate uncertainty by calculating what the maximum possible area would be of this measurement within its uncertainty and then the minimum area value and then check the difference between the two, divide by 2 to get this spread or the absolute uncertainty in our area measurement and then go from there. So this is the straightforward calculation to begin with is πr squared to calculate the area of the circle and we get 3.01907 times 10 to the 9 centimeter squared. And then we are not actually given what the uncertainty is and so we just have to make an assumption at least it's going to be the very minimum plus or minus 1 of the most precise digit and that's the tenths place, you know, in this number here— it's actually the ten thousandths place when you expand it with times 10 to the 4 but never mind— it's just this digit here is going to be plus or minus 1 and so the minimum possible area with an uncertainty would be π times 3.0 times 10 to the 4 centimeters and then we get this number. And then the area could be as high as π times 3.2 times 10 to the 4 centimeter squared and you get that and then the difference of the spread—the ΔA— is gonna be this maximum minus the minimum and then take half of that total difference to get 1.9473 times 10 to the 8 centimeter squared which can just be 2; you only have one significant figure in any error estimate because it is just an estimate there's no precision in how we came up with this number, it was just an assumption that we used to get it in the first place and at most, you can have two significant figures in a error estimate but in this case, we have no basis other than just our regular assumptions here that this is plus or minus 1 and so yeah one significant figure is all we are entitled to here. And then we are gonna actually change this to 2 or 0.2 times 10 to the 9 that just makes things a little cleaner in our final answer because our final answer is times 10 to the 9, it's gonna be 3.0 times 10 to the 9 and it's nice to have this 3.0 in brackets with the uncertainty of 0.2 and in order to put the uncertainty within the brackets, we have to have it also be multiplied by 10 to the 9 and so we'll and so if we are gonna have this multiplied by 10 in order to make it 10 to the 8 times 10 makes 10 to the 9, we'll have to divide this thing by 10 as well so that in fact we are not changing the quantity and then 2 divided by 10 is how we come up with the 0.2. So there we go. 3.0 times plus or minus 0.2 all times 10 to the 9 centimeter squared is our area measurement.
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http://scitation.aip.org/content/aip/journal/rsi/84/8/10.1063/1.4818913?showFTTab=true&containerItemId=content/aip/journal/rsi
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math
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Sketch of the Faraday imaging system and a resulting image (right) at T = 1.5 μK and N = 106 atoms. The polarization of the light is indicated (top left) by displaying cross sections of the imaging beam obtained from a rotation around the dashed-dotted lines.
Characterization of DFFI. (a) Faraday coefficient c F and photon scattering rate as a function of the detuning Δ. (b) Absorption image calibration factor α depending on the number of measurements with (dots) and without (triangles) Faraday scaling. (c) and (d) Temperature and atom number obtained from DFFI compared to results from absorption imaging. Arrows in the figures indicate the appropriate axes. In (a), (c), and (d), the data have been binned and the shaded regions indicate the standard error for each bin; in (b), the shaded region is the standard error of the measured value of α (see text).
Single-run magnetometry in an optical lattice. DFFI signal as a function of the applied magnetic field along the z axis. (Open circles) Magnetic field sweep over 0.93 G at |B r | = 0.053 G. (Full dots) Magnetic field sweep over 6.2 G at |B r | = 1.03 G. The inset shows the sensitivity of the offset field extraction for the two realizations as a function of number of included data points (centered around the signal minimum). The sensitivity is estimated as the error of the fit times the square root of time taken to record the included data points.
Monitoring of spatial dynamics. (a) Non-destructive measurement of the cloud position during a damped oscillation. (b) Non-destructive measurement of the cloud position during a decompression of the magnetic trap. The cloud position and oscillation frequency are shown within three time intervals during the decompression.
Signal-to-noise ratio in the measured phase shift. (a) SNRθ/Π for common dispersive imaging techniques (DFSI blue solid; PCI green dashed; DFFI red dotted-dashed; DPFI black dotted) as a function of the respective phase shift. (b) Relative SNRθ ratio for DFFI and DPFI vs. Faraday rotation angle for two values of cube suppression .
Far-detuned vector to scalar phase shift ratio |θF/θS| for hydrogen-like atoms with nuclear spin I.
Signal properties of common dispersive imaging methods.
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https://cadcamengineering.net/unfolding-in-visi-progress/
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math
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Basically a sheet metal part is a formed part by drawings and bends. There are two kinds of bend: linear and not linear.
In both cases the unfolding process is based on neutral fibre, and it can be set by user for linear unfolding or fixed (0.5 of thickness) for the blank/flange unfolding.
Actually, in case of no-linear bend unfolding, we must use the flange unfolding functions based on (FTI) engine. In case of linear bends, is possible to apply in the same time: linear flange and flange unfolding, but the advice is to prefer linear flange, because is based on analytic algorithm and thus more precise.
The linear unfolding algorithm is used on:
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https://rhettallain.com/2008/10/04/what-is-energy/
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math
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I think it is time for me to talk about energy. My ultimate goal is to give some insight into the many stories about perpetual motion. To do this, I will first talk about the fundamentals of energy.
**What is Energy**
I started thinking about this, and at first I realized that I did not have a good, short explanation of energy. The most commonly used definition in science text books is:
*Energy: the ability to do work (or something dreadfully vague like this).*
But what is work? It may be no surprise to find that many college level physics texts avoid defining energy. After some serious contemplation, I think I have this energy figured out.
**There are only two types of energy**
I don’t need a general definition of energy, since there are only two types I can just describe those two. ALL energy is either:
- Particle Energy: Energy of particles (obviously). I was originally going to just say kinetic energy (energy of things that move) but I forgot about mass (of course you remember E=mc2). This is sort of complicated, so I can perhaps summarize it by saying a particle can have energy because of its mass and because of its motion (really this is just one thing). So, particle energy can be an electron moving, a water molecule moving, or a car (a car is a collection of atoms that are mostly moving in the same direction). For the rotational kinetic energy of the Earth, this is really the same thing. Imagine all the pieces of the Earth (atoms) they are moving and thus have kinetic energy. The idea of rotational kinetic energy is to simplify the calculation. Instead of summing the kinetic energy of each of the atoms of the Earth, one can use the radius, mass, and the angular velocity of the Earth to do the same thing. But realize this is mostly just a short cut.
- Field Energy: Energy in the fields associated with the fundamental forces – gravity, electric, magnetic, strong nuclear and weak nuclear. Suppose I hold a ball above the Earth, it has particle energy (because of its mass) and there is also energy in the gravitational field associated with the ball and Earth. A chemical battery has energy stored in the electric field due to the configuration of atoms. A final example of energy in fields would be the energy from electromagnetic radiation.
But wait! What about ….. What about …. (insert some energy). All these other energies you read about are one of the above two. Other energies (for example thermal energy) are short cuts. They allow us to deal with large collections of particles without having to calculate ALL the particle energies and the field energies.
**Conservation of Energy**
There have been many many experiments in the history of science. In all of these experiments, the total energy of the situation as been conserved. Well, this is to say that there has not been an experiment where clearly the total energy before something happened was different than the total energy after something happened. Most experiments don’t look at this “energy accounting” directly. Energy conservation isn’t the law, its just what we see. How about a couple of examples of everyday things and I explain where all the energy is?
**Example: A cup of hot tea sitting on a table**
First, where is all the energy in this hot cup of tea? The cup and the tea both have particle energy. The particles (carbon and stuff) have mass energy. If I somehow annihilated this cup and tea it would turn all this mass into field energy. In this case that energy would be in the form of electromagnetic radiation. In fact, this would be so much energy in electromagnetic radiation that it would create pairs of particles (matter and antimatter pairs).
The particles also have energy because of their motion. If we assume the cup is stationary, the particles in the cup are still moving. The hotter something is, the more they move. For the particles that make up the cup, these particles are essentially just vibrating and staying in the same general area. For the tea, the particles are moving around and mostly staying in the cup (but some are leaving at the surface through evaporation). This energy is generally called thermal energy.
The cup also has energy in fields. There is energy associated with the gravitational field of the Earth-Cup(and tea) system. This would be called gravitational potential energy. There is also energy associated with the electric field is the interactions between the electrons and protons in the atoms of both the tea and the cup. People usually call this chemical energy, you could see this energy change forms if you burned the cup or had some other chemical reaction.
As the cup is sitting in the room, it gets cooler. That corresponds to lower particle energies. Where does the energy go? In this case, the stuff surrounding the cup gains energy. The table gets a little warmer (particle energy) and so does the air. This energy transfer takes place by the higher energy particles of the cup and tea interacting (through the electric field) with the particles of the air and the table. You might ask, why is it that the table gains energy and the cup loses energy? Couldn’t it happen the other way and energy would still be conserved? Yes, it would. But the probability of this happening (remember that there are on the order of 1025 particles in this cup) is so near to zero that you have a much greater chance of winning the lottery.
What if the cup were in outer space with nothing touching it? It would still cool (unless the sun was shinning on it). The particles in the cup still radiate electromagnetic energy (usually in the Infra Red region). This IR radiation could causes something else to increase in energy, but the cup still loses energy. The tea would all evaporate and lose energy to IR radiation.
I didn’t think it would be possible to take a simple thing and make it so boring, but I did it. I know that was painful (and likely in some places technically wrong) but it was necessary. Don’t make me do it again. Hopefully, you have an idea of conservation of energy and of the fundamental ideas of energy.
Main entry continued
One thought on “What is Energy?”
Perpetual motion machines
Time is homogeneous; Noether’s theorem; mass-energy is locally conserved.
Noether’s theorem demands any continuous symmetry is coupled to a conserved observable, and vice-versa. External symmetries couple to translation and rotation, internal symmetries do not. Diddle a non-Noetherian external symmetry to get in physics’ face.
There is one non-Neotherian external symmetry: parity – opposite chirality along all directions. (Charge conjugation is an internal symmetry, so no antimatter anomalies.) Quantitative mass distribution parity divergence is ab initio calculated on a scale of CHI = zero (achiral) to one (perfectly parity divergent) in mathematician Michel Petitjean’s QCM software. Geometrically right- or left-handed quartz single crystals in enantiomorphic space groups P3(1)21 or P3(2)21 respectively are CHI = 1 atomic mass distributions. Chirality emergent scale is a 0.304 nm diameter sphere.
Do centimeter diameter solid single crystal spheres (no direction bias!) of space groups P3(1)21 and P3(2)21 quartz vacuum free fall identically? There is no compelling reason why they should. A parity Eötvös experiment is the relevant observation. Somebody should look for symmetry breaking.
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http://answers.northerntool.com/answers/0394/product/12470/emp-tractor-draw-bar-stabilizer-for-category-0-tractors-model-7350-0-questions-answers/questions.htm?sort=affiliationaStaff
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Please explain the difference between category 0 and the category 1 draw bar stabilizer?
there are two numbers on the draw bar stabilizer
12470 category 0 and 12471 category 1
what is the difference between the o and 1?
asked 1 year, 1 month ago
on "EMP Tractor Draw Bar Stabilizer for Category 0 Tractors, Model# 7350-0"
1out of 1found this question helpful.
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http://www.education.com/reference/article/monomials-binomials-polynomials/
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math
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Algebra and Probability Study Guide 2 for McGraw-Hill's ASVAB
Practice questions for this study guide can be found at:
Monomials, Binomials, and Polynomials
You can guess by the prefix mono- that a monomial has something to do with "one." A monomial is a mathematical expression consisting of only one term. Examples include 12x, 3a2, and 9abc.
- A binomial (the prefix bi- means "two") has exactly two terms: 12z + j.
- A polynomial, as indicated by the prefix poly-, meaning "many," has two or more terms.
Examples include x + y, x + y + z, and y2 – 2z + 12.
Multiplying Monomials When multiplying monomials, multiply any numbers, then multiply unknowns. Add any exponents. Keep in mind that in a term like x or 2x, the x is understood to have the exponent 1 even though the 1 is not shown.
Dividing Monomials To divide monomials, divide the numbers and subtract any exponents (the exponent of the divisor from the exponent of the number being divided).
Adding and Subtracting Polynomials Arrange the expressions in columns with like terms in the same column. Add or subtract like terms.
Multiplying Polynomials To multiply polynomials, multiply each term in the first polynomial by each term in the second polynomial. The process is just like regular multiplication. For example, if you multiply 43 times 12, the problem looks like this:
Dividing a Polynomial by a Monomial Just divide the monomial into each term of the polynomial.
Dividing a Polynomial by a Polynomial To divide a polynomial by another polynomial, first make sure the terms in each polynomial are in descending order (i.e., cube → square → first power).
For example, 6c + 3c2 + 9 should be written 3c2 + 6c + 9.
10 + 2c + 5c2 should be written 5c2 + 2c + 10.
Then use long division to solve the problem.
Factoring a Polynomial A factor is a number that is multiplied to get a product. Factoring a mathematical expression is the process of finding out which numbers, when multiplied together, produce the expression.
To factor a polynomial, follow these two steps:
- Find the largest common monomial in the polynomial. This is the first factor.
- Divide the polynomial by that monomial. The result will be the second factor.
Special Case: Factoring the Difference between Two Squares
Factor the following expression:
y2 – 100
In this expression, each term is a perfect square; that is, each one has a real-number square root. The square root of y2 is y, and the square root of 100 is 10.
When an expression is the difference between two squares, its factors are the sum of the squares (y +10) and the difference of the squares (y – 10). Multiplying the plus sign and the minus sign in the factors gives the minus sign in the original expression.
Factoring Polynomials in the Form ax2 + em>bx + c, where a, b, and c are numbers Remember that you want to find two factors that when multiplied together produce the original expression.
Factor the expression:
- x2 + 5x + 6
First, you know that x times x will give x2, so it is likely that each factor is going to start with x.
- (x ) (x )
Now you need to find two factors of 6 that when added together give the middle term of 5. Some options are 1 and 6 and 2 and 3. 2 and 3 add to 5, so add those numbers to your factors. Now you have
- (x 2) (x 3)
Finally, deal with the sign. Since the original expression is all positive, both signs in the factors must be positive. So the two factors must be
- (x + 2)(x + 3)
Check your work by multiplying the two factors to see if you come up with the original expression.
- (x + 2)(x + 3) = x2 + 5x + 6
Factor the expression:
- 6x2 + 8x = 8
6x2 can be factored into either (6x)(x) or (2x)(3x). Using the latter, the first terms in our factors are as follows:
- (2x ) (3x )
Now let's consider the –8. Factors of 8 can be (8)(1) or (2)(4). Let's try 2 and 4, so our factors are now:
- (2x 2)(3x 4)
Now for the signs. In order to get a minus 8 in the original expression, one of the numbers must be a negative and the other a positive. Let's try making the 4 negative, making the factors:
- (2x + 2)(3x – 4) = 6x2 – 2x – 8
That's close, but the original expression was 6x2 + 8x – 8, not 6x2 – 2x – 8. What if we switched the numbers 2 and 4?
- (2x + 4)(3x – 2) = 6x2 + 8x – 8
Now the factors give the original expression when multiplied together, so this is the correct answer.
- Coats and Car Seats: A Lethal Combination?
- Kindergarten Sight Words List
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- 10 Fun Activities for Children with Autism
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development
- The Homework Debate
- First Grade Sight Words List
- Social Cognitive Theory
- GED Math Practice Test 1
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http://www.stat.ucla.edu/~sabatti/statarray/textr/node4.html
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math
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How one measures the outcome of the experiment depends mainly on the available technology. However, there is often room for choices and it is important to keep an eye on the way in which data will be analyzed to optimize the results. In general, the way in which measurements are taken determines the nature of the error: because statistical analysis is all about recovering the true signal from the error, this is obviously very important.
Measuring the outcome of a microarray experiment is a complex matter; one step of the process involves transforming a two-dimensional picture in an array of numbers. The image segmentation technique used here, for example, can bias the results in different directions, so that it is important to discuss it with the statistician. It is important to correct for background intensities and the appropriate background values vary across the array. Attention should be paid that the background measurements are not excessively noisy: a robust and stable method for background estimation is needed (presumably using a smoothing procedure).
If the measurements are taken as ratio, it is often convenient to take the Log of these numbers making their distribution symmetric, so that a departure from the average in each direction has the same meaning. Excessively high or low expression values should be probably considered as outliers. Dye-renormalization is often needed to take into account differential incorporating power of the dyes. As this interact with the sensitivity of the scanner, the difference in luminosity between the dyes has been noted to follow a non-linear pattern. So that normalization should be non linear. Having a quadrant-dependent normalization is also a good idea. Care should also be taken in deciding which genes to use for normalization.
References The following groups have analyzed in detail the imaging, background correction and normalization of the signals.
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https://www.physicsforums.com/threads/tight-binding-model-help-please.369872/
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math
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1. The problem statement, all variables and given/known data In the tight binding model, (E=Eo + alpha + (2gamma coska)), how are parameters alpha and gamma affected if the lattice spacing, a, is increased? I know the equations for gamma and alpha but can't type them in here because they involve integrals and I don't know how to put those in on this forum. Thanks in advance! Sorry, didn't realise how the forums worked, so posted in this area instead. Its taking far too long to figure out how to post the equations in latex on here so i apologise but this isnt technically homework help. My final year solid state physics 2 exam is tomorrow and this wasn't in our notes but is in the revision questions handed out so please don't delete this one!
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https://stereoelectronics.org/webSC/SC307.html
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math
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Applying (R,S) descriptors to allenes
1,3-Disubstituted allenes have fixed perpendicular dissymmetric planes, another type of structure that is naturally 'handed.' If the two groups in one of the planes (i.e. at one end) of the allene are the same, then the other plane of the allene will be a plane of symmetry and the molecule will be the same as its mirror image. However, if neither of the two planes of an allene is a plane of symmetry (i.e. neither end has two identical groups), then the mirror images are non-superimposable and the structure is chiral. One of the simplest examples is 2,3-pentadiene, which is shown in the interactive model below.
The procedure for assigning (R,S) stereochemical descriptors to allenes is similar to that for biaryls. The molecule is viewed along its axis and the relevant substituents are projected on to a plane at right angles to the axis. The groups are prioritised using the Sequence Rules with the extra proximity rule, which requires that near groups precede far groups.
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http://www.typophile.com/node/46724
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math
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I am having an issue with TTF fonts in PCs being displayed too small in MS Word and Bistream's font manager . (see attach)
In the beginning I thought I've made a mistake in the TrueType Metrics, but after reading in the FL Forum "Tips and Tricks: Setting Font Family Metrics in FontLab" carefully I realized this was not the case.
The fonts has 1000 units UPM and only if I set it with 2048 they are displayed correctly.
I thought 2048 units were not mandatory for TTF fonts but apparently this is the case here.
Or am I doing something else wrong?
Have anyone had the same problem with TTF set with 1000 units? And which was the solution?
Thanks in advance.
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https://www.transtutors.com/homework-help/statistics/geometric-mean/geometric-mean.aspx
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math
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What is Geometric Mea N?
Geometric mean is the nth root of the product of n items of a series. Thus if the geometric mean of 3, 6 and 8 is to be calculated it would be equal to the cube root of the product of these figures. Similarly the geometric mean of 8, 9, 12 and 16 would be the 4th root of the product of these four figures.
Symbolically GM =√(n&x1..x2..x3………xn )
where GM stands for the geometric mean, N for the number of items and X for the values of the variable.
The calculation of the geometric mean by this process is possible only if the number of items is very few. If the number of items is large and their size is big, this method is more or less out of question. In such cases calculations have to be done with the help of logs. In terms of logs.
Thus geometric mean is the anti-log of the arithmetic average of the logs, of the values of a variable. It is also possible to assume a log mean and to find out the deviations from it and then calculate the geometric mean. It should be noted that the value of the geometric mean is always less than the value of the arithmetic average unless all the items have equal value in which case the geometric mean and arithmetic average have identical values.
Email Based, Online Homework Assignment Help in Geometric Mean
Transtutors is the best place to get answers to all your doubts regardinggeometric mean. Transtutors has a vast panel of experienced statistics tutorswho can explain the different concepts to you effectively. You can submit your school, college or university level homework or assignment to us and we will make sure that you get the answers related to geometric mean.
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https://www.jiskha.com/display.cgi?id=1290472414
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posted by Carl .
1.) A lawnmower engine claims to provide 18 hp of power and 6.25 ft lb of torque. If this is true, what must be the rotational speed in RPM?
2.) The conversion of the shaft power of a wind turbine to electricity is the function of the generator. Assume it has an efficiency of 0.45. If the wind turns the shaft of a generator at 26.2 rad/s (about 250 RPM) and the generator produces 300 W of electricity, what torque is produced by the wind on the shaft?
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https://lwgportlandharbor.org/how-to-convert-moles-to-ml
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math
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MOLES FROM VOLUME OF PURE LIQUID OR SOLID Multiply the volume by the density to get the mass. Divide the mass by the molar mass to get the number of moles.
What is the formula for calculating moles?
The unit is denoted by mol.
The formula for the number of moles formula is expressed as.
Number of moles formula is.
Number of moles = Mass of substance / Mass of one mole.
Number of moles = 95 / 86.94.
What is the formula for moles?
Avogadro's number is a very important relationship to remember: 1 mole = 6.022×1023 6.022 × 10 23 atoms, molecules, protons, etc. To convert from moles to atoms, multiply the molar amount by Avogadro's number. To convert from atoms to moles, divide the atom amount by Avogadro's number (or multiply by its reciprocal).
How do you calculate moles to grams?
One mole consists of Avogadro number of atoms. If you know the quantity of mole, it can be converted into grams and vice versa. You have three steps to convert mole values to grams.
Calculate how many moles are mentioned in the question.
Find the molar mass of the substance.
Multiply both the values.
What is a mole equal to?
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12 (12 g C = 1 mol C atoms = 6.022 × 1023 C atoms). Here some of the language of the current SI definition is incorporated.
How many moles is NaOH?
The molar mass of the compound NaOH is 40 g/mol.
How many moles are in 4 grams of NaOH?
How many moles are in 5g of NaOH?
Moles = mass / molar mass So, 0.125 moles are present in 5g of NaOH.
How many moles are in 100g of NaOH?
The answer is 39.99711. We assume you are converting between grams NaOH and mole.
How many moles are in 2 grams of NaOH?
Solution : Number of moles `= ("Mass ")/( "Molecular mass ")` ` (2)/(40 )=0.05.
How many moles are in 40 grams of NaOH?
Thus 40 grams of NaOH equals one mole of NaOH, and a 1 molar solution of NaOH will contain 40 grams of NaOH.
How many grams are there in 1.00 mole of NaOH?
39.997 g The molar mass tells you the mass of exactly one mole of a given substance. In this case, sodium hydroxide has a molar mass of 39.997 g mol−1 , which implies that one mole of sodium hydroxide has a mass of 39.997 g .
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http://mathcentral.uregina.ca/QQ/database/QQ.02.06/spencer1.html
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math
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To answer this, you must break it into two parts: the cost of digging and the cost of finishing. The digging cost is based on cubic yards, which is a volume measurement. How many cubic yards of dirt do you need dug? If you convert your measurements in feet to yards, then multiply them together, you will get the number of cubic yards. Multiply that by the cost per cubic yard and you have your digging cost.
The finishing cost is easier: you don't care about the depth of the cellar and the cost is in square feet - an area measurement. Multiply the width and length of the cellar (in feet) together to get the area in square feet. Then multiply that by the finishing rate.
When you add the two parts together, you get your final quote.
Stephen La Rocque>
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https://brainly.com/question/317517
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math
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Slope is equal to the change in y over the change in x, or 'rise over run'.
m=change in y/change in x
The change in x is equal to the difference in x-coordinates (also called run), and the change in y is equal to the difference in y-coordinates (also called rise).
Substitute in the values of x and y into the equation to find the slope.
m= 1/2 −(− 3/2 )/ 3/5 −( 2/10 )
The slope is 5
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https://www.coursehero.com/tutors-problems/Finance/19520619-ABC-Company-sells-3755-chairs-a-year-at-an-average-price-per-chair-of/
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math
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ABC Company sells 3,755 chairs a year at an average price per chair of $191. The carrying cost per unit is $20.71.
The company orders 543 chairs at a time and has a fixed order cost of $127 per order. The chairs are sold out before they are restocked. How many orders will company place if it follows the economic order quantity model?
Enter your answer rounded off to two decimal points. Do not enter comma in the answer box. For example, if your answer is 12.345 then enter as 12.35 in the answer box.
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https://aas.org/archives/BAAS/v25n4/aas183/abs/S4205.html
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We present a sample of Lick Image Dissector Scanner (Lick-IDS) nuclear spectral indices of elliptical galaxies. Galaxies which show noticeable [OIII] emission have been culled from the sample, as the presence of [OIII] may indicate filling-in of H$\beta$ stellar absorption by emission. The final, culled sample contains 87 ellipticals, of which 61 are galaxies not in the emission-culled sample of Gonzalez (1993). Of these, a sub-sample of 23 Virgo-cluster ellipticals over range of seven absolute magnitudes is identified, a six-fold increase over Gonzalez's (1993) study. The Lick-IDS metallicity indicators Mg and Fe plus the key H$\beta$ age index were measured. These were combined with model predictions of Worthey (1993) for co-eval generations of stars to estimate ages and metallicities of the stellar populations in each elliptical galaxy. In both the full sample of 87 galaxies and the Virgo sub-sample, ellipticals cover a wide range of ages, from $\sim 3\ Gyr$ to greater than $15\ Gyr$. Furthermore, weak-lined galaxies (Mgb$ < 4$ \AA) are statistically younger and more metal-poor than strong-lined galaxies, confirming Gonzalez's (1993) results from a smaller sample. This is also true of the Virgo ellipticals, although this sub-sample is small. Comparing galaxy ages with various structural parameters, we find no trend in absolute magnitude with age, but low-velocity-dispersion ellipticals are statically younger than those with high velocity-dispersions. Some of these ellipticals are young enough that their stellar populations will look quite different even at moderate lookback times.
Thursday program listing
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https://www.physicsforums.com/threads/vectors-adding-subtracting-and-velocity-questions.372660/
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So I have a few problems that I need help with. Any help would be greatly appreciated as I am new here and stumped. 1. A woman walks 220 m in the direction 27 degrees east of north, then 205 m directly east. (a) find the magnitude (finished it). 362.5 m (b) Find the angle of her final displacement from the starting point. ??? (c) Find the distance she walks (finished it). 425 2. Suppose a-b=2c, a+b=9c, and c=5i+3j. (***NOTE: all "a","b", and "c" have directional arrows above them I just didn't know how to put them there. The i and j have ^ above them as well). (a). What is a? ??? (b). What is b? 5. At one instant a bicyclist is 35 m due east of a park's flagpole, going due south with a speed of 11 m/s. Then, 12 s later, the cyclist is 50 m due north of the flagpole, going due east with a speed of 12 m/s. For the cyclist in this 12 s interval, find each of the following. (a) displacement (solved) magnitude = 61 m direction = 55 degrees north of west (b) average velocity (solved) magnitude =5.08 m/s direction = 55 degrees north of west (c) average acceleration magnitude = ??? m/s^2 direction = ??? degrees north of east 6. A ball rolls horizontally off the edge of a tabletop that is 2.00 m high. It strikes the floor at a point 1.57 m horizontally away from the table edge. (Neglect air resistance.) (a) How long was the ball in the air? ??? s (b) What was its speed at the instant it left the table. ??? I am sorry there are so many, I got the majority of them but was very confused on these. At first I thought I new how to do 6 but then I realized that I didn't have the instant velocity in the x direction. All help is greatly appreciated. Thanks again.
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https://nike-huarache.com/book-254
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We'll provide some tips to help you select the best Linear equation calculator two variables for your needs.
Is very friendly and easy to learn, it help with any math problems I have trouble with. The best app ever, it's really helpful for maths solutions and well explaination.
To find the derivative of say sinhx, the app rather calculates the 'h' separately. Efficient, just need to say amazing, amazing, amazing, you also need to try it, it will be the perfect app if it can solve other areas of math.
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Linear Equations in Two Variables Calculator is a free online tool that displays the value of the variables for the given linear equation. BYJU’S online linear equations in two variables
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CC-MAIN-2023-06
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https://community.zemax.com/got-a-question-7/index4.html
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math
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Got a question?
Can't find the answer you need? Ask your peers!
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Hello, I am having issues with ZOS CAD export function in Ansys ZOS 2023 R2.02 where an exported STEP file, when imported into Solidworks (2023), lacks some of the bodies and rays. Has anyone encountered this problem before? I tried to export the same design file to a STEP file from Ansys ZOS 2022 R2.02 and the imported STEP file contained no missing part. So, perhaps it is related to the newer Ansys ZOS version? Any help would be appreciated.Thanks,Hanshin
Tolerancing of achromatic doublet results in 2 element front surface moving in front of first element back surface
Hello Zemax Community!I have been looking at the tolerancing for an achromatic doublet design, and had a question regarding the positioning of the elements post-tolerancing. In some cases, the front surface of the 2nd element of the lens will move in front of the back surface of the second element of the lens. Is there a way to control thicknesses such that the doublet surfaces are physically in sequence? The first image below highlights the position of the back surface of the first lens element, and the second image highlights the position of the front surface of the second lens element.Additionally, before tolerancing the lens was modelled with 4 surfaces: Any ideas/help would be appreciated.Thank you,Musab
Hello,For a silicon photodiode detector, there’s a given Directivity plot that describes the amplitude change in amount of light that the detector sees at a specific angle relative to the normal axis.I would like to know how to apply this Directivity for a Detector Rectangle object in Non-Sequential Component mode. The results of the simulation are meant to show how much light the photodiode fails to detect due to non-collimated light.I’ve attached ZAR of a simple source gaussian non-collimated light onto a detector rectangle.
We use zemax to simulate a galvo scanner geometry in mixed mode NSC. Almost all seqential analyses work in mixed mode. However the paraxial magnification gets messed up. I am pretty certain this is due to the beam beeing turned in the nsc component. In the depicted example an object point (0,1) gets imaged to (-1,0). Apperently this can be corrected by setting the “Exit Tilt Z” for the non sequential component to a value that object points on the x-axis are imaged to image points on the x-axis.However the documentation says “The radial magnification, being the ratio of paraxial image height to object height”. My impression is, that zemax internally uses vector components instead of “heights”.
I don’t understand some results generated by the Power Pupil Map analysis feature. Perhaps someone here in the forum can provide insight.For a general lens system, consider the wavefront leaving the exit pupil. This wavefront is typically not spherical because of aberrations, or perhaps because the lens is a freeform optic. In either event, the min and max values of the local curvature at any point on the wavefront in the pupil determine the min and max focusing power associated with that pupil location. Equivalently, we can look at the max and min local focal lengths, which are just the reciprocals of the powers. The Power Pupil Map is a nice tool to help the user visual this effect. Several options are available for the form of the data. Also, a surface must be selected. Consider a simple test case using a plano-convex lens that focuses an on-axis collimated beam. Spherical aberration causes the exit pupil wavefront to have local curvature that varies over the pupil. Presuma
Dear ZEMAX Staff and fellow Zemaxers!My question might be very silly/simple, but I have limited experience in non-sequential mode. I noticed this issue for few days, and I have tried all the tricks that I could think of. I guess it's now a good time to check with the experienced users. Maybe my struggles are quite straightforward to you.I have this optical system that light from a filament source was collected by a condenser. After passing through the slit (I made using an Air gap nested in absorber). The beam will be reflected through prism mirror then travel upwards. Here is the issue: When I use the filament source even it emits randomly, the layout always captures/displays the beam that follows the optical axis of the system. However, when I changed the light source to a Cree LED that I created based on supplier datasheet. Even the beams are still random, but there is never a beam will follow the optical axis anymore. Please refer to the diagrams below... I understand maybe the bea
Hi, I set up the vortex phase metasurface and would like to generate a circular beam. I have the correct phase distribution at the metasurface position, but I am getting a very strange tracing image that is off center. The POP propagation shows a round bright spot rather than a ring. What am I doing wrong?Kind regards,Mery
I've been performing a tolerance analysis on some of my designs. In my tolerance results, I ran into a bug where the output is always the nominal value. I verified that the output should change by adjusting the values in the LDE. However, when I run the Monte Carlo or sensitivity analysis there is no change. Example: My TDE contains TTHI to change the distance from object to lens. My merit function contains operand RSRE to find the RMS radius of the image. Nominal is 8.092mm, when I change the thickness to the extreme of the tolerances in the LDE, the spot changes from 10.322mm to 5.773mm. So, I know that the thickness changes the spot. If I run the monte carlo, or sensitivity analysis with the criterion set to my merit function. The sensitivity says there is no change with TTHI, and the monte carlo only outputs 8.092mm (nominal). I recreated my file multiple times and sometimes operands work and sometimes they don't. Any ideas on what is happening?
Hi Experts, I want to check the OPD different of different configurations. But I found that I can only check the OPD within one configuration only. I just need the OPD different of the chief rays of different configuration.Any ideas of it will be appreciated. Thanks!
Hi I am curious if anybody in the community knows what this means, I’ve never seen it before…. After optimizing the mechanical semi-diamter and chip zones on this surface disappeared and an asterisk appeared next to the clear semi-diameter value. I’m not sure what’s going on here… I had a pickup solve for mechanical semi-diameter on the following surfaces and when this happened them seem to have automatically changed to picking up the clear semi-diameter. Thanks EDIT: The radii for the two adjacent surfaces (of the surface in question) in the image were getting very small, almost hyper-hemispherical, and when I manually changed them to a larger value the mechanical semi-diameter came back and the asterisk went away…. still curious what happened if anybody knows.
Hi,A friend recently asked me how to get highly aspheric “freeform” surfaces made, and specifically how to create jobfiles for various CNC machines given an OpticStudio prescription. I advised him to look at Diffsys, as I once knew the developer of the program (long since lost contact) and as it’s owned by Precitech I thought it was a no-brainer choice.He basically laughed at me (we’re good friends) so I downloaded the demo from the Precitech website and I see what he means. It’s very much a 1990s product still available for purchase.Does anyone here use it, and if so with what success? Is there a better program that I should be recommending?Mark
I’m facing a weird problem while exporting .igs files from both sequential and non-sequential designs.When importing into Solidworks 2017 all doublets have one of the two components transformed into a full sphere.Moreover, an empty part is imported togheter with the design assembly.I don’t know if its a matter of lack of compatibility with old Solidworks version, or just me having the sw configured in a bad way.
Hi zemaxers I am trying to do a line scan optical system image simulation. And the first thing I thought is multi-configuration. It makes different layouts to show as one visually, but I am wondering is it possible to show different image surfaces to show as one in sequential mode or different detectors as one in non-sequential mode? I checked the zemax help but did not find the solution. Could you kindly give me some advice on it ?Thanks Yang
Hello, I am setting up an optical system with a CAD object with unknown curvature and I want to determine where a baseline ray and a ray at some other field angle crosses. I would like to do this to generate a “best focus range” so I may determine the best position for a reflective flat surface. I can manually find these points, but is there a way to find these points using an operand or some other feature?Method:I first define a point where three Source Ray sources are positioned incident on the CAD object.I use one of the rays to be the baseline ray to find the focus.I then want to find where the baseline ray crosses with the other two rays as these will be two different points. Thanks for the help.
Hi ZemaxersI am learning to do an image simulation in non-sequential mode.but not sure whether I am right, could you give me some advice?what I did1) set the slide object after the source2) use detector rectangle to analysis3) ray tracing3) show the radiant intensity in detector viewerMy question1) is the step above all right?2) the result the detector viewer shows includes aberrations?Thank you in advance!
Is there a way to plot/save the complex part of the refraction index of a (multilayer) coating? I can find the complex refraction index for a single layer of a coating in the COATING.DAT, but not for multilayer coatings. Moreover, encrypted coating do not have details on their single layers, so I cannot reconstruct anyway the complex refraction index.
Hi, simple system: SM fiber, lens, SM fiber. I want to offset the receiving fiber by a certain amount using a coordinate break. Doing so, I expect the POPD beam profile to be 'off-center' by exactly this offset. However, it looks as if the beam profile is not considering the Coordinate break. Coupling is calculated correctly. Any solution for that? Best, Jörg
hello. I am a beginner zemax user.I am studying.Obtain a sample source file that demonstrates the laser going out as a line using a laser diode and Powell lens.I would like to know how to set up a laser diode.I would like to request a simple sample file with a structure similar to the attached image.thank you
I have the latest version of OpticStudio and am trying to run an example https://optics.ansys.com/hc/en-us/articles/18254409091987When I open the.zprj file from step 6 of the manual, I get the following error (Image attached). What is wrong with it? How I can fix it?
Hi all, I tried to figured out the difference between the chief ray definition by the classical books and the convention by ZEMAX. chief ray definition from Arizona Universitychief ray difinition on SPIE website chief ray convention from ZEMAX manual the chief ray, by SPIE and Arizona University, is defined to pass through the center of the aperture stop. whereas, is defined to pass through the center of the entrance pupil. the difference causes a lot of troubles using ZEMAX. Can anybody explain this to me ?
It is learned from zemaxmanual that FFT MTF is calculated from FFT PSF or FFT LSF, while FFT PSF/LSF is calculated from pupil data. As pupil sampling 2(64x64) is concerned, we can get 128x128 PSF data or 1x128 LSF data on image. When I deal these PSF/LSF data with FFT method directly, I find two problems: Only 64 MTF data with different frequency can be get. However, there are more MTF data in FFT MTF vs frequency curve. Does Zemax perfrom a curve fitting with PSF/LSF data before FFT or a curve fitting with MTF data after FFT?At the same frequency, MTF value get by directly FFT with FFT PSF/LSF data is always higher than MTF value get by FFT MTF curve. And with frequency rising, the gap will be larger. What’s the reason?
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http://cooking.stackexchange.com/questions/15969/milk-vs-home-made-kefir
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math
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I got a bit confused when looking up nutrition values for milk versus kefir made from that milk. For some reason many tables list higher values of some nutrients (e.g. potassium and magnesium) than is in the source form, i.e. the milk.
Can someone shed light on how this would happen?
I realize that lactose from the milk gets turned into carbon dioxide and alcohol, but where does it take the additional nutrients from if the only nutrition for the kefir grains is the milk and air?
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http://mathforum.org/library/drmath/view/55306.html
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math
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Origami Equilateral Triangle
Date: 04/26/2001 at 00:06:02 From: Oliver Subject: Folding Construction: Equilateral Triangle Hi, I have spent quite a long time trying to figure out how to create an equilateral triangle by folding; not using the sides of a piece of paper and using nothing but your hands and, of course, your brain. I have tried to make a hexagon from a piece of paper by folding, because that would give you six equilateral triangles, but I have had no luck in creating a hexagon from only folding. I would appreciate your help and I am thankful you took the time to read this. Oliver
Date: 04/26/2001 at 12:36:34 From: Doctor Peterson Subject: Re: Folding Construction: Equilateral Triangle Hi, Oliver. I'm not quite sure what you mean by "not using the sides"; does that mean the edge of the paper can't form one of the sides of the triangle, or that you can't assume the sides are parallel, or what? I think what I've done probably is allowed, regardless of the interpretation. Origami books teach a standard way to make a 60-degree angle by folding, which you can use to accomplish the task. If you have two parallel lines and a perpendicular to them, form a third parallel midway between the first two, and then fold the perpendicular AB at an angle so that point A lies on the midline (A'), while the fold passes through point B. Then the fold BC will be at the desired angle: | ---B----------------------------- |\ \ | \ \ | \ \ | \ \ | \ \ ---D------\--------------A'------ | \ / | \ / | \ / | \ / | \ / ---A-------------C--------------- | From the fact that A'B = AB, you can prove easily that ABA' is an equilateral triangle, so that angle ABC measures 30 degrees. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
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| 1,851 | 6 |
https://link.springer.com/chapter/10.1007/978-1-4419-1270-1_8
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math
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Categorical Dependent Variables
In this chapter, we consider statistical models to analyze variables where the numbering does not have any meaning and, in particular, where there is no relationship between one level of the variable and another level. In these cases, we are typically trying to establish whether it is possible to explain with other variables the level observed of the criterion variable. The chapter is divided in two parts. The first part presents discriminant analysis , which is a traditional method in multivariate statistical analysis. The second part introduces quantal choice statistical models. The models are described, as well as their estimation. Their measures of fit are also discussed.
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| 716 | 2 |
https://www.cakecentral.com/forum/t/609192/dolphin-cake
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math
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I have recd a query for a dolphin cake. Can someone please tell how much cake is required to make a dolphin cake. customers wants a 2 kg cake that is roughly around 4 pounds. is that sufficient to make a dolphin cake.
ALso does anyone have any instructions on how this is done. Please help
Couldn't you make 2 11 x 15 cakes stack them then carve a dolphin? Just a thought!
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http://apk4free.mobi/apps/engineering-math-formulas-free-download-for-android
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math
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Engineering math formulas for Android
Engineering math formulas app for android description: Mumbai university engineering math formulas.
All chapters are sorted according to semesters.
Ready reference for chapterwise formulas.
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The app is a complete free handbook of automobile engineering which covers important topics notes materials news blogs on the course. Download the app as a..
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The app is a complete free handbook of control systems engineering which covers important topics notes materials news blogs on the course. Download the app..
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| 2,157 | 24 |
https://usupdates.net/know-your-mutual-fund-sip-calculator/
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math
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A Mutual fund SIP calculator is a tool which helps you to Calculate the SIP return at the end of a specific term for a specific amount which is invested periodically. It also enables you to calculate how much money you have to invest for a particular time so that you can achieve the objective of your financial goal.
How does a SIP calculator work?
SIP Plans or returns are calculated based on compound interest. You have to enter the amount you want to invest every month, choose several years you want to continue the investment and the calculator will automatically calculate the amount of return.
On these frameworks, an investor calculates his SIP investment.
- Investment amount per month
Investment amount per month is the amount that will be invested every month. It is a fixed amount. For example, if a sip is of rupees interval 5,000 per month is fixed, then the mutual fund SIP calculator will assume that every month at the amount of rupees 5000 will be invested in the given fund.
- Total number of payments
The total number of payments means the term mutual fund sip; the total number of payments you are going to invest in a SIP.
For example, an investor wants to make 24 payments for a fixed amount for about 2 years.
- Expected Annual return
Annual returns mean the investors are expecting a return from their investment in the yearly basis. However, it is not possible to know the exact return, but just for a reference, we can use the past returns percentage. The fluctuation can be there in numbers only to see the total value in a different scenario.
What is Systematic Investment Plans (SIP)?
Systematic Investment Plans are the abbreviation of SIP. It is an investment strategy where an investor does the investment to build up their wealth over a long period. In this type of investment, a fixed amount is invested for over a particular period at a regular interval. SIP plans do not require a huge investment. Through a SIP, an investor can invest with just a small amount of ₹500 on a monthly, quarterly or annually basis on a Mutual fund d schemes.
What are the benefits of mutual fund SIP calculator?
SIP calculator helps investors make their payment for Mutual Fund Investment plans easily and quickly. SIP calculator is useful for both experienced and new investors. You need three variables to calculate the details of your SIP plans very quickly.
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https://www.physicsforums.com/threads/two-demensional-motion.375659/
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math
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1. The problem statement, all variables and given/known data a ball is thrown up to the top of a building 6 m tall. It is thrown at a 53 degree angle above the horizontal at a point 24m from the base of the building. The ball takes 2.20s to reach a point vertically above the wall. a) find the vertical distance by which the ball clears the wall. b) find the distance from the wall to the point on the roof where the ball lands 2. Relevant equations Kinematics and other trig equations. 3. The attempt at a solution I know that Vx is 18.1 m/s and Vy is 3.42 m/s. From there I am not sure where to go. I have tried several different things and I am not coming up with the answer. Please tell me how to go about this problem to see if I am headed in the right direction. Thanks.
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https://jwcn-eurasipjournals.springeropen.com/articles/10.1155/2010/189157
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math
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- Research Article
- Open Access
Resource Allocation in MU-OFDM Cognitive Radio Systems with Partial Channel State Information
EURASIP Journal on Wireless Communications and Networking volume 2010, Article number: 189157 (2010)
In wireless communications, the assumption that the transmitter has perfect channel state information (CSI) is often unreasonable, due to feedback delays, estimation errors, and quantization errors. In order to accurately assess system performance, a more careful analysis with imperfect CSI is needed. In this paper, the impact of partial CSI due to feedback delays in a multiuser Orthogonal Frequency Division Multiplexing (MU-OFDM) cognitive radio (CR) system is investigated. The effect of partial CSI on the bit error rate (BER) is analyzed. A relationship between the transmit power and the number of bits loaded on a subcarrier is derived which takes into account the target BER requirement. With this relationship, existing resource allocation schemes which are based on perfect CSI being available can be applied when only partial CSI is available. Simulation results are provided to illustrate how the system performance degrades with increasingly poor CSI.
In performance analyses of wireless communication systems, it is often assumed that perfect channel state information (CSI) is available at the transmitter. This assumption is often not valid due to channel estimation errors and/or feedback delays. To ensure that the system can satisfy target quality of service (QoS) requirements, a careful analysis which takes into account imperfect CSI is required .
Cognitive radio (CR) is a relatively new concept for improving the overall utilization of spectrum bands by allowing unlicensed secondary users (also referred to as CR users or CRUs) to access those frequency bands which are not currently being used by licensed primary users (PUs) in a given geographical area. In order to avoid causing unacceptable levels of interference to PUs, CRUs need to sense the radio environment and rapidly adapt their transmission parameter values [2–6].
Orthogonal frequency division multiplexing (OFDM) is a modulation scheme which is attractive for use in a CR system due to its flexibility in allocating resources among CRUs. The problem of optimal allocation of subcarriers, bits, and transmit powers among users in a multiuser-(MU-) OFDM system is a complex combinatorial optimization problem. In order to reduce the computational complexity, the problem is solved in two steps by many suboptimal algorithms [7–10]: determine the allocation of subcarriers to users and determine the allocation of bits and transmit powers to subcarriers. Resource allocation algorithms for MU-OFDM systems have been studied in [11–14]. These algorithms are designed for non-CR MU-OFDM systems in which there are no PUs.
In an MU-OFDM CR system, mutual interference between PUs and CRUs needs to be considered. The problem of optimal allocation of subcarriers, bits, and transmit powers among users in an MU-OFDM CR system is more complex. It is commonly assumed that perfect CSI is available at the transmitter [15, 16]. As noted earlier, this assumption is often not reasonable. In this paper, we investigate the problem of resource allocation in an MU-OFDM CR system when only partial CSI is available at the CR base station (CRBS). We assume that CSI is acquired perfectly at the CRUs and fed back to the CRBS with a delay of seconds. The channel experiences frequency-selective fading. The objective is to maximize the total bit rate while satisfying BER, transmit power, and mutual interference constraints.
The rest of the paper is organized as follows. The system model is described in Section 2. Based on the system model, a constrained multiuser resource allocation problem is formulated in Section 3. A suboptimal algorithm for solving the problem is discussed in Section 4. Simulation results are presented in Section 5 and the main findings are summarized in Section 6.
2. System Model
We consider the problem of allocating resources on the downlink of an MU-OFDM CR system with one base station (BS) serving one PU and CRUs. The basic system model is the same as that described in and is summarized here for the convenience of the reader.
The PU channel is Hz wide and the bandwidth of each OFDM subchannel is Hz. On either side of the PU channel, there are OFDM subchannels. The BS has only partial CSI and allocates subcarriers, transmit powers, and bits to the CRUs once every OFDM symbol period. The channel gain of each subcarrier is assumed to be constant during an OFDM symbol duration.
Suppose that is the transmit power allocated on subcarrier and is the channel gain of subcarrier from the BS to the PU. The resulting interference power spilling into the PU channel is given by
represents the interference factor for subcarrier , is the spectral distance between the center frequency of subcarrier and that of the PU channel, and denotes the normalized baseband power spectral density (PSD) of each subcarrier.
Let be the channel gain of subcarrier from the BS to CRU , and let be the baseband PSD of the PU signal. The interference power to CRU on subcarrier is given by
Let denote the transmit power allocated to CRU on subcarrier . For QAM modulation, an approximation for the BER on subcarrier of CRU is
where is the one-sided noise PSD and is given by (3). Rearranging (4), the maximum number of bits per OFDM symbol period that can be transmitted on this subcarrier is given by
where and denotes the floor function.
Equation (4) shows the relationship between the transmit power and the number of bits loaded on the subcarrier for a given BER requirement when perfect CSI is available at the transmitter. We now establish an analogous relationship when only partial CSI is available.
The imperfect CSI that is available to the BS is modeled as follows. We assume that perfect CSI is available at the receiver. The channel gain, hnk, for subcarrier n and CRU k is the outcome of an independent complex Gaussian random variable, that is, , corresponding to Rayleigh fading. For clarity, we will denote random variables and their outcomes by uppercase and lowercase letters, respectively.
For notational simplicity, we will use to denote an arbitrary channel gain. The BS receives the CSI after a feedback delay , where is the OFDM symbol duration. We assume that the noise on the feedback link is negligible. Suppose that is the channel gain information that is received at the BS, then . From , the correlation between and is given by
where the correlation coefficient, , is given by
In (6) and (7), denotes the zeroth-order Bessel function of the first kind, is the Doppler frequency, is the expectation operator, and denotes the complex conjugate of .
The minimum mean square error (MMSE) estimator of based on is given by
From (6), the actual channel gain can be written as follows:
where with .
3. Formulation of the Multiuser Resource Allocation Problem
Based on the partial CSI available at the BS, we wish to maximize the total CRU transmission rate while maintaining a target BER performance on each subcarrier and satisfying PU interference and total BS CRU transmit power constraints. Let denote the average BER on subcarrier , and let represent the prescribed target BER. The optimization problem can be expressed as follows:
where is the total power budget for all CRUs, is the maximum interference power that can be tolerated by the PU, and is a subcarrier assignment indicator, that is, if and only if subcarrier is allocated to CRU . The term represents the nominal bit rate weight (NBRW) for CRU , and
denotes the total bit rate achieved by CRU . Constraint (11) ensures that the average BER for each subcarrier is below the given BER target. Constraint (12) states that the total power allocated to all CRUs cannot exceed , while constraint (14) ensures that the interference power to the PU is maintained below an acceptable level . Constraint (15) results from the assumption that each subcarrier can be assigned to at most one CRU. Constraint (17) ensures that the bit rate achieved by a CRU satisfies a proportional fairness condition.
Based on (9), we calculate the average of the right-hand side (RHS) of (4), treating as an outcome of an independent complex Gaussian variable. For an arbitrary vector , we have the following:
where denotes the identity matrix. Applying (19) to (4), we obtain
where , , and denotes the channel gain that is fedback to the BS.
From (20), an explicit relationship between minimum transmit power and number of transmitted bits cannot be easily derived. However, since in (20) is a monotonically decreasing function of , we obtain the minimum power requirement while satisfying the constraint in (11) by setting .
We now derive a simpler, albeit approximate, relationship between the required transmit power, , and the number of loaded bits.
When setting , , , and , the RHS of (20) has the form
with . The function is Rician distributed with Rician factor . A Rician distribution with can be approximated by a Nakagami- distribution as follows:
with , where . Therefore, we approximate the RHS of (20) by
Then, from (23), we obtain
where . From (24), we obtain
4. Resource Allocation with Partial Csi
Note that the joint subcarrier, bit, and power allocation problem in (10)–(17) belongs to the mixed integer nonlinear programming (MINP) class . For brevity, we use the term "bit allocation" to denote both bit and power allocation. Since the optimization problem in (10)–(17) is generally computationally complex, we first use a suboptimal algorithm, which is based on a greedy approach, to solve the subcarrier allocation problem in Section 4.1. After subcarriers are allocated to CRUs, we apply a memetic algorithm (MA) to solve the bit allocation problem in Section 4.2.
4.1. Subcarrier Allocation
From (17), it can be seen that the subcarrier allocation depends not only on the channel gains, but also on the number of bits allocated to each subcarrier. Moreover, allocation of subcarriers close to the PU band should be avoided in order to reduce the interference power to the PU to a tolerable level. Therefore, we use a threshold scheme to select subcarriers for CRUs.
Suppose that subcarriers are available for allocating to CRUs. We assume equal transmit power for each subcarrier. Let
If a subcarrier is assigned to CRU , the maximum number of bits which can be loaded on the subcarrier is given by
Using (26)–(28), we can determine the number of subcarriers assigned to each CRU as follows. Let be the number of subcarriers allocated to CRU . Assuming that the same number of bits is loaded on every subcarrier assigned to a given CRU, the objective in (10) is equivalent to finding a set of subcarriers to maximize
where is the total transmit power allocated to all subcarriers and is the total interference power experienced by the PU due to CRU signals. The subcarrier allocation problem in (29)–(32) can be solved using the SA algorithm proposed in . Note that we need to make use of (24) in the SA algorithm if only partial CSI is available. A pseudocode listing for the SA algorithm is shown in Pseudocode 1. The algorithm has a relatively low computational complexity . After subcarriers are allocated to CRUs, we then determine the number, , of bits allocated to subcarrier .
4.2. Bit Allocation
Memetic algorithm (MAs) are evolutionary algorithms which have been shown to be more efficient than standard genetic algorithms (GAs) for many combinatorial optimization problems [25–27]. Using (24), the bit allocation problem can be solved using the MA algorithm proposed in . It should be noted that the chosen genetic operators and local search methods greatly influence the performance of MAs. The selection of these parameters for the given optimization problem is based on the results in . A pseudocode listing of the proposed memetic algorithm is shown in Pseudocode 2.
Let be the chromosome of member in a population, expressed as
where denotes the population size. A brief description of the MA algorithm in is now provided.
The function selects a set, , of chromosomes from in a roulette wheel fashion, that is, selection with replacement.
Crossover: suppose that . Let denote the crossover probability, and let , denote the outcome of an independent random variable which is uniformly distributed in , then is selected as a candidate for crossover if and only if . Suppose that we have such candidates, we then form disjoint pairs of candidates (parents).
For each pair of parents and ,(34)
we first generate a random integer , then we obtain the (possibly identical) chromosomes of two children as follows:(35)
Mutation: let denote the mutation probability. For each chromosome in , we generate , where denotes the outcome of an independent random variable which is uniformly distributed in . Then for each component for which , we substitute the value with a randomly chosen admissible value.
Selection of surviving chromosomes: we select the chromosomes of parents and offsprings with the best fitness values as input for the next generation.
In this section, performance results for the proposed algorithm described in Section 4 are presented. In the simulation, the parameters of the MA algorithm were chosen as follows: population size, ; number of generations = 20; crossover probability, ; mutation probability, .
We consider a system with one PU and CRUs. The total available bandwidth for CRUs is 5 MHz and supports 16 subcarriers with MHz. We assume that and an OFDM symbol duration, of s. In order to understand the impact of the fair bit rate constraint in (17) on the total bit rate, three cases of user bit rate requirements with were considered. In addition, three cases of partial CSI with were studied. It is assumed that the subcarrier gains and , for are outcomes of independent identically distributed (i.i.d.) Rayleigh-distributed random variables (rvs) with mean square value . The additive white Gaussian noise (AWGN) PSD, , was set to W/Hz. The PSD, , of the PU signal was assumed to be that of an elliptically filtered white noise process. The total CRU bit rate, , results were obtained by averaging over 10,000 channel realizations. The confidence intervals for the simulated results are within of the average values shown.
Figure 1 shows the average total bit rate, , as a function of the total CRU transmit power, , for , and 1 with , W, and a PU transmit power, , of 5 W. As expected, the average total bit rate increases with the maximum transmit power budget . It can be seen that the average total bit rate, , varies greatly with . For example, at W, increases by a factor of 2 as increases from 0.7 to 0.9. This illustrates the big impact that inaccurate CSI may have on system performance. The curves level off as increases due to the fixed value of the maximum interference power that can be tolerated by the PU.
Corresponding results for and are plotted in Figures 2 and 3, respectively. The average total bit rate, , decreases as the NBRW distribution becomes less uniform; the reduction tends to increase with .
Figure 4 shows as a function of for three different cases of with , W, and W. As to be expected, increases with . It can be seen that for is larger than for , and for is larger than for . When the bit rate requirements for CRUs become less uniform, decreases due to a decrease in the benefits of user diversity. With W, increases by about when changes from to . Results for are shown in Figure 5 and are qualitatively similar to those in Figure 4.
The average total bit rate, , is plotted as a function of the maximum PU tolerable interference power, , with W and W, for and 0.7 in Figures 6 and 7, respectively. As expected, increases with and decreases as the CRU bit rate requirements become less uniform. The curves level off as increases due to the fixed value of the total CRU transmit power, .
The assumption of perfect CSI being available at the transmitter is often unreasonable in a wireless communication system. In this paper, we studied an MU-OFDM CR system in which the available partial CSI is due to a delay in the feedback channel. The effect of partial CSI on the BER was investigated; a relationship between transmit power, number of bits loaded, and BER was derived. This relationship was used to study the performance of a resource allocation scheme when only partial CSI is available. It is found that the performance varies greatly with the quality of the partial CSI.
Pseudocode 1: Pseudocode for subcarrier allocation algorithm.
for to number of subcarriers do
find which maximizes
Using (25), calculate the number of bits loaded on
as with ;
initialize to 0;
subcarrier is available; increment by 1;
subcarrier is not available;
For each , initialize the number, , of
subcarriers allocated to CRU to 0
calculate using (28);
for to do
find the value, , of which minimizes
allocate subcarrier to CRU ;
increment by one.
Pseudocode 2: Pseudocode for the memetic algorithm.
initialize Population ;
for to Number_of_Generatio do
add to ;
add to ;
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Huang, D., Shen, Z., Miao, C. et al. Resource Allocation in MU-OFDM Cognitive Radio Systems with Partial Channel State Information. J Wireless Com Network 2010, 189157 (2010). https://doi.org/10.1155/2010/189157
- Orthogonal Frequency Division Multiplex
- Cognitive Radio
- Channel State Information
- Channel Gain
- Orthogonal Frequency Division Multiplex Symbol
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http://www.thoughtsfromatvgeek.com/2016/12/episode-7.html
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math
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It’s also a good suggestion to check out|to take a glance at} the casino’s web site to see what games and options it offers. Live Roulette games are suggested to incorporate between two and 6 gamers. The majority of casino tables accommodate little greater than seven gamers concurrently. To present a stage enjoying in} area, one of 로스트아크 many online roulette individuals ought to act as the vendor; we urge that this individual stay anonymous. This brings the edge for the home method down to} a negligible 1.35%.
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https://www.vedantu.com/physics/motion
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math
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What is Motion?
A book that falls from a table, water flowing from the tap, or rattling windows, all these are the examples of motion. Let us see some important terms that will help us in determining motion.
Both distance as well as displacement help in describing the change in the object’s position. The distance covered by an object from point A to point B depends on the type of path it has taken. So, if an object takes a circular path, then the distance covered by it on this path will be different from the distance covered in case of a linear path.
In the case of displacement, it is defined as the shortest distance that connects the two points, A and B (basically initial and final point).
So, let us say A and B are two cities. The distance between them is ‘d’. Now, when a person moves from city A to city B and then again returns back to city A, then
Distance traveled = d (From A to B) + d (From B to A)
Displacement = 0, as there is no difference in the initial and final position of the person. He started from A and returned back to A. So, in this case, displacement is nil.
Speed is the rate at which the position of the object changes with respect to its origin. It is measured as distance (in meters)/ Time (in seconds).
Types of Motion
Now, let us talk about the different types of motions that can be seen in an object. These are Linear, Rotary, Oscillating motion, and Periodic motion. Each of these types is achieved with a different mechanical means. Let us learn more about them in detail.
1) Linear Motion
In linear motion, the object moves from one position to another in either a curved path or a straight line. On the basis of the type of path taken by an object, linear motion is further classified as:
Rectilinear Motion – Here, the path taken by an object is a straight line.
Curvilinear Motion – Here, the path taken by the object is curved.
One of the best examples of linear motion is linear actuators where you can find cars, cycles, trains and other vehicles travelling in one straight direction. But it will not be called as a linear motion when the road or rail track is perfectly circular. You can even find linear cylinders that display linear motion in pneumatic, hydraulic and electric options. A linear motion has a lot of significance in the field of manufacturing, automation, robotics, etc.
2) Rotary Motion
Rotary motion is a kind of motion where the object moves in a circle. This type of motion occurs when an object rotates at its own place or axis. The rotary motion was the first type of motion that was invented by scientists in primitive times.
Some of the examples that would help you understand about the rotatory motion are:
Earth rotating on its own axis about the sun is the best example of rotary motion.
Another example is the movement of the wheels and steering wheel of the car in the driving state. You will find that both are rotating around their own axis. The same thing goes with the engine of the car as it also moves on its own place.
Similar to linear cylinders, nowadays rotary actuators are extensively used in various industries. These cylinders come in pneumatic, electric, and hydraulic options.
3) Oscillatory Motion
This is the third type of motion that is characterized by the movement of the object in the form of front and back oscillation. In other words, an oscillatory motion is defined as the movement of an object around its mean position. If an object repeats the cycle of its motion after a specific time period, then it is regarded as an oscillating motion.
One of the best examples of oscillating motion is the pendulum of a clock. It repeats its motion after a certain time frame. In actual sense, the pendulum is not displacing from its position. It is stationary in one position, yet it displays motion. Such type of recurring motion after a certain time period is called an oscillating motion.
In this kind of motion, the movement of the object is termed as oscillation. This is because of the repeated nature of the motion that takes place after a set time period. A few other illustrations of an oscillating motion are:
When a child oscillates to and fro around its fixed position on the swing.
Table fan is another example of periodic motion.
Both linear, as well as a rotary actuator, has oscillating motion.
Waves Sound waves are the result of oscillation of particles.
When the cords of the sitar or guitar are struck, then they move to and fro about their mean position (i.e, oscillatory motion).
4) Periodic Motion:
Periodic motion is a type of motion that is repeated at equal time intervals. A few examples of this motion is a bouncing ball, a rocking chair, a swing in motion, a water wave, a vibrating tuning fork etc.
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http://math.stackexchange.com/questions/232203/ring-of-invariants-for-the-action-of-rotation-groups-in-tensors
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math
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Consider the component-wise action of the group $SO(p)\times SO(q)$ in the tensor product of two real vector spaces $S^2(R^p)\otimes R^q$. How to parametrize orbits of this action ? For $q=1$ we obtain a nice answer: the parameters are eigenvalues of the bilinear form.
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http://earthdoc.eage.org/publication/publicationdetails/?publication=32817
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math
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ω-k filter design
Ralph A. Wiggins
Journal name: Geophysical Prospecting
Issue: Vol 14, No 4, December 1966 pp. 427 - 440
Info: Article, PDF ( 688.44Kb )
Two-dimensional band-pass filters can be constructed by a simple extension of the theory of one-dimensional band-pass filters. Similarly to the one-dimensional analogue the shape of the two-dimensional filter is important in determining its effectiveness. The band-pass filter formulation can be further refined so that the filter will concentrate its rejection energies in certain areas of the ω, k plane. Such band-pass, band-reject filters are found by solving a set of simultaneous equations.
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Wolfram. They Tried to Outsmart Wall Street. Basics - Gut Instinct’s Surprising Role in Math. Tikhonov regularization. When the following problem is not well posed (either because of non-existence or non-uniqueness of where may be ill-conditioned or singular).
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In the late nineteen-seventies, when Yau was in his twenties, he had made a series of breakthroughs that helped launch the string-theory revolution in physics and earned him, in addition to a Fields Medal—the most coveted award in mathematics—a reputation in both disciplines as a thinker of unrivalled technical power. Yau had since become a professor of mathematics at Harvard and the director of mathematics institutes in Beijing and Hong Kong, dividing his time between the United States and China. His lecture at the Friendship Hotel was part of an international conference on string theory, which he had organized with the support of the Chinese government, in part to promote the country’s recent advances in theoretical physics. Annals of Mathematics: Manifold Destiny. A First Course in Linear Algebra (A Free Textbook) Ulam spiral. Ulam spiral of size 200×200. Black dots represent prime numbers. Diagonal, vertical, and horizontal lines with a high density of prime numbers are clearly visible.
MimeTeX quickstart. Announcement: If latex is installed on your server then see mathTeX , the successor to mimeTeX. It's based on real LaTeX and renders higher quality images than mimeTeX. If latex is not available on your server, continue using mimeTeX. Copyright © 2002-2011, John Forkosh Associates, Inc. email: [email protected] —— Important Notice —— LatexRender & TeX Converter. TEX Converter is for 32 bit Windows 95/98/Me/NT/2000/XP Programs that you'll need MiKTEX TeX4ht Useful book The Latex Web Companion by Michael Goosens and Sebastian Rahtz Amazon USA Amazon UK Internet Bookshop TEX Converter is freeware written by Steve Mayer © Steve Mayer 1999 LatexRender for PHP Steve's Maths Pages Introduction.
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If you know a distinctive fact about a number not listed here, please e-mail me. Welcome to CalcEnstein - The Ultimate Web. The Mystery Of The Ale. TNT Documentation Page. Theme and Variations (Netscape version) Starmaze Intro. Before I can explain my decades-long quest to map the starmaze I must acquaint you with a small puzzle.
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Scilab Home Page. Play with PI. Untitled. [P] failure functions - another example by akdevarajApr 14 Let our definition of a failure be a non - Devarajnumber which is not a Carmichael number ( see A104017 on OEIS ).Let the mother function be 2^n + 3113.
Then n = 16 + 42*k is a failure function ). [P] Failure functions - role of by akdevarajApr 13 failure functions play an important role in proving a conjecture indirectly. See sketch proof. [P] failure functions - another example by akdevarajApr 12 Here k belongs to N.When n= 10, f(n) = 1105, a Carmichael number; however when n is generated by the failure function 18 + 20*k we get values of f(n) which are not square free and hence incapable of being Carmichael numbers i.e. failures. Numerical Recipes in C.
NumberSpiral.com. Mathematical Symbol. By Douglas Weaver Mathematics Coordinator, Taperoo High School with the assistance of Anthony D. Smith Computing Studies teacher, Taperoo High School. Introduction On the topic of mathematical symbols..... Math Link. Modern Algebra. BEATCALC. Main Page - Knot Atla. LAPACK Users Guide. How to Read Mathematic. Golden ratio - Wikipedia, the free encyclo. Line segments in the golden ratio In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities.
The figure on the right illustrates the geometric relationship. Expressed algebraically, for quantities a and b with a > b > 0, The golden ratio is also called the golden section (Latin: sectio aurea) or golden mean. Other names include extreme and mean ratio, medial section, divine proportion, divine section (Latin: sectio divina), golden proportion, golden cut, and golden number. Geometric Paper Folding: Dr. David Huffman. Nov 1996 David Huffman has been creating some very complex and original folded structures.
He works with both straight and curved folds, using mathematical techniques that he has developed over many years. Mr. Huffman teaches at the University of California at Santa Cruz. Folding Paper in Half Twelve Time. Folding Paper in Half 12 Times: The story of an impossible challenge solved at the Historical Society office Alice laughed: "There's no use trying," she said; "one can't believe impossible things. " "I daresay you haven't had much practice," said the Queen. Through the Looking Glass by L. College Algebra Homepage. If you need help in college algebra, you have come to the right place.
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| 6,756 | 14 |
https://merithub.com/tutorial/problem-solving-strategies-c8os8tqckrg6007ptt6g
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math
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Problem Solving Strategies
Counting All- Students who use this strategy are not yet able to add on from either addend. They cannot visualize and
hold a number in their mind; instead they must mentally build every number quantity.
8+9 The student literally starts with 1 and counts up to 17 using
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Using models to help the student keep track of their
11, 12, 13, 14, 15, 16, 17 location when counting is helpful.
Counting On-The student starts with on of the numbers and counts on from this point. When students are able to
conceptualize a number, they will transition from Counting All to Counting On.
8+9 It is not unusual to hear students say, “I put the eight in
8…9, 10, 11, 12, 13, 14, 15, 16, 17 head and counted up nine more.”
9…10, 11, 12, 13, 14, 15, 16, 17 From an efficiency standpoint, it is important to note
whether the student counts on from the smaller or larger
Doubles/Near Doubles-This strategy capitalizes on this strength by adjusting one or both numbers to make a double or
8+9 The student could choose from several doubles/near
doubles combinations to solve this problem: 8+8
8+(8+1) Using this double requires the student to decompose 9 into
+1 Using this double requires the student to add an extra 1 and
9+9=18 then subtract it from the total.
Making Tens-The focus of this strategy is to be able to utilized fluency with ten to expedite adding. Being able to take
numbers apart with ease or fluency is the key to using this strategy.
8+9 By changing the 8 to 7+1 the student can restructure the
(7+1) +9 problem to create a combination of 10 with 1+9.
8+9 The student could also choose to make a 10 by breaking
8+(2+7) apart the 9 into 7+2 and combining the 2 with the 8 to
(8+2)+7 create 10.
Making Landmark or Friendly Numbers-Landmark or friendly numbers are numbers that are easy to use in mental
computation. Fives, multiples of ten, as well as monetary amounts such as twenty-five and fifty are examples of
numbers that fall into this category. Students may adjust one or all addends by adding or subtracting amounts to make a
23+(48+2) In this example only the 48 is adjusted to make an easy
23+50=73 landmark number.
73-2=71 The extra 2 that was added on must be subtracted.
Compensation-When compensating, students will remove a specific amount from one addend and give that exact
amount to the other addend to make friendlier numbers. Taking from one addend and giving the same quantity to the
other addend to maintain the total sum is a big mathematical idea in addition.
A. 8 + 6 Example A demonstrates a first grader’s Compensation strategy
-1 +1 for making a double.
7 + 7 = 14
In example B, the student changes 18 to the friendly number of
B. 18 + 23
+2 -2 20. Notice how 2 was subtracted from the 23 and added to the
20 + 21 = 41 18.
C. 36 + 9 Example C demonstrates that Compensation can be used to make
-1 +1 an easy 10. Choosing which number to adjust is an important
35 + 10 = 45 student decision that is linked to the student’s thinking about
Breaking Each Number into Its Place Value-Once students begin to understand place value, this is one of the first
strategies they utilize. Each addend is broken into expanded form and like place value amounts are combined. When
combining quantities, children typically work left to right because it maintains the magnitude of the numbers.
24+38 Each addend is broken into its place value.
(20+4) + (30+8)
20+30=50 Tens are combined.
4+8=12 Ones are combined.
50+12=62 Totals are added from the previous sums.
Adding Up in Chunks-The strategy is similar to the Breaking Each Number into Its Place Value strategy except for the
focus is on keeping one addend whole and adding the second number in easy-to-use chunks. This strategy is slightly
more efficient than the Breaking Each Number into Its Place Value strategy, since you are not breaking apart every
A. 45+28 In Example A, 45 is kept whole while 28 is broken into its place
45+(20+8) value and added in parts to the 45.
B. 45+28 Example B demonstrates that either number can be kept whole.
(40+5)+28 This time the 28 is kept intact while the 45 is broken into
40+28=68 combinations that make the problem easier to solve.
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CC-MAIN-2023-40
| 4,209 | 60 |
https://www.teachme2.co.za/additional-mathematics-tutors-wilgespruit-190-iq-roodepoort
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math
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Highest Quality Additional Mathematics Tutors in Wilgespruit 190-Iq, Roodepoort. Get Additional Mathematics Lessons in your home with Teach Me 2
I tutor 1st year Calculus and I have covered Add Maths concepts and beyond. Furthermore I have been tutoring this subject since 2009. *I cover all Syllabi*
I managed to pass all my modules (Statistics, Applied Maths, Calculus and Computer Science) with distinctions (with an average of 85%).
I have a degree in Actuarial Science and still studying towards my Fellowship in Actuarial Science.
Good day; as a recent young B.A graduate; I possess excellent problem-solving and numeracy skills which I acquired especially whilst taking Maths courses throughout my degree.
Taught up to Grade 10; diploma in mathematics and applied mathematics
I've found that maths can be very enjoyable when you understand how to do it, and practice is the best way to understand. When helping others with Maths I try to focus on the joy you can get out of it, as this helped me to achieve better marks.
I excelled in Advanced Level Mathematics and I can easily transfer this knowledge.
Mathematics (as well as Financial Mathematics) is one of my majors for Actuarial Science.
I did Additional Mathematics until grade 10 and I possess the required skills to teach high-school graders. Moreover,I majored in Maths in high school which makes it easier to tutor Additional Maths.
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s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00314-ip-10-31-129-80.ec2.internal.warc.gz
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CC-MAIN-2016-50
| 1,400 | 10 |
http://www.solipsys.co.uk/new/MathematicalMovingChairs_edit.html
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math
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File: MathematicalMovingChairs One of the topics on which ColinWright has given a MathematicsTalk. Suppose we have 8 people in chairs numbered 0 to 7. Each person multiplies their seat number by 5, divides by 8 (that being the number of people) and keeps the remainder. That's their new seat. Do they all go to different seats? Yes they do! * What about with 9 people ... does it still work? * What about 10 people? * What about 11 people? * What if we multiply by 4? * ... or by 5? * ... or by 6? When does it work? | Number of people -> ---- Multiply by ... | 6 | 7 | 8 | 9 | 10 | ... | | 2 | . ? . | . ? . | . ? . | . ? . | . ? . | | 3 | . ? . | . ? . | . ? . | . ? . | . ? . | | 4 | . ? . | . ? . | . ? . | . ? . | . ? . | | 5 | . ? . | . ? . | #Yes# | . ? . | . ? . | | 6 | . ? . | . ? . | . ? . | . ? . | . ? . | | ... | And why? This has connections with cryptography, juggling, computer algorithms, telling the time, and loads of other mathematics.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00586.warc.gz
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CC-MAIN-2018-09
| 1,282 | 11 |
https://www.mathworksheetscenter.com/mathskills/geometry/ExploringFigures/
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math
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Exploring Similar Figures Worksheets
What Characteristics Do Similar Figures Share? Polygons are defined as the two-dimensional closed figures that have straight sides. Every figure or shape that has finite sides and exhibits straight is included as polygons. Circles are not categorized as polygons as they have curved sides. Similar polygons are those that share the exact same shape but of different sizes. Even if one of the two polygons is flipped or rotated, they still will be called similar polygons. At first, the flipped or rotated shape might look different, but they still share the same structure and sometimes even size. In similar polygons, the measurement of the angle remains the same. No matter how much you increase or decrease, enlarge, or shrink the polygon, the angles do not change. It one angle of a triangle measures 70 degrees, then the same angle in a similar triangle will also measure 70 degrees.
Golden rule for math teachers:
You must tell the truth, and nothing but the truth, but not the whole truth.
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CC-MAIN-2023-50
| 1,033 | 4 |
http://abs.gov.au/websitedbs/CaSHome.nsf/4a256353001af3ed4b2562bb00121564/1923d044ab9a237eca2573fa001e61af!OpenDocument
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math
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Census data are being delivered in a number of new formats which are free online. Articles in Education News (cat.no 1330.0) for 2007 illustrate how to search for Census data.
The activity uses one of the Census products, namely QuickStats, which provides summary data in tabular form, with comparative data for Australia. QuickStats are available for areas as large as Australia or as small as a Collection District (about 230 households) and a range of geographic units in between. QuickStats is one of a number of census products available free on the web.
1. Identify the line of transect. It may be from the central business district outwards to the rural urban fringe. Identify the geographical localities to be sampled along the line of transect. For the transect in urban areas use statistics from suburbs, postal areas, Local Government Areas and Statistical Local Areas, whilst in rural districts Statistical Local Areas or Statistical Divisions may be more useful. The examples provided are for suburbs in Melbourne and Statistical Local Areas in Sydney.
2. Provide students with a map of the area. Calculate the distance along the line of transect for each of the chosen geographic areas to be sampled.
3. Ask students to create their own copy of the diagram shown in Figure 1. They should create a scale for the horizontal axis representing the distances in kilometres of each sample location along the line of transect. Begin with zero where the y axis intercepts the horizontal or ‘x’ axis.
4. Students may work in groups or individually. Each group could be allocated a different line of transect to investigate.
5. If working in groups, allocate each member of the group one of the characteristics from the QuickStats table (e.g. people aged 0-4 years, median household income) to investigate. The group size will depend on the number of characteristics being studied. The examples in Tables 1a and 1b provides eight characteristics.
6. Ask the students to use QuickStats to obtain the value of their assigned characteristic for the chosen suburbs. In addition, ask the student to find the comparative statistic for the larger representative geographic region e.g. the Statistical Division or the value for the State.
7. Ask students to calculate the difference between the values for the suburb and comparative values for the representative area, by subtracting the values. Use a table similar to Table 2 to record the results.
Some characteristics will be above the representative region (e.g. more 5-14 year olds in the chosen suburb compared with the Statistical Division) and will therefore be positive. Some values will be below and will therefore be represented by a negative value on the ‘y’ axis in Figure 1.
8. Ask students to review the differences between the suburb and the representative area and choose an appropriate scale for the ’y’ axis. For each suburb students draw a column graph at the appropriate distance along the transect, which represents the percentage by which the suburb varies either positively or negatively from the representative region for that city or state.
9. The group comes together to see how each characteristic varies along the line of transect.
1. Using the data created, describe the changes in the chosen demographic and socio-economic characteristics along the line of transect.
2. For the line of transect, explain the marked variations from the representative area values.
3. If several transects of different localities have been undertaken, compare transects for different parts of the city or State. Explain any differences identified.
4. Find a diagram of an urban transect showing changes in land use and building height. Relate this diagram to the completed transect. Is it possible to identify links between land use and socio-economic data for the chosen study area?
5. Prior to commencing the activity, ask students to suggest the demographic and socio-economic changes expected along the line of transect. On completion of the activity, assess the extent to which the expectations were accurate.
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CC-MAIN-2017-30
| 4,084 | 17 |
http://mercalli.cl/lib/algebra-i-textbook-for-students-of-mathematics
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math
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By Alexey L. Gorodentsev (auth.)
This publication is the 1st quantity of a radical “Russian-style” two-year graduate path in summary algebra, and introduces readers to the fundamental algebraic buildings – fields, jewelry, modules, algebras, teams, and different types – and explains the most ideas of and techniques for operating with them.
The direction covers sizeable components of complicated combinatorics, geometry, linear and multilinear algebra, illustration conception, classification conception, commutative algebra, Galois conception, and algebraic geometry – subject matters which are usually missed in common undergraduate courses.
This textbook is predicated on classes the writer has performed on the self reliant college of Moscow and on the college of arithmetic within the larger university of Economics. the most content material is complemented by means of a wealth of routines for sophistication dialogue, a few of which come with reviews and tricks, in addition to difficulties for self sustaining study.
Read Online or Download Algebra I: Textbook for Students of Mathematics PDF
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Extra resources for Algebra I: Textbook for Students of Mathematics
Otherwise, we say that K has positive characteristic and define the characteristic char K to be the minimal m 2 N such that „ 1C1C 1 D 0. 1 „C 1 C ƒ‚ m C… 1/ n implies that the characteristic of an integral domain is either zero or a prime number p 2 N. 30) takes all multiples of p to zero and therefore can be factorized into the composition ~p ∘ p of ring homomorphisms p WZ Fp ; z 7! Œzp and ~p W Fp ,! K ; Œzp 7! 2 on p. 34, because Fp is a field. Thus, the smallest subring with unit in an integral domain K of positive characteristic p is a field isomorphic to Fp D Z=.
13) 1 Logical “exclusive OR ” (XOR ): a C b is true if and only if precisely one of a, b is true (and not both). Logical AND : a b is true if and only if both a and b are true. 2 Logical NOT x evaluates to true if and only if x is false. Nonexclusive x OR y is true if and only if at least one of x and y is true. 5 on p. 9. 10) are satisfied. 10) requires some intellectual effort, which is generally undertaken in the first chapters of a course in real analysis. I hope that you have taken such a course.
Fx . ax / 2 Kx , with not all the ax equal to zero, either equals zero or divides zero in K , then the family is a zero divisor in x Q Kx . 0; 0; : : : ; 0/ : Thus, the direct product of more than one ring cannot be a field. 13 Show that for two given prime numbers p; q 2 N, the product Fp Fq consists of the zero element, . 0; b/, where a ¤ 0 and b ¤ 0. Note that Fp Fq ' Fp Fq . ax / such Q that each ax is invertible in Kx . 1 Homomorphisms of Abelian Groups A map of abelian groups ' W A ! 27) holds in the group B.
Algebra I: Textbook for Students of Mathematics by Alexey L. Gorodentsev (auth.)
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CC-MAIN-2019-13
| 4,575 | 19 |
http://wlcourseworkbywy.jordancatapano.us/percent-yield-of-a-chemical-reaction.html
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math
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Percent yield of a chemical reaction
Base the percent yield on the please note that the theoretical yield is never the expected yield as we never expect 100% yield in a chemical reaction. Sheffield chemputer reaction yield calculator to use the yield calculator: select whether you want to work in grammes or millimoles select a yield multiplier (must. Limiting reagents and percent yield test your understanding of chemical reactions and stoichiometry with these 9 questions start test about this unit. Video on how to find the percent yield for chemical experiments we represent the amount of product created in a chemical reaction as the percent yield since chemical. Calculate the theoretical yield to determine the % yield in a chemical reaction percent yield. Start studying chem retake in reality the reaction only produces 19l of oxygen gas the percent yield of the reaction is in a chemical reaction between two. Experiment 7 – reaction stoichiometry and percent yield write the balanced chemical equation for the reaction step 2: calculate the moles of given substance. Guided practice in calculating product formation and percent yield is also provided we begin by balancing the chemical reaction.
Determining percentage yield of a chemical reaction aim (research question) related international baccalaureate chemistry essays disappearing cross experiment. What experimental information do you need in order to calculate both the theoretical and percent yield of any chemical reaction (select all that apply. This video shows you how to calculate the theoretical and percent yield in chemistry the theoretical yield is the maximum amount of product that can be. We will also explore limiting reagents and percent yield to address practical aspects of chemical reactions because there may be 505 theoretical & percent. Chemical reactions of copper and percent yield key pre-lab (review questions) 1 give an example, other than the ones listed in this experiment, of redox and.
Purpose the purpose of this investigation is to explore the percent yield of the precipitate in the reaction introduction for known amounts of reactants, theoretical. How to calculate percent yield of a chemical reaction socratic subjects science anatomy & physiology astronomy astrophysics biology chemistry earth science. Calculating the percent yield of alum last update: april 25, 2011 to determine the percent yield of a product in a chemical reaction we need to know the amount of. Percent yield is the amount of a product made in the chemical reaction measured by dividing the amount of product by the amount of theoretical yield which could be.
Chemical reactions of copper and percent yield objective to gain familiarity with basic laboratory procedures, some chemistry of a typical transition element, and. Chemical reactions of copper and percent yield lab experiment. Experiment 5 experiment 5 – determining percent yield in a chemical reaction -1. Start studying chemistry chapter 12 test learn vocabulary what does the percent yield of a reaction measure the chemical reaction does not stop until the.
Percent yield of a chemical reaction
Learn what the theoretical yield, actual yield and percent yield are given the limiting reactant, learn how to calculate the theoretical reaction. The first step in finding theoretical and percentage yield is to balance the relevant chemical equation if you are unsure how to do.
Via wikipedia the ideal or theoretical yield of a chemical reaction would be 100%, a value that is impossible to achieve due to limitations in measurement accuracy. Purpose:to find out the percent yield of copper in the reaction between copper sulfate (cuso4) and iron (fe)materials:balance100-ml beaker250-ml beakerbunsen. An explanation of why chemical reactions do not always produce the maximum yield and how to calculate percentage yield. This worksheet and answer sheet/mark scheme is aimed at chemistry students calculating percentage yields in chemical reactions. Percentage yield a balanced chemical equation tells what the theoretical or ideal yield of the reaction should be.
Why there is a percent yield reactions rarely produce the predicted amount of product from the masses of reactants in the reaction an. In chemistry, yield, also referred to as reaction yield, is the amount of product obtained in a chemical reaction the absolute yield can be given as the weight in. Theoretical yield is the mass of product predicted by the balanced chemical equation for the reaction percentage yield yield 100% for a chemical reaction at.
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CC-MAIN-2018-43
| 4,568 | 9 |
http://www.statemaster.com/encyclopedia/Carsten-Thomassen
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math
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Carsten Thomassen (b. August 22, 1948 in Grindsted) is a danishmathematician, Technical University of Denmark. He is member of the Royal Danish Academy of Sciences and Letters since 1990. His research concerns discrete mathematics and more specifically graph theory. August 22 is the 234th day of the year in the Gregorian calendar (235th in leap years), with 131 days remaining. ... 1948 (MCMXLVIII) was a leap year starting on Thursday (the link is to a full 1948 calendar). ... Grindsted is a municipality in south-west Denmark, in the county of Ribe on the peninsula of Jutland. ... To meet Wikipedias quality standards, this article or section may require cleanup. ... Discrete mathematics, also called finite mathematics, is the study of mathematical structures that are fundamentally discrete, in the sense of not supporting or requiring the notion of continuity. ... A pictorial representation of a graph In mathematics and computer science, graph theory is the study of graphs, mathematical structures used to model pairwise relations between objects from a certain collection. ...
He is editor-in-chief of the Journal of Graph Theory and the Electronic Journal of Combinatorics, and editor of Combinatorica, The Journal of Combinatorial Theory, Ser B, Discrete Mathematics, and the European Journal of Combinatorics.
He was awarded the Dedicatory Award of the 6th International Conference on the Theory and Applications of Graphs by the Western Michigan University in May 1988, the Lester R. Ford Award by the Mathematical Association of America in 1993 and the Faculty of Mathematics Alumni Achievement Medal by the University of Waterloo in 2005. He is included on the ISI Web of Knowledge list (http://www.isihighlycited.com/) of the 250 most cited mathematicians.
Carsten Thomassen's home page
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CC-MAIN-2019-39
| 2,336 | 12 |
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