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8: Nucleotides and Nucleic Acids
Fundamentals of Biochemistry Vol. I - Structure and Catalysis
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Wed, 19 Mar 2025 10:20:47 GMT
8.3: Nucleic Acids - Comparison of DNA and RNA
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Henry Jakubowski
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[ "article:topic", "DNA", "RNA", "mutations", "nucleic acids", "deamination", "epigenetics", "showtoc:no", "license:ccbysa", "autonumheader:yes2", "licenseversion:40", "authorname:jakubowski-flatt", "avatar@ "molecular dynamics", "genetic mutations", "base pairing", "chemical modifications", "post-translation modifications", "replication errors", "double-stranded helix", "intermediary molecule", "phosphodiester link", "catalytic activities", "genetic information", "molecular flexibility", "structural comparison", "Watson-Crick", "Hoogsteen", "tertiary structures", "catalytic sites", "evolutionary adaptation" ]
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Contents
1. Home
2. Bookshelves
3. Biochemistry
4. Fundamentals of Biochemistry (Jakubowski and Flatt)
5. Fundamentals of Biochemistry Vol. I - Structure and Catalysis
6. 8: Nucleotides and Nucleic Acids
7. 8.3: Nucleic Acids - Comparison of DNA and RNA
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Fundamentals of Biochemistry Vol. I - Structure and Catalysis
Front Matter
1: The Foundations of Biochemistry
2: Water and its Role in Life
3: Amino Acids, Peptides, and Proteins
4: The Three-Dimensional Structure of Proteins
5: Protein Function
6: Enzyme Activity
7: Carbohydrates and Glycobiology
8: Nucleotides and Nucleic Acids
9: Investigating DNA
10: Lipids
11: Biological Membranes and Transport
Back Matter
8.3: Nucleic Acids - Comparison of DNA and RNA
Last updated Mar 19, 2025
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8.2: Nucleic Acids - RNA Structure and Function
8.4: Chromosomes and Chromatin
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Page ID 72660
Henry Jakubowski and Patricia Flatt
College of St. Benedict/St. John's University and Western Oregon University
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Table of contents
1. Learning Goals (ChatGPT o3-mini, 2/1/25)
2. Chemical modifications of DNA and RNA
3. Mutations
4. Why DNA and RNA - A Chemical Perspective
5. A Structural Comparison
6. Summary
7. References
Search Fundamentals of Biochemistry
Learning Goals (ChatGPT o3-mini, 2/1/25)
Explain Chemical Modifications:
Describe how intentional chemical modifications—such as methylation and hydroxymethylation—alter the structure and function of DNA and RNA, including their roles in regulating transcription and translation.
Distinguish Epigenetic Versus Epitranscriptomic Changes:
Compare how DNA methylation (and histone modifications) contribute to epigenetic regulation, while analogous modifications in RNA contribute to the emerging field of epitranscriptomics.
Analyze Mutation Mechanisms:
Explain how spontaneous chemical reactions, such as the hydrolytic deamination of cytosine, lead to point mutations and discuss the role of repair enzymes (e.g., uracil-DNA glycosylases) in maintaining genomic integrity.
Evaluate the Impact of Chemical Agents:
Assess how external mutagens (like nitrous acid and alkylating agents) cause point mutations and structural rearrangements in DNA, and predict their potential consequences for genomic stability.
Compare Uracil and Thymine Roles:
Discuss the chemical rationale behind using uracil in RNA and thymine in DNA, emphasizing the stabilizing effect of the methyl group in thymine to prevent erroneous repair of deaminated cytosine.
Examine Backbone Linkage Chemistry:
Evaluate why DNA and RNA employ phosphodiester bonds instead of alternative linkages (e.g., carboxylic acid esters or amides) and how the chemical properties of these bonds contribute to the stability and overall structure of nucleic acids.
Understand Sugar Structure and Flexibility:
Describe how the presence or absence of a 2'-OH group in ribose versus deoxyribose influences molecular stability, backbone flexibility, and the ability to form extended double-helical structures.
Interpret Sugar Puckering and Helical Conformations:
Analyze the effects of ribose puckering (C3'-endo vs. C2'-endo) on the formation of A-RNA versus B-DNA helices, and explain how these conformational differences affect nucleic acid-protein interactions.
Differentiate Base Pairing Modes:
Contrast Watson-Crick and Hoogsteen base pairing in DNA, discuss how their dynamic equilibrium is influenced by structural modifications, and understand the implications for protein recognition and damaged DNA repair.
Integrate Structure-Function Relationships:
Synthesize the comparative structural features of DNA, RNA, and proteins to explain how these differences have been evolutionarily selected to optimize genetic information storage, replication fidelity, and catalytic activity.
These goals are intended to guide your study of how subtle chemical differences in nucleic acids impact their structure, stability, and biological functions, thereby deepening your understanding of molecular biology at a fundamental level.
Now that we understand the structures of DNA and the structures and various functions of RNA, we can more fully explore how their chemical similarities and differences contribute to different functions.
Chemical modifications of DNA and RNA
Post-translation modifications of proteins alter their structural/functional properties. Likewise, intentional chemical modifications of nucleic acid bases alter both their structures and potentially their transcriptional and translational status. Figure 8.3.1 8.3.1 shows common modifications of bases in DNA.
Figure 8.3.1 8.3.1: Common modifications of bases in DNA. Matthew K.Bilyard et al. Current Opinion in Chemical Biology. Volume 57, August 2020, Pages 1-7. Under a Creative Commons license
Likewise, RNA is chemically modified. Figure 8.3.2 8.3.2 shows common modifications of bases in RNA. Methylation and subsequent hydroxylation to hydroxymethyl are common to both DNA and RNA. Methylation of DNA often represses the transcription of the DNA into RNA. Hence, it has huge potential to alter gene transcription. Such changes to the DNA are called epigenetic modifications. These changes can also be passed down to future generations and affect a cell's phenotype. Histone proteins involved in DNA packing into nucleosomes can also be methylated and acetylated, altering the interaction of the DNA with the nucleosome core and further packing, again affecting transcription.
Figure 8.3.2 8.3.2:Common modifications of bases in RNA
Chemical modification to RNA can also change the reading of the genome. The epitranscriptomerefers to the collective chemical modifications to RNA, and its understanding is part of a new field,epitranscriptomics.
Mutations
Mutation can arise from the chemical modification of bases. Uracil in RNA is a demethylated form of thymine in DNA. In RNA, AU base pairs replace AT base pairs. Why the need for uracil in RNA? The question could be rephrased as to why there is a need for thymine, with its extra methyl group, in DNA. It's useful to think about the consequence of replacing a single H in a molecule with a -CH 3. Take HOH (i.e., water) as an example. Our bodies are over 60% water. We drink liters of water with a concentration of 55 M each day. Yet if we drink 0.07 L of methanol (CH 3 OH), half of us would die! Let's probe some consequences of the U (no -CH 3) and T (with -CH 3) changes in DNA. It can get confusing, but remember that the normal base pairs in DNA are AT, but AU base pairs also form (they are the norm in RNA). The -CH 3 substituent on thymine does not affect its base pairing.
a. Spontaneous deamination of cytosine in DNA
Why are we now discussing cytosine in DNA? One reason is that the most common mutation in DNA is a C to T replacement. One way that happens is through the spontaneous hydrolytic deamination of cytosine in DNA to uracil, which we have presumed to be found only in RNA. The mechanism for this deamination and subsequence conversion of a GCto an ATbase pair is shown in Figure 8.3.3 8.3.3. The inset box shows a simplified mechanism for spontaneous deamination.
Figure 8.3.3 8.3.3: GC to AT base pair mutation on spontaneous hydrolytic deamination of cytosine in DNA.
Hence, a possible consequence of the deamination reaction is a GC to AT base pair mutation if the uracil in DNA is not removed before DNA replication. Fortunately, the enzyme uracil-DNA glycosylases can remove any uracils found in DNA, leaving an abasic site, which can be fixed with DNA repair enzymes.
We can now ask why T and not U in DNA. Pretend you are a DNA repair enzyme and see a UA base pair in DNA. How can you tell if the UA base pair is correct and intended to be there or if it should be a CG base pair that underwent deamination? The most common uracil-DNA glycosylases remove the uracil whether it is across from guanine, the correct base but which can not hydrogen bond with uracil (in the green oval in Figure 8.3.3 8.3.3), or if it is across from adenine, the wrong base (in red oval), which is present after a round of replication. Evolution has addressed this problem by adding a methyl group to uracil to form thymine and using that base, which forms a base pair with adenine. Now, no decision on which base across from a uracil (guanine if the uracil arose from deamination) or across from a "uracil-like" thymine (adenine) is correct.
b. Other mutations
Since we are considering chemical modifications to DNA and mutations, giving a more expanded background on them is appropriate. In addition to mutations caused by spontaneous hydrolytic cytosine deamination, mutations can also arise by adding a wrong base during DNA replication, by chemical damages caused by radiation or chemical modifying agents. How many mistakes in replication are made? You would be ecstatic if you received a 99% on an examination. That's not good enough for DNA replication. In Cell Biology by the Numbers, they calculate it this way. Assume the replication /repair is so good that it takes 10 8 replications to make a mistake (an error rate of 10-8/BP). Assume also there are 3 x 10 9 base pairs in the human genome. This leads to a mutation rate 10-100 mutations/genome/generation or about 0.1-1 mutations/genome/replication. Not bad!
Figure 8.3.4 8.3.4 shows how common point mutations might randomly arise.
Figure 8.3.4 8.3.4: How common point mutations might arise randomly
Chemical agents also can cause point mutations. Figure 8.3.5 8.3.5 shows point mutations arising from oxidative deaminations (not hydrolytic) by nitrous acid/nitrosamines and from alkylating agents.
Figure 8.3.5 8.3.5: Nitrous acid/nitrosamines and alkylating agent point mutations
Figure 8.3.6 8.3.6 shows a variety of alkylating agents with mutagenic potential.
Figure 8.3.6 8.3.6: Alkylating agents with mutagenic potential.
Finally, large-scale changes in chromosome structure can also occur, as shown in Figure 8.3.7 8.3.7, usually with profound consequences.
Figure 8.3.7 8.3.7: Large scale structural rearrangements in DNA
Why DNA and RNA - A Chemical Perspective
Asking a "why" question (like above) in the sciences is inappropriate as teleological questions are more philosophical or religious. Yet we will in this section, in part, to be in the company of Alexander Rich, who wrote a very cool article entitled "Why RNA and DNA have different Structures".
Given that RNA expresses catalytic activities and can carry genetic information (some viruses have ds and ss RNA as their genome), it has been suggested that early life might have been based on RNA. DNA would evolve later as a more secure carrier of genetic information. Inspecting the chemical properties of DNA, RNA, and proteins shows them to have attributes needed for their expressed function. Let's examine each for structural features that might be important for function.
a. Why does DNA lack a 2' OH group (found in RNA), which has been replaced with hydrogen? This required the evolutionary creation of a new enzyme, ribonucleotide reductase, to catalyze the replacement of the OH in a ribonucleotide monomer to form the deoxyribonucleotide form. One possible explanation is offered in the figure below. DNA, the main carrier of genetic information, must be an extremely stable molecule. An OH present on C'2 could act as a nucleophile and attack the proximal P in the phosphodiester bond, leading to a nucleophilic substitution reaction and potential cleavage of the bond. RNA, an intermediary molecule whose concentration (at least as mRNA) should rise and fall based on the need for a potential transcript, should be more labile to such hydrolysis. Figure 8.3.8 8.3.8 shows a possible reaction diagram for the internal cleavage of RNA. (The reaction would probably proceed with no actual intermediate but just a transition state.
Figure 8.3.8 8.3.8: Internal cleave of RNA using the C'2-OH as an intramolecular nucleophile
b. Why do both DNA and RNA contain a phosphodiester link between adjacent monomers instead of more "traditional" links such as carboxylic acid esters, amides, or anhydrides? One possible explanation is given below. Nucleophilic attack on the sp 3 hybridized P in a phosphodiester is much more difficult than for a more open sp 2 hybridized carboxylic acid derivative. In addition, the negative charge on the O in the phosphodiester link would decrease the likelihood of a nucleophilic attack. The negative charges on both strands in ds-DNA probably help keep the strands separated, allowing the traditional base pairing and double-stranded helical structure to be observed. The cleavage of the phosphodiester link in DNA and a hypothetic ester link is shown in Figure 8.3.9 8.3.9. Again, the reaction of the phosphodiester shows a pentavalent intermediate, but most likely, the reaction proceeds directly from the transition state.
Figure 8.3.9 8.3.9: cleavage of the phosphodiester link in DNA and a hypothetic ester link
c. Why is DNA found as a repetitive double-stranded helix, but RNA is usually found as a single-stranded molecule that can form complicated tertiary structures with some ds-RNA motifs?
Another reason for the absence of the 2' OH in DNA is that it allows the deoxyribose ring in DNA to pucker just the right way to allow extended ds-DNA helices (B type). The pucker in deoxyribose and ribose can be visualized by visualizing a single plane in the sugar ring defined by the ring atoms C1', O, and C4'. If a ring atom points in the same direction as the C4'-C5' bond, the ring atom is defined as endo. If it points in the opposite direction, it is defined as exo. In the most common form of double-stranded DNA, B-DNA, the iconic extended double helix you know, C2' is in the endo form. It can also adopt the C3' endo form, forming another less common helix, a more open ds-A helix. In contrast, steric interference prevents ribose in RNA from adopting the 2'endo conformation. It allows only the 3'endo form, precluding the occurrences of extended ds-B-RNA helices but allowing more open, A-type helices.
Figure 8.3.10 8.3.10 shows another comparison between the A-RNA and B-DNA double helices and the C'3 and C'2 endo forms of the ribose
Figure 8.3.10 8.3.10: after Zhou et al. Nature Structural and Molecular Biology. doi:10.1038/nsmb.3270
Figure 8.3.5 8.3.5 shows interactive iCn3D models of the pentoses in a strand of A-RNA (413D), double-stranded, left, and B-DNA (1BNA), double-stranded, right.
C'3-endo ribose, A-RNA (413D, double stranded)C'2 endo ribose, B-DNA (1BNA, double stranded)
Click the image for a popup or use this external link:
Click the image for a popup or use this external link:
d. What about the molecular dynamics of A-RNA and B-DNA?
The information above suggests that the sugar ring of DNA is conformationally more flexible than the ribose ring of RNA. This can be inferred from the observation that dsDNA can adopt B and A forms, which requires a switch from the 2' endo in the B form to the 3'endo form in the A form. The smaller H on the 2'C would offer less steric interference with such flexibility. The rigidity in ribose is associated with a smaller 5'O to 3'O distance in RNA, leading to a compression of the nucleotides into a helix with a smaller number of base pairs/turn.
The increased flexibility in DNA allows rotation around the C1'-N glycosidic bond connecting the deoxyribose and base in DNA, allowing different orientations of AT and GC base pairs with each other. The normal "anti" orientation allows "Watson-Crick" (WC) base pairing between AT and GC base pairs, while the altered rotation allows "Hoogsteen" (Hoog) base pairs. Figure 8.3.11 8.3.11 shows the different orientations for an AT base pair.
Figure 8.3.11 8.3.11: Xu, Y., McSally, J., Andricioaei, I. et al. Modulation of Hoogsteen dynamics on DNA recognition. Nat Commun9,1473 (2018). Commons Attribution 4.0 International License.
The Watson-Crick (WC) and Hoogsteen (HG) base pairs in B-DNA are in a dynamic equilibrium, with the equilibrium greatly favoring the WC form, as indicated by the arrows in the figure above. In a DNA:protein complex, the WC ↔ HG equilibrium can favor the WG form for AT and GC+ forms (in the latter, the C is protonated) when those base pairs are also involved in protein recognition. They can also occur more frequently in damaged DNA. In contrast, molecular dynamic studies show that the HG base pairs A-U and GC+ are strongly disfavored in ds A-RNA.
One type of DNA damage is methylation on N1-adenosine and N1-guanosine. This modification prevents normal Watson-Crick base pairing, but for DNA, these modified bases can still engage in Hoogsteen base pairing, preserving the overall structure of dsDNA and its ability to carry genetic information stably. This same methylation occurs normally in post-transcriptional modified RNA. Hence, N1 adenosine and N1 guanosine methylation prevent any base pairing in the modified RNA. These properties make DNA a better carrier of molecular information and offer another way to regulate RNA's structural and functional properties.
Hoogsteen base pairs can be found in distorted dsDNA structures (caused by protein:DNA interactions) and normal B-DNA. Figure 8.3.12 8.3.12 shows a Hoogsteen base pair between dA7 and dT37 in the MAT α 2 homeodomain:DNA complex (pdb 1K61). Note that the dA base in the Hoogsteen base pair is rotated syn (with respect to the deoxyribose ring) instead of the usual anti, allowing the Hoogsteen base pair.
Figure 8.3.12 8.3.12: Hoogsteen base pair between dA7 and dT37 in the MAT α 2 homeodomain:DNA complex (pdb 1K61)
A Structural Comparison
Now, let's review the structures adopted by the three major macromolecules: DNA, RNA, and proteins. DNA predominately adopts the classic ds-BDNA structure, although it is wound around nucleosomes and "supercoiled" in eukaryotic cells since it must be packed into the nucleus. Prokaroytic DNA is typically packed into a more amorphous nuclear region, the nucleoid, through interactions with other proteins that also facilitated supercoiling. It is, in effect, a dynamic molecular condensate.
The extended ds-BDNA helical form arises partly from the significant electrostatic repulsions of two strands of this polyanion (even in counter-ions). Given its high charge density, it is unsurprising that it forms complexes with positive proteins and does not adopt complex tertiary structures. RNA, conversely, can not form long B-type double-stranded helices (due to steric constraints of the 2'OH and the resulting 3'endo ribose pucker). Rather, it can adopt complex tertiary conformations (albeit with significant counter-ion binding to stabilize the structure) and, in doing so, can form regions of secondary structure (ds-A RNA) in the form of stem/hairpin forms. Proteins, with their combination of polar charged, polar uncharged, and nonpolar side chains, offer little electrostatic hindrance in adopting secondary and tertiary structures. RNA and proteins can adopt tertiary structures with potential binding and catalytic sites, making them ideal catalysts for chemical reactions. Given its four nucleotide alphabet, RNA can also carry genetic information, making it an ideal candidate for the first evolved macromolecules enabling the development of life. Proteins with abundant organic functionalities eventually supplanted RNA as a better choice for life's catalyst. DNA, with its greater stability, would supplant RNA as the choice for the primary carrier of genetic information (Figure 8.3.13 8.3.13):
Figure 8.3.13 8.3.13: Summary comparison of DNA, RNA, and Protein structures
A final note on the simplicity of the dsDNA structure. A mutation causing a single base pair change in DNA does notchange the iconic ds-stranded DNA structure. If it did, DNA would not be a reliable molecule to store and read out the genetic blueprint. In contrast, a single mutation in the DNA leading to a single amino acid substitution may lead to a protein with altered structure and function. This could be deleterious or even fatal to the organism. On the other hand, the new protein structure might have new functionalities that allow adaptation to new environments or allow new types of reactions. Evolution would favor the latter.
Summary
This chapter examines how subtle chemical variations and modifications in nucleic acids—DNA and RNA—lead to distinct structural and functional outcomes that are essential for life. Designed for junior and senior biochemistry majors, the chapter integrates concepts from chemical modifications and mutation mechanisms to explain why DNA and RNA serve different roles in the cell.
Chemical Modifications and Regulation:
Both DNA and RNA undergo intentional chemical modifications that can alter their physical structure and influence gene expression. In DNA, common modifications such as methylation (and subsequent hydroxymethylation) are central to epigenetic regulation. These modifications can repress transcription and are heritable, impacting cell phenotype over generations. RNA modifications, collectively referred to as the epitranscriptome, similarly affect RNA stability, processing, and translation, highlighting the dynamic regulatory potential of these molecules.
Mutation Mechanisms and DNA Repair:
The chapter discusses how chemical modifications can also lead to mutations. For example, the spontaneous hydrolytic deamination of cytosine to uracil in DNA represents a major source of point mutations, potentially converting GC to AT base pairs if not corrected. DNA repair enzymes, such as uracil-DNA glycosylases, play a critical role in identifying and excising these aberrant bases. Additionally, errors during DNA replication and damage induced by chemical agents (e.g., nitrous acid, alkylating agents) contribute to mutations ranging from single base changes to large-scale chromosomal rearrangements.
Structural Basis for Functional Differences:
A significant portion of the chapter is devoted to understanding why DNA and RNA, despite their chemical similarities, adopt very different structures and fulfill distinct roles:
Sugar Chemistry and Backbone Stability:
DNA’s deoxyribose lacks a 2'-OH group, making it more chemically stable and less prone to self-cleavage compared to RNA, which contains ribose with a reactive 2'-OH. This difference is crucial for DNA’s role as a long-term, reliable repository of genetic information, whereas RNA’s inherent lability suits its function in transient information transfer and catalysis.
Phosphodiester Linkages:
Both nucleic acids use phosphodiester bonds to link nucleotides. The inherent chemical resistance of these bonds to nucleophilic attack—enhanced by their negative charge—contributes to the overall stability of the DNA double helix and the more dynamic secondary and tertiary structures found in RNA.
Sugar Puckering and Helical Forms:
The conformational flexibility of the sugar ring is a key determinant of nucleic acid structure. In DNA, the deoxyribose can adopt different puckers (C2'-endo for B-DNA, C3'-endo for A-DNA), which influences the overall helix geometry. In contrast, the ribose in RNA is more restricted, favoring the C3'-endo conformation, which precludes extended double-stranded helices and promotes the formation of complex tertiary structures.
Base Pairing Dynamics:
The chapter also explores the equilibrium between Watson-Crick and Hoogsteen base pairing in DNA. Although Watson-Crick pairing predominates in stable B-DNA, dynamic shifts to Hoogsteen pairings can occur—particularly in protein-DNA complexes or damaged regions—affecting recognition and repair mechanisms. In RNA, such alternative pairing is less common due to structural constraints imposed by the ribose.
Comparative Structural Overview:
In a broader context, the chapter compares the structural attributes of DNA, RNA, and proteins. DNA’s robust double-helical structure, which is maintained despite mutations, underscores its reliability as a genetic archive. RNA’s versatility in folding into complex structures enables it to perform catalytic and regulatory roles, while proteins, with their diverse side chains, are well-suited to function as dynamic catalysts and structural components. This comparison reinforces the evolutionary specialization of these macromolecules for their respective biological roles.
In summary, the chapter provides a comprehensive exploration of how chemical modifications and intrinsic structural features govern the stability, function, and evolutionary utility of DNA and RNA. Understanding these principles is fundamental for appreciating the molecular basis of gene regulation, mutation, and the overall maintenance of genetic integrity in biological systems.
References
Börner, R., Kowerko, D., Miserachs, H.G., Shaffer, M., and Sigel, R.K.O. (2016) Metal ion induced heterogeneity in RNA folding studied by smFRET. Coordination Chemistry Reviews 327 DOI: 10.1016/j.ccr.2016.06.002 Available at:
Hardison, R. (2019) B-Form, A-Form, and Z-Form of DNA. Chapter in: R. Hardison’s Working with Molecular Genetics. Published by LibreTexts. Available at:
Lenglet, G., David-Cordonnier, M-H., (2010) DNA-destabilizing agents as an alternative approach for targeting DNA: Mechanisms of action and cellular consequences. Journal of Nucleic Acids 2010, Article ID: 290935, DOI: 10.4061/2010/290935 Available at:
Mechanobiology Institute (2018) What are chromosomes and chromosome territories? Produced by the National University of Singapore. Available at:
National Human Genome Research Institute (2019) The Human Genome Project. National Institutes of Health. Available at:
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Wikipedia contributors. (2019, July 17). Nucleosome. In Wikipedia, The Free Encyclopedia. Retrieved 17:17, July 26, 2019, from en.Wikipedia.org/w/index.php?title=Nucleosome&oldid=906654745
Wikipedia contributors. (2019, July 26). Human genome. In Wikipedia, The Free Encyclopedia. Retrieved 06:12, July 27, 2019, from en.Wikipedia.org/w/index.php?title=Human_genome&oldid=908031878
Wikipedia contributors. (2019, July 19). Gene structure. In Wikipedia, The Free Encyclopedia. Retrieved 06:16, July 27, 2019, from en.Wikipedia.org/w/index.php?title=Gene_structure&oldid=906938498
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4301 | https://www.origoeducation.com.au/blog/number-line/ | Published Time: 2018-09-20T21:08:16+00:00
Visual Models in Mathematics: The Importance of a Number Line | ORIGO Education
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Home/ Blogs / Visual Models in Mathematics: The Importance of a Number Line
20 Sep
Visual Models in Mathematics: The Importance of a Number Line
The number line: A powerful tool in the primary maths classroom
Simple tools have been used to help make giant leaps to understand or explain new ideas in science, mathematics, and technology. One of the more important “inventions” for mathematics has been the number line. About 2,500 years ago the Ancient Egyptians could be credited with making a very practical number line. They tied knots equal distance along a rope that was used to measure length by counting the number of spaces. In the picture, Egyptian rope measurers have formed a triangle with 3 spaces, 4 spaces, and 5 spaces respectively along each side. The rope was one single length so it could be used to measure buildings, walls, or fields.
Photo courtesy of Key Curriculum Press 1
The rope was a visual tool that worked well with real things. It was good for measurement as well as calculations that involved “straight forward” operations such as addition and multiplication. However, by the 1600s, mathematicians were aware of numbers that they could not easily explain with “real things.” It is interesting to note that Fahrenheit chose 32 as the number for the freezing point of water when he created the first accurate thermometer. He seemed to know about using a number line for temperatures, but did not have a way to describe a reading less than 0. His choice of 32 insured, at least for his time, that all temperatures would be numbers greater than 0.
Representing Negative Numbers
The credit for creating a number line to show numbers less than 0 is usually given to the Englishman John Wallis. In his first illustrations, he used a solid line for 0 and the positive numbers and a dotted line for the negative numbers.
About the same time, the Dutch mathematician Rene´ Descartes used two lines to create the x-y coordinate system which served to reinforce the approach that Wallis used. The use of number lines provided ways to show negative numbers, and more importantly, these tools were used to link all numbers. With these tools, huge strides where made to explore new mathematical ideas because there was a way to work with numbers using just a single visual model!
Number Lines in the Classroom
When a number line is first used in the classroom it is a good idea to show a line extending in both directions without referring to the left side of 0 (the origin). If 0 is shown as an end-point, the example is a ray and not a true line.
In the first experiences, it is important to focus on the distance from 0 to 1. This is called the unit (or one-unit) length that gives meaning to all numbers along the line. For example, knowing the length of one-unit means that doubling or tripling that distance gives a way to interpret 2 or 3 units. Marking a distance from 0 to a point less than 1 gives a distance that can be written as a fraction or decimal less than 1. The distance from 0 to 1 can be any length, but once it is determined, the value of all other numbers is decided.
Why Are Number Lines Important?
Number lines are important because they present numbers in real life. Primarily, because they enable negative numbers to be represented in a way that made sense. A secondary and equally important outcome was a way to show all real numbers, including the mysterious irrational numbers such as π, – π, √2, -√2, etc. This result meant that the number line became a versatile and powerful visual tool to help students understand numbers.
Click on the ORIGO ONE video for more about how number lines can be used to represent all real numbers.
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1:22
Use the support page shown below to facilitate a productive conversation about number lines in your classroom. (Click on the link below to download the support resource.)
The next blog will discuss the special features of visual models that are useful in the mathematics classroom. This will include tools that might be used in special ways (e.g. equivalent fraction strips). The discussion will include visual models that are called “graphic organisers” that are used with familiar manipulatives such as counters and cubes to illustrate powerful ideas.
Reference
1 Serra, Michael. Discovering Geometry an Inductive Approach. Key Curriculum Press, 1997.
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Calvin Irons cofounded ORIGO Education with James Burnett in 1995. Cal has been involved in mathematics education for over 50 years. He started his career as a specialist teacher of mathematics in Iowa after completing his BA and MA at the University of Northern Iowa in 1967. Dr. Irons received his PhD from Indiana University in 1975 (the dissertation topic was the teaching of division). In 1975, he accepted a position at the Queensland University of Technology, Brisbane, Australia where he has been involved in the teaching and the development of mathematics curricula for elementary schools. He has received outstanding achievement awards from the university for his work and in 2014 was the student’s nominee for University Outstanding Teacher of the Year from a university faculty of over 3000 professors. He is the author/co-author of over 600 books or articles including the award-winning ORIGO Stepping Stones mathematics program.
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4302 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10603284/ | Improving Accuracy and Reliability of Hearing Tests: An Exploration of International Standards - PMC
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. 2023 Oct 10;27(4):169–180. doi: 10.7874/jao.2023.00388
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Improving Accuracy and Reliability of Hearing Tests: An Exploration of International Standards
Michelle J Suh
Michelle J Suh
1 Department of Otorhinolaryngology, Jeju National University College of Medicine, Jeju, Korea
Find articles by Michelle J Suh
1, Jihyun Lee
Jihyun Lee
2 Department of Otorhinolaryngology, Yonsei University Wonju College of Medicine, Wonju, Korea
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2, Wan-Ho Cho
Wan-Ho Cho
3 Division of Physical Metrology, Korea Research Institute of Standards and Science, Daejeon, Korea
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3, In-Ki Jin
In-Ki Jin
4 Division of Speech Pathology and Audiology, Research Institute of Audiology and Speech Pathology, College of Natural Sciences, Hallym University, Chuncheon, Korea
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4, Tae Hoon Kong
Tae Hoon Kong
2 Department of Otorhinolaryngology, Yonsei University Wonju College of Medicine, Wonju, Korea
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2, Soo Hee Oh
Soo Hee Oh
5 Department of Audiology and Speech Language Pathology, Hallym Univesity of Graduate Studies, Seoul, Korea
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5, Hyo-Jeong Lee
Hyo-Jeong Lee
6 Department of Otorhinolaryngology-Head and Neck Surgery, Hallym University College of Medicine, Anyang, Korea
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6, Seong Jun Choi
Seong Jun Choi
7 Department of Otorhinolaryngology-Head and Neck Surgery, Soonchunhyang University Cheonan Hospital, Soonchunhyang University College of Medicine, Cheonan, Korea
Find articles by Seong Jun Choi
7, Dongchul Cha
Dongchul Cha
8 Healthcare Lab, Naver Corporation, Seongnam, Korea
9 Healthcare Lab, Naver Cloud Corporation, Seongnam, Korea
Find articles by Dongchul Cha
8,9, Kyung-Ho Park
Kyung-Ho Park
10 Department of Otorhinolaryngology, Seoul St. Mary’s Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea
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10, Young Jun Seo
Young Jun Seo
2 Department of Otorhinolaryngology, Yonsei University Wonju College of Medicine, Wonju, Korea
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2,✉
Author information
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Copyright and License information
1 Department of Otorhinolaryngology, Jeju National University College of Medicine, Jeju, Korea
2 Department of Otorhinolaryngology, Yonsei University Wonju College of Medicine, Wonju, Korea
3 Division of Physical Metrology, Korea Research Institute of Standards and Science, Daejeon, Korea
4 Division of Speech Pathology and Audiology, Research Institute of Audiology and Speech Pathology, College of Natural Sciences, Hallym University, Chuncheon, Korea
5 Department of Audiology and Speech Language Pathology, Hallym Univesity of Graduate Studies, Seoul, Korea
6 Department of Otorhinolaryngology-Head and Neck Surgery, Hallym University College of Medicine, Anyang, Korea
7 Department of Otorhinolaryngology-Head and Neck Surgery, Soonchunhyang University Cheonan Hospital, Soonchunhyang University College of Medicine, Cheonan, Korea
8 Healthcare Lab, Naver Corporation, Seongnam, Korea
9 Healthcare Lab, Naver Cloud Corporation, Seongnam, Korea
10 Department of Otorhinolaryngology, Seoul St. Mary’s Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea
✉
Address for correspondence Young Joon Seo, MD, PhD Department of Otorhinolaryngology, Yonsei University Wonju College of Medicine, 20 Ilsan-ro, Wonju 26426, Korea Tel +82-33-741-0644 Fax +82-33-732-8287 E-mail okas2000@hanmail.net
Received 2023 Aug 23; Revised 2023 Sep 13; Accepted 2023 Sep 13; Issue date 2023 Oct.
Copyright © 2023 The Korean Audiological Society and Korean Otological Society
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC10603284 PMID: 37872752
Abstract
This study explores the internal standards for hearing tests and benefits of implementing international standard protocols, including the International Organization for Standardization (ISO) and International Electrotechnical Commission (IEC), and discusses how ISO and IEC standards provide a framework for designing, calibrating, assessing hearing test instruments and methods, and exchanging and comparing data globally. ISO and IEC standards for hearing tests improve accuracy, reliability, and consistency of test results by applying standardized methods and environments. Moreover, they promote international harmonization and data interoperability, enabling information exchange and research collaboration. Those standards for hearing tests are beneficial but have challenges and limitations, such as variation in equipment and calibration, lag in updating standards, variation in implementation and compliance, and lack of coverage of clinical aspects, cultural diversity, and linguistic diversity. These affect the quality and interpretation of test results. Adapting ISO or IEC standards locally would improve their applicability and acceptability, while balancing customization and compatibility with global standards.
Keywords: Hearingtests, Standardization
Introduction
Hearing tests with International Organization for Standardization (ISO) standards focus on accuracy and reliability in assessing individuals’ hearing abilities . These standards provide a standardized framework for conducting hearing tests, making it more reliable to compare results across different clinics, researchers, and countries. Obtaining accurate test results is one of the important parts of audiology and contributes to diagnosing hearing loss and determining appropriate interventions.
Hearing tests play a pivotal role in assessing individuals’ auditory capabilities and identifying potential hearing impairments. Evaluation of individual hearing threshold levels across different frequencies with hearing tests provides valuable insights into the functioning of the auditory system. Early detection of hearing loss is essential for timely intervention, as untreated hearing impairments can have significant social, emotional, and cognitive implications . Furthermore, hearing tests are not only relevant for clinical purposes but also play a vital role in occupational health and safety, ensuring that individuals are fit for specific job requirements that involve auditory communication or exposure to high noise levels [3-5].
The standardized hearing tests facilitate effective communication and collaboration among professionals in the field, enabling the exchange of data, research findings, and best practices. Moreover, the standards provide a foundation for benchmarking and monitoring trends in hearing health on a broader scale, contributing to the development of evidencebased interventions, public health policies, and quality assurance in audiological care. Ultimately, the implementation of standard in hearing testing ensures that individuals receive consistent and reliable evaluations of their hearing abilities, enabling appropriate interventions and support tailored to their specific needs.
Development of International Standards for Hearing Tests
The Western Electric 2-A developed by Harvey Fletcher and R.L. Wegel in 1923 was the first widely used commercial audiometer and it was operated with limited frequency and intensity scales without standard reference levels . Fletcher also contributed to promoting speech audiometry and developed a method in 1929 . The unstandardized calibration of audiometers led to the “Beasley Survey” conducted by the United States Public Health Service (USPHS) in 1935, providing data on average normal hearing thresholds across frequencies from 128 Hz to 8,192 Hz. In 1951, the American Standards Association (ASA) published the ASA-1951 standard, establishing the 0 dB threshold line in sound pressure level (SPL) at each frequency, gaining global recognition .
The ISO-64 standard was introduced in 1964, providing a zero-line reference for audiometer calibration and aligning with the ASA-1951. In 1969, the American National Standards Institute (ANSI) refined the ISO-64 and released the ANSI-69 standard, resolving the transition period between ASA-1951 and ISO-64. The current ISO standard for pure-tone audiometry is ISO 8253-1, first published in 1983, specifying general requirements, calibration procedures, and verification methods for pure-tone audiometers . In the United Kingdom, the British Society of Audiology (BSA) publishes recommended procedures for pure-tone audiometry, aligned with ISO standards. The American Speech-Language-Hearing Association (ASHA) published “Guidelines for Manual Pure-Tone Threshold Audiometry” in 2005 in the United States.
ISO 8253 is a series of standards for audiometric test methods. ISO 8253-1 specifies procedures and requirements for pure-tone air conduction and bone conduction threshold audiometry, including the maximum permissible ambient noise levels in audiometric rooms (ISO 8253-1: first edition in 1989, second edition in 2010) . ISO 8253-2 specifies sound field audiometry with pure-tone and narrow-band test signals presented by means of one or more loudspeakers, including the maximum permissible ambient noise levels in audiometric rooms (ISO 8253-2: first edition in 1989, second edition in 2009) . ISO 8253-3 specifies speech audiometry (ISO 8253-3: first edition in 1998, second edition in 2012, third editionin 2022) . Both ISO 8253-1 (2010) and ISO 8253-2 (2009) are standards for audiometric test methods. The establishment of standardized calibration protocols, such as ASA-1951, ISO-64, ANSI-69, and ANSI-96, has contributed to consistent and comparable hearing test results, ensuring accurate assessments and facilitating effective patient care.
Other ISO standards related to pure-tone audiometry include ISO 389 (1976), which focused on audiometer calibration. ISO 389 is a series of standards that deal with several essential aspects of hearing tests including reference of hearing threshold scales, testing environments, and calibration [12-20]. ISO 389-1 and ISO 389-2 specify the standard reference zeros for the scale of hearing threshold level applicable to pure-tone air conduction audiometers with supra-aural and insert earphones, respectively [12,13]. ISO 389-3 specifies the standard reference zero for the scale of hearing threshold level applicable to pure-tone bone-conduction audiometry . ISO 389-4 specifies the reference levels of narrow-band masking noise . ISO 389-5 specifies the reference equivalent threshold sound pressure levels for pure tones in the frequency range 8 kHz to 16 kHz . ISO 389-6 specifies the reference threshold of hearing for short duration test signal . ISO 389-7 specifies a reference threshold of hearing for the calibration of audiometric equipment used under specific listening conditions . Finally, ISO 389-8 specifies a reference equivalent threshold sound pressure levels for pure tones and circumaural earphones . In addition, ISO 389-9 specifies the preferred test conditions for determination of reference hearing threshold levels .
The International Electrotechnical Commission (IEC) focuses on standardizing various aspects of electrical and electronic technologies, including medical equipment used in audiology and hearing assessment [21-26]. While the ISO 389 and 8253 series of standards primarily address hearing measurement and assessment methods, the IEC standards deal with requirements and calibration methods of equipments such as audiometer [21-23], impedance audiometers , otoacoustic emissions (OAE) , and auditory brainstem response (ABR) test systems . The requirements and specification of devices to simulate the human ear and head, which are applied to calibrate the hearing test equipment, are also specified in IEC 60318 series [27-32]. The list of international standards related to the hearing test is summarized in Table 1.
Table 1.
International standards for hearing tests
| Category | No. | Title | Contents | Korean standard (KS) |
:---: :---:
| Audiometric test method | ISO 8253-1 | Acoustics — Audiometric test methods | • Pure-tone bone conduction audiometry methods (threshold determination, masking, automatic audiometry, screening) | KSIISO8253-1 |
| | Part 1: Pure-tone air and bone conduction audiometry | • Maintenance and calibration methods | |
| | • Qualifications for examiners, environmental conditions (location of subject and examiner, temperature, permissible ambient noise range) | |
| | • Preparation and education of subjects, headphone/bone vibrator use | |
| | • Hearing measurement, uncertainty evaluation methods | |
| ISO 8253-2 | Acoustics — Audiometric test methods | • Methods for sound field audiometry | KSIISO8253-2 |
| | Part 2: Sound field audiometry with pure-tone and narrow-band test signals | • Maintenance and calibration methods | |
| | • Explanation of signal sounds that can be used | |
| | • Examination environment (characteristics, settings, permissible ambient noise range), preparation and education of subjects, and reporting of results | |
| ISO 8253-3 | Acoustics — Audiometric test methods | • Methods for speech audiometry (threshold determination, masking, etc.) | KSIISO8253-3 |
| | Part 3: Speech audiometry | • Maintenance and calibration methods | |
| | • Recording methods for speech data (recording equipment, standard recording, speech data, verification, documentation) | |
| | • Examination environment (settings, permissible ambient noise range), subject preparation and education | |
| | • Reporting of results, measurement uncertainty evaluation methods | |
| Reference zero to the calibration of audiometric equipment | ISO 389-1 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the reference threshold level (RETSPL) for calibrating headphones | KSIISO389-1 |
| | Part 1: Reference equivalent threshold sound pressure levels for pure tones and supra-aural earphones | • Types of headphones | |
| | • Method for fixing headphones to a simulated ear canal | |
| | Simulated ear canal: an acoustic characteristic that simulates the average person’s ear acoustic characteristics and is generally used to calibrate headphones | |
| ISO 389-2 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the RETSPL for calibrating insert earphones (ER-3A) | KSIISO389-2 |
| | Part 2: Reference equivalent threshold sound pressure levels for pure tones and insert earphones | • Method for connecting insert earphones to a coupler | |
| ISO 389-3 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the RETSPL for calibrating bone vibrators | KSIISO389-3 |
| | Part 3: Reference equivalent threshold vibratory force levels for pure tones and bone vibrators | | |
| ISO 389-4 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the standard level for calibrating masking noise | KSIISO389-4 |
| | Part 4: Reference levels for narrow-band masking noise | | |
| ISO 389-5 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the RETSPL for calibrating headphones or earphones | KSIISO389-5 |
| | Part 5: Reference equivalent threshold sound pressure levels for pure tones in the frequency range 8 kHz to 16 kHz | • Types of headphones and earphones | |
| ISO 389-6 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the standard level for calibrating short signals such as click sounds | N/A |
| | Part 6: Reference threshold of hearing for test signals of short duration | | |
| ISO 389-7 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the standard level for calibrating loudspeakers | KSIISO389-7 |
| | Part 7: Reference threshold of hearing under free-field and diffuse-field listening conditions | | |
| ISO 389-8 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the reference threshold level (RETSPL) for calibrating circumaural earphones | N/A |
| | Part 8: Reference equivalent threshold sound pressure levels for pure tones and circumaural earphones | | |
| ISO 389-9 | Acoustics — Reference zero for the calibration of audiometric equipment | • Specifies the preferred test condition including environmental and ambient condition for the determination of reference hearing threshold levels | N/A |
| | Part 9: Preferred test conditions for the determination of reference hearing threshold levels | | |
| Audiometric equipment | IEC 60645-1 | Electroacoustics — Audiometric equipment | • Requirements and specifications for each type of audiometer and verification criteria | KSCIEC60645-1A |
| | Part 1: Equipment for pure-tone and speech audiometry | • Measured signals, transducers (headphones, bone vibrators, etc.), level adjustment, reference sound | |
| | • List of standards for calibration | |
| | • Audiogram format displayed on the equipment | |
| IEC 60645-2 (withdrawn and replaced by IEC 60645-1:2017) | Audiometers — Part 2: Equipment for speech audiometry | • Requirements and specifications for each type of audiometer | KSCIEC60645-2A |
| IEC 60645-3 | Electroacoustics — Audiometric equipment | • Types and characteristics of short signals | KSCIEC60645-3A |
| | Part 3: Test signals of short duration | • Calibration and measurement methods for short signals | |
| IEC 60645-5 | Electroacoustics — Audiometric equipment | • General specifications and verification criteria, items that manufacturers must specify | KSCIEC60645-5A |
| | Part 5: Instruments for the measurement of aural acoustic impedance/admittance | • Calibration methods (cavity specifications, probe connection method, expanded uncertainty) | |
| | • Equipment usage and environmental conditions | |
| | • Audiogram format displayed on the equipment | |
| IEC 60645-6 | Electroacoustics — Audiometric equipment | • General specifications, essential functions, verification criteria, items that manufacturers must specify | KSCIEC60645-6 |
| | Part 6: Instruments for the measurement of otoacoustic emissions | • Routine calibration methods, equipment usage and environmental conditions | |
| IEC 60645-7 | Electroacoustics — Audiometric equipment | • Auditory brainstem response (ABR) device types, general specifications, quality assurance system, verification criteria | KSCIEC60645-7 |
| | Part 7: Instruments for the measurement of auditory brainstem responses | • Calibration parameters, environmental conditions | |
| Simulators of human head and ear | IEC 60318-1 | Electroacoustics — Simulators of human head and ear | • Specifies the requirements of ear simulator having the overall acoustic impedance of the device approximates that of the normal human ear for calibrating supra-aural earphones | KSCIEC60318-1 |
| | Part 1: Ear simulators for the calibration of supra-aural earphones | • Structure, headphone connection method | |
| | • Calibration method of its acoustical transfer impedance and its expanded uncertainty | |
| IEC 60318-3 | Electroacoustics — Simulators of human head and ear | • Specifies the acoustic coupler for the measurement of supra-aural audiometric earphones | KSCIEC60318-3 |
| | Part 3: Acoustic coupler for the calibration of supra-aural earphones used in audiometry | • Structure, headphone connection method | |
| IEC 60318-4 | Electroacoustics — Simulators of human head and ear | • Specifies the simulated ear canal used as a coupler for calibrating ear insert earphones | KSCIEC60318-4 |
| | Part 4: Occluded-ear simulator for the measurement of earphones coupled to the ear by means of ear inserts | • Structure, headphone connection method | |
| IEC 60318-5 | Electroacoustics — Simulators of human head and ear | • Specifies the 2 cm 3 coupler used for calibrating insert earphones or verifying hearing aid output | KSCIEC60318-5 |
| | Part 5: 2 cm 3 coupler for the measurement of hearing aids and earphones coupled to the ear by means of ear inserts | • Structure, insert earphone or hearing aid connection method, and expanded uncertainty | |
| IEC 60318-6 | Electroacoustics — Simulators of human head and ear | • Specifies the artificial mastoid used as a coupler for calibrating bone vibrators | KSCIEC60318-6 |
| | Part 6: Mechanical coupler for the measurement of bone vibrators | • Structure, bone vibrator connection method | |
| | • Calibration method of its mechanical impedance and its expanded uncertainty | |
| IEC/TS 60318-7 | Electroacoustics — Simulators of human head and ear | • Describes a head and torso simulator, or manikin, intended for the measurement of air-conduction hearing aids | KSCIECTS60318-7 |
| | Part 7: Head and torso simulator for the measurement of air-conduction hearing aids | • Specifies the manikin in terms of both its geometrical dimensions and its acoustical properties | |
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Overview of International Standards for Hearing Tests
Pure-tone audiometry in ISO and IEC standards
ISO 8253-1 consists of procedures and requirements for pure-tone audiometry . Air conduction audiometry involves presenting the test signal through earphones, while bone conduction audiometry uses a bone vibrator placed on the mastoid or forehead. It is recommended to start with air conduction measurements followed by bone conduction measurements. Threshold levels can be determined using fixedfrequency audiometry or sweep-frequency audiometry. Both ears’ hearing threshold levels should be determined separately, and masking noise may be applied to the non-test ear under specified conditions.
The calibration of audiometric equipment follows the standard reference zero in ISO 389 series [12-20], and equipment requirements are outlined in IEC 60645-1 . Qualified testers or those under their supervision should conduct the tests, and care should be taken to prevent subject fatigue. The test environment conditions, including ambient sound pressure level, are specified, and the uncertainty of measurement results should be evaluated according to ISO/IEC Guide 98-3 .
ISO 8253-1 focuses on the preparation and instruction of test subjects before audiometric testing and the correct positioning of transducers. It emphasizes avoiding recent noise exposure and allowing subjects to arrive early to minimize errors. An otoscopic examination and preliminary tuning fork tests are recommended to assess hearing loss and masking requirements. Clear and appropriate instructions should be given to the subjects, specifying the response task and the importance of remaining still. Proper placement of transducers, such as earphones and bone vibrators, is crucial for accurate testing.
The procedure for determining air conduction hearing threshold levels using fixed-frequency audiometry involves presenting test tones manually or with an automatic-recording audiometer. The specific order of test tones should be followed, starting from 1,000 Hz and going upwards, followed by the lower frequency range. Threshold measurements are performed with and without masking noise. Two methods are specified for threshold measurements without masking: ascending method and bracketing method. For threshold measurements with masking, masking noise is applied to the nontest ear if necessary.
Estimation of hearing threshold level is done by determining the lowest level at which responses occur in more than half of the ascents for the ascending method and averaging the lowest levels of responses in ascents and descents separately, then calculating the mean of these two averages for the bracketing method. ISO 8253-1 also briefly mentions procedures for automatic recording audiometry, computer-controlled threshold determination, sweep-frequency audiometry for air conduction threshold measurements, and bone conduction hearing threshold audiometry. It provides guidelines for bone conduction testing and masking procedures in bone conduction audiometry. Additionally, it covers guidelines for screening audiometry, both manually controlled and computer-controlled, for determining pass or fail results in screening tests.
Speech audiometry in ISO 8253-3
Speech audiometry is used for diagnostic evaluation and audiological rehabilitation. In order to ensure minimum requirements of precision and comparability between different test procedures including speech recognition tests in different languages, ISO 8253-3 specifies requirements for the composition, validation and evaluation of speech test materials, and the realization of speech recognition tests. It provides procedures for presenting recorded speech test material through earphones or loudspeakers. Methods for using noise for masking or as competing sound are described.
ISO 8253-3 also provides guidelines for test preparation, instructions to the subject, response modes, and intervals between test items in speech audiometry. Prior to speech audiometry, pure-tone audiometry is assumed to have been conducted. Preparation includes an otoscopic examination and confirming the subject’s understanding and ability to reproduce the test material. Clear instructions of the tester with an appropriate language are necessary and the subject’s response can be spoken, written, or indicated through a keyboard.
ISO 8253-3 specifies the procedure for determining the speech detection threshold level during monaural testing. The procedure involves using connected speech as the speech signal and starting with a high level, approximately 30 dB above the average of the subject’s pure-tone hearing threshold levels at 500 Hz, 1,000 Hz, and 2,000 Hz. The level is then decreased in 20 dB steps until the subject no longer responds and then increased in 5 dB steps until the subject responds. To determine the speech recognition threshold level, the procedure involves using complete test lists of single words, phrases, or sentences and descending procedures with step sizes of 5 dB and 2 dB. Adaptive procedures using fixed step sizes are also described.
In speech audiometry, avoid repeating test items in the same session and present a complete test list. Determine the speech recognition threshold beforehand or familiarize the subject with test items at an audible level. Choose appropriate test levels based on the purpose of the assessment (maximum speech recognition score, speech recognition score, half-optimum speech level). Express scores as percentages and record the achieved level. Use contralateral masking to prevent speech signals from reaching the non-test ear during monaural speech audiometry, adjusting masking level as needed for accuracy. During testing speech audiometry with competing sound, the recommended speech level is 65 dB, which corresponds to normal speech level in conversation. The level of the competing sound can be fixed or variable, with a recommended fixed noise level of 60 dB or variable levels changed in steps of 5 dB or less. Additionally, this standard discusses two methods for assessing speech recognition scores with competing sounds. The first method involves determining the score at a fixed speech-to-noise ratio, where the equipment is set to the required speech level, and the subject is familiarized with test items at a low competing sound level. The score is calculated as a percentage based on the desired competing sound level. The second method is for measuring the speech recognition threshold with a competing sound of varying loudness. It includes increasing the competing sound level gradually while presenting test items until the subject incorrectly recognizes one item. The threshold is determined using specific criteria or linear interpolation. Equations are provided for calculating the competing sound level for a 50% correct (for example, if you answer 2 out of 3 or 3 out of 5 times on a test, you meet the criterion of “50% or more”) speech recognition score based on the starting level and step size.
Tympanometry in IEC 60645-5
Middle ear examination or aural acoustic admittance measurement including tympanometry, acoustic reflex test, acoustic reflex decay test, and Eustachian test does not require a patient’s behavioral response, is inexpensive and has a very short measurement time. It is an important component of the audiologic test battery and is a physiological measure that must be included in any comprehensive audiologic assessment. The purpose of tympanometry is to indirectly evaluate the function of middle ear by measuring aural acoustic admmittances with a probe to the ear cannel attached. This test can be evaluated with an aural acoustic immittance instrument, and in particular, the eardrum motility test and the acoustic reflex test can be performed using a single pure tone, multi-frequency stimulation, or broadband stimulation.
IEC 60645-5 specifies the calibration of aural acoustic immittance instruments. The standard includes procedures for calibrating instruments that measure admittance, impedance, reflectance, and absorbance. The standard also specifies procedures for calibrating instruments that measure acoustic reflex thresholds and decay times. According to the newest international and national standards (IEC 60645-5 , ANSI S3.39-1987 [R2012]), the aural acoustic immittance instrument basically includes six components: 1) calibration cavity, 2) acoustic immittance analysis system, 3) probe assembly/unit and signal, 4) pneumatic air-pressure pump system, 5) acoustic reflex activator system, and 6) tympanogram and acoustic reflex plotting system, each of these components should meet set standards. This standard covers instruments designed primarily for the measurement of acoustic impedance/admittance in the human external acoustic meatus using a stated probe tone. It is recognized that other probe signals may also be used. The standard defines the characteristics to be specified by the manufacturer, lays down performance specifications for three types of instruments, and specifies the facilities to be provided on these types. This standard describes methods of test to be used for approval testing and guidance on methods for undertaking routine calibration. The purpose of this standard is to ensure that measurements made under comparable test conditions with different instruments complying with the standard will be consistent.
Otoacoustic emissions in IEC 60645-6
OAE testing is a non-invasive and objective method used to assess the function of the inner ear. It measures sound pressure levels in eardrum representing the outer hair cells’ responses with or without sound stimulation. OAE testing is widely used for various purposes for early screening, accurate diagnosis, and monitoring of hearing health in various populations. It includes testing newborn hearing screening, assessing individual’s hearing loss (especially for in infants and individuals with communication difficulties), monitoring hearing health, identifying cochlear pathologies, and identifying auditory neuropathy/dys-synchrony [34,35].
IEC 60645-6 pertains to instruments primarily designed for measuring OAE, elicited by acoustic probe stimuli. The standard defines the characteristics to be specified by the manufacturer, specifies minimum mandatory functions for two types of instruments, and provides performance specifications applicable to both instrument types. The standard describes methods to be used to demonstrate conformance with the specifications in this document and guidance on methods for periodic calibration.
Recent notable technical changes in IEC 60645-6:2022 include defining the nominal test frequency for distortion product optoacoustic emissions (DPOAE) as the higher of the two frequencies, f2, allowable deviation of the stimulus signal for TEOAE, the frequency range for DPOAE stimulus signals, the stimulus level requirements for TEOAE/DPOAE, harmonic distortion requirements for DPOAE, and a minimum measurement range for DPOAE.
Auditory brainstem response in IEC 60645-7
ABR test is objective electrophysiological method used to assess hearing function and neural responses to sound stimuli. ABR measures the electrical activity of the auditory nerve and brainstem in response to sounds, providing information about hearing thresholds and neural integrity . It is used in newborn hearing screening, threshold estimation, and diagnosing auditory neuropathy. ABR test serves as a critical tool in universal newborn hearing screening programs, aiding in the early identification of hearing loss in infants. Estimating hearing thresholds with ABR is particularly valuable for patients who cannot provide consistent behavioral responses or are difficult to test, such as young children or those with developmental disabilities. Additionally, ABR test plays a crucial role in diagnosing auditory neuropathy/dys-synchrony, a condition characterized by impaired neural transmission despite normal cochlear function. Furthermore, during certain surgeries involving the brainstem, ABR is utilized for intraoperative monitoring, ensuring the safety and protection of the auditory pathway. With its objective and reliable measurements of neural responses to sound stimuli, ABR significantly contributes to the accurate diagnosis, monitoring, and intervention decisions in the field of audiology.
IEC 60645-7 applies to instruments designed for the measurement of evoked potentials from the inner ear, the auditory nerve and the brainstem, evoked by acoustic and/or vibratory stimuli of short duration. IEC 60645-7 defines the characteristics to be specified by the manufacturer, specifies performance requirements for two types of instrument, screening and diagnostic, and specifies the functions to be provided on these types. The purpose of IEC 60645-7 is to ensure that measurements made under comparable test conditions with different instruments complying with this standard will be consistent. It is not intended to restrict development or incorporation of new features, nor to discourage innovative approaches. The application of electric stimuli for special purposes is beyond the scope of this standard.
Devices to simulate the auditory system in IEC 60318
IEC 60318 series specify the devices to simulate the response of human auditory system and these devices are employed to calibrate the output level of audiometer with each transducer. For the air-conduction audiometry, two types of devices, ear simulators and acoustic coupler, are defined.
Ear simulator is the device designed to have the overall acoustic impedance of the device approximates that of the normal human ear . IEC 60318-1 specifies the requirement and its calibration method of ear simulators for supraaural earphones . The requirement of occluded-ear simulators for earphones coupled to the ear by means of ear inserts is specified in IEC 60318-4 . The basic requirement is given in terms of the acoustical transfer impedance and its permissible deviation is also specified in the related standards.
Acoustic coupler is a device designed to have a cavity of predetermined shape and volume, which does not necessarily approximate the acoustical impedance of the normal human ear. IEC 60318-3 and IEC 60318-5 specify the required dimensions and connection method for supra-aural and earinsert earphones, respectively [28,30].
For calibrating the bone conduction stimuli, the mechanical coupler simulating the mechanical impedance of mastoid position is employed and the reference mechanical impedance and permissible deviation are specified in IEC 60318-6 . The basic method and condition for calibration of bone vibrator are also described in this standard.
IEC TS 60318-7 describes the requirement of the head and torso to simulate the auditory response including the effect of head and shoulder in free-field or specific room condition . This standard specifies the geometrical dimensions and acoustical properties of system. These types of devices are not directly required to calibrate the audiometric devices for clinical purpose; however, it can be applied for the investigation concerning various practical situations.
Benefits and Challenges of International Standards
Benefits of using international standards in hearing testing
Improving test accuracy and reliability
International standards for various hearing tests establish guidelines for equipment calibration, test procedures, and verification methods. By adhering to these standards, healthcare professionals can ensure the accuracy and reliability of their testing equipment, leading to high-quality and trustworthy results.
Ensuring consistency and comparability of results
ISO and IEC standards ensure that hearing tests are conducted in a consistent manner, regardless of the different testing places or healthcare providers. Measurement consistency in hearing tests allows for improved comparability of test results across different clinics, researchers, and countries, accurate assessments, and treatment decisions.
Improved patient care
By following standards, healthcare providers can enhance the quality of patient care. Standardized protocols help ensure that hearing tests are administered correctly, minimizing errors and inconsistencies. Accurate and reliable test results aid in making appropriate diagnoses, developing personalized treatment plans, and monitoring the effectiveness of interventions over time.
Facilitating international collaboration and research
ISO and IEC standards are globally recognized and accepted. This recognition promotes interoperability and harmonization of hearing test practices across different countries and healthcare systems. It facilitates collaboration, research, and the exchange of information among professionals working in the field of audiology. Those standards provide a foundation for research and development in the field of hearing testing. They provide a common language and framework for researchers to compare and analyze data, fostering advancements in diagnostic techniques, treatment modalities, and technological innovations. These standards may be required or recommended by regulatory bodies and accreditation organizations. Adhering to these standards helps healthcare facilities and professionals comply with regulatory requirements, ensuring adherence to best practices and quality standards in hearing testing.
Challenges and limitations associated with ISO/IEC standards
Potential variations in equipment and calibration
While ISO/IEC standards for hearing testing bring numerous benefits, there are also challenges and limitations associated with their implementation. One significant challenge is the potential for variations in equipment and calibration across different clinics, manufacturers, and regions. Despite standardization efforts, there may still be differences in audiometric equipment and calibration procedures used by different providers. These variations can impact the accuracy and comparability of test results, leading to discrepancies in diagnoses and treatment decisions. Another limitation is the need for ongoing updates and revisions of ISO/IEC standards to keep pace with advancements in technology and research. As new equipment and testing methods emerge, standards may need to be updated to ensure their relevance and effectiveness. However, the process of updating standards can be time-consuming and may lag behind technological advancements, leading to potential gaps between current practice and standard requirements. Additionally, the adoption and adherence to ISO/IEC standards may vary across different healthcare settings and regions. Some clinics or countries may have limited resources or awareness of the standards, resulting in inconsistent implementation. This variation can affect the consistency and comparability of test results, hindering international collaboration and data exchange.
Moreover, ISO standards primarily focus on technical aspects of hearing testing and may not address all clinical considerations. Audiologists and healthcare providers need to consider individual patient factors, such as medical history, communication needs, and cognitive abilities, which may not be explicitly covered in the standards. Clinical judgment and expertise are still essential in the interpretation and application of test results.
Furthermore, ISO standards may not fully address cultural and linguistic diversity in patient populations. Different languages, dialects, and cultural norms can influence the administration and interpretation of hearing tests. Adapting the standards to accommodate these diversities can be challenging and may require additional guidelines or recommendations.
Consideration of regional and cultural factors
Regional and cultural factors are crucial to consider when implementing ISO standards in hearing testing. These factors can affect how hearing tests are conducted, interpreted, and accepted in different populations. Here are some key considerations.
Language barriers can affect the administration and interpretation of hearing tests. Healthcare providers should ensure effective communication with patients who have limited proficiency in the dominant language. They should use translation services or interpreters, and provide test instructions, materials, and communication that are culturally sensitive and accessible to diverse linguistic backgrounds.
Cultural norms and beliefs may influence individuals’ attitudes towards hearing health and their willingness to participate in testing. Understanding cultural perspectives on hearing loss, stigma, and help-seeking behavior can help healthcare providers tailor their approach to testing and counseling. Sensitivity to cultural practices, taboos, and religious beliefs is essential to establish trust and foster open communication.
Environmental factors, such as ambient noise levels or testing room characteristics, may vary across different regions and healthcare settings. Adhering to ISO standards for ambient noise control is essential, but additional considerations may be needed to account for regional variations. Local norms and practices regarding test conditions and patient positioning should also be taken into account to ensure patient comfort and accurate test results.
Consideration should be given to the appropriateness of test materials and stimuli for specific cultural contexts. For example, certain frequency ranges or speech sounds may have different significance or salience in different languages or cultural groups. Adapting test materials to reflect the language, dialect, or specific sound characteristics of the population being tested can improve the validity and reliability of results. Healthcare providers should receive training on cultural competence and diversity to better understand and address the needs of diverse patient populations. This training can enhance their ability to adapt testing procedures, communicate effectively, and provide culturally sensitive counseling and support.
Conclusion
Following ISO and IEC standards in hearing tests improves accuracy and reliability of test results by applying consistent methods and environments. They provide a comprehensive framework for designing, calibrating, and assessing hearing test instruments, as well as conducting both subjective and objective hearing tests. Adherence to these standards allows hearing professionals to confidently measure and interpret results, mitigating the impact of various complex factors. Regular calibration checks and instrument maintenance ensure precise results, fostering trust in hearing test outcomes.
Moreover, these globally recognized standards promote international harmonization and data interoperability, facilitating information exchange and research collaboration across borders. Comparing findings from different locations enhances understanding of hearing-related issues and improves treatment strategies.
Additionally, adapting ISO or IEC guidelines to a country’s context and language enhances their applicability and local acceptance. Customization addresses cultural factors and unique healthcare challenges while preserving the core principles of the original standards. Striking a balance between customization and maintaining compatibility with international benchmarks ensures alignment with global best practices and supports collaboration in research and advancements.
Therefore, following international standards in hearing tests is essential for ensuring the quality and reliability of hearing assessment and treatment across different settings and populations.
Acknowledgments
This research was supported by “Regional Innovation Strategy (RIS)” through the National Research Foundation of Korea (NRF) funded by the Ministry of Education(MOE) (2022RIS-005).
Footnotes
Conflicts of Interest
The authors have no financial conflicts of interest.
Author Contributions
Conceptualization: Kyung-Ho Park, Young Jun Seo. Investigation: Michelle J. Suh. Methodology: Jihyun Lee, Wan-Ho Cho, In-Ki Jin. Project administration: Young Jun Seo. Resources: Jihyun Lee, Young Jun Seo. Supervision: Young Jun Seo. Writing—original draft: Michelle J. Suh, Young Jun Seo. Writing—review & editing: Wan-Ho Cho, In-Ki Jin, Tae Hoon Kong, Soo Hee Oh, Hyo-Jeong Lee, Seong Jun Choi, Dongchul Cha. Approval of final manuscript: all authors.
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Abstract
Introduction
Development of International Standards for Hearing Tests
Overview of International Standards for Hearing Tests
Benefits and Challenges of International Standards
Conclusion
Acknowledgments
Footnotes
REFERENCES
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4303 | https://www.ck12.org/flexi/chemistry/determining-molecular-formulas/determine-the-molar-mass-of-choh/ | Determine the molar mass of CH₃OH.
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The molar mass of CH3OH (methanol) can be calculated by adding up the molar masses of its constituent atoms:
Carbon (C): 12.01 g/mol
Hydrogen (H): 1.01 g/mol × 4 = 4.04 g/mol
Oxygen (O): 16.00 g/mol
So, the molar mass of CH3OH can be calculated as follows:
@$\begin{align}12.01 \, \text{g/mol} + 4.04 \, \text{g/mol} + 16.00 \, \text{g/mol} = 32.05 \, \text{g/mol}\end{align}@$
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How can one calculate mmol from molecular weight?Maleic acid is an organic compound composed of 41.39% C, 3.47% H, and the rest oxygen. If 0.129 mole of maleic acid has a mass of 15.0 g, what are the empirical and molecular formulas of maleic acid?What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 g/mol?
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Calculus Cal
Ariyan R.
asked • 04/24/22
Find the absolute extrema of 𝑓(𝑥) = 2 sin 𝑥 − cos 2𝑥 on the interval [0, 2pi]
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f'(x) = 2cosx + 2sin2x = 0; 2cosx(1 + 2sinx) = 0; cosx = 0; x = π/2 or x = 3π/2
sinx = - 1/2; x = 7π/6 or x = 11π/6
f(0) = - 1
f(π/2) = 3 abs max
f(7π/6) = - 1 - 1/2 = - 3/2 abs min
f(3π/2) = - 1
f(11π/6) = - 3/2 abs min
f(2π) = - 1
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4305 | https://www.youtube.com/watch?v=v_Trr06YMEo | Solve x³ - 3x - 2 = 0
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x cube - 3x - 2 equal to zero
x³ - 3x - 2 = 0
How to solve x^3-3x-2=0
Solve by Factoring x^3-3x^2-2=0
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Transcript:
hi kid in this video lecture I'm going to find the root of this equation so first writing the given xq - 3x - 2 = 0 so we can easily achieve the output if I write there xq and here x² + x² and subtracting with - x² and - 3x - 2 = 0 so this is your addition and subtraction so x² - x² the value comes zero okay so there is no effect with this equation now thereafter or - 3x we can write so first writing xq x² - x² so - 3x we can write there - x - 2x or here - 3x we can write - 2x - x so because this place 2 is there so that we can write here - 3 x - x - 2x - 2 so because two is there uh so that uh making go pair with uh this twos after that the next is let's take the common so between these twos x² is the common so we can write x + 1 and between these twos - x is the common so we can write x + 1 and between these twos we can or or take the common min-2 so this is now x + 1 = 0 now the next is among these threes x + 1 is the common so x + 1 is the common let's take that so we can write here x² - x - 2 and this is equal to zero after that the next is so let's write there x + 1 and this we can write here x² so this we can write or Min - x we can write here - 2 - 2x + x so if you do the calculation this is equal to - x okay so I did like this because we need to take the common again and do the calculation so Min - x we can write here or let's write X and this place - 2 x - 2 = 0 then we can take the common again between these two so between these twos X is the common so let's write here x + 1 and between these twos Min - 2 is the common this is now x + 1 = 0 so after that uh here you can see x + 1 is the again common between these twos so let's take the common so we can write x + 1 and here this is now x - 2 equal to Zer so let's uh take this common is there now or we can write after calculating this x + 1 a² and this is here x - 2 = 0 or we can write this one x + 1 s² = 0 making the relation with this one with zero and that one with zero so this we can write x - 2 = 0 right we can write this one x + 1 whole s we can write x + 1 and x + 1 tce time equal to 0 so this is twice St so let's make the Rel so x + 1 = to 0 and again we can write here x + 1 = 0 so here x = -1 that's come x = -1 and this place we can write two moving on the right direction so minus that change to plus so we can write here 0 + 2 and x = + 2 that's come there so here you can see or directly add this so there I'm going to write x + 1 = 0 so X this moving on the right so 0 - 1 that come X = to -1 that's come there okay minus one so directly then I write so now you can see the value that come of X is there xal to that come - one again minus one and 1 is two so this is the answer or you can write 1 - one or two this also you can write the answer if you have option like this then you can write this and either you can write this so one time I'm going to repeat this all so this is the equation there so we can easily achieve the output if I add there plus x² and subtract there - x² then we can take the common x² is the common take that then you can see this term this terms x + 1 is there x + 1 you can take the common then you can get here x² - x - 2 = 0 after that one part that has been done so between these twos we have to solve this one so x² - x - 2 equal to Z so we can easily achieve the output if I write there Min - x + 2x - 2 = 0 so you can see plus 2X okay I did the mistake so this place right there plus then you can write there minus so - 2x + x that is equal to here - x so this is equal to of this one after that or between these twos again you need take the common X is the common then you can get x +1 x + 1 so x + 1 x + 1 or there is the there is so this is now X + 1 sare come and x - 2 = 0 after that x + 1 sare = to 0 so we can write T terms so here the value come X = to -1 twice times and x value that come so zero if I make down on the right then x = 2 that's good so this is the way we can solve this question so I hope guys this video is the helpful for you so now this video it's over so thanks for watching see next thank you |
4306 | https://www.chegg.com/homework-help/questions-and-answers/nearsighted-eye-corrected-placing-diverging-lens-front-eye-lens-create-virtual-image-dista-q12721273 | Solved A nearsighted eye is corrected by placing a diverging | Chegg.com
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A nearsighted eye is corrected by placing a diverging lens in front of the eye. The lens will create a virtual image of a distant object at the far point (the farthest an object can be from the eye and still be in focus) of the myopic viewer where it will be clearly seen. In the traditional treatment of myopia, an object at infinity is focused to the far
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Question: A nearsighted eye is corrected by placing a diverging lens in front of the eye. The lens will create a virtual image of a distant object at the far point (the farthest an object can be from the eye and still be in focus) of the myopic viewer where it will be clearly seen. In the traditional treatment of myopia, an object at infinity is focused to the far
A nearsighted eye is corrected by placing a diverging lens in front of the eye. The lens will create a virtual image of a distant object at the far point (the farthest an object can be from the eye and still be in focus) of the myopic viewer where it will be clearly seen. In the traditional treatment of myopia, an object at infinity is focused to the far point of the eye. If an individual has a far point of 31.5 cm, prescribe the correct power of the lens that is needed. Assume that the distance from the eye to the lens is negligible.
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4307 | https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_General_Chemistry/02%3A_Atoms_Molecules_and_Ions/2.3%3A_Atomic_Structure_and_Symbolism | Skip to main content
2.3: Atomic Structure and Symbolism
Last updated
: Jun 5, 2019
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2.2: Evolution of Atomic Theory
2.4: Chemical Formulas
Page ID
: 105546
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Skills to Develop
Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
Define the atomic mass unit and average atomic mass
Calculate average atomic mass and isotopic abundance
The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom’s mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom’s volume. The diameter of an atom is on the order of 10−10 m, whereas the diameter of the nucleus is roughly 10−15 m—about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure ).
Figure : If an atom could be expanded to the size of a football stadium, the nucleus would be the size of a single blueberry. (credit middle: modification of work by “babyknight”/Wikimedia Commons; credit right: modification of work by Paxson Woelber).
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 10−23 g, and an electron has a charge of less than 2 10−19 C (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu. (This isotope is known as “carbon-12” as will be discussed later in this module.) Thus, one amu is exactly of the mass of one carbon-12 atom: 1 amu = 1.6605 10−24 g. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 10−19 C.
A proton has a mass of 1.0073 amu and a charge of 1+. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1− and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton. The properties of these fundamental particles are summarized in Table . (An observant student might notice that the sum of an atom’s subatomic particles does not equal the atom’s actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu, slightly larger than the 12.00 amu of an actual carbon-12 atom. This “missing” mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)
Table : Properties of Subatomic Particles
| Name | Location | Charge (C) | Unit Charge | Mass (amu) | Mass (g) |
| electron | outside nucleus | | 1− | 0.00055 | |
| proton | nucleus | | 1+ | 1.00727 | |
| neutron | nucleus | 0 | 0 | 1.00866 | |
The number of protons in the nucleus of an atom is its atomic number (Z). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of neutrons is therefore the difference between the mass number and the atomic number: A – Z = number of neutrons.
Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:
Atomic charge = number of protons − number of electrons
As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+). A neutral oxygen atom (Z = 8) has eight electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).
Example : Composition of an Atom
Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure ).
Figure : (a) Insufficient iodine in the diet can cause an enlargement of the thyroid gland called a goiter. (b) The addition of small amounts of iodine to salt, which prevents the formation of goiters, has helped eliminate this concern in the US where salt consumption is high. (credit a: modification of work by “Almazi”/Wikimedia Commons; credit b: modification of work by Mike Mozart)
The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.
Solution
The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].
Exercise
An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?
Answer
: 78 protons; 117 neutrons; charge is 4+
Chemical Symbols
A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure ). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).
Figure : The symbol Hg represents the element mercury regardless of the amount; it could represent one atom of mercury or a large amount of mercury. Image used with permission from Wikipedia (user: Materialscientist).
The symbols for several common elements and their atoms are listed in Table . Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Symbols have one or two letters, for example, H for hydrogen and Cl for chlorine. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table.
Table : Some Common Elements and Their Symbols
| Element | Symbol | Element | Symbol |
| aluminum | Al | iron | Fe (from ferrum) |
| bromine | Br | lead | Pb (from plumbum) |
| calcium | Ca | magnesium | Mg |
| carbon | C | mercury | Hg (from hydrargyrum) |
| chlorine | Cl | nitrogen | N |
| chromium | Cr | oxygen | O |
| cobalt | Co | potassium | K (from kalium) |
| copper | Cu (from cuprum) | silicon | Si |
| fluorine | F | silver | Ag (from argentum) |
| gold | Au (from aurum) | sodium | Na (from natrium) |
| helium | He | sulfur | S |
| hydrogen | H | tin | Sn (from stannum) |
| iodine | I | zinc | Zn |
Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists or locations; for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements.
Isotopes
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure ). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as 24Mg, 25Mg, and 26Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, 24Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” 25Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.
Figure : The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.
Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table . Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized 2H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized 3H, is also called tritium and sometimes symbolized T.
Table : Nuclear Compositions of Atoms of the Very Light Elements
| Element | Symbol | Atomic Number | Number of Protons | Number of Neutrons | Mass (amu) | % Natural Abundance |
| hydrogen | (protium) | 1 | 1 | 0 | 1.0078 | 99.989 |
| (deuterium) | 1 | 1 | 1 | 2.0141 | 0.0115 |
| (tritium) | 1 | 1 | 2 | 3.01605 | — (trace) |
| helium | | 2 | 2 | 1 | 3.01603 | 0.00013 |
| | 2 | 2 | 2 | 4.0026 | 100 |
| lithium | | 3 | 3 | 3 | 6.0151 | 7.59 |
| | 3 | 3 | 4 | 7.0160 | 92.41 |
| beryllium | | 4 | 4 | 5 | 9.0122 | 100 |
| boron | | 5 | 5 | 5 | 10.0129 | 19.9 |
| | 5 | 5 | 6 | 11.0093 | 80.1 |
| carbon | | 6 | 6 | 6 | 12.0000 | 98.89 |
| | 6 | 6 | 7 | 13.0034 | 1.11 |
| | 6 | 6 | 8 | 14.0032 | — (trace) |
| nitrogen | | 7 | 7 | 7 | 14.0031 | 99.63 |
| | 7 | 7 | 8 | 15.0001 | 0.37 |
| oxygen | | 8 | 8 | 8 | 15.9949 | 99.757 |
| | 8 | 8 | 9 | 16.9991 | 0.038 |
| | 8 | 8 | 10 | 17.9992 | 0.205 |
| fluorine | | 9 | 9 | 10 | 18.9984 | 100 |
| neon | | 10 | 10 | 10 | 19.9924 | 90.48 |
| | 10 | 10 | 11 | 20.9938 | 0.27 |
| | 10 | 10 | 12 | 21.9914 | 9.25 |
Atomic Mass
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.
The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.
For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are 10B with a mass of 10.0129 amu, and the remaining 80.1% are 11B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:
It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.
Example : Calculation of Average Atomic Mass
A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu), 0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?
Solution
The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)
Exercise
A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25Mg atoms (mass 24.99 amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.
Answer
: 24.31 amu
We can also do variations of this type of calculation, as shown in the next example.
Example : Calculation of Percent Abundance
Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
Solution
The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37Cl times the mass of 37Cl.
If we let x represent the fraction that is 35Cl, then the fraction that is 37Cl is represented by 1.00 − x.
(The fraction that is 35Cl + the fraction that is 37Cl must add up to 1, so the fraction of 37Cl must equal 1.00 − the fraction of 35Cl.)
Substituting this into the average mass equation, we have:
So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% 35Cl and 24.24% 37Cl.
Exercise
Naturally occurring copper consists of 63Cu (mass 62.9296 amu) and 65Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes?
Answer
: 69.15% Cu-63 and 30.85% Cu-65
Figure : Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr.
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure ), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.
Video : Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.
Summary
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly of the mass of a carbon-12 atom and is equal to 1.6605 10−24 g.
Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.
Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.
Key Equations
Glossary
anion
: negatively charged atom or molecule (contains more electrons than protons)
atomic mass
: average mass of atoms of an element, expressed in amu
atomic mass unit (amu)
: (also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to of the mass of a 12C atom
atomic number (Z)
: number of protons in the nucleus of an atom
cation
: positively charged atom or molecule (contains fewer electrons than protons)
chemical symbol
: one-, two-, or three-letter abbreviation used to represent an element or its atoms
Dalton (Da)
: alternative unit equivalent to the atomic mass unit
fundamental unit of charge
: (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 10−19 C
ion
: electrically charged atom or molecule (contains unequal numbers of protons and electrons)
mass number (A)
: sum of the numbers of neutrons and protons in the nucleus of an atom
unified atomic mass unit (u)
: alternative unit equivalent to the atomic mass unit
Contributors
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at
2.2: Evolution of Atomic Theory
2.4: Chemical Formulas |
4308 | https://journal.ecust.edu.cn/cn/article/pdf/preview/10.14135/j.cnki.1006-3080.20210506010.pdf | 基于CFD-DEM的固液分级过滤模拟 赵钟杰 张建鹏 唐艳玲 肖桐 黄子宾 程振民 Simulation of Solid-Liquid Cascade Filtration Based on CFD-DEM ZHAO Zhongjie, ZHANG Jianpeng, TANG Yanling, XIAO Tong, HUANG Zibin, CHENG Zhenmin 在线阅读 View online: 您可能感兴趣的其他文章 Articles you may be interested in 颗粒床内固液过滤的三维CFD-DEM模拟 Three-Dimensional CFD-DEM Simulation of Solid-Liquid Filtration in Granular Bed 华东理工大学学报（自然科学版）. 2020, 46(2): 164-172 废催化剂气流加速度分选CFD-DEM模拟 CFD-EDM Simulation of Discharged Catalyst by Acceleration Air Classification 华东理工大学学报（自然科学版）. 2019, 45(2): 293-300 固液混合过程的数值模拟及实验研究 Numerical Simulation and Experimental Study of Solid-Liquid Mixing Process 华东理工大学学报（自然科学版）. 2019, 45(4): 675-680 螺旋管强化传热的CFD模拟与优化 CFD Simulation and Optimization of Spiral Tube Heat Transfer Enhancement 华东理工大学学报（自然科学版）. 2018(3): 296-302 工业级MIP提升管反应器气固两相流动特性的数值模拟 Numerical Simulation of the Gas-Solid Two-Phase Flow Characteristics in the Industrial Grade MIP Riser Reactor 华东理工大学学报（自然科学版）. 2019, 45(6): 860-867 鼓泡塔反应器中两相流动CFD-PBM耦合数值模拟 CFD-PBM Coupled Numerical Simulation of Two Phase Flow in Bubble Column Reactor 华东理工大学学报（自然科学版）. 2021, 47(1): 1-10 扫码关注公众号，获取更多信息！ 文章编号：1006-3080(2022)05-0591-09 DOI: 10.14135/j.cnki.1006-3080.20210506010 基于CFD-DEM 的固液分级过滤模拟 赵钟杰， 张建鹏， 唐艳玲， 肖 桐， 黄子宾， 程振民 （华东理工大学化学工程联合国家重点实验室，上海 200237） 摘要：采用计算流体力学(CFD) 和离散单元法(DEM) 耦合的方法，在不同滤层结构的三维 随机堆积颗粒层过滤器内进行固液分级过滤的数值模拟研究。实验结果表明，过滤效率的模拟 计算值与实验值吻合良好，压降值的偏差在Ergun 方程允许误差范围内。过滤器的容垢能力用 计算模型的颗粒沉积均匀度表示，并拟合得到沉积均匀度的关联式。颗粒沉积分布的模拟结果 显示：单层细滤料过滤器的颗粒沉积主要发生在近入口处，容垢能力较低；分级过滤器的细滤 料层保证了高过滤效率，粗滤料层则提供了较大的容垢能力。 关键词：CFD-DEM；分级过滤；固液分离；容垢量；沉积分布 中图分类号：TQ 028.5; TQ 015.9 文献标志码：A 颗粒层过滤作为一种低成本的分离方式被人们 广泛关注，并应用于烟气除尘、水处理、油浆净化等 领域[1-3]。研究者们总结了包括惯性碰撞、拦截和扩 散等在内的基本颗粒层过滤机理，得出了过滤性 能与流速、粒径、床层深度等因素有关的结论[4-6]。 Zamani 等 归纳了颗粒层过滤的微观和宏观模型， 这些模型的应用范围是非普适性的，均存在各自的 缺点：如宏观经验模型无法揭示内部过滤机理；随机 模型无法计算过滤效率；迹线模型无法计算过滤压 降等。 由于上述模型应用的局限性，计算流体力学 (Computational Fluid Dynamics，CFD) 和离散单元法 (Discrete Element Method，DEM) 耦合模拟的方法被 许多研究者所重视，并被应用于过滤研究。Qian 等 利用随机算法构建了三维纤维过滤器模型，并采用 CFD-DEM 耦合方法研究不同气速和空隙率下的气 溶胶过滤性能。Yue 等 基于CFD-DEM 的耦合模 型对含尘气体的过滤进行了模拟，全面分析了颗粒 的运动以及颗粒所受接触力对过滤性能的影响。肖 桐等 采用CFD-DEM 耦合方法，构建三维随机堆 积的颗粒床模型，研究不同时刻和表观过滤速率下 颗粒沉积的变化，并考虑了颗粒沉积对于过滤效率 和压降的影响。 颗粒层过滤器中杂质颗粒的沉积分布通常是不 均匀的，当滤料粒径较小时，颗粒更容易沉积在过滤 器前端，床层深处的滤料几乎没有被利用。滤料粒 径越小，过滤效率越高，然而这也意味着过滤器容垢 能力的降低。为了同时提高过滤效率和容垢量，分 级过滤器的概念被提出。Yang 等 进行了分级过 滤器的除尘实验，上、下层分别填装轻而大和重而小 的滤料。实验结果表明，在相同的床层压降下，分级 过滤器的容尘能力是单层过滤器的10 倍。Shi 等 通过实验研究了上层膨胀珍珠岩和下层细砂组成的 分级过滤器的过滤性能，结果发现分级过滤器的压 降增长率与单层粗滤料过滤器基本相同，且远低于 单层细滤料过滤器的压降增长率。 综上所述，可发现前人对于分级过滤的研究主 要集中在实验层面，而杂质颗粒在过滤器内的沉积 分布无法通过实验方法准确获得。本文利用CFD-DEM 耦合方法，构建不同滤层结构的三维过滤器模 收稿日期： 2021-05-06 基金项目： 国家重点研发计划（2019YFC1906705）；国家自然科学基金资助项目（21676085） 作者简介： 赵钟杰(1996—)，男，浙江人，硕士生，主要研究方向为多相反应器的模拟与计算。E-mail：1012925517@qq.com 通信联系人： 程振民，E-mail：zmcheng@ecust.edu.cn 引用本文： 赵钟杰, 张建鹏, 唐艳玲, 等. 基于CFD-DEM 的固液分级过滤模拟[J]. 华东理工大学学报（自然科学版）, 2022, 48(5): 591-599.
Citation： ZHAO Zhongjie, ZHANG Jianpeng, TANG Yanling, et al. Simulation of Solid-Liquid Cascade Filtration Based on CFD-DEM[J]. Journal of East China University of Science and Technology, 2022, 48(5): 591-599.
Vol. 48 No. 5 华 东 理 工 大 学 学 报（自 然 科 学 版） 2022-10 Journal of East China University of Science and Technology 591 型，获取颗粒沉积分布结果以辅助解释实验结果。 1 实验描述 图1 为颗粒层过滤实验装置简图，包括原料液 供应和控制系统、输送原料液的泵系统、颗粒过滤系 统。充分搅拌的原料液通过泵系统输送至滤层总高 为40 mm 的过滤器的入口，随后原料液流经床层，未 被收集的颗粒从过滤器出口逃逸。 Agitator Raw material tank Control valve Pump Flow meter Differential manometer Random packed granular filter 图 1 颗粒层过滤实验装置 Fig. 1 Experimental apparatus of granular filtering 实验中使用的滤料为α-Al2O3，并采用分级法进 行填装。上层填充的粗滤料(记为D1) 粒径为4 mm， 下层填充的细滤料(记为D2) 粒径为2 mm。图2 所 示为5 种不同滤层结构的颗粒层过滤器，其中Ⅰ、 Ⅴ为单层过滤器，Ⅱ、Ⅲ、Ⅳ为分级过滤器。利用流 量计测量表观过滤速率(u)，并分别控制表观过滤速 率为50、100、150、200、250 mm/s。原料罐内料液的 密度近似为1 000 kg/m3，动力黏度为0.001 Pa·s。常 规的固液颗粒层过滤器中颗粒的体积分数在0.01% 以下，本文中颗粒体积分数为0.005 6%，质量浓度 控制在180 mg/L。进入过滤器的杂质颗粒通过细筛 分获得，直径约为0.4 mm。每次实验在20 ℃的室温 下重复3 次，以尽量消除偶然性。 2 数值方法和模型描述 2.1 数值方法 2.1.1 流体相控制方程 选取Eulerian 模型描述流体 相的行为变化，并在控制方程中引入了过滤器的床 层空隙率ε，实现流体相和离散相的双向耦合，以考 察颗粒沉积对于过滤的影响。流体相的控制方程 如下： ∂ ρfε ∂ t +∇·(ρfεu) = 0 (1) ∂ ∂ t (ρfεu)+∇·(ρfεµu) = −∇ρf −S +∇·(µε∇u)+ρfεg (2) S = n ∑ i FD/∆V (3) ∆V 其中：ρf 是流体密度；g 是重力加速度；μ 是流体黏度； S 是作用在网格单元体积内的阻力FD 的总和， 表示网格单元体积。 流体在颗粒床层内的流动状态主要取决于床层 雷诺数Re： Re = ρvD 6αµ(1−ε) (4) 其中：ρ 为颗粒密度，D 是颗粒直径，v 是水流平均速 率，α 为滤料形状系数。当Re>2 时，流体流动状态为 湍流。 根据式(4)，若视颗粒为球体，则层流临界速率 (Re=2 时的水流平均速率) 为0.003 43 m/s。由于在文 中最小表观过滤速率为50 mm/s，因此选择湍流模 型。采用SST k-ω 模型描述介质空隙间湍流的影响， 湍流动能k 和比耗散率ω 的方程分别为[16-17]： ∂ ∂ t (ρk)+ ∂ ∂ xi (uiρk) = ∂ ∂ xj ( Γk ∂ k ∂ xj ) +Gk −Yk +S k (5) L = 40 mm L1 = 8 mm L2 = 32 mm L1 = 16 mm L2 = 24 mm L1 = 24 mm L2 = 16 mm L = 40 mm Direction of fluid flow (a) CaseⅠ (b) CaseⅡ (c) CaseⅢ (d) Case Ⅳ (e) CaseⅤ D1 D2 图 2 不同滤料填装方式的过滤器 Fig. 2 Filters with different granular packing methods 592 华 东 理 工 大 学 学 报（自 然 科 学 版） 第 48 卷 ∂ ∂ t (ρω)+ ∂ ∂ xi (uiρω) = ∂ ∂ xj ( Γω ∂ ∂ xj ω ) +Gω−Yω+Dω+S ω (6) 其中：xi、xj 分别为i、j 方向上的坐标值；Γk 和Γω 为湍 流模型的有效扩散项；Gk 为湍流动能；Gω 代表ω 方 程；Yk 和Yω 表示湍流模型的发散项；Dω 表示正交散 度项；Sk 和Sω 由用户自定义。 2.1.2 固体离散相模型 在过滤过程中，由于颗粒与 颗粒、滤料和流体相间作用力的存在，使得颗粒发生 迁移，迁移轨迹通过Lagrange 方法求解。通常微米 级颗粒才需要考虑非接触力，而本文中模拟涉及的 颗粒粒径远大于微米级，所以不考虑非接触力的影 响。颗粒间的接触力则利用Hertz-Mindlin 模型 计 算，并表示为非线性的软球模型。在流体相中，颗粒 的迁移运动方程如式(7)、式(8) 所示： mp dup dt = Fg + fD (u−up )+ fn + ft (7) Ip dωp d p = Tp (8) 其中：mp、up、ωp、fn、ft 和Tp 分别表示颗粒的质量 (g)、速度(m/s)、角速度(rad/s)、法向碰撞力(N)、切 向碰撞力(N)、碰撞力矩(N·m)；Ip 是颗粒的惯性矩 (m4)；Fg 和fD(u−up) 分别是颗粒的重力(N) 和流体对 于颗粒的阻力(N)，fD 表示为： fD = 3µCDRep 4ρpd2 p (9) 其中：ρp 是颗粒密度；CD 是非线性阻力系数，表达式 如下： CD = 24 Rep Rep ⩽0.5 24 ( 1+0.15Re0.687 p ) Rep 0.5 < Rep ⩽1 000 0.43 Rep > 1 000 (10) Rep = ρ u−up dp µ (11) 有关颗粒之间力的详细计算，可参阅文献。 2.1.3 耦合计算方法 在CFD-DEM 双平台数值模 拟过程中采用Eulerian 耦合法，其不仅考虑液固两 相间的动量交互过程，同时也考虑颗粒对流场的影 响。DEM 平台选用EDEM (V 2018)，CFD 平台选用 Fluent (V 16.0)，耦合流程由编译的UDF (User-Defined Function，用户定义函数) 加载。 图3 为CFD-DEM 耦合流程图，首先在Fluent 中 初始化流场并进行零时刻流场数据的计算，随后将 流场数据传输至EDEM 中，并利用Hertz-Mindlin 模 型计算颗粒受到的曳力等所有接触力，由此计算颗 粒速度等信息；将作用力等信息通过耦合程序传输 至Fluent 中进行下一时间步长流场数据的计算，直至 达到目标模拟时间。计算流场的时间步长取为3×10−4 s， 为准确获取颗粒接触力的信息，EDEM 时间步长设 置为3×10−6 s。本文的计算在Intel(R) Core(TM) i7-9750H CPU @ 2.60 GHz (6 核心，16 GB 内存) 上进 行，计算平均时长为60 h。 Start CFD Define boundary conditions Initialize fluid field Compute liquid field No DEM End Yes t = tend?
Update forces, positions and speed of particles Compute contact force Initialize solid field and geometry 图 3 CFD-DEM 耦合流程图 Fig. 3 Flow chart of the CFD-DEM coupling 2.2 过滤模型 图4 所示为过滤器的几何模型(以过滤器Ⅳ为 例) 及其边界条件，模型整体是半径为25 mm 的三维 圆柱，DEM 计算域的高度为40 mm。滤料由EDEM 中的“颗粒工厂”在DEM 计算域内随机生成并快速 堆积。计算域底部的交织滤网在支撑滤料的同时保 证杂质颗粒都能通过。流体的入口和出口分别采用 速度进口和压力出口，进口速率与实验保持一致。 为了保持进口速度的稳定，并避免流体从出口回流， 分别对模型的进口和出口区域进行了加长处理。颗 粒由颗粒入射面进入，并从压力出口边界离开模拟 区域。模型的壁面设置为静止且无滑移，模拟参数 如滤料粒径和杂质颗粒粒径等均与实验条件相同。 表1 所示为模拟所用物性参数。结构化网格被应用 于划分Fluent 中过滤器模型的网格，网格数量为 2 261。为了保证数值结果的准确性，EDEM 中过滤器 模型的网格数量分别依次划分为982 600、1 706 256、 3 342 336 和7 860 800。模拟结果显示，网格数量为 第 5 期 赵钟杰，等：基于CFD-DEM的固液分级过滤模拟 593 3 342 336 和7 860 800 的模型所得到的初始压降值和 过滤效率均非常接近。因此，从数值结果的精度和 计算机资源两方面考虑，EDEM 中网格数量划分选 取3 342 336。 (η) 在颗粒层过滤实验中，过滤效率 定义为： η=Cin −Cout Cin (12) 其中：Cin 为进口颗粒质量浓度，mg/L；Cout 为出口颗 粒质量浓度，mg/L。假设所有颗粒的物性参数相同 并视为球体，则可采用颗粒数目代替颗粒质量浓度 的方法计算过滤效率。而初始过滤效率定义为滤料 处于干净状态下的过滤效率。 ∆p Ergun 方程 可用于预测颗粒床的初始压降（ ）： ∆p L = 150 (1−ε)2 ε3(ψde)2 µu+1.75 1−ε ε3ψde ρu2 (13) ψ 其中： 为颗粒球形度；de 为颗粒体积等效直径，m。 虽然不存在计算容垢量的公式，但很显然，颗粒 沉积均匀度(λ) 越大，则相同质量的颗粒沉积引起的 压降增量越小，过滤床的容垢能力就越高。因此，本 文采用沉积均匀度来评价过滤器的容垢能力。 Velocity-inlet Particle injecting surface DEM solution domain Wall Pressure-outlet Y X Symmetry Z 4 mm 40 mm R = 25 mm 80 mm 图 4 过滤器模型及其边界条件 Fig. 4 Filter model and its boundary conditions 表 1 模拟物性参数 Table 1 Physical parameters in the simulation Item Density/ (kg·m−3) Poisson’s ratio Shear modulus/Pa Restitution coefficient Static friction coefficient Rolling friction coefficient Particle 3 200 0.25 1.0×108 — — — Wall 1 500 0.25 1.1×109 — — — Particle to particle — — — 0.5 0.154 0.05 Particle to wall — — — 0.3 0.154 0.01 图5 所示为颗粒层过滤器分节示意图。将过滤 器沿着过滤方向分为20 节，颗粒沉积均匀度由式 (14) 定义： λ= 1−1 2n n ∑ i=1 √ (Ni −Nave)2 Nave (14) 其中：n 为床层节数，Ni 为每一节床层中沉积的颗粒 ϕ 数目，Nave 为沉积颗粒数目的算术平均数。颗粒沉积 越均匀，λ 越趋向于1。每一节床层中的颗粒沉积分 数( ) 定义为： ϕ= Ni n ∑ i=1 Ni (15) N1 N2 N3 N18 N19 N20 … 图 5 颗粒层过滤器分节示意图 Fig. 5 Schematic diagram of granular bed filter segmentation 594 华 东 理 工 大 学 学 报（自 然 科 学 版） 第 48 卷 3 结果与讨论 3.1 初始压降与过滤效率 图6(a) 所示为过滤器Ⅰ、Ⅳ和Ⅴ的初始压降结 果。从图6(a) 可知，压降模拟结果与实验结果基本 一致，但后者数值略大，这可能是因为在实验过程 中，流体流过过滤器壁面时，由于摩擦而产生了微小 的压降。在相同过滤速率下，过滤器Ⅳ的压降仅为 过滤器Ⅰ和Ⅴ压降之和的一半左右。 图6(b) 所示为表观过滤速率为50 mm/s 时，不 同床层深度下过滤器V 的初始压降模拟结果和 Ergun 方程结果之间的比较。Ergun 方程允许的误差 范围为±25%，虽然模拟计算得到的压降值稍大，但两 者之间的偏差仍处于有效范围内。 图7(a) 所示为不同表观过滤速率下过滤器Ⅰ、 Ⅱ和Ⅴ的初始过滤效率变化，黑点和红点分别代表 实验和计算结果。由图可得，模拟结果与实验结果 具有良好的一致性。过滤器Ⅰ的过滤效率明显小于其 他滤层结构的过滤器。如当表观过滤速率为50 mm/s 时，过滤器Ⅰ的过滤效率约为70%，而其他滤层结构 过滤器的过滤效率均约为90%，说明采用分级填料 可以大幅提高过滤效率。 很显然，通过模拟分析可便捷地得到过滤效率 随床层深度的连续变化。图7(b) 所示为表观过滤速 率为50 mm/s 时，不同床层结构和不同床层深度下的 过滤效率模拟值。可以看出，当过滤器Ⅴ的床层深 度约为24 mm(称为“转折点”) 时，其过滤效率达到最 大值并保持稳定。过滤器Ⅲ和Ⅳ的转折点分别约为 32 mm 和28 mm；而过滤器Ⅰ和Ⅱ的转折点大于40 mm。 综上分析可知，下层细滤料层深度的增加使转折点 提前出现。 50 100 150 200 250 0 5 10 15 20 25 30 Δp/kPa (a) Simulation of caseⅠ Experiment of caseⅠ Simulation of case Ⅳ Experiment of case Ⅳ Simulation of case Ⅴ Experiment of case Ⅴ 4 8 12 16 20 200 400 600 800 1 000 1 200 (b) Δp/Pa L/mm Results of Ergun’s equation Simulation results u/(mm·s−1) 图 6 (a) 不同过滤器初始压降的实验值和计算值比较；(b) 不同床层深度下过滤器Ⅴ的初始压降 Fig. 6 (a) Comparison of initial pressure drop of different filters between experiment and simulation; (b) Initial pressure drop under different bed depths of filter with case Ⅴ 50 100 150 200 250 50 60 70 80 90 100 (a) η/% u/(mm·s−1) Simulation of caseⅠ Experiment of caseⅠ Simulation of case Ⅱ Experiment of case Ⅱ Simulation of case Ⅴ Experiment of case Ⅴ 4 8 12 16 20 24 28 32 36 40 15 30 45 60 75 90 (b) Bed structure: η/% L/mm Case Ⅰ Case Ⅱ Case Ⅲ Case Ⅳ Case Ⅴ 图 7 (a) 不同过滤器过滤效率的实验值和模拟值比较；(b) 不同过滤器不同床层深度下的过滤效率模拟值 Fig. 7 (a) Comparison of filtration efficiency of different filters between experiment and simulation; (b) Simulation results of filtration efficiency under different bed depths of different filters 过滤器的综合性能由过滤效率和压降共同决 定，不同滤层结构过滤器的综合评价指标(Y) 根据 式(16) 定义，计算得到的Y 值如图8 所示，可见在不 同表观过滤速率下，分级过滤器相比于单层过滤器 第 5 期 赵钟杰，等：基于CFD-DEM的固液分级过滤模拟 595 具有更高的Y 值，即分级过滤的综合过滤性能最佳。 Y=−ln(1−η) ∆p (16) 50 100 150 200 250 0 0.001 0.002 0.003 0.004 Bed structure: Y Y u/(mm·s−1) u/(mm·s−1) 0.000 4 0.000 3 0.000 2 0.000 1 150 200 250 Case Ⅰ Case Ⅱ Case Ⅲ Case Ⅳ Case Ⅴ 图 8 不同床层结构过滤器的Y 值 Fig. 8 Y Value of filters with different bed structures 3.2 颗粒沉积形貌 杂质颗粒沉积形貌的结果更加直观，可加深对 过滤现象的理解。图9 所示为当表观过滤速率为 200 mm/s、过滤时间为10 s 时，过滤器Ⅱ和Ⅴ的细滤 料层表面的颗粒沉积形貌，可发现两者的颗粒沉积 图是近似的，过滤器Ⅴ的细滤料层表面沉积的杂质 颗粒数目大于分级过滤器Ⅱ的细滤料层表面杂质颗 粒数。由此可以推断，分级过滤可以使杂质颗粒部 分容纳在粗滤料层中，提高了床层容垢量。 图10 所示为当表观过滤速率为50 mm/s，过滤 时间为10 s 时，模拟所得同一纵截面下过滤器Ⅰ、 Ⅱ和Ⅴ的颗粒沉积图。可以看出，过滤器Ⅴ近入口 处及过滤器Ⅱ上、下层滤料的分界处形成明显的团 聚沉积结构。这可以从惯性机制 的角度进行解释， 当滤料粒径为2 mm、过滤速率为50 mm/s 时，斯托 克斯数(St) 约为1.5，颗粒的惯性发挥作用，使得颗粒 容易偏离流体流向，其运动轨迹与滤料表面相交的 几率增大，颗粒发生高频率碰撞并被快速捕集。较 高的捕集几率促成颗粒间的架桥现象，从而形成团 聚结构，其可视为新的“过滤器”来截留颗粒，因而大 幅降低了滤层表面的空隙率，也解释了图6(a) 中过 滤器Ⅴ的压降远高于其他过滤器的原因。 由图10 可知，过滤器Ⅰ及过滤器Ⅱ的上层滤料 表面并未出现明显的团聚结构。相比于过滤器Ⅴ， 过滤器Ⅰ较大的滤料粒径使得惯性作用明显减弱， 颗粒更容易随着流体流线运动而被夹带至床层的深 处，使得颗粒的分布较为均匀。此时可考虑拦截机 制，而只有当颗粒的轨迹恰好落在滤料表面到滤料 半径的流线范围内时，拦截机制才对颗粒迁移产生影 响，颗粒才会被滤料表面截留，因此难以形成团聚结构。 3.3 颗粒沉积均匀度 从杂质颗粒沉积形貌来分析颗粒沉积分布仅是 定性分析，通过EDEM 的“group bin”后处理功能可 得到不同床层节数上滤料表面及间隙所沉积颗粒的 数量，从而定量分析颗粒沉积均匀度。当表观过滤 速率为200 mm/s、过滤时间为10 s 时，不同床层节数 (a) CaseⅡ (b) CaseⅤ Simulation image Experimental image 图 9 细滤料层表面的颗粒沉积形貌 Fig. 9 Deposition morphology on the surface of fine granular layer 596 华 东 理 工 大 学 学 报（自 然 科 学 版） 第 48 卷 上的杂质颗粒沉积分数如图11(a) 所示。对于过滤 器Ⅴ，杂质颗粒的沉积分数随着床层节数的增大而 明显减小。可以看出，前5 节床层的颗粒沉积分数之 和达到了71.2%；当床层节数达到10 左右时，颗粒沉 积分数就几乎降低为0，后半段床层的颗粒沉积分数 之和仅为8.4%，这意味着过滤器Ⅴ收集的颗粒几乎 都容纳在过滤器的前端部分。过滤器Ⅰ中每一节床 层的颗粒沉积分数基本都为0.05~1.00。对于过滤器 Ⅱ，当床层节数小于16 时，每一节床层的颗粒沉积分 数相差不大，提供较高容垢量；而当在床层节数为 17 时，颗粒沉积分数显著增大，表明颗粒收集量快速 增加，保证了分级过滤器的过滤效率。 0 2 4 6 8 10 12 14 16 18 20 0 0.05 0.10 0.15 0.20 0.25 ϕ n Bed structure: CaseⅠ CaseⅡ CaseⅤ (a) 50 100 150 200 250 0.45 0.55 0.65 0.75 0.85 (b) Bed structure: λ u/(mm·s−1) Case Ⅰ Case Ⅱ Case Ⅲ Case Ⅳ Case Ⅴ 图 11 颗粒的沉积分布 Fig. 11 Particle deposition distribution 图11(b) 定量分析了不同滤层结构下颗粒沉积 均匀度的变化。结果表明，沉积均匀度随表观过滤 速率的增加而增加。如当过滤速率从50 mm/s 增加 至250 mm/s 时，过滤器II 的沉积均匀度由0.73 增加 到0.84。这是由于过滤速率的增大使滤料表面已沉 积颗粒的剥离行为更显著。过滤器V 的沉积均匀度 明显低于其他4 种滤层结构过滤器。当表观过滤速 率为200 mm/s时，过滤器II 的沉积均匀度比过滤器 V 高59.4%；当滤层结构由V 变化至IV，沉积均匀度 增加0.217，而从IV 变化到I，沉积均匀度仅增加 0.103。 引入分级过滤器的平均滤料直径(Dave)，并定义为： Dave = D1L1 L1 + L2 + D2L2 L1 + L2 = D1L1 + D2L2 L1 + L2 (17) 式中：D1、D2 分别为上、下层滤料直径；L1、L2 分别为 上、下层滤料层厚度。 对影响沉积均匀度的各个参数进行量纲分析， 通过白金汉定理和量纲一致性，沉积均匀度可表示为： λ = 1−exp K ( ρpud2 p 9µDave )n1(Daveuρ µ )n2( dp Dave )n3 (18) 其中，(ρpudp 2)/(9μDave) 为斯托克斯数St；Daveuρ/μ 为雷 诺数Re；dp/Dave=NR 代表拦截机制的影响。 利用1st Opt 软件进行多元非线性回归分析，得 到在50 mm/s <u< 250 mm/s, 2 mm <Dave< 4.0 mm 范 围内，沉积均匀度的关系式如下： λ = 1−exp [ −0.61×10−2St−1.98Re2.14N2.79 R ] (19) λpre.
λcal.
图12 对比了沉积均匀度的预测结果（ ）与计 算结果（ ）。由图可知，沉积均匀度的模拟结果与 回归方程值吻合良好，偏差小于5%，说明拟合具有 足够的精度。 (a) CaseⅠ (b) CaseⅡ (c) CaseⅤ 图 10 不同过滤器同一纵截面的颗粒沉积形貌 Fig. 10 Particle deposition morphology in the same longitudinal section of different filters 第 5 期 赵钟杰，等：基于CFD-DEM的固液分级过滤模拟 597 0.5 0.6 0.7 0.8 0.9 0.5 0.6 0.7 0.8 0.9 −5% λpre.
λcal.
+5% 图 12 沉积均匀度的预测结果和计算结果的对比 Fig. 12 Comparison of deposition uniformity between predicted and calculated results 4 结 论 采用CFD-DEM 耦合方法，对不同滤层结构的 颗粒层过滤器进行固液分级过滤的数值模拟研究， 主要结论如下： (1) 过滤效率的模拟计算值与实验值吻合良好， 压降值的偏差在Ergun 方程允许误差范围内。过滤 效率和压降随床层深度的增加而增加，当床层深度 增大至转折点时，过滤效率基本不变，且随着下层细 滤料层深度的增加，转折点提前出现。 (2) 过滤器的容垢能力由颗粒沉积均匀度表 示。颗粒沉积均匀度随表观过滤速率的增大而增 大，滤层的分级填装使得杂质颗粒在床层中的沉积 均匀度大幅提升。通过拟合回归得到不同滤层结构 过滤器沉积均匀度的关联式，且偏差小于5%。 (3) 杂质颗粒的沉积分布表明：单层细滤料过滤 器由于较强的颗粒惯性作用而形成颗粒团聚结构， 颗粒沉积主要发生在近入口处，容垢量较低；分级过 滤器的上层粗滤料截留了部分的杂质颗粒，在保证 过滤效率的前提下，降低了过滤器的压降，同时提高 了容垢量。 参考文献： ALTMANN J, REHFELD D, TRAEDER K, et al. Combin-ation of granular activated carbon adsorption and deep-bed filtration as a single advanced wastewater treatment step for organic micropollutant and phosphorus removal[J]. Water Research, 2016, 92: 131-139.
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Simulation of Solid-Liquid Cascade Filtration Based on CFD-DEM ZHAO Zhongjie, ZHANG Jianpeng, TANG Yanling, XIAO Tong, HUANG Zibin, CHENG Zhenmin (State Key Laboratory of Chemical Engineering, East China University of Science and Technology, Shanghai 200237, China) Abstract: The three-dimensional model for a randomly packed bed filter was established by the combination of computational fluid dynamics (CFD) and discrete element method (DEM). To obtain more reliable simulation results, the liquid-solid, particle-granule and particle-particle interactions were all taken into consideration. The filtration performances of the system including filtration efficiency, pressure drop and impurity holding capacity were systematically analyzed, and the particle deposition distribution and morphology were quantitatively studied. The simulated results of filtration efficiency agreed well with those obtained from experimental operations. The deviation of the pressure drop fell in the allowable error range of the Ergun equation. The impurity holding capacity could be represented by the deposition uniformity obtained by simulation, which increased with superficial velocity. Correlation of deposition uniformity for granular bed filters was also simulated with a good prediction accuracy. The results show that cascade filtration has both a high filtration efficiency and a low pressure drop by combining deep bed filtration and surface filtration. The quality factor of the cascade filters is greater than those with single-layer filters. Simulation of particle deposition morphology and distribution suggests that the particles mainly deposit on the surface of single-layer filter packed with fine granules, resulting in a small holding capacity. As for the cascade filter, the fine and coarse granular layer produces a high filtration efficiency and large impurity holding capacity, respectively Key words: CFD-DEM；cascade filtration；solid-liquid separation；particle holding capacity；deposition distribution 第 5 期 赵钟杰，等：基于CFD-DEM的固液分级过滤模拟 599 |
4309 | https://brainly.in/question/20871574 | what is the relation between Pascal and N/m2 - Brainly.in
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Explanation:
The SI unit of the pressure is the pascal with the formula sign Pa.
1 Pascal is equal to the pressure of 1 newton per square meter. 1 Pa = 1 N / m2 ≡ 1 kg / m · s2.
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Answer:
1Pascal = 1N/1m²
Explanation:
the pressure of one Newton force on an area of one meter² is equal to one pascal
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4310 | https://embryology.med.unsw.edu.au/embryology/index.php/Musculoskeletal_System_-_Muscle_Development | Musculoskeletal System - Muscle Development
From Embryology
Jump to:navigation, search
| ExpandEmbryology - 27 Aug 2025 Expand to Translate |
| Google Translate - select your language from the list shown below (this will open a new external page) |
| العربية | català | 中文 | 中國傳統的 | français | Deutsche | עִברִית | हिंदी | bahasa Indonesia | italiano | 日本語 | 한국어 | မြန်မာ | Pilipino | Polskie | português | ਪੰਜਾਬੀ ਦੇ | Romnă | русский | Español | Swahili | Svensk | ไทย | Türkçe | اردو | ייִדיש | Tiếng Việt These external translations are automated and may not be accurate. (More? About Translations) |
Contents
1 Introduction
2 Some Recent Findings
3 Myogenesis
4 Muscle Groups
4.1 Epaxial Muscle
4.2 Hypaxial Muscle
4.3 Head Muscle
5 Skeletal Muscle Stages
6 Muscle Fibre Types
7 Muscle Contraction
8 Adult Myotome
9 Mouse Limb Muscle
10 Histology Images
10.1 Electron Microscopy Virtual Slides
11 Puberty
12 Satellite Cells
13 Abnormalities
13.1 Duchenne Muscular Dystrophy
13.2 Becker Muscular Dystrophy
13.3 Autosomal Recessive Muscular Dystrophy
13.4 Myotonic Dystrophy
13.5 Facioscapulohumeral muscular dystrophy (FSHD)
14 References
14.1 Reviews
14.2 Articles
14.3 Search PubMed
15 Additional Images
16 External Links
17 Glossary Links
Introduction
There are 3 main different types of muscle: skeletal, cardiac and smooth. This page describes skeletal muscle development, descriptions of cardiac muscle and smooth muscle development can be found in other notes. Skeletal muscle forms by fusion of mononucleated myoblasts to form mutinucleated myotubes.
: There are more than 640 skeletal muscles in the adult human body.
Differentiation/determination of mesoderm into muscle cells is thought to involve a family of basic Helix-Loop-Helix transcription factors, the first of which discovered was MyoD1. MyoD1 needs to form a dimer to be active and is maintained in an inactive state by binding of an inhibitor, Id.
Specific Skeletal Muscles: tongue | diaphragm
| |
| Musculoskeletal Links: Introduction | mesoderm | somitogenesis | limb | cartilage | bone | bone timeline | bone marrow | shoulder | pelvis | axial skeleton | skull | joint | skeletal muscle | muscle timeline | tendon | diaphragm | Lecture - Musculoskeletal | Lecture Movie | musculoskeletal abnormalities | limb abnormalities | developmental hip dysplasia | cartilage histology | bone histology | Skeletal Muscle Histology | Category:Musculoskeletal |
| | ExpandHistoric Embryology - Musculoskeletal | | 1853 Bone | 1885 Sphenoid | 1902 - Pubo-femoral Region | Spinal Column and Back | Body Segmentation | Cranium | Body Wall, Ribs, and Sternum | Limbs | 1901 - Limbs | 1902 - Arm Development | 1906 Human Embryo Ossification | 1906 Lower limb Nerves and Muscle | 1907 - Muscular System | Skeleton and Limbs | 1908 Vertebra | 1908 Cervical Vertebra | 1909 Mandible | 1910 - Skeleton and Connective Tissues | Muscular System | Coelom and Diaphragm | 1913 Clavicle | 1920 Clavicle | 1921 - External body form | Connective tissues and skeletal | Muscular | Diaphragm | 1929 Rat Somite | 1932 Pelvis | 1940 Synovial Joints | 1943 Human Embryonic, Fetal and Circumnatal Skeleton | 1947 Joints | 1949 Cartilage and Bone | 1957 Chondrification Hands and Feet | 1968 Knee | |
Some Recent Findings
| |
| Hitherto unknown detailed muscle anatomy in an 8-week-old embryo "Human embryo at Carnegie embryo 950 stage 23 (8 weeks of development, crown-rump length of 23.8 mm), using Amira reconstruction software. Reconstructed muscles, tendons, bones and nerves were exported in a 3D-PDF file to permit interactive viewing. Almost all adult skeletal muscles of the trunk and limbs could be individually identified in their relative adult position. The pectoralis major muscle was divided in three separate muscle heads. The reconstructions showed remarkable highly developed extraocular, infrahyoid and suprahyoid muscles at this age but surprisingly also absence of the facial muscles that have been described to be present at this stage of development. The overall stage of muscle development suggests heterochrony of skeletal muscle development. Several individual muscle groups were found to be developed earlier and in more detail than described in current literature." Review - Making muscle: skeletal myogenesis in vivo and in vitro "In this Review, we provide a comprehensive overview of skeletal myogenesis from the earliest premyogenic progenitor stage to terminally differentiated myofibers, and discuss how this knowledge has been applied to differentiate PSCs into muscle fibers and their progenitors in vitro." Development of the ventral body wall in the human embryo "Initially, vertebrae and ribs had formed medially, and primordia of sternum and hypaxial flank muscle primordium laterally in the body wall at Carnegie Stage (CS)15 (5.5 weeks). The next week, ribs and muscle primordium expanded in ventrolateral direction only. At CS18 (6.5 weeks), separate intercostal and abdominal wall muscles differentiated, and ribs, sterna, and muscles began to expand ventromedially and caudally, with the bilateral sternal bars fusing in the midline after CS20 (7 weeks) and the rectus muscles reaching the umbilicus at CS23 (8 weeks). The near-constant absolute distance between both rectus muscles and approximately fivefold decline of this distance relative to body circumference between 6 and 10 weeks identified dorsoventral growth in the dorsal body wall as determinant of the 'closure' of the ventral body wall. Concomitant with the straightening of the embryonic body axis after the 6th week, the abdominal muscles expanded ventrally and caudally to form the infraumbilical body wall. Our data, therefore, show that the ventral body wall is formed by differential dorsoventral growth in the dorsal part of the body." |
| ExpandMore recent papers |
| This table allows an automated computer search of the external PubMed database using the listed "Search term" text link. This search now requires a manual link as the original PubMed extension has been disabled. The displayed list of references do not reflect any editorial selection of material based on content or relevance. References also appear on this list based upon the date of the actual page viewing. References listed on the rest of the content page and the associated discussion page (listed under the publication year sub-headings) do include some editorial selection based upon both relevance and availability. More? References | Discussion Page | Journal Searches | 2019 References | 2020 References Search term: Muscle Development | Skeletal Muscle Development | Myogenesis |
| ExpandOlder papers |
| These papers originally appeared in the Some Recent Findings table, but as that list grew in length have now been shuffled down to this collapsible table. See also the Discussion Page for other references listed by year and References on this current page. Tbx15 controls skeletal muscle fibre-type determination and muscle metabolism "Skeletal muscle is composed of both slow-twitch oxidative myofibers and fast-twitch glycolytic myofibers that differentially impact muscle metabolism, function and eventually whole-body physiology. Here we show that the mesodermal transcription factor T-box 15 (Tbx15) is highly and specifically expressed in glycolytic myofibers. Ablation of Tbx15 in vivo leads to a decrease in muscle size due to a decrease in the number of glycolytic fibres, associated with a small increase in the number of oxidative fibres. This shift in fibre composition results in muscles with slower myofiber contraction and relaxation, and also decreases whole-body oxygen consumption, reduces spontaneous activity, increases adiposity and glucose intolerance. Mechanistically, ablation of Tbx15 leads to activation of AMPK signalling and a decrease in Igf2 expression. Thus, Tbx15 is one of a limited number of transcription factors to be identified with a critical role in regulating glycolytic fibre identity and muscle metabolism." TBX Notch regulation of myogenic versus endothelial fates of cells that migrate from the somite to the limb "Multipotent Pax3-positive (Pax3(+)) cells in the somites give rise to skeletal muscle and to cells of the vasculature. We had previously proposed that this cell-fate choice depends on the equilibrium between Pax3 and Foxc2 expression. In this study, we report that the Notch pathway promotes vascular versus skeletal muscle cell fates. ...We now demonstrate that in addition to the inhibitory role of Notch signaling on skeletal muscle cell differentiation, the Notch pathway affects the Pax3:Foxc2 balance and promotes the endothelial versus myogenic cell fate, before migration to the limb, in multipotent Pax3(+) cells in the somite of the mouse embryo." limb | }}Notch}} Jamb and jamc are essential for vertebrate myocyte fusion "Cellular fusion is required in the development of several tissues, including skeletal muscle. In vertebrates, this process is poorly understood and lacks an in vivo-validated cell surface heterophilic receptor pair that is necessary for fusion. Identification of essential cell surface interactions between fusing cells is an important step in elucidating the molecular mechanism of cellular fusion. We show here that the zebrafish orthologues of JAM-B and JAM-C receptors are essential for fusion of myocyte precursors to form syncytial muscle fibres. Both jamb and jamc are dynamically co-expressed in developing muscles and encode receptors that physically interact." (mammalian orthologues JAM-B/Jam2a = JAM2 JAM-C/Jam3b = JAM3) Origin of vertebrate limb muscle: the role of progenitor and myoblast populations7 "Muscle development, growth, and regeneration take place throughout vertebrate life. In amniotes, myogenesis takes place in four successive, temporally distinct, although overlapping phases. Understanding how embryonic, fetal, neonatal, and adult muscle are formed from muscle progenitors and committed myoblasts is an area of active research. In this review we examine recent expression, genetic loss-of-function, and genetic lineage studies that have been conducted in the mouse, with a particular focus on limb myogenesis." The histone methyltransferase Set7/9 promotes myoblast differentiation and myofibril assembly "Together, our experiments define a biological function for Set7 in muscle differentiation and provide a molecular mechanism by which Set7 modulates myogenic transcription factors during muscle differentiation. The expression pattern of myogenic regulatory factors MyoD, Myf6 and Pax7 in postnatal porcine skeletal muscles "The MyoD, Myf6 genes, which belong to the family of muscle regulatory factors (MRFs) play a major role in muscle growth and development. These basic helix-loop-helix (bHLH) transcription factors regulate myogenesis: they initiate the formation of muscle fibres and regulate the transcription of muscle specific genes. The paired-box transcription factor Pax7 plays critical roles during fetal development and this protein is essential for renewal and maintenance of muscle stem cells. In particular, expression of Pax7 and MyoD is correlated with presence of active satellite cells, important in hyperplastic and hypertrophic growth in skeletal muscle." Expression of Gα(z) in C2C12 cells restrains myogenic differentiation "The recent identification of Gα(z) expression in C2C12 myoblasts and its demonstrated interaction with the transcription factor Eya2 inferred an unanticipated role of Gα(z) in muscle development. In the present study, endogenous Gα(z) mRNA and protein expressions in C2C12 cells increased upon commencement of myogenesis and peaked at around 4-6days after induction but were undetectable in adult skeletal muscle. " |
Myogenesis
Somites in human embryo (Carnegie stage 11)
Three different types of muscle form in the body.
Skeletal muscle - cells originate from the paraxial mesoderm, forming somites, then dermamyotome and finally the myotome. Myoblasts undergo frequent divisions and coalesce with the formation of a multinucleated, syncytial muscle fibre or myotube. The nuclei of the myotube are still located centrally in the muscle fibre. In the course of the synthesis of the myofilaments/myofibrils, the nuclei are gradually displaced to the periphery of the cell.
Cardiac muscle - cells originate from the prechordal splanchnic mesoderm.
Smooth muscle - cells originate from undifferentiated mesenchymal cells. These cells differentiate first into mitotically active cells, myoblasts, which contain a few myofilaments. Myoblasts give rise to the cells which will differentiate into mature smooth muscle cells.
Muscle Groups
Cartoon showing myotome fate forming epaxial and hypaxial muscle groups.
In humans, body muscles lying dorsal to the vertebral column form the epaxial muscles. The body muscles lying ventral (anterior) to the vertebral column form the hypaxial muscles.
Epaxial Muscle
Anatomical term describing skeletal muscles which lie dorsal (posterior) to the vertebral column developing from the somite. Epaxial muscles are only a small muscle group formed by the transversospinalis, longissimus, and iliocostalis muscles. At the ribcage level, the levatores costarum muscles (transverse processes of C7 to T11 vertebrae) are also involved with rib elevation during respiration.
A recent study has determined the developmental sequence of epaxial muscles in the human embryo between week 5 to 10 (see summary below).
Carnegie stage 15 - epaxial portions of myotomes become identifiable laterally to developing vertebrae.
Carnegie stage 16 - epaxial portions fusion starting cranially to form a longitudinal muscle column,
become innervated by spinal nerve dorsal branches.
Carnegie stage 17 - longitudinal muscle mass segregated into medial and lateral columns (completed at CS18).
Carnegie stage 18 - medial column segregated again into intermediate and medial columns (completed at CS20).
lateral and intermediate columns did not separate in the lower lumbar and sacral regions.
Carnegie stage 20 to Carnegie stage 23 - cervical portions of the three columns segregated again from lateral to medial resulting ventrolaterally in rod-like continuations of the caudal portions of the columns and dorsomedially in spade-like portions.
lateral column - iliocostalis and splenius
intermediate column - longissimus
medial column - (semi-)spinalis
Early fetal - medial (multifidus) group gain the transversospinal course during vertebral arch closure.
Hypaxial Muscle
(hypomere) Anatomical term describing skeletal muscles which lie ventral (anterior) to the vertebral column developing from the somite myotome. These muscles contribute both body (trunk) and limb skeletal muscle.
In the trunk, these form the three anterior body muscle layers.
In the limb, these form the extensor and flexor muscle groups.
Ventral body wall timeline
Carnegie stage 15 - hypaxial flank muscle primordium laterally in the body wall.
Carnegie stage 18 - separate intercostal and abdominal wall muscles differentiated.
Carnegie stage 20 - muscles expand ventromedially and caudally bilateral sternal bars fusing in the midline
Carnegie stage 23 - rectus muscles reach the umbilicus
6 and 10 weeks - dorsal body wall growth closes the ventral body wall.
Head Muscle
jaw associated muscles mainly from cranial mesoderm.
jaw, connective tissues and tendons from neural crest cells.
Head muscle precursor myoblast summary from a review.
myoblasts for the tongue muscle, migrate like those seen in the limb.
myoblasts for extraocular muscles, condense within paraxial mesoderm, then cross the mesoderm:neural crest interface en route to periocular regions.
myoblasts for branchial muscle, establish contacts with neural crest populations before branchial arch formation and maintain these relations through subsequent stages of development.
See also for head muscle and connective tissue.
Skeletal Muscle Stages
3D virtual muscle model
3D virtual muscle model
Myoblast - individual progenitor cells
Myotube - multinucleated, but undifferentiated contractile apparatus (sarcomere)
Myofibre (myofiber, muscle cell) - multinucleated and differentiated sarcomeres
primary myofibres - first-formed myofibres, act as a structural framework upon which myoblasts proliferate, fuse in linear sequence
secondary myofibers - second later population of myofibres that form surrounding the primary fibres.
Muscle Fibre Types
Human skeletal muscle generally consists of individual fibres with different contractile and other properties, this is the basis of classification into "types".
Muscle fiber types
type IIB, IIA, IIX, and I fibres - based only on the myosin ATPase activity.
Type I fibres appear red, due to the presence of myoglobin.
Type II fibres appear white, due to the absence of myoglobin and their glycolytic nature.
A group of individual myofibres within a muscle will be innervated by a single motor neuron (motor unit).
The electrical properties of the motor neuron will regulate the contractile properties of all associated myofibres.
| Fibre Type | Type I fibres | Type II a fibres | Type II x fibres | Type II b fibres |
--- ---
| Contraction time | Slow | Moderately Fast | Fast | Very fast |
| Size of motor neuron | Small | Medium | Large | Very large |
| Resistance to fatigue | High | Fairly high | Intermediate | Low |
| Activity Used for | Aerobic | Long-term anaerobic | Short-term anaerobic | Short-term anaerobic |
| Maximum duration of use | Hours | <30 minutes | <5 minutes | <1 minute |
| Power produced | Low | Medium | High | Very high |
| Mitochondrial density | High | High | Medium | Low |
| Capillary density | High | Intermediate | Low | Low |
| Oxidative capacity | High | High | Intermediate | Low |
| Glycolytic capacity | Low | High | High | High |
| Major storage fuel | Triglycerides | Creatine phosphate, glycogen | Creatine phosphate, glycogen | Creatine phosphate, glycogen |
| Myosin heavy chain, human genes | MYH7 | MYH2 | MYH1 | MYH4 |
| ExpandMuscle Fibre Type | | | | |
--- ---
| Fibre Type | Type I fibres | Type II a fibres | Type II x fibres | Type II b fibres |
| Contraction time | Slow | Moderately Fast | Fast | Very fast |
| Size of motor neuron | Small | Medium | Large | Very large |
| Resistance to fatigue | High | Fairly high | Intermediate | Low |
| Activity Used for | Aerobic | Long-term anaerobic | Short-term anaerobic | Short-term anaerobic |
| Maximum duration of use | Hours | <30 minutes | <5 minutes | <1 minute |
| Power produced | Low | Medium | High | Very high |
| Mitochondrial density | High | High | Medium | Low |
| Capillary density | High | Intermediate | Low | Low |
| Oxidative capacity | High | High | Intermediate | Low |
| Glycolytic capacity | Low | High | High | High |
| Major storage fuel | Triglycerides | Creatine phosphate, glycogen | Creatine phosphate, glycogen | Creatine phosphate, glycogen |
| Myosin heavy chain, human genes | MYH7 | MYH2 | MYH1 | MYH4 |
| Links: Muscle Development | Muscle Development Timeline | | | | |
Muscle Contraction
Skeletal muscle sarcomeres
Individual myoblasts in the developing muscle bed initial fuse together to form multi-nucleated myotubes. These myotubes then express the contractile proteins, that are organized into sarcomeres in series along the length of the myotube.
This animation shows the molecular interactions that occur within the skeletal muscle sarcomere between actin and myosin during skeletal muscle contraction.
| | |
--- |
| Legend Moving blob and stick - myosin complex. Moving blob and stick - myosin complex with ATPase activation. Ball binding myosin and splitting - ATP losing a phosphate to form ADP. Twisted string of beads - actin helix. Blue string - tropomyosin. Beads stacked on large bead on blue string - troponin. Small ball binding troponin - Calcium ion (Ca2+). Grey pyramid - Magnesiun ion (Mg2+). | |
Adult Myotome
In both development and the adult, the group of skeletal muscles supplied by a specific segmental spinal nerve is also referred to as a "myotome". The muscle arises from a specific somite and the spinal nerve arises from a specific level of the spinal cord (identified by veretebral column).
In humans this corresponds to the following spinal nerves (from top to bottom) and muscular functions:
Spinal Nerve Muscle Innervation Pattern
| Spinal Nerve | Acronym | Innervation |
| Cervical spinal nerve 3 | C3 | supply the diaphragm for breathing. |
| Cervical spinal nerve 4 | C4 | |
| Cervical spinal nerve 5 | C5 | supply the diaphragm for breathing and supply shoulder muscles and muscles to bend our elbow. |
| Cervical spinal nerve 6 | C6 | for bending the wrist back. |
| Cervical spinal nerve 7 | C7 | for straightening the elbow. |
| Cervical spinal nerve 8 | C8 | bends the fingers. |
| Thoracic spinal nerve 1 | T1 | spreads the fingers and supplies the chest wall and abdominal muscles. |
| Thoracic spinal nerve 2 | T2 | supplies the chest wall and abdominal muscles. |
| Thoracic spinal nerve 3 | T3 | |
| Thoracic spinal nerve 4 | T4 | |
| Thoracic spinal nerve 5 | T5 | |
| Thoracic spinal nerve 6 | T6 | |
| Thoracic spinal nerve 7 | T7 | |
| Thoracic spinal nerve 8 | T8 | |
| Thoracic spinal nerve 9 | T9 | |
| Thoracic spinal nerve 10 | T10 | |
| Thoracic spinal nerve 11 | T11 | |
| Thoracic spinal nerve 12 | T12 | |
| Lumbar spinal nerve 1 | L1 | |
| Lumbar spinal nerve 2 | L2 | bends the hip. |
| Lumbar spinal nerve 3 | L3 | straightens the knee. |
| Lumbar spinal nerve 4 | L4 | pulls the foot up. |
| Lumbar spinal nerve 5 | L5 | wiggles the toes. |
| Sacral spinal nerve 1 | S1 | pulls the foot down. |
| Sacral spinal nerve 2 | S2 | |
| Sacral spinal nerve 3 | S3 | supply the bladder, bowel, sex organs, anal and other pelvic muscles. |
| Sacral spinal nerve 4 | S4 | |
| Sacral spinal nerve 5 | S5 | |
| Links: Spinal Nerve Myotomes | Spinal Nerve Dermatomes | skeletal muscle | | |
Mouse Limb Muscle
Change in cell types and tissue formation as a function of mouse developmental stage.{#pmid:22174793|PMID22174793}}
: Links: mouse
Histology Images
Human HE x4 longitudinal and transverse
Human HE x40 transverse
Human HE x40 longitudinal
Human HE x40 longitudinal
: Muscle Histology: Muscle Development | Human HE x4 longitudinal and transverse | Human HE x40 transverse | Human HE x40 longitudinal | Human HE x40 longitudinal | Human HE x4 longitudinal and transverse | Muscle Spindle HE x40 | Human HE x40 | Human HE x40 | Human HE x40 | Human HE x100 | Human HE x100 | Fetal human muscle | Myotendinous junction label | Myotendinous junction HE x40 | Whipf 1 | Whipf 2 | Whipf 3 | Tongue HE x10 transverse | Tongue x100 | Muscle spindle HE x20 | Muscle spindle HE x40
Electron Microscopy Virtual Slides
Electron micrographs below are thin longitudinal section cut through adult human skeletal muscle tissue.
| | | | | | | | | | | | |
--- --- --- --- --- --- |
| | | | Skeletal Muscle EM1 | | Mobile | Desktop | Original | | Skeletal Muscle | EM Slides | | | | | Skeletal Muscle EM2 | | Mobile | Desktop | Original | | Skeletal Muscle | EM Slides | | | | | Skeletal Muscle EM3 | | Mobile | Desktop | Original | | Skeletal Muscle | EM Slides | |
| | | | Skeletal Muscle EM4 | | Mobile | Desktop | Original | | Skeletal Muscle | EM Slides | | | | | Skeletal Muscle EM5 | | Mobile | Desktop | Original | | Skeletal Muscle | EM Slides | |
Image Source: Contributed by Dartmouth College Electron Microscope Facility special thanks to Chuck Daghlian and Louisa Howard. Gallery. Original images may have been altered in size contrast and labelling. (These images are in the public domain)
: Links: Electron Microscopy Virtual Slides
Puberty
Musculoskeletal mass doubles by the end of puberty
regulated growth by - sex steroid hormones, growth hormone, insulin-like growth factors
accumulation of (peak) bone mass during puberty relates to future osteoporosis in old age
Satellite Cells
These cells remain as muscle stem cells under the basal lamina around each skeletal muscle fibre. They have a role in postnatal growth and also regeneration of muscle fibres. (see review)
| ExpandMore recent papers |
| This table allows an automated computer search of the external PubMed database using the listed "Search term" text link. This search now requires a manual link as the original PubMed extension has been disabled. The displayed list of references do not reflect any editorial selection of material based on content or relevance. References also appear on this list based upon the date of the actual page viewing. References listed on the rest of the content page and the associated discussion page (listed under the publication year sub-headings) do include some editorial selection based upon both relevance and availability. More? References | Discussion Page | Journal Searches | 2019 References | 2020 References Search term: Satellite Cell Satellite Cell |
: Links: stem cells
Abnormalities
There can be abnormalities associated directly with muscle differentiation and function as well as those mediated indirectly by abnormalities of innervation or skeletal development and other associated systems.
Duchenne Muscular Dystrophy
The most common occuring in Boys and in Duchenne Muscular Dystrophy (DMD). This cause of the disease was discovered in 1988 as a mutation in dystrophin, a protein that lies under the muscle fiber membrane and maintains the cell's integrity. As skeletal muscles have little prenatal load or use it is not until postnatally that muscle wasting occurs, usually in the anti-gravity muscles first. This is a progressive disease usually detected between 3-5 years old.
X-linked dystrophy
large gene encoding cytoskeletal protein - Dystrophin
progressive wasting of muscle, die late teens
Becker Muscular Dystrophy
(BMD) Similar to DMD but allows muscles to function better than in DMD, slower progression, make a shortened form of the mutated protein. Named after Peter Emil Becker, a German doctor who first described this variant in the 1950s.
Autosomal Recessive Muscular Dystrophy
Dystroglycan, a protein that associates with both dystrophin and membrane molecules, is a candidate gene for the site of the mutation in autosomal recessive muscular dystrophies. A knockout mouse has been generated that has early developmental abnormalities.
Myotonic Dystrophy
An inherited disorder in which the muscles contract but have decreasing power to relax. With this condition, the muscles also become weak and waste away. The myotonic dystrophy gene, found on chromosome 19, codes for a protein kinase that is found in skeletal muscle, where it likely plays a regulatory role. The disease is "amplified" through generations probably by a similar GC expansion associated with Huntington disease.
Facioscapulohumeral muscular dystrophy (FSHD)
characterized by the progressive weakness and atrophy of a specific subset of skeletal muscles.
mostly affects the muscles of the face, scapula, and upper arms.
involvement of specific muscles that it is often used clinically to distinguish FSHD from other forms of muscular dystrophy.
: Links: musculoskeletal abnormalities | PLoS Currents - Muscular Dystrophy
References
↑ Warmbrunn MV, de Bakker BS, Hagoort J, Alefs-de Bakker PB & Oostra RJ. (2018). Hitherto unknown detailed muscle anatomy in an 8-week-old embryo. J. Anat. , , . PMID: 29726018 DOI.
↑ Chal J & Pourquié O. (2017). Making muscle: skeletal myogenesis in vivo and in vitro. Development , 144, 2104-2122. PMID: 28634270 DOI.
↑ Jump up to: 3.0 3.1 Mekonen HK, Hikspoors JP, Mommen G, Köhler SE & Lamers WH. (2015). Development of the ventral body wall in the human embryo. J. Anat. , 227, 673-85. PMID: 26467243 DOI.
↑ Lee KY, Singh MK, Ussar S, Wetzel P, Hirshman MF, Goodyear LJ, Kispert A & Kahn CR. (2015). Tbx15 controls skeletal muscle fibre-type determination and muscle metabolism. Nat Commun , 6, 8054. PMID: 26299309 DOI.
↑ Mayeuf-Louchart A, Lagha M, Danckaert A, Rocancourt D, Relaix F, Vincent SD & Buckingham M. (2014). Notch regulation of myogenic versus endothelial fates of cells that migrate from the somite to the limb. Proc. Natl. Acad. Sci. U.S.A. , 111, 8844-9. PMID: 24927569 DOI.
↑ Powell GT & Wright GJ. (2011). Jamb and jamc are essential for vertebrate myocyte fusion. PLoS Biol. , 9, e1001216. PMID: 22180726 DOI.
↑ Murphy M & Kardon G. (2011). Origin of vertebrate limb muscle: the role of progenitor and myoblast populations. Curr. Top. Dev. Biol. , 96, 1-32. PMID: 21621065 DOI.
↑ Tao Y, Neppl RL, Huang ZP, Chen J, Tang RH, Cao R, Zhang Y, Jin SW & Wang DZ. (2011). The histone methyltransferase Set7/9 promotes myoblast differentiation and myofibril assembly. J. Cell Biol. , 194, 551-65. PMID: 21859860 DOI.
↑ Ropka-Molik K, Eckert R & Piórkowska K. (2011). The expression pattern of myogenic regulatory factors MyoD, Myf6 and Pax7 in postnatal porcine skeletal muscles. Gene Expr. Patterns , 11, 79-83. PMID: 20888930 DOI.
↑ Mei H, Ho MK, Yung LY, Wu Z, Ip NY & Wong YH. (2011). Expression of Gα(z) in C2C12 cells restrains myogenic differentiation. Cell. Signal. , 23, 389-97. PMID: 20946953 DOI.
↑ Mekonen HK, Hikspoors JP, Mommen G, Eleonore KÖhler S & Lamers WH. (2016). Development of the epaxial muscles in the human embryo. Clin Anat , 29, 1031-1045. PMID: 27571325 DOI.
↑ Noden DM & Francis-West P. (2006). The differentiation and morphogenesis of craniofacial muscles. Dev. Dyn. , 235, 1194-218. PMID: 16502415 DOI.
↑ Grenier J, Teillet MA, Grifone R, Kelly RG & Duprez D. (2009). Relationship between neural crest cells and cranial mesoderm during head muscle development. PLoS ONE , 4, e4381. PMID: 19198652 DOI.
↑ Jump up to: 14.0 14.1 Waardenberg AJ, Reverter A, Wells CA & Dalrymple BP. (2008). Using a 3D virtual muscle model to link gene expression changes during myogenesis to protein spatial location in muscle. BMC Syst Biol , 2, 88. PMID: 18945372 DOI.
↑ Baghdadi MB & Tajbakhsh S. (2018). Regulation and phylogeny of skeletal muscle regeneration. Dev. Biol. , 433, 200-209. PMID: 28811217 DOI.
↑ BECKER PE & KIENER F. (1955). [A new x-chromosomal muscular dystrophy]. Arch Psychiatr Nervenkr Z Gesamte Neurol Psychiatr , 193, 427-48. PMID: 13249581
Reviews
Baghdadi MB & Tajbakhsh S. (2018). Regulation and phylogeny of skeletal muscle regeneration. Dev. Biol. , 433, 200-209. PMID: 28811217 DOI.
Michailovici I, Eigler T & Tzahor E. (2015). Craniofacial Muscle Development. Curr. Top. Dev. Biol. , 115, 3-30. PMID: 26589919 DOI.
Abmayr SM & Pavlath GK. (2012). Myoblast fusion: lessons from flies and mice. Development , 139, 641-56. PMID: 22274696 DOI.
Murphy M & Kardon G. (2011). Origin of vertebrate limb muscle: the role of progenitor and myoblast populations. Curr. Top. Dev. Biol. , 96, 1-32. PMID: 21621065 DOI.
Mok GF & Sweetman D. (2011). Many routes to the same destination: lessons from skeletal muscle development. Reproduction , 141, 301-12. PMID: 21183656 DOI.
Albini S & Puri PL. (2010). SWI/SNF complexes, chromatin remodeling and skeletal myogenesis: it's time to exchange!. Exp. Cell Res. , 316, 3073-80. PMID: 20553711 DOI.
Buckingham M & Vincent SD. (2009). Distinct and dynamic myogenic populations in the vertebrate embryo. Curr. Opin. Genet. Dev. , 19, 444-53. PMID: 19762225 DOI.
Cossu G & Biressi S. (2005). Satellite cells, myoblasts and other occasional myogenic progenitors: possible origin, phenotypic features and role in muscle regeneration. Semin. Cell Dev. Biol. , 16, 623-31. PMID: 16118057 DOI.
Articles
Wang S, Zhang B, Addicks GC, Zhang H, J Menzies K & Zhang H. (2018). Muscle Stem Cell Immunostaining. Curr Protoc Mouse Biol , 8, e47. PMID: 30106515 DOI.
Tzahor E. (2015). Head muscle development. Results Probl Cell Differ , 56, 123-42. PMID: 25344669 DOI.
Tao Y, Neppl RL, Huang ZP, Chen J, Tang RH, Cao R, Zhang Y, Jin SW & Wang DZ. (2011). The histone methyltransferase Set7/9 promotes myoblast differentiation and myofibril assembly. J. Cell Biol. , 194, 551-65. PMID: 21859860 DOI.
Philipot O, Joliot V, Ait-Mohamed O, Pellentz C, Robin P, Fritsch L & Ait-Si-Ali S. (2010). The core binding factor CBF negatively regulates skeletal muscle terminal differentiation. PLoS ONE , 5, e9425. PMID: 20195544 DOI.
Bhatnagar S, Kumar A, Makonchuk DY, Li H & Kumar A. (2010). Transforming growth factor-beta-activated kinase 1 is an essential regulator of myogenic differentiation. J. Biol. Chem. , 285, 6401-11. PMID: 20037161 DOI.
Grenier J, Teillet MA, Grifone R, Kelly RG & Duprez D. (2009). Relationship between neural crest cells and cranial mesoderm during head muscle development. PLoS ONE , 4, e4381. PMID: 19198652 DOI.
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June 2010 " Skeletal Muscle Development" All (19316) Review (2515) Free Full Text (5587)
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Mouse E11.5 Myog PMID 23236180
Mouse E12.5 Myog PMID 23236180
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4311 | https://www.geogebra.org/m/fjtrj8Cw | Google Classroom
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4312 | https://chemistrytalk.org/ionic-radius-trends/ | Skip to content
Ionic Radius Trends
Core Concepts
In this tutorial, you will be introduced to ionic radius trends on the periodic table of elements. You will also be introduced to the concepts that contribute to ionic radius, including how to find it.
Topics Covered in Other Articles
Periodic Table Metals and Non-Metals
Atomic Radius Trends
How to Write an Electron Configuration
What are Valence Electrons?
What is Lattice Energy?
Vocabulary
Atomic radius: The atomic radius of a chemical element is a measure of the size of its atom, usually the mean or typical distance from the center of the nucleus to the outermost isolated electron.
Ionic radius: the radius of a monatomic ion in an ionic crystal structure. Although neither atoms nor ions have sharp boundaries, they are treated as if they were hard spheres with radii such that the sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
Valence electrons: the electrons in the outermost shell, or energy level, of an atom.
What is an Ion?
An ion is an atom that carries a positive or negative electric charge. When an atom on the periodic table loses an electron it becomes a cation. In contrast, when an atom in the periodic table gains an electron it becomes an anion.
Most metals on the periodic table tend to form cations. For example, the metals in group 1A (Alkali Metals) all have a +1 charge meaning they want to give away an electron. The non-metals tend to become anions.
Ionic Radius
By definition, an ionic radius is the distance of the outermost shell of electrons (valence shell) from the nucleus of an ion. So in the picture below the ionic radius would be found by measuring the distance between the yellow and blue protons in the nucleus. To the red electrons at the outermost orbital. This is a good visualization of how the radius may be measured but this is unlikely to be accurate because the borders of orbitals are quite fuzzy.
The ionic radius of an ion is measured when the atom is in a crystal lattice structure. The ionic radius is half of the distance between two ions that is just touching one another. As mentioned above, the shell boundaries of electron shells are difficult to read. So the ions of an atom are treated as if they were spheres. The ionic radius can easily be a little smaller or larger than the atomic radius, which is the radius a neutral atom of the element possesses.
Finding Ionic Radius
Since ions tend to exist in bonds, the ionic radius can be found via the ionic bond between two atoms. Moreover, the ionic radius is tricky to measure since it depends on the varying factors of the environment in which the ion is located. It depends on the coordination number, or the number of atoms, ions, or molecules that a central atom or ion holds as its nearest neighbors in a complex or coordination compound. It also depends on the spin state of the ion. The ionic radius is generally calculated by estimating the distance between the two nuclei and dividing it according to the atomic sizes. Ionic radius is generally measured in picometer (pm) or nanometer (nm).
Cations vs. Anions
Because cations lose electrons from the valence shell they often have smaller ionic radii than their parent neutral atom. Vice Versa for anions since anions gain electrons to the outermost shell they usually will have bigger radii than their parent neutral atom.
Ex. The atomic radius of sodium (Na) is 190 pm, but the ionic radius of sodium (Na+1) is only 116 pm.
Ex. The atomic radius of the chlorine atom (Cl) is 79 pm, and the ionic radius of the chlorine ion (Cl-1) is 167 pm.
Ionic Radius Trends in the Periodic Table
Ionic Radius and Group
As you move down the periodic table additional electrons are being added, which in turn causes the ionic radius to increase.
Look at group 2 on the periodic table (Alkaline Earth Metals)
Beryllium ionic radius – 31 pm
Magnesium ionic radius – 65 pm
Calcium ionic radius – 99 pm
Strontium ionic radius – 113 pm
Barium ionic radius – 135 pm
Congruent with the trend of increasing atomic radius as you move down the periodic table.
Ionic Radius and Period
As you move across the periodic table more electrons, protons, and neutrons are being added. So it would make sense for the radius to increase, but it does not. This is because as you move over a row on the periodic table the ionic radius decreases for metals forming cations. The ionic radius increases for nonmetals as the effective nuclear charge decreases.
Moving across the periodic table from potassium (K)
Potassium (K) ionic radius – 137 pm
Calcium ionic (Ca) radius – 99 pm
Scandium ionic (Sc) radius – 87 pm
Notice as you move to the right on the periodic table the ionic radius size is decreasing. |
4313 | https://www.youtube.com/watch?v=nNT4Q--Zkho | Solve a Linear Inequality & Show the Solution on a Number Line
Maths with Jay
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A linear inequality solved & checked, then the solution represented on a number line
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Transcript:
hello welcome to maths with Jay here we're going to solve an inequality and then we're going to represent that solution on the number line so the inequality is that 8 X plus 5 is greater than or equal to 29 now if you haven't seen something like this before it may be a good idea if you think about the related equation so we would simply write down instead of greater than or equal to just equals so that would mean that the equation would be 8 X plus 5 is equal to 29 so if you think about what you would do to solve that you would subtract 5 from both sides so you'd have 4 8 X is 24 and then you would divide both sides by 8 so X is equal to 3 and you would check your answer by multiplying 8 by your answer so 8 times 3 24 plus 5 is 29 so that's the right answer now when you're looking at an inequality you do more or less the same kind of thing you can do the same to both sides but you do need to be careful if you're multiplying or dividing by a negative if you do want to do that then you would need to change the direction of the inequality anyway let's write down this inequality so we've got 8 X plus 5 is greater than or equal to 29 so we can subtract 5 from both sides so 8 X must be greater than or equal to 29 minus 5 so that's 24 and then dividing both sides by 8 we're dividing by a positive number so that's not going to change the direction of the inequality we've got X is greater than or equal to 3 so you can see that if you can solve the related equation solving the inequality is just a simple really but when it comes to checking it there are two checks to do he would check for 3 so 8 times 3 plus 5 is equal to 29 so let's just write that down so when x equals three we've got that eight times three plus five equals 29 and then because we found that X is either equal to three or greater than three we're going to choose any value that we like greater than three so you could choose for nine point 210 million whatever you like but it's easier obviously to choose a relatively small number so let's just choose four and see what we get so eight times four plus five putting the the value for X into the original left-hand side so eight X plus five is eight times four plus five so that's 32 plus five so that's going to be 37 and that is greater than 29 so that's showing us that it looks like we have got the correct answer whereas if we checked x equals two for example we would expect that not to work because two is less than three and in fact that would be the case wouldn't it because eight times two is only 16 and when we add 5 to that we get 21 which is less than 29 so we're doing two checks to check that we've got the inequality of the right way round you've got greater than or equal to three rather than less than or equal to three okay so that's the solution and then we not we want to show the answer on the number line so we're starting off at x equals three so we draw a circle at x equals three and because X can be equal to three we want to shade in that circle and then we want the rest of the number line to be shaded in if you're using pencil and your pencil mark can't be seen on the number line what you could do is draw just above the number line so shading your three and then draw your line like that so that's all you need to do to represent all the values that solve this inequality anything to the right of three on that number line and we've included three |
4314 | https://wayground.com/library/videos/math/geometry/circles-and-conic-sections/circle-properties/equation-derivation-of-parabola-from-focus-directrix | Equation Derivation Of Parabola From Focus & Directrix Interactive Videos Kindergarten to 12th Grade Math | Wayground (formerly Quizizz)
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4315 | https://igcsemathstutor.com/calculating-height-using-trigonometry/ | Can you calculate height using Trigonometry?
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Using Trigonometry to calculate height
Using Trigonometry to Calculate the Height of Objects:
Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. By applying trigonometric concepts, such as sine, cosine, and tangent, we can determine the height of an object or distance using angle measurements and known lengths. Understanding how to use trigonometry for height calculations is essential in various fields, including engineering, surveying, and physics.
The SOHCAHTOA Method
The SOHCAHTOA acronym is a helpful mnemonic to remember the three primary trigonometric ratios:
Sine (sin): The ratio of the length of the side opposite an angle to the length of the hypotenuse in a right triangle.
Cosine (cos): The ratio of the length of the side adjacent to an angle to the length of the hypotenuse in a right triangle.
Tangent (tan): The ratio of the length of the side opposite an angle to the length of the side adjacent to the angle in a right triangle.
These trigonometric ratios are fundamental in trigonometry and are used to solve various types of problems involving angles and sides of triangles.
Steps for Calculating the Height of Objects using Trigonometry:
To calculate the height of objects using trigonometry, follow these steps:
Identify the right triangle formed by the object and the observer’s line of sight.
Measure the angle of elevation or depression from the observer’s line of sight to the object. Ensure that the angle is with respect to the horizontal line.
Determine the length of a known side or distance in the triangle. This can be the distance from the observer to the object or any other relevant length.
Choose the appropriate trigonometric ratio (sine, cosine, or tangent) based on the known side and the angle.
Set up the trigonometric equation and solve for the unknown height.
Height Calculation Examples Using Trigonometry
Example 1: Calculating Height Using Trigonometry
Example 1: Calculating Height Using Trigonometry
Imagine you’re standing 20 meters away from a tree. You measure the angle of elevation to the top of the tree as 30 degrees. To find the height of the tree, you can use the tangent ratio:
Given:
Distance from the tree: 20 meters
Angle of elevation: 30 degrees
To find the height, use the formula:
height = distance × tan(angle of elevation)
Substitute the values into the formula:
height = 20 × tan(30°)
height ≈ 20 × 0.577
height ≈ 11.54 meters
Therefore, the height of the tree is approximately 11.54 meters.
Example 2: Calculating Building Height Using Trigonometry
Example 2: Calculating Building Height Using Trigonometry
Suppose you’re looking at the top of a building from a distance of 50 meters. The angle of elevation to the top of the building is 45 degrees. To determine the height of the building, you can once again use the tangent ratio:
Given:
Distance from the building: 50 meters
Angle of elevation: 45 degrees
To find the height, use the formula:
height = distance × tan(angle of elevation)
Substitute the values into the formula:
height = 50 × tan(45°)
height ≈ 50 × 1
height ≈ 50 meters
Therefore, the height of the building is approximately 50 meters.
Example 3: Calculating Mountain Height Using Trigonometry
Example 3: Calculating Mountain Height Using Trigonometry
Imagine you’re at a viewing point 500 meters away from the base of a mountain. The angle of elevation to the peak of the mountain is 20 degrees. To estimate the height of the mountain, you can use the tangent ratio:
Given:
Distance from the mountain: 500 meters
Angle of elevation: 20 degrees
To find the height, use the formula:
height = distance × tan(angle of elevation)
Substitute the values into the formula:
height = 500 × tan(20°)
height ≈ 500 × 0.364
height ≈ 182 meters
Therefore, the estimated height of the mountain is approximately 182 meters.
Video guide on how to calculate height using Trigonometry
This YouTube video provides a detailed explanation of using trigonometry to calculate the height of objects. It covers the SOHCAHTOA method, explains the trigonometric ratios, and demonstrates how to apply them in practical scenarios. By watching the video, you will gain a clear understanding of using trigonometry for height calculations and be able to solve problems with confidence.
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4316 | https://www.rxlist.com/urokinase/generic-drug.htm | Urokinase: Side Effects, Uses, Dosage, Interactions, Warnings
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Urokinase
Brand Name: Kinlytic
Drug Class: Thrombolytics
Medical Author: Divya Jacob, Pharm. D.
Medical Reviewer: Sarfaroj Khan, BHMS, PGD Health Operations
generic drugs
Uses
What Is Urokinase and How Does It Work?
Side Effects
What Are Side Effects Associated with Using Urokinase?
Dosages
What Are Dosages of Urokinase?
Drug Interactions
What Other Drugs Interact with Urokinase?
Warnings and Precautions
What Are Warnings and Precautions for Urokinase?
What Is Urokinase and How Does It Work?
Urokinase is a prescription medication used to treat blood clots in the lungs.
Urokinase are available under various brand names: Abbokinase, Kinlytic
What Are Side Effects Associated with Using Urokinase?
Common side effects of Urokinase include:
easy bruising, and
bleeding
Serious side effects of Urokinase include:
hives,
difficult breathing,
swelling of the face, lips, tongue, or throat.
risk of bleeding,
easy bruising or bleeding (nosebleeds, bleeding gums, bleeding from a wound, incision, catheter, or needle injection);
bloody or tarry stools, coughing up blood or vomit that looks like coffee grounds;
red or pink urine; or
sudden numbness or weakness (especially on one side of the body), sudden severe headache, slurred speech, problems with vision or balance,
chest pain or heavy feeling, pain spreading to the jaw or shoulder, nausea, sweating, general ill feeling;
swelling, rapid weight gain, little or no urinating;
severe stomach pain, nausea, and vomiting;
darkening or purple discoloration of your fingers or toes;
very slow heartbeats, shortness of breath, feeling light-headed;
sudden severe back pain, muscle weakness, numbness or loss of feeling in your arms or legs;
dangerously high blood pressure--severe headache, blurred vision, pounding in your neck or ears, nosebleed, anxiety, confusion, severe chest pain, shortness of breath, irregular heartbeats; or
pancreatitis--severe pain in your upper stomach spreading to your back, nausea and vomiting, fast heart rate.
Rare side effects of Urokinase include:
none
Seek medical care or call 911 at once if you have the following serious side effects:
Severe headache, confusion, slurred speech, arm or leg weakness, trouble walking, loss of coordination, feeling unsteady, very stiff muscles, high fever, profuse sweating, or tremors.
Serious eye symptoms such as sudden vision loss, blurred vision, tunnel vision, eye pain or swelling, or seeing halos around lights.
Serious heart symptoms such as fast, irregular, or pounding heartbeats; fluttering in the chest; shortness of breath; sudden dizziness, lightheartedness, or passing out.
This is not a complete list of side effects and other serious side effects or health problems that may occur because of the use of this drug. Call your doctor for medical advice about serious side effects or adverse reactions. You may report side effects or health problems to FDA at 1-800-FDA-1088.
What Are Dosages of Urokinase?
Adult and pediatric dosage
Sterile lyophilized white powder
250,000 international units urokinase per vial
Pulmonary embolism
Adult and pediatric dosage
The loading dose of 4,400 international units per kilogram of urokinase injection is given at a rate of 90 mL per hour over a period of 10 minutes.
This is followed with a continuous infusion of 4,400 international units per kilogram per hour at a rate of 15 mL for 12 hours.
Administration of urokinase injection may be repeated as necessary.
A dosing and preparation chart for patients who weigh 37 to 114 kilograms (81 to 250 pounds) is provided as a guide in the Preparation Section that follows below. If the patient is outside of these weights, calculate with dosing information provided above.
Dosage Considerations – Should be Given as Follows:
See “Dosages”
What Other Drugs Interact with Urokinase?
If your medical doctor is using this medicine to treat your pain, your doctor or pharmacist may already be aware of any possible drug interactions and may be monitoring you for them.Do not start, stop, or change the dosage of any medicine before checking with your doctor, health care provider, or pharmacist first.
Urokinase has severe interactions with the following drugs:
a blood thinner (heparin, warfarin, Coumadin, Jantoven);
NSAIDs (nonsteroidal anti-inflammatory drugs)--aspirin, ibuprofen (Advil, Motrin), naproxen (Aleve), celecoxib, diclofenac, indomethacin, meloxicam, and others; or
medication used to prevent blood clots--abciximab, eptifibatide, tirofiban, vorapaxar.
Urokinase has serious interactions with no other drugs.
Urokinase has moderate interactions with no other drugs.
Urokinase has minor interactions with no other drugs.
This information does not contain all possible interactions or adverse effects. Visit the rxlist Drug Interaction Checker for any drug interactions. Therefore, before using this product, tell your doctor or pharmacist about all your products.Keep a list of all your medications with you and share this information with your doctor and pharmacist.Check with your health care professional or doctor for additional medical advice, or if you have health questions or concerns.
What Are Warnings and Precautions for Urokinase?
Contraindications
allergic reaction to urokinase
active bleeding inside your body;
a brain tumor or blood vessel disorder;
a brain aneurysm (dilated blood vessel);
a bleeding or blood clotting disorder (such as hemophilia);
severe or uncontrolled high blood pressure;
if you have had a recent medical emergency requiring CPR (cardiopulmonary resuscitation); or
if you have had a stroke, brain surgery, or spinal surgery within in the past 2 months.
Effects of drug abuse
None
Short-Term Effects
See “What Are Side Effects Associated with Using Urokinase?”
Long-Term Effects
See “What Are Side Effects Associated with Using Urokinase?”
Cautions
If possible before you receive urokinase, tell your doctor if you have a brain tumor or aneurysm, hemophilia or other bleeding disorder, high blood pressure, or if you have recently had a stroke, brain or spinal surgery, or medical emergency requiring CPR (cardiopulmonary resuscitation).
In an emergency situation it may not be possible to tell your caregivers about your health conditions. Make sure any doctor caring for you afterward knows you have received this medicine..
Ask your doctor before taking aspirin or ibuprofen (Motrin, Advil) shortly after you have received urokinase. These medications can increase your risk of bleeding.
Avoid activities that may increase your risk of bleeding or injury. Use extra care to prevent bleeding while shaving or brushing your teeth.
Urokinase is made from human plasma (part of the blood) which may contain viruses and other infectious agents. Donated plasma is tested and treated to reduce the risk of it containing infectious agents, but there is still a small possibility it could transmit disease. Talk with your doctor about the risks and benefits of using this medication.
Pregnancy and Lactation
Reproduction studies have been performed in mice and rats at doses up to 1,000 times the human dose and have revealed no evidence of impaired fertility or harm to the fetus due to urokinase injection. There are, however, no adequate and well-controlled studies in pregnant women. Because animal reproduction studies are not always predictive of human response, this drug should be used during pregnancy only if clearly needed.
Lactation
It is not known whether this drug is excreted in human milk. Because many drugs are excreted in human milk, caution should be exercised when urokinase injection is administered to a nursing woman.
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4317 | https://discussion.tiwariacademy.com/question/what-does-the-area-under-the-stress-strain-curve-represent/ | What does the area under the stress-strain curve represent? – Discussion Forum
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Asked:January 16, 20252025-01-16T07:03:16+00:00 2025-01-16T07:03:16+00:00 In: Class-11-Physics
What does the area under the stress-strain curve represent?
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0%Elastic modulus 50%Work done on the material per unit volume ( 1 voter ) 0%Strain energy density 50%Modulus of rigidity ( 1 voter )
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Work done on the material per unit volume
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The area under the stress-strain curve provides insight into the mechanical behavior of the material.
The CBSE syllabus for 2024-2025 includes Chapter 8 “Mechanical Properties of Solids” from the Class 11 NCERT Physics textbook. This chapter introduces key concepts such as stress, strain, elasticity, Young’s modulus, bulk modulus, shear modulus and Poisson’s ratio. It explains how solids respond to different forces and the factors that affect their mechanical properties. Multiple-choice questions are used to test the understanding of these principles like Hooke’s Law and the stress-strain relationship. These concepts are essential for academic learning and practical applications in fields like engineering and material science.
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manish12 2025-01-16T09:30:36+00:00 Added an answer on January 16, 2025 at 9:30 am
The area under the stress-strain curve represents the energy per unit volume absorbed by the material during deformation.
This question related to Chapter 8 physics Class 11th NCERT. From the Chapter 8. Mechanical Properties of Solids. Give answer according to your understanding.
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4318 | https://www.slideshare.net/slideshow/solution-manual-vector-mechanics-for-engineersstatics-beer-johnston-mazurek-12th-edition/148995960 | Uploaded byMichaelLeigh25
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solution manual Vector Mechanics for Engineers:Statics Beer Johnston Mazurek 12th edition
The document contains 18 practice problems involving determining the magnitude and direction of the resultant force of two or more applied forces using trigonometric methods like the law of sines and cosines. The problems involve forces applied to structures like hooks, brackets, stakes pulled from the ground, and tugboats pulling barges. The student must use trigonometry to calculate values like the unknown force magnitude, angle between forces, or magnitude of the resultant force.
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In this document
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These slides detail graphical methods like parallelogram law and triangle rule for determining the resultant of two forces in various scenarios, with numerical values such as 1391 N, 906 lb, and 125 lb.
Slides focus on calculating tensions in cables and resultant forces in various contexts, with examples using forces of 120 N and components yielding results such as tension in cables and forces in newtons.
These slides outline problems involving calculating x, y, z components of forces and resultant forces in multiple dimensions, including forces like 600 N and 820 N, and their respective angles.
Focuses on tension in cables under various loads, including calculations of weight forces, equilibrium conditions, and systems involving multiple supports with forces ranging from 60 lb to 600 lb.
This section discusses the force and tension analysis of mechanical systems, determining required forces, tensions in cables, and resultant forces using free-body diagrams.
Encompasses calculations of equilibrium in systems with multiple cables or forces acting, determining tensions, weights of containers, and conditions needed for static equilibrium in force systems.
Explores advanced problems in cable tensions, including force components and resolutions regarding weights, tensions, and vertical forces acting from various load applications.
Includes free-body diagrams for various experimental setups assessing load resistance, including systems with multiple force directions, tensions, and the resulting force conditions.
Concludes with various mechanical and tension problems discussing real-world applications such as supporting systems under various load conditions.
solution manual Vector Mechanics for Engineers:Statics Beer Johnston Mazurek 12th edition
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1 CHAPTER 2 www.solutions-guides.com
2.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 2 PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 1391 kN, 47.8R a= = 1391 N=R 47.8 ◀ 45° 30° 900 N 600 N
3.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 3 PROBLEM 2.2 Two forces are applied as shown to a bracket support. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 906 lb,R = 26.6a = 906 lbR = 26.6 ◀ 500 lb 800 lb 35° 60°
4.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 4 PROBLEM 2.3 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P 75 N= and 125 N,Q = determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 179 N, 75.1R a= = R 179 N= 75.1
5.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 5 PROBLEM 2.4 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P 60 lb= and 25 lb,Q = determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 77.1lb, 85.4R a= = R 77.1 lb= 85.4
6.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 6 PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the triangle rule and the law of sines: (a) 120 N sin30 sin25 P = 101.4 NP = ◀ (b) 30 25 180 180 25 30 125 b b + + = = - - = 120 N sin30 sin125 R = 196.6 NR = ◀ 120 N P α25°
7.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 7 PROBLEM 2.6 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1= 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75 40 180 180 75 40 65 a a + + = = - - = 2 800 lb sin65 sin75 T = 2 853 lbT = ◀ (b) 800 lb sin65 sin40 R = 567 lbR = ◀ A B 25°15° T1 T2
8.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 8 PROBLEM 2.7 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75 40 180 180 75 40 65 b b + + = = - - = 11000 lb sin 75° sin65 T = 1 938 lbT = ◀ (b) 1000 lb sin 75° sin 40 R = 665 lbR = ◀ A B 25°15° T1 T2
9.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 9 PROBLEM 2.8 A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A. SOLUTION Using the law of sines: 2.2 kN sin30° sin125 sin25 ACT R = = 2.603 kN 4.264 kN ACT R = = (a) 2.60 kNA CT = ◀ (b) 4.26 kNR = ◀ 30° B C A α
10.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 10 PROBLEM 2.9 A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile. SOLUTION Using the law of cosines: 2 2 2 (3 kN) (4.8 kN) 2(3 kN)(4.8 kN)cos 30° 2.6643 kN AC AC T T = + - = Using the law of sines: sin sin30 3 kN 2.6643 kN 34.3 a a = = 2.66 kNA C =T 34.3 ◀ 30° B C A α
11.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 11 PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle a if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: (a) sin sin 25 50 N 35 N sin 0.60374 a a = = 37.138a = 37.1a = ◀ (b) 25 180 180 25 37.138 117.862 a b b + + = = - - = 35 N sin117.862 sin25 R = 73.2 NR = ◀ 50 N 25° P α
12.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 12 PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that a = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 50 60 180 180 50 60 70 b b + + = = - - = 425 lb sin 70 sin60 P = 392 lbP = ◀ (b) 425 lb sin 70 sin50 R = 346 lbR = ◀ 425 lb A P 30° α
13.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 13 PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle a if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) ( 30 ) 60 180 180 ( 30 ) 60 90 sin(90 ) sin60 425 lb 500 lb a b b a b a a + + + = = - + - = - - = 90 47.402a- = 42.6a = ◀ (b) 500 lb sin(42.598 30 ) sin60 R = + 551 lbR = ◀ 425 lb A P 30° α
14.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 14 PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) (425 lb)cos30P = 368 lb=P ◀ (b) (425 lb)sin30R = 213 lbR = ◀ 425 lb A P 30° α
15.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 15 PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) (50 N)sin 25P = 21.1 N=P ◀ (b) (50 N)cos 25R = 45.3 NR = ◀ 50 N 25° P α
16.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 16 PROBLEM 2.15 The barge B is pulled by two tugboats A and C. At a given instant the tension in cable AB is 4500 lb and the tension in cable BC is 2000 lb. Determine by trigonometry the magnitude and direction of the resultant of the two forces applied at B at that instant. SOLUTION Using the law of cosines: 180 30 45 105 b b = - - = ( ) ( ) ( )( ) 2 22 4500 lb 2000 lb 2 4500 lb 2000 lb cos 105 5380 lb R R = + - = Using the law of sines: ( ) ( ) 2000 lb sin sin 30 5380 lb 2000 lb sin105 sin 30 R b a a = - = - 8.94a = 5380 lb=R 8.94 ◀
17.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 17 PROBLEM 2.16 Solve Prob. 2.1 by trigonometry. PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the law of cosines: 2 2 2 (900N) (600N) 2(900N)(600 N)cos(135 ) 1390.57N R R = + - = Using the law of sines: sin(135 )sin( 30 ) 600N 1390.57N 30 17.7642 47.764 a a a - = - = = 1391N=R 47.8 ◀ 45° 30° 900 N 600 N
18.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 18 PROBLEM 2.17 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that 60 lbP = and 25 lb,Q = determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the triangle rule and the law of cosines: 2 2 2 2 2 2 2 20 35 180 125 2 cos (60 lb) (25 lb) 2(60 lb)(25 lb)cos125 3600 625 3000(0.5736) 77.108 lb R P Q PQ R R R a a a + + = = = + - = + - = + + = Using the law of sines: sin125sin 25 lb 77.108 lb 15.402 b b = = 70 85.402b+ = R 77.1 lb= 85.4 ◀
19.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 19 PROBLEM 2.18 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the laws of cosines and sines: 2 2 2 (120 N) (160 N) 2(120 N)(160 N)cos25 72.096 N P P = + - = And sin sin25 120 N 72.096 N sin 0.70343 44.703 a a a = = = 72.1 N=P 44.7 ◀ 120 N P α25°
20.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 20 PROBLEM 2.19 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 10 kN in member A and 15 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines We have 180 (40 20 ) 120 g = - + = Then 2 2 2 2 (10 kN) (15 kN) 2(10 kN)(15 kN)cos120 475 kN 21.794 kN R R = + - = = and 21.794 kN15 kN sin sin120 15 kN sin sin120 21.794 kN 0.59605 36.588 a a a = æ ö÷ç= ÷ ç ÷ç ÷è ø = = Hence: 50 86.588f a= + = R 21.8 kN= 86.6°
21.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 21 PROBLEM 2.20 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines: We have 180 (40 20 ) 120 g = - + = Then 2 2 2 2 (15 kN) (10 kN) 2(15 kN)(10 kN)cos120 475 kN 21.794 kN R R = + - = = and 21.794 kN10 kN sin sin120 10 kN sin sin120 21.794 kN 0.39737 23.414 a a a = æ ö÷ç= ÷ ç ÷ç ÷è ø = = Hence: 50 73.414f a= + = R 21.8 kN= 73.4°
22.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 22 PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (84) (80) 116 in. (28) (96) 100 in. (48) (90) 102 in. OA OB OC = + = = + = = + = 29-lb Force: 84 (29 lb) 116xF = + 21.0 lbxF = + ◀ 80 (29 lb) 116yF = + 20.0 lbyF = + ◀ 50-lb Force: 28 (50 lb) 100xF =- 14.00 lbxF = - ◀ 96 (50 lb) 100yF = + 48.0 lbyF = + ◀ 51-lb Force: 48 (51 lb) 102xF = + 24.0 lbxF = + ◀ 90 (51 lb) 102yF =- 45.0 lbyF = - ◀ 29 lb 51 lb O x y 90 in. 96 in. 28 in. 84 in. 80 in. 48 in. 50 lb
23.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 23 PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (600) (800) 1000 mm (560) (900) 1060 mm (480) (900) 1020 mm OA OB OC = + = = + = = + = 800-N Force: 800 (800 N) 1000xF = + 640 NxF = + ◀ 600 (800 N) 1000yF = + 480 NyF = + ◀ 424-N Force: 560 (424 N) 1060xF =- 224 NxF = - ◀ 900 (424 N) 1060yF =- 360 NyF = - ◀ 408-N Force: 480 (408 N) 1020xF = + 192.0 NxF = + ◀ 900 (408 N) 1020yF =- 360 NyF = - ◀ O Dimensions in mm 424 N 408 N 800 N x y 900 800 600 560 480
24.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 24 PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION 350-N Force: (350 N)cos25xF =+ 317 NxF = + ◀ (350 N)sin25yF = + 147.9 NyF = + ◀ 800-N Force: (800 N)cos70xF =+ 274 NxF = + ◀ (800 N)sin70yF = + 752 NyF = + ◀ 600-N Force: (600 N)cos60xF =- 300 NxF = - ◀ (600 N)sin60yF = + 520 NyF = + ◀
25.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 25 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION 80-lb Force: (80 lb)cos30xF =+ 69.3 lbxF = + ◀ (80 lb)sin30yF =- 40.0 lbyF = - ◀ 120-lb Force: (120 lb)cos75xF =+ 31.1 lbxF = + ◀ (120 lb)sin75yF =- 115.9 lbyF = - ◀ 150-lb Force: (150 lb)cos40xF =- 114.9 lbxF = - ◀ (150 lb)sin40yF =- 96.4 lbyF = - ◀
26.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 26 PROBLEM 2.25 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION 2 2 (650 mm) (720 mm) 970 mm BC = + = (a) 650 970xP P æ ö÷ç= ÷ç ÷ç ÷è ø or 970 650 970 325 N 650 485 N xP P æ ö÷ç= ÷ç ÷ç ÷è ø æ ö÷ç= ÷ç ÷ç ÷è ø = 485 NP = ◀ (b) 720 970 720 485 N 970 360 N yP P æ ö÷ç= ÷ç ÷ç ÷è ø æ ö÷ç= ÷ç ÷ç ÷è ø = 970 NyP = ◀ A C B 720 mm 650 mm
27.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 27 PROBLEM 2.26 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION (a) sin35 300 lbP = 300 lb sin35 P = 523 lbP = ◀ (b) Vertical component cos35vP P= (523 lb)cos35= 428 lbvP = ◀ A B C D 35° Q
28.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 28 PROBLEM 2.27 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB. SOLUTION 180 45 90 30 180 45 90 30 15 a a = + + + = - - - = (a) cos cos 600 N cos15 621.17 N x x P P P P a a = = = = 621 NP = ◀ (b) tan tan (600 N)tan15 160.770 N y x y x P P P P a a = = = = 160.8 NyP = ◀ 45° 30° B A M C
29.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 29 PROBLEM 2.28 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) cos 55 yP P = 350 lb cos 55 610.21 lb = = 610 lbP = ◀ (b) sin 55xP P= (610.21 lb)sin 55 499.85 lb = = 500 lbxP = ◀ A B C 55° Q
30.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 30 PROBLEM 2.29 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. SOLUTION (a) 750 N sin 20P= 2192.9 NP = 2190 NP = ◀ (b) cos20A BCP P= (2192.9 N)cos20= 2060 NA BCP = ◀ 60° 50° B C D A Q
31.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 31 PROBLEM 2.30 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) 37 12 37 (720 N) 12 2220 N xP P= = = 2.22 kNP = ◀ (b) 35 12 35 (720 N) 12 2100 N y xP P= = = 2.10 kNyP = ◀ A B C D 7 m 2.4 m
32.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 32 PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: Force x Comp. (lb) y Comp. (lb) 29 lb +21.0 +20.0 50 lb -14.00 +48.0 51 lb +24.0 -45.0 31.0xR = + 23.0yR = + (31.0 lb) (23.0 lb) tan 23.0 31.0 36.573 23.0 lb sin(36.573 ) x y y x R R R R R a a = + = + = = = = R i j i j 38.601 lb= 38.6 lb=R 36.6 ◀ 29 lb 51 lb O x y 90 in. 96 in. 28 in. 84 in. 80 in. 48 in. 50 lb
33.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 33 PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.23: Force x Comp. (N) y Comp. (N) 350 N +317 +147.9 800 N +274 +752 600 N -300 +520 291xR = + 1419.9yR = + (291 N) (1419.9 N) tan 1419.9 N 291 N 78.418 1419.9 N sin(78.418 ) x y y x R R R R R a a = + = + = = = = R i j i j 1449 N= 1449 N=R 78.4 ◀
34.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 34 PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.24: Force x Comp. (lb) y Comp. (lb) 80 lb +69.3 -40.0 120 lb +31.1 -115.9 150 lb -114.9 -96.4 14.50xR = - 252.3yR =- (14.50 lb) (252.3 lb) tan 14.50 252.3 3.289 252.3 lb cos(3.289 ) x y x y R R R R R b b = + = - - = - = - = = R i j i j 253 lb= 253 lb=R 86.7 ◀
35.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 35 PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.22: Force x Comp. (N) y Comp. (N) 800 lb +640 +480 424 lb -224 -360 408 lb +192 -360 608xR = + 240yR = - (608 lb) ( 240 lb) tan 240 608 21.541 240 N sin(21.541°) 653.65 N x y y x R R R R R a a = + = + - = = = = = R i j i j 654 N=R 21.5 ◀ O Dimensions in mm 424 N 408 N 800 N x y 900 800 600 560 480
36.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 36 PROBLEM 2.35 Knowing that a = 35°, determine the resultant of the three forces shown. SOLUTION 100-N Force: (100 N)cos35 81.915 N (100 N)sin35 57.358 N x y F F = + = + = - = - 150-N Force: (150 N)cos65 63.393 N (150 N)sin65 135.946 N x y F F = + = + = - = - 200-N Force: (200 N)cos35 163.830 N (200 N)sin35 114.715 N x y F F = - = - = - = - Force x Comp. (N) y Comp. (N) 100 N +81.915 -57.358 150 N +63.393 -135.946 200 N -163.830 -114.715 18.522xR = - 308.02yR = - ( 18.522 N) ( 308.02 N) tan 308.02 18.522 86.559 x y y x R R R R a a = + = - + - = = = R i j i j 308.02 N sin86.559 R = 309 N=R 86.6 ◀ 200 N 150 N 100 N 30° αα
37.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 37 PROBLEM 2.36 Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at Point B of beam AB. SOLUTION Cable BC Force: 840 (725 N) 525 N 1160 840 (725 N) 500 N 1160 x y F F = - = - = = 500-N Force: 3 (500 N) 300 N 5 4 (500 N) 400 N 5 x y F F =- =- =- = - 780-N Force: 12 (780 N) 720 N 13 5 (780 N) 300 N 13 x y F F = = =- =- and 2 2 105 N 200 N ( 105 N) ( 200 N) 225.89 N x x y y R F R F R = S =- = S =- = - + - = Further: 1 200 tan 105 200 tan 105 62.3 a a - = = = Thus: R 226 N= 62.3
38.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 38 PROBLEM 2.37 Knowing that a = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos20 56.382 lb (60 lb)sin20 20.521 lb x y F F = = = = 80-lb Force: (80 lb)cos60 40.000 lb (80 lb)sin60 69.282 lb x y F F = = = = 120-lb Force: (120 lb)cos30 103.923 lb (120 lb)sin30 60.000 lb x y F F = = = - = - and 2 2 200.305 lb 29.803 lb (200.305 lb) (29.803 lb) 202.510 lb x x y y R F R F R = S = = S = = + = Further: 29.803 tan 200.305 a = 1 29.803 tan 200.305 8.46 a - = = 203 lb=R 8.46 ◀ 120 lb 80 lb 60 lb a a' α α 20°
39.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 39 PROBLEM 2.38 Knowing that a = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos 20 56.382 lb (60 lb)sin 20 20.521 lb x y F F = = = = 80-lb Force: (80 lb)cos 95 6.9725 lb (80 lb)sin 95 79.696 lb x y F F = = - = = 120-lb Force: (120 lb)cos 5 119.543 lb (120 lb)sin 5 10.459 lb x y F F = = = = Then 168.953 lb 110.676 lb x x y y R F R F = S = = S = and 2 2 (168.953 lb) (110.676 lb) 201.976 lb R = + = 110.676 tan 168.953 tan 0.65507 33.228 a a a = = = 202 lb=R 33.2 ◀ 120 lb 80 lb 60 lb a a' α α 20°
40.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 40 PROBLEM 2.39 A collar that can slide on a vertical rod is subjected to the three forces shown. Determine (a) the required value of α if the resultant of the three forces is to be horizontal, (b) the corresponding magnitude of the resultant. SOLUTION (85 N)cos (170 N)sin( ) x xR F a a = S = + (1) (110 N) (85 N)sin( ) (170 N)cos y yR F a a = S = + + - (2) (a) For R to be horizontal, we must have 0.yR = We make 0yR = in Eq. (2): ( ) 2 2 2 2 110 85sin 170cos 0 22 17sin 34cos 0 17sin 22 34 1 sin 289sin 748sin 484 1156 1 sin 1445sin 748sin 672 0 a a a a a a a a a a a + - = + - = + = - - + + = - + - = Solving by use of the quadratic formula: sin 0.47059 28.072 a a = = 28.1a = ◀ (b) Since xR R= using Eq. (1): 85cos28.072 170sin28.072 155.0 N R = + = 155.0 NR = ◀
41.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 41 PROBLEM 2.40 For the beam of Problem 2.36, determine (a) the required tension in cable BC if the resultant of the three forces exerted at Point B is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION 840 12 3 (780 N) (500 N) 1160 13 5 21 420 N 29 x x BC x BC R F T R T = S = - + - = - + (1) 800 5 4 (780 N) (500 N) 1160 13 5 20 700 N 29 y y BC y BC R F T R T = S = - - = - (2) (a) For R to be vertical, we must have 0xR = Set 0xR = in Eq. (1) 21 420 N 0 29 BCT- + = 580 NBCT = (b) Substituting for BCT in Eq. (2): 20 (580 N) 700 N 29 300 N y y R R = - =- | | 300 NyR R= = 300 NR =
42.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 42 PROBLEM 2.41 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. SOLUTION Using the x and y axes shown: sin10 (50 lb)cos35 (75 lb)cos60 sin10 78.458 lb x x AC AC R F T T = S = + + = + (1) (50 lb)sin35 (75 lb)sin60 cos10 93.631 lb cos10 y y AC y AC R F T R T = S = + - = - (2) (a) Set 0yR = in Eq. (2): 93.631 lb cos10 0 95.075 lb AC AC T T - = = 95.1 lbA CT = ◀ (b) Substituting for ACT in Eq. (1): (95.075 lb)sin10 78.458 lb 94.968 lb x x R R R = + = = 95.0 lbR = ◀ 75 lb 50 lb 25° 65° 35° A B C
43.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 43 PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of a if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. SOLUTION Select the x axis to be along a a¢. Then (60 lb) (80 lb)cos (120 lb)sinx xR F a a= S = + + (1) and (80 lb)sin (120 lb)cosy yR F a a= S = - (2) (a) Set 0yR = in Eq. (2). (80 lb)sin (120 lb)cos 0a a- = Dividing each term by cosa gives: (80 lb)tan 120 lb 120 lb tan 80 lb 56.310 a a a = = = 56.3a = ◀ (b) Substituting for a in Eq. (1) gives: 60 lb (80 lb)cos56.31 (120 lb)sin 56.31 204.22 lbxR = + + = 204 lbxR = ◀ 120 lb 80 lb 60 lb a a' α α 20°
44.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 44 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 400 lb sin60 sin 40 sin80 AC BCT T = = (a) 400 lb (sin60 ) sin80 ACT = 352 lbACT = ◀ (b) 400 lb (sin40 ) sin80BCT = 261 lbBCT = ◀ A B C 400 lb 50° 30°
45.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 45 PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 6 kN sin60 sin35 sin85 AC BCT T = = (a) 6 kN (sin60 ) sin85ACT = 5.22 kNA CT = ◀ (b) 6 kN (sin35 ) sin85BCT = 3.45 kNBCT = ◀ A B C 6 kN55° α
46.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 46 PROBLEM 2.45 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle ( )( )2 120 kg 9.81 m/s 1177 NW mg= = = Law of sines: 1.177 kN sin130 sin30 sin 20 AC BCT T = = (a) 1.177 kN (sin130 ) sin20ACT = 2.64 kNACT = ◀ (b) 1.177 kN (sin30 ) sin20BCT = 1.721 kNBCT = ◀
47.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 47 PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and a = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 500 N sin35 sin 75 sin 70° AC BCT T = = (a) 500 N sin35 sin 70ACT = 305 NA CT = ◀ (b) 500 N sin75 sin70BCT = 514 NBCT = ◀ 45° A B C P 25° α
48.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 48 PROBLEM 2.47 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle ( )( )2 50 kg 9.81 m/s 490 NW mg= = = Law of sines: 490 N sin110 sin30 sin 40° AC BCT T = = (a) 490 N sin110 sin40ACT = 716 NACT = ◀ (b) 490 N sin30 sin40BCT = 381 NBCT = ◀
49.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 49 PROBLEM 2.48 Knowing that 20 ,a = determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 1200 lb sin 110 sin 5 sin 65 AC BCT T = = (a) 1200 lb sin 110 sin 65ACT = 1244 lbA CT = ◀ (b) 1200 lb sin 5 sin 65BCT = 115.4 lbBCT = ◀ 5° A C B α 1200 lb
50.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 50 PROBLEM 2.49 Two cables are tied together at C and are loaded as shown. Knowing that P = 300 N, determine the tension in cables AC and BC. SOLUTION Free-Body Diagram 0 sin30 sin30 cos45 200N 0x CA CBF T T PS = - + - - = For 200NP = we have, 0.5 0.5 212.13 200 0CA CBT T- + + - = (1) 0yFS = cos30 cos30 sin 45 0CA CBT T P- - = 0.86603 0.86603 212.13 0CA CBT T+ - = (2) Solving equations (1) and (2) simultaneously gives, 134.6 NCAT = ◀ 110.4 NCBT = ◀ A B C 45° 30°30° 200 N P
51.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 51 PROBLEM 2.50 Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free-Body Diagram 0 sin30 sin30 cos45 200N 0x CA CBF T T PS = - + - - = For 0CAT = we have, 0.5 0.70711 200 0CBT P+ - = (1) 0yFS = cos30 cos30 sin45 0CA CBT T P- - = ; again setting 0CAT = yields, 0.86603 0.70711 0CBT P- = (2) Adding equations (1) and (2) gives, 1.36603 200CBT = hence 146.410NCBT = and 179.315NP = Substituting for 0CBT = into the equilibrium equations and solving simultaneously gives, 0.5 0.70711 200 0 0.86603 0.70711 0 CA CA T P T P - + - = - = And 546.40N,CAT = 669.20NP = Thus for both cables to remain taut, load P must be within the range of 179.315 N and 669.20 N. 179.3 N < < 669 NP ◀ A B C 45° 30°30° 200 N P
52.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 52 PROBLEM 2.51 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 600 lb and Q = 800 lb, determine the tension in rods A and B. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R P Q T T 0A B= + + + = Substituting components: R j i j i i j (600 lb) [(800 lb)cos30 ] [(800 lb)sin30 ] ( cos60 ) ( sin60 ) 0B A AT T T =- - + + + + = Summing forces in the y-direction: 600 lb (800 lb)sin30 sin60 0AT- + + = 230.94 lbAT = 231 lbAT = ◀ Summing forces in the x-direction: (800 lb)cos30 cos60 0B AT T- + + = Thus, (230.94 lb)cos60 (800 lb)cos30BT = - + 577.35 lb= 577 lbBT = ◀
53.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 53 PROBLEM 2.52 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the tensions in rods A and B are TA = 240 lb and TB = 500 lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R P Q T T 0A B= + + + = Substituting components: R j i j i j i cos 30 sin 30 [(240 lb)cos 60 ] [(240 lb)sin 60 ] (500 lb) P Q Q=- - + + + + Summing forces in the x-direction: cos 30 (240 lb)cos 60 500 lb 0Q- + + = 715.91 lbQ = Summing forces in the y-direction: sin 30 (240 lb)sin 60 0P Q- + + = sin 30 (240 lb)sin 60 (715.91 lb)sin 30 (240 lb)sin 60 565.80 lb P Q= + = + = 566 lb; 716 lbP Q= = ◀
54.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 54 PROBLEM 2.53 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 8AF = kN and 16BF = kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 0: 0 5 5x B C AF F F FS = - - = With 8 kN 16 kN A B F F = = 4 4 (16 kN) (8 kN) 5 5CF = - 6.40 kNCF = ◀ 3 3 0: 0 5 5y D B AF F F FS = - + - = With FA and FB as above: 3 3 (16 kN) (8 kN) 5 5DF = - 4.80 kNDF = ◀ FD FC FA FB B A D C 3 4
55.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 55 PROBLEM 2.54 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 5AF = kN and 6DF = kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 0: 0 5 5y D A BF F F FS = - - + = or 3 5B D AF F F= + With 5 kN, 8 kNA DF F= = 5 3 6 kN (5 kN) 3 5BF é ù ê ú= + ê úë û 15.00 kNBF = ◀ 4 4 0: 0 5 5x C B AF F F FS = - + - = 4 ( ) 5 4 (15 kN 5 kN) 5 C B AF F F= - = - 8.00 kNCF = ◀ FD FC FA FB B A D C 3 4
56.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 56 PROBLEM 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that a = 30° and b = 10° and that the combined weight of the boatswain’s chair and the sailor is 200 lb, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. SOLUTION Free-Body Diagram 0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T TS = - - = 0.137158CD A CBT T= (1) 0: sin 10 sin 30 sin 30 200 0y ACB ACB CDF T T TS = + + - = 0.67365 0.5 200A CB CDT T+ = (2) (a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 200ACB ACBT T+ = 269.46 lbA CBT = 269 lbA CBT = ◀ (b) From (1): 0.137158(269.46 lb)CDT = 37.0 lbCDT = ◀ A B C α β D
57.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 57 PROBLEM 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and b = 15° and that the tension in cable CD is 20 lb, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) the tension in the support cable ACB. SOLUTION Free-Body Diagram 0: cos 15 cos 25 (20 lb)cos 25 0x ACB ACBF T TS = - - = 304.04 lbA CBT = 0: (304.04 lb)sin 15 (304.04 lb)sin 25yFS = + (20 lb)sin 25 0 215.64 lb W W + - = = (a) 216 lbW = ◀ (b) 304 lbA CBT = ◀ A B C α β D
58.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 58 PROBLEM 2.57 For the cables of prob. 2.44, find the value of α for which the tension is as small as possible (a) in cable bc, (b) in both cables simultaneously. In each case determine the tension in each cable. SOLUTION Free-Body Diagram Force Triangle (a) For a minimum tension in cable BC, set angle between cables to 90 degrees. By inspection, 35.0a = ◀ (6 kN)cos35A CT = 4.91 kNA CT = ◀ (6 kN)sin 35BCT = 3.44 kNBCT = ◀ (b) For equal tension in both cables, the force triangle will be an isosceles. Therefore, by inspection, 55.0a = ◀ 6 kN (1 / 2) cos35AC BCT T= = 3.66 kNAC BCT T= = ◀ A B C 6 kN55° α
59.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 59 PROBLEM 2.58 For the cables of Problem 2.46, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of a. SOLUTION Free-Body Diagram Force Triangle (a) Law of cosines 2 2 2 (600) (750) 2(600)(750)cos(25 45 )P = + - + 784.02 NP = 784 NP = ◀ (b) Law of sines sin sin(25 45 ) 600 N 784.02 N b + = 46.0b = 46.0 25a = + 71.0a = ◀ 45° A B C P 25° α
60.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 60 PROBLEM 2.59 For the situation described in Figure P2.48, determine (a) the value of a for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. SOLUTION Free-Body Diagram Force Triangle To be smallest, BCT must be perpendicular to the direction of .A CT (a) Thus, 5.00a = 5.00a = ◀ (b) (1200 lb)sin 5BCT = 104.6 lbBCT = ◀ 5° A C B α 1200 lb
61.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 61 PROBLEM 2.60 Two cables tied together at C are loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free-Body Diagram 3 0: sin30 0 5x BC ACF P T TS = - - = ( )5 sin30 3 BC ACP T T= + (1) 4 0: cos30 120 0 5y ACF P TS = + - = ( )5 120 cos30 4 ACP T= - (2) Requirement: 0:ACT = From Eq. (2): 150.0 lbP = Requirement: 0:BCT = From Eq. (1): ( )5 sin 30 3 ACP T= From Eq. (2): ( )5 120 cos30 4 ACP T= - Solving simultaneously yields: 78.294 lbACT = and 65.2 lbP = 65.2 lb< 150.0 lbP < ◀
62.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 62 PROBLEM 2.61 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram tan 0.6 m h a = (1) Isosceles Force Triangle Law of sines: 1 2 1 2 (2.8 kN) sin 5 kN (2.8 kN) sin 5 kN 16.2602 AC AC T T a a a = = = = From Eq. (1): tan16.2602 0.175000 m 0.6 m h h = = Half-length of chain 2 2 (0.6 m) (0.175 m) 0.625 m AC= = + = Total length: 2 0.625 m= ´ 1.250 m ◀ A C 0.7 m B 1.2 m
63.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 63 PROBLEM 2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium. SOLUTION Free-Body Diagram 800 NA C BCT T= = ( )2 2 AC BC h d= = + 2 2 0: 2(800 N) 0y h F P h d S = - = + 2 800 1 2 P d h æ ö÷ç= + ÷ç ÷ç ÷è ø Data: 200 N, 600 mmP d= = and solving for h 2 200 N 600 mm 800 N 1 2 h æ ö÷ç= + ÷ç ÷ç ÷è ø 75.6 mmh = ◀ P W d h d
64.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 64 PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) 4.5 in.,x = (b) 15 in.x = SOLUTION (a) Free Body: Collar A Force Triangle 50 lb 4.5 20.5 P = 10.98 lbP = ◀ (b) Free Body: Collar A Force Triangle 50 lb 15 25 P = 30.0 lbP = ◀ 50 lb x C B A P 20 in.
65.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 65 PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb. SOLUTION Free Body: Collar A Force Triangle 2 2 2 (50) (48) 196 14.00 lb N N = - = = Similar Triangles 48 lb 20 in. 14 lb x = 68.6 in.x = ◀ 50 lb x C B A P 20 in.
66.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 66 PROBLEM 2.65 A cable loop of length 1.5 m is placed around a crate. Knowing that the mass of the crate is 300 kg, determine the tension in the cable for each of the arrangements shown. SOLUTION Free-Body Diagram Isosceles Force Triangle ( )( ) ( ) 2 1 300 kg 9.81 m/s 2943.0 N 1 1500 mm 400 mm 300 mm 300 mm 2 250 mm 200 mm cos 36.87 250 mm W EB EB a - = = = - - - = æ ö÷ç= ÷= ç ÷ç ÷è ø ( ) ( ) 1 sin 2943.0 N 2 1 sin36.87 2943.0 N 2 2452.5 N AE BE AE AE AE T T T T T a = = = =
67.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 67 (a) 2450 NAET = ◀ Problem 2.65 (Continued) Isosceles Force Triangle Free-Body Diagram ( ) 1 1 1500 mm 300 mm 400 mm 400 mm 2 250 mm 150 mm cos 41.41 200 mm EB EB a - = - - - = æ ö÷ç= ÷= ç ÷ç ÷è ø ( ) ( ) 1 sin 2943.0 N 2 1 sin41.41 2943.0 N 2 2224.7 N AE BE AE AE AE T T T T T a = = = = (b) 2220 NAET = ◀
68.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 68 PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.) SOLUTION Free-Body Diagram: Pulley A 5 0: 2 cos 0 281 cos 0.59655 53.377 xF P P a a a æ ö÷ç ÷S = - + =ç ÷ç ÷çè ø = = For 53.377 :a = + 16 0: 2 sin53.377 1962 N 0 281 yF P P æ ö÷ç ÷S = + - =ç ÷ç ÷çè ø 724 N=P 53.4 ◀ For 53.377 :a = - 16 0: 2 sin( 53.377 ) 1962 N 0 281 yF P P æ ö÷ç ÷S = + - - =ç ÷ç ÷çè ø 1773=P 53.4 ◀ 2.4 m P A α 200 kg 0.75 m B
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 69 PROBLEM 2.67 A 600-lb crate is supported by several rope- and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley (a) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T S = - = = 300 lbT = ◀ (b) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T S = - = = 300 lbT = ◀ (c) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T S = - = = 200 lbT = ◀ (d) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T S = - = = 200 lbT = ◀ (e) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T S = - = = 150.0 lbT = ◀ T T T T T (a) (b) (c) (d) (e)
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 70 PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley and Crate (b) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T S = - = = 200 lbT = ◀ (d) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T S = - = = 150.0 lbT = ◀ T T T T T (a) (b) (c) (d) (e)
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 71 PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that 750 N,P = determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C (a) 0: (cos25 cos55 ) (750 N)cos55° 0x ACBF TS = - - = Hence: 1292.88 NA CBT = 1293 NA CBT = ◀ (b) 0: (sin25 sin55 ) (750 N)sin55 0 (1292.88 N)(sin25 sin55 ) (750 N)sin55 0 y ACBF T Q Q S = + + - = + + - = or 2219.8 NQ = 2220 NQ = ◀ A D B C P 25° 55° Q
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 72 PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P. SOLUTION Free-Body Diagram: Pulley C 0: (cos25 cos55 ) cos55 0x ACBF T PS = - - = or 0.58010 ACBP T= (1) 0: (sin25 sin55 ) sin55 1800 N 0y ACBF T PS = + + - = or 1.24177 0.81915 1800 NACBT P+ = (2) (a) Substitute Equation (1) into Equation (2): 1.24177 0.81915(0.58010 ) 1800 NA CB A CBT T+ = Hence: 1048.37 NA CBT = 1048 NA CBT = ◀ (b) Using (1), 0.58010(1048.37 N) 608.16 NP = = 608 NP = ◀ A D B C P 25° 55° Q
73.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 73 PROBLEM 2.71 Determine (a) the x, y, and z components of the 500-N force, (b) the angles qx, qy, and qz that the force forms with the coordinate axes. SOLUTION (a) (500 N)sin 40 cos30xF = 278.34 NxF = 278 NxF = ◀ (500 N)cos40° 383.02 N y y F F = = 383 NyF = ◀ (500 N)sin40 sin30 160.697 N z z F F = = 160.7 NzF = ◀ (b) 278.34 N cos 500 N x x F F q = = 56.2xq = ◀ 383.02 N cos 500 N y y F F q = = 40.0yq = ◀ 160.697 N cos 500 N z z F F q = = 71.3zq = ◀
74.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 74 PROBLEM 2.72 Determine (a) the x, y, and z components of the 800-N force, (b) the angles qx, qy, and qz that the force forms with the coordinate axes. SOLUTION (a) (800 N)cos 70 sin25xF =- 115.635 NxF = - 115.6 NxF = - ◀ (800 N)sin 70° 751.75 N y y F F = = 752 NyF = ◀ (800 N)cos70 cos25 247.98 N z z F F = = 248 NzF = ◀ (b) 115.635 N cos 800 N x x F F q - = = 98.3xq = ◀ 751.75 N cos 800 N y y F F q = = 20.0yq = ◀ 247.98 N cos 800 N z z F F q = = 71.9zq = ◀ Note: From the given data, we could have computed directly 90 35 55 , which checks with the answer obtained.yq = - =
75.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 75 PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force 400 NF = (400 N)cos40 306.42 N HF = = (a) sin35 (306.42 N)sin35 x HF F=- =- 175.755 N=- 175.8 NxF = - ◀ sin40 (400 N)sin 40 257.12 N yF F=- =- =- 257 NyF = - ◀ cos35 (306.42 N)cos35 251.00 N z HF F= + = + = + 251 NzF = + ◀ (b) 175.755 N cos 400 N x x F F q - = = 116.1xq = ◀ 257.12 N cos 400 N y y F F q - = = 130.0yq = ◀ 251.00 N cos 400 N z z F F q = = 51.1zq = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 76 PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force 400 NF = (400 N)cos25 362.52 N HF = = (a) cos15 (362.52 N)cos15 x HF F=+ =+ 350.17 N= + 350 NxF = + ◀ sin25 (400 N)sin25 169.047 N yF F= - = - = - 169.0 NyF =- ◀ sin15 (362.52 N)sin15 93.827 N z HF F= + = + = + 93.8 NzF = + ◀ (b) 350.17 N cos 400 N x x F F q + = = 28.9xq = ◀ 169.047 N cos 400 N y y F F q - = = 115.0yq = ◀ 93.827 N cos 400 N z z F F q + = = 76.4zq = ◀
77.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 77 PROBLEM 2.75 The angle between the guy wire AB and the mast is 20°. Knowing that the tension in AB is 300 lb, determine (a) the x, y, and z components of the force exerted on the boat at B, (b) the angles qx, qy, and qz defining the direction of the force exerted at B. SOLUTION sin 20 (300 lb)sin 20 102.606 lb h AB h F F F = = = cos40 cos20 sin 40 ( 102.606 lb)cos40 (300 lb)cos20 ( 102.606 lb)sin 40 78.601 lb 281.91 lb x h y AB z h x y z x y F F F F F F F F F F F = - = = - = - = = - = - = 65.954 lbzF = - (a) 78.6 lbxF = - ◀ 282 lbyF = ◀ 66.0 lbzF = - ◀ (b) 78.601 lb cos 300 lb x x F F q - = = 105.2xq = ◀ 281.91 lb cos 300 lb y y F F q = = 20.0yq = ◀ 65.954 lb cos 300 lb z z F F q - = = 102.7zq = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 78 PROBLEM 2.76 The angle between the guy wire AC and the mast is 20°. Knowing that the tension in AC is 300 lb, determine (a) the x, y, and z components of the force exerted on the boat at C, (b) the angles qx, qy, and qz defining the direction of the force exerted at C. SOLUTION sin 20 (300 lb)sin 20 102.606 lb h AC h F F F = = = cos40 cos20 sin 40 (102.606 lb)cos40 (300 lb)cos20 ( 102.606 lb)sin 40 78.601 lb 281.91 lb x h y AC z h x y z x y F F F F F F F F F F F = = = - = = = - = = 65.954 lbzF = - (a) 78.6 lbxF = ◀ 282 lbyF = ◀ 66.0 lbzF = - ◀ (b) 78.601 lb cos 300 lb x x F F q = = 74.8xq = ◀ 281.91 lb cos 300 lb y y F F q = = 20.0yq = ◀ 65.954 lb cos 300 lb z z F F q - = = 102.7zq = ◀
79.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 79 PROBLEM 2.77 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles ,xq ,yq and zq defining the direction of that force. SOLUTION From triangle AOB: 56 ft cos 65 ft 0.86154 30.51 y y q q = = = (a) sin cos20 (3900 lb)sin30.51 cos20 x yF F q=- =- 1861 lbxF = - ◀ cos (3900 lb)(0.86154)y yF F q= + = 3360 lbyF =+ ◀ (3900 lb)sin 30.51° sin 20°zF = + 677 lbzF = + ◀ (b) 1861 lb cos 0.4771 3900 lb x x F F q = =- =- 118.5xq = ◀ From above: 30.51yq = 30.5yq = ◀ 677 lb cos 0.1736 3900 lb z z F F q = = + = + 80.0zq = ◀ x D A y 56 ft α O 50° 20° B Cz
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 80 PROBLEM 2.78 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles qx, qy, and qz defining the direction of that force. SOLUTION In triangle AOB: 70 ft 56 ft 5250 lb AC OA F = = = 56 ft cos 70 ft 36.870 sin (5250 lb)sin36.870 3150.0 lb y y H yF F q q q = = = = = (a) sin50 (3150.0 lb)sin50 2413.0 lbx HF F=- =- =- 2410 lbxF = - ◀ cos (5250 lb)cos36.870 4200.0 lby yF F q= + = + = + 4200 lbyF =+ ◀ cos50 3150cos50 2024.8 lbz HF F=- =- =- 2025 lbzF = - ◀ (b) 2413.0 lb cos 5250 lb x x F F q - = = 117.4xq = ◀ From above: 36.870yq = 36.9yq = ◀ 2024.8 lb 5250 lb z z F F q - = = 112.7zq = ◀ x D A y 56 ft α O 50° 20° B Cz
81.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 81 PROBLEM 2.79 Determine the magnitude and direction of the force F = (260 N)i – (320 N)j + (800 N)k. SOLUTION 2 2 2 2 2 2 (260 N) ( 320 N) (800 N) x y zF F F F F = + + = + - + 900 NF = ◀ 260 N cos 900 N x x F F q = = 73.2xq = ◀ 320 N cos 900 N y y F F q - = = 110.8yq = ◀ 800 N cos 900 N z y F F q = = 27.3zq = ◀
82.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 82 PROBLEM 2.80 Determine the magnitude and direction of the force F = (700 N)i – (820 N)j + (960 N)k. SOLUTION 2 2 2 2 2 2 (700 N) (820 N) (960 N) 1443.61 N x y zF F F F F F = + + = - + = 1444 NF = ◀ 700 N cos 1443.61 N x x F F q = = 61.0xq = ◀ 820 N cos 1443.61 N y y F F q - = = 124.6yq = ◀ 960 N cos 1443.61 N z y F F q = = 48.3zq = ◀
83.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 83 PROBLEM 2.81 A force F of magnitude 250 lb acts at the origin of a coordinate system. Knowing that qx = 65, qy = 40°, and Fz > 0, determine (a) the components of the force, (b) the angle qz . SOLUTION 2 2 2 2 2 2 cos cos cos 1 cos (65 ) cos (40 ) cos 1 cos 0.48432 x y z z z q q q q q + + = + + = = (b) Since 0,zF > we choose cos 0.48432zq = 61.0zq = ◀ (a) cos (250 lb)cos65x xF F q= = 105.7 lbxF = ◀ cos (250 lb)cos40y yF F q= = 191.5 lbyF = ◀ cos (250 lb)cos 61z zF F q= = 121.2 lbzF = ◀
84.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 84 PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles qx = 70.9 and qy = 144.9. Knowing that the z component of the force is s-52.0 lb, determine (a) the angle qz, (b) the other components and the magnitude of the force. SOLUTION 2 2 2 2 2 2 cos cos cos 1 cos 70.9 cos 144.9 cos 1 cos 0.47282 x y z z z q q q q q + + = + + = = (a) Since 0,zF < we choose cos 0.47282zq = - 118.2zq = ◀ (b) cos 52.0 ( 0.47282) z zF F lb F q= - = - 110.0 lbF = 110.0 lbF = ◀ cos (110.0 lb)cos70.9x xF F q= = 36.0 lbxF = ◀ cos (110.0 lb)cos144.9y yF F q= = 90.0 lbyF =- ◀
85.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 85 PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, qz= 151.2, and Fy< 0, determine (a) the components Fy and Fz, (b) the angles qx and qy. SOLUTION (a) cos (210 N)cos151.2z zF F q= = 184.024 N=- 184.0 NzF =- ◀ Then: 2 2 2 2 x y zF F F F= + + So: 2 2 2 2 (210 N) (80 N) ( ) (184.024 N)yF= + + Hence: 2 2 2 (210 N) (80 N) (184.024 N)yF = - - - 61.929 N=- 62.0 lbyF =- ◀ (b) 80 N cos 0.38095 210 N x x F F q = = = 67.6xq = ◀ 61.929 N cos 0.29490 210 N y y F F q = = =- 107.2yq = ◀
86.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 86 PROBLEM 2.84 A force acts at the origin of a coordinate system in a direction defined by the angles qy = 120 and qz = 75. Knowing that the x component of the force is +40 N, determine (a) the angle qx, (b) the magnitude of the force. SOLUTION 2 2 2 2 2 2 cos cos cos 1 cos cos 120 cos 75 1 cos 0.82644 x y z x x q q q q q + + = + + = = (b) Since 0,xF > we choose cos 0.82644xq = 34.3xq = ◀ (a) cos 40 N cos34.3 x xF F F q= = 48.4 NF = ◀
87.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 87 PROBLEM 2.85 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BG is 540 N, determine the components of the force exerted by cable BG on the frame at B. SOLUTION i j k T λ i j k i j k 2 2 2 (0.8 m) (1.48 m) (0.64 m) ( 0.8 m) (1.48 m ) ( 0.64 m) 1.8 m 540 N [( 0.8 m) (1.48 m) ( 0.64 m) ] 1.8 m ( 240 N) (444 N) (192.0 N) BG BG BG BG BG BG BG T BG T BG T =- + - = - + + - = = = = - + + - = - + - 240 N, 444 N, 192.0 Nx y zF F F= - = + = + ◀
88.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 88 PROBLEM 2.86 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BH is 750 N, determine the components of the force exerted by cable BH on the frame at B. SOLUTION i j k T λ i j k i j k 2 2 2 (0.6 m) (1.2 m) (1.2 m) (0.6 m) (1.2 m) (1.2 m) 1.8 m 750 N [ 2 2 ] (m) 3 m (250 N) (500 N) (500 N) BH BH BH BH BH BH BH T BH T BH T = + - = + - = = = = + - = + - 250 N, 500 N, 500 Nx y zF F F= + = + =- ◀
89.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 89 PROBLEM 2.87 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: i j k λ i j k T λ ( 46.765 ft) (45 ft) (36 ft) 74.216 ft 46.765 45 36 74.216 AB AB AB AB AB AB T - + + = = - + + = = ( ) 1.260 kipsAB xT =- ◀ ( ) 1.213 kipsAB yT = + ◀ ( ) 0.970 kipsAB zT = + ◀ 36 ft 28.8 ft 18 ft 45 ft 54 ft 30° A B C
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 90 PROBLEM 2.88 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: i j k λ i j k T λ ( 46.765 ft) (55.8 ft) ( 45 ft) 85.590 ft 46.765 55.8 45 (1.5 kips) 85.590 AC AC AC AC AC AC T - + + - = = - + - = = ( ) 0.820 kipsAC xT =- ◀ ( ) 0.978 kipsAC yT = + ◀ ( ) 0.789 kipsAC zT =- ◀ 36 ft 28.8 ft 18 ft 45 ft 54 ft 30° A B C
91.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 91 PROBLEM 2.89 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B. SOLUTION We have: i j k(320 mm) (480 mm) (360 mm) 680 mmBA BA= + + - = Thus: λ i j k 8 12 9 F 17 17 17B BA BA BA BA BA T T T BA æ ö÷ç= = = + - ÷ç ÷ç ÷è ø 8 12 9 0 17 17 17BA BA BAT T T æ ö æ ö æ ö÷ ÷ ÷ç ç ç÷ + ÷ - ÷ =ç ç ç÷ ÷ ÷ç ç ç÷ ÷ ÷è ø è ø è ø i j k Setting 408 NBAT = yields, 192.0 N, 288 N, 216 Nx y zF F F= + = + =- ◀ x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm
92.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 92 PROBLEM 2.90 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D. SOLUTION We have: i j k(250 mm) (480 mm) (360 mm) 650 mmDA DA=- + + = Thus: λ i j k 5 48 36 F 13 65 65D DA DA DA DA DA T T T DA æ ö÷ç= = = - + + ÷ç ÷ç ÷è ø 5 48 36 0 13 65 65DA DA DAT T T æ ö æ ö æ ö÷ ÷ ÷ç ç ç- ÷ + ÷ + ÷ =ç ç ç÷ ÷ ÷ç ç ç÷ ÷ ÷è ø è ø è ø i j k Setting 429 NDAT = yields, 165.0 N, 317 N, 238 Nx y zF F F=- = + = + ◀ x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm
93.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 93 PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION (300 N)[ cos30 sin15 sin30 cos30 cos15 ] (67.243 N) (150 N) (250.95 N) (400 N)[cos50 cos20 sin50 cos50 sin20 ] (400 N)[0.60402 0.76604 0.21985] (241.61 N) (306.42 N) (87.939 N) (174. = - + + =- + + = + - = + - = + - = + = P i j k i j k Q i j k i j i j k R P Q 2 2 2 367 N) (456.42 N) (163.011 N) (174.367 N) (456.42 N) (163.011 N) 515.07 N R + + = + + = i j k 515 NR = ◀ 174.367 N cos 0.33853 515.07 N x x R R q = = = 70.2xq = ◀ 456.42 N cos 0.88613 515.07 N y y R R q = = = 27.6yq = ◀ 163.011 N cos 0.31648 515.07 N z z R R q = = = 71.5zq = ◀ z x y 30° 20° 15° 50°P Q
94.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 94 PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N. SOLUTION (400 N)[ cos30 sin15 sin30 cos30 cos15 ] (89.678 N) (200 N) (334.61 N) (300 N)[cos50 cos20 sin50 cos50 sin20 ] (181.21 N) (229.81 N) (65.954 N) (91.532 N) (429.81 N) (268.66 N) (91.5R = - + + =- + + = + - = + - = + = + + = P i j k i j k Q i j k i j k R P Q i j k 2 2 2 32 N) (429.81 N) (268.66 N) 515.07 N + + = 515 NR = ◀ 91.532 N cos 0.177708 515.07 N x x R R q = = = 79.8xq = ◀ 429.81 N cos 0.83447 515.07 N y y R R q = = = 33.4yq = ◀ 268.66 N cos 0.52160 515.07 N z z R R q = = = 58.6zq = ◀ z x y 30° 20° 15° 50°P Q
95.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 95 PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION i j k i j k i j k T λ 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (425 lb) 85 in.AB AB AB AB AB AB AC AC AB T T AB = - + = + + = = - + = + + = é - + = = = ë T i j k i j k T λ T i j k R T T i j k (200 lb) (225 lb) (300 lb) (100 in.) (45 in.) (60 in.) (510 lb) 125 in. (408 lb) (183.6 lb) (244.8 lb) (608) (408.6 lb) (544.8 lb) AB AC AC AC AC AC AB AC AC T T AC ù ê ú ê ú ê úû = - + é ù- +ê ú= = = ê ú ê úë û = - + = + = - + Then: 912.92 lbR = 913 lbR = ◀ and 608 lb cos 0.66599 912.92 lb xq = = 48.2xq = ◀ 408.6 lb cos 0.44757 912.92 lb yq = =- 116.6yq = ◀ 544.8 lb cos 0.59677 912.92 lb zq = = 53.4zq = ◀ y xz A B C D O 40 in. 60 in. 60 in. 45 in.
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 96 PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION i j k i j k i j k T λ 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (510 lb) 85 in.AB AB AB AB AB AB AC AC AB T T AB = - + = + + = = - + = + + = é - + = = = ë T i j k i j k T λ T i j k R T T i j k (240 lb) (270 lb) (360 lb) (100 in.) (45 in.) (60 in.) (425 lb) 125 in. (340 lb) (153 lb) (204 lb) (580 lb) (423 lb) (564 lb) AB AC AC AC AC AC AB AC AC T T AC ù ê ú ê ú ê úû = - + é ù- +ê ú= = = ê ú ê úë û = - + = + = - + Then: 912.92 lbR = 913 lbR = ◀ and 580 lb cos 0.63532 912.92 lb xq = = 50.6xq = ◀ 423 lb cos 0.46335 912.92 lb yq - = =- 117.6yq = ◀ 564 lb cos 0.61780 912.92 lb zq = = 51.8zq = ◀ y xz A B C D O 40 in. 60 in. 60 in. 45 in.
97.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 97 PROBLEM 2.95 For the frame of Prob. 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tension is 540 N in cable BG and 750 N in cable BH. PROBLEM 2.85 Two cables BG and BH are attached to frame ACD as shown. Knowing that the tension in cable BG is 540 N, determine the components of the force exerted by cable BG on the frame at B. SOLUTIO Then: and ON R = cos xθ = cos yθ = cos 1zθ = T (0 BG B B BG BG BH BH T T = − = = = = = T T R T ( (2 BH B B BH B T T = − = = = = ( 2 2 10 944+ 10 N 1170.51 N 944 N 1170.51 N 692 N 1170.51 N − i i λ 2 2 (0.8 m) (1.4 ( 0.8 m) ( 0.6 m) (1.2 (0.6 m) (1. 540 1.8 m BG BG BG BG BG − + − + + + = i λ i T 240 N) (4 750 3 m 250 N) (500 (1 BH BH BH BG BH BH BH − + = + + = ( ) )2 692+ − = j j i 2 2 48 m) (0.64 (1.48 m ) ( m) (1.2 m) .2 m) (1.2 m N [( 0.8 m) m − + − − − − j i j k j i 444 N) (192 N [ 2 2 m 0 N) (500 N 10 N) (944 − + − − + 1170.51 N= k k j 2 2 4 m) 0.64 m) 1 ) m) 1.8 m (1.48 m) − = = + + k k k j 2.0 N) ] (m) N) N) (692 N− 1.8 m ( 0+ − k k .64 m) ] ) 1171NR = 89.5xθ = ° 36.2yθ = ° 126.2zθ = ° ] ◀ ◀ ◀ ◀
98.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 98 PROBLEM 2.96 For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resultant of the forces exerted by the three cables at A must be vertical. SOLUTION We have: ( ) i j k i j k i j k (320 mm) (480 mm) (360 mm) 680 mm (450 mm) (480 mm) (360 mm) 750 mm (250 mm) (480 mm) 360 mm 650 mm AB AB AC AC AD AD =- - + = = - + = = - - = Thus: ( ) ( ) ( ) T λ i j k T λ i j k T λ i j k 320 480 360 680 54 450 480 360 750 250 480 360 650 AB AB AB AB AB AC AC AC AC AD AD AD AD AD TAB T T AB AC T T AC TAD T T AD = = = - - + = = = - + = = = - - Substituting into the Eq. = SR F and factoring , , :i j k 320 250 32.40 680 650 480 480 34.560 680 650 360 360 25.920 680 650 AB AD AB AD AB AD T T T T T T æ ö÷ç= - + + ÷ç ÷ç ÷è ø æ ö÷ç+ - - - ÷ç ÷ç ÷è ø æ ö÷ç+ + - ÷ç ÷ç ÷è ø R i j k x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm
99.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 99 PROBLEM 2.96 (Continued) Since R is vertical, the coefficients of i and k are zero: :i 320 250 32.40 0 680 650AB ADT T- + + = (1) :k 360 360 25.920 0 680 650AB ADT T+ - = (2) Multiply (1) by 3.6 and (2) by 2.5 then add: 252 181.440 0 680 ABT- + = 489.60 NA BT = 490 NA BT = ◀ Substitute into (2) and solve for :ADT 360 360 (489.60 N) 25.920 0 680 650 ADT+ - = 514.80 NA DT = 515 NA DT = ◀
100.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 100 PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION Cable AB: 183 lbABT = i j k T λ T i j k ( 48 in.) (29 in.) (24 in.) (183 lb) 61 in. (144 lb) (87 lb) (72 lb) AB AB AB AB AB AB T T AB - + + = = = =- + + Cable AC: i j k T λ T i j k ( 48 in.) (25 in.) ( 36 in.) 65 in. 48 25 36 65 65 65 AC AC AC AC AC AC AC AC AC AC T T T AC T T T - + + - = = = =- + - Load P: P=P j For resultant to be directed along OA, i.e., x-axis 36 0: (72 lb) 0 65 z z ACR F T¢ = S = - = 130.0 lbACT = ◀ z 24 in. 29 in. 25 in. 48 in. A C B O y 36 in. x P
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 101 PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write 25 0: (87 lb) 0 65y y ACR F T P= S = + - = 130.0 lbACT = from Problem 2.97. Then 25 (87 lb) (130.0 lb) 0 65 P+ - = 137.0 lbP = ◀ z 24 in. 29 in. 25 in. 48 in. A C B O y 36 in. x P
102.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 102 PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION Free-Body Diagram at A: The forces applied at A are: , , , andAB AC ADT T T W where .W=W j To express the other forces in terms of the unit vectors i, j, k, we write i j j k i j k (450 mm) (600 mm) 750 mm (600 mm) (320 mm) 680 mm (500 mm) (600 mm) (360 mm) 860 mm AB AB AC AC AD AD =- + = = + - = = + + + = and i j T λ ( 450 mm) (600 mm) 750 mmAB AB AB AB AB AB T T T AB - + = = = 45 60 75 75 ABT æ ö÷ç= - + ÷ç ÷ç ÷è ø i j i j T λ j k i j k T λ i j k (600 mm) (320 mm) 680 mm 60 32 68 68 (500 mm) (600 mm) (360 mm) 860 mm 50 60 36 86 86 86 AC AC AC AC AC AC AD AD AD AD AD AD AC T T T AC T AD T T T AD T - = = = æ ö÷ç= - ÷ç ÷ç ÷è ø + + = = = æ ö÷ç= + + ÷ç ÷ç ÷è ø y x z 450 mm 500 mm 360 mm 320 mm 600 mm A C D B
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 103 PROBLEM 2.99 (Continued) Equilibrium condition: 0: 0AB AC ADFS = + + + =T T T W Substituting the expressions obtained for , , and ;AB AC ADT T T factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 45 50 0 75 86AB ADT T- + = (1) From j: 60 60 60 0 75 68 86AB AC ADT T T W+ + - = (2) From k: 32 36 0 68 86AC ADT T- + = (3) Setting 6 kNABT = in (1) and (2), and solving the resulting set of equations gives 6.1920 kN 5.5080 kN AC AC T T = = 13.98 kNW = ◀
104.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 104 PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: 45 50 0 75 86AB ADT T- + = (1) 60 60 60 0 75 68 86AB AC ADT T T W+ + - = (2) 32 36 0 68 86AC ADT T- + = (3) Setting 4.3 kNADT = into the above equations gives 4.1667 kN 3.8250 kN AB AC T T = = 9.71 kNW = ◀ y x z 450 mm 500 mm 360 mm 320 mm 600 mm A C D B
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 105 PROBLEM 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N. SOLUTION FREE-BODY DIAGRAM AT A The forces applied at A are: , , , andAB AC ADT T T P where .P=P j To express the other forces in terms of the unit vectors i, j, k, we write i j i j k j k (4.20 m) (5.60 m) 7.00 m (2.40 m) (5.60 m) (4.20 m) 7.40 m (5.60 m) (3.30 m) 6.50 m AB AB AC AC AD AD = - - = = - + = = - - = and T λ i j T λ i j k T λ j k ( 0.6 0.8 ) (0.32432 0.75676 0.56757 ) ( 0.86154 0.50769 ) AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD = = = - - = = = - + = = = - - A B C D O 4.20 m 4.20 m 3.30 m 5.60 m 2.40 m x y z
106.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 106 PROBLEM 2.101 (Continued) Equilibrium condition: 0: 0AB AC ADF PS = + + + =T T T j Substituting the expressions obtained for , , andAB AC ADT T T and factoring i, j, and k: ( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 ) (0.56757 0.50769 ) 0 AB AC AB AC AD AC AD T T T T T P T T - + + - - - + + - = i j k Equating to zero the coefficients of i, j, k: 0.6 0.32432 0AB ACT T- + = (1) 0.8 0.75676 0.86154 0AB AC ADT T T P- - - + = (2) 0.56757 0.50769 0AC ADT T- = (3) Setting 481 NADT = in (2) and (3), and solving the resulting set of equations gives 430.26 N 232.57 N AC AD T T = = 926 N=P ◀
107.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 107 PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). 0.6 0.32432 0AB ACT T- + = (1) 0.8 0.75676 0.86154 0AB AC ADT T T P- - - + = (2) 0.56757 0.50769 0AC ADT T- = (3) From Eq. (1): 0.54053AB ACT T= From Eq. (3): 1.11795AD ACT T= Substituting for ABT and ADT in terms of ACT into Eq. (2) gives 0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0AC AC ACT T T P- - - + = 2.1523 ; 800 N 800 N 2.1523 371.69 N AC AC T P P T = = = = Substituting into expressions for ABT and ADT gives 0.54053(371.69 N) 1.11795(371.69 N) AB AD T T = = 201 N, 372 N, 416 NAB AC ADT T T= = = ◀ A B C D O 4.20 m 4.20 m 3.30 m 5.60 m 2.40 m x y z
108.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 108 PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight W of the crate, knowing that the tension in cable AD is 924 lb. SOLUTION The forces applied at A are: T T T W, , andAB AC AD where P j.P= To express the other forces in terms of the unit vectors i, j, k, we write (28 in.) (45 in.) 53 in. (45 in.) (24 in.) 51 in. (26 in.) (45 in.) (18 in.) 55 in. AB AB AC AC AD AD = + = = = = - + = i j j - k i j + k
109.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 109 and T λ i j T λ j k T λ i j k (0.5283 0.84906 ) (0.88235 0.47059 ) ( 0.47273 0.81818 0.32727 ) AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T AB T AC T T AC T AD T T AD T = = = + = = = - = = = - + + Equilibrium Condition with W jW= - T T T j0: 0AB AC ADF WS = + + - = PROBLEM 2.103 (Continued) Substituting the expressions obtained for T T T, , andAB AC AD and factoring i, j, and k: i j k (0.5283 0.47273 ) (0.84906 0.88235 0.81818 ) ( 0.47059 0.32727 ) 0 AB AD AB AC AD AC AD T T T T T W T T - + + + - + - + = Equating to zero the coefficients of i, j, k: 0.5283 0.47273 0 (1) 0.84906 0.88235 0.81818 0 (2) 0.47059 0.32727 0 (3) AB AD AB AC AD AC AD T T T T T W T T - = + + - = - + = Substituting 924 lbADT = in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: 826.81 lb 642.59 lb AB AC T T = = 2030 lbW = ◀
110.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 110 PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight W of the crate, knowing that the tension in cable AB is 1378 lb. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.5283 0.47273 0 (1) 0.84906 0.88235 0.81818 0 (2) 0.47059 0.32727 0 (3) AB AD AB AC AD AC AD T T T T T W T T - = + + - = - + = Substituting 1378 lbABT = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: 1539.99 lb 1070.98 lb AD AC T T = = 3380 lbW = ◀
111.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 111 PROBLEM 2.105 A 12-lb circular plate of 7-in. radius is supported as shown by three wires, each of 25-in. length. Determine the tension in each wire, knowing that α = 30°. SOLUTION ( )2 2 Let be angle between the vertical and any wire. 24 = 25 7 24 in. thus cos 25 By symmetry AB AC OA T T q q- = = = 0:xFS = 2( sin )(cos ) (sin ) 0AB ADT Tq a q- + = For 30a = : ( )2 cos30 1.73205AD AB ABT T T= = 0: 12 lb 2 (cos ) (cos ) 0y AB ADF T Tq qS = - - = or ( )12 lb 2 cosAB ADT T q= + ( ) 24 12 lb 2 1.73205 25AB ABT T æ ö÷ç= + ÷ç ÷ç ÷è ø 3.35 lbAB ACT T= = ◀ 5.80 lbADT = ◀
112.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 112 PROBLEM 2.106 Solve Prob. 2.105, knowing that α = 45°. PROBLEM 2.105 A 12-lb circular plate of 7-in. radius is supported as shown by three wires, each of 25-in. length. Determine the tension in each wire, knowing that α = 30°. SOLUTION ( )2 2 Let be angle between the vertical and any wire. 24 = 25 7 24 in. thus cos 25 By symmetry AB AC OA T T q q- = = = 0:xFS = 2( sin )(cos ) (sin ) 0AB ADT Tq a q- + = For 45a = : ( )2 cos45 1.41421AD AB ABT T T= = 0: 12 lb 2 (cos ) (cos ) 0y AB ADF T Tq qS = - - = or ( )12 lb 2 cosAB ADT T q= + ( ) 24 12 lb 2 1.41421 25AB ABT T æ ö÷ç= + ÷ç ÷ç ÷è ø 3.66 lbAB ACT T= = ◀ 5.18 lbADT = ◀
113.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 113 PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 0,Q = find the value of P for which the tension in cable AD is 305 N. SOLUTION 0: 0A AB AC ADS = + + + =F T T T P where P=P i i j k i j k i j k (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm (960 mm) (720 mm) (220 mm) 1220 mm AB AB AC AC AD AD =- - + = =- - - = =- + - = T λ i j k T λ i j k T λ i j k i j k 48 12 19 53 53 53 12 3 4 13 13 13 305 N [( 960 mm) (720 mm) (220 mm) ] 1220 mm (240 N) (180 N) (55 N) AB AB AB AB AB AC AC AC AC AC AD AD AD AB T T T AB AC T T T AC T æ ö÷ç= = = - - + ÷ç ÷ç ÷è ø æ ö÷ç= = = - - - ÷ç ÷ç ÷è ø = = - + - =- + - Substituting into 0,AS =F factoring , , ,i j k and setting each coefficient equal to f gives: 48 12 : 240 N 53 13AB ACP T T= + +i (1) :j 12 3 180 N 53 13AB ACT T+ = (2) :k 19 4 55 N 53 13AB ACT T- = (3) Solving the system of linear equations using conventional algorithms gives: 446.71 N 341.71 N AB AC T T = = 960 NP = ◀ y x z 220 mm 240 mm 960 mm Q P A B C D O 380 mm 320 mm 960 mm
114.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 114 PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 1200 N,P = determine the values of Q for which cable AD is taut. SOLUTION We assume that 0ADT = and write 0: (1200 N) 0A AB AC QS = + + + =F T T j i i j k i j k (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm AB AB AC AC =- - + = =- - - = T λ i j k T λ i j k 48 12 19 53 53 53 12 3 4 13 13 13 AB AB AB AB AB AC AC AC AC AC AB T T T AB AC T T T AC æ ö÷ç= = = - - + ÷ç ÷ç ÷è ø æ ö÷ç= = = - - - ÷ç ÷ç ÷è ø Substituting into 0,AS =F factoring , , ,i j k and setting each coefficient equal to f gives: 48 12 : 1200 N 0 53 13AB ACT T- - + =i (1) 12 3 : 0 53 13AB ACT T Q- - + =j (2) 19 4 : 0 53 13AB ACT T- =k (3) Solving the resulting system of linear equations using conventional algorithms gives: 605.71 N 705.71 N 300.00 N AB AC T T Q = = = 0 300 NQ£ < ◀ Note: This solution assumes that Q is directed upward as shown ( 0),Q ³ if negative values of Q are considered, cable AD remains taut, but AC becomes slack for 460 N.Q = - y x z 220 mm 240 mm 960 mm Q P A B C D O 380 mm 320 mm 960 mm
115.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 115 PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A: 0: 0AB AC ADF PS = + + + =T T T j We have: ( ) i j k i j k i j k (320 mm) (480 mm) (360 mm) 680 mm (450 mm) (480 mm) (360 mm) 750 mm (250 mm) (480 mm) 360 mm 650 mm AB AB AC AC AD AD =- - + = = - + = = - - = Thus: ( ) T λ i j k T λ i j k T λ i j k 8 12 9 17 17 17 0.6 0.64 0.48 5 9.6 7.2 13 13 13 AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD æ ö÷ç= = = - - + ÷ç ÷ç ÷è ø = = = - + æ ö÷ç= = = - - ÷ç ÷ç ÷è ø Substituting into the Eq. 0FS = and factoring , , :i j k 8 5 0.6 17 13 12 9.6 0.64 17 13 9 7.2 0.48 0 17 13 AB AC AD AB AC AD AB AC AD T T T T T T P T T T æ ö÷ç- + + ÷ç ÷ç ÷è ø æ ö÷ç+ - - - + ÷ç ÷ç ÷è ø æ ö÷ç+ + - ÷ =ç ÷ç ÷è ø i j k x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm Dimensions in mm
116.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 116 PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: :i 8 5 0.6 0 17 13AB AC ADT T T- + + = (1) :j 12 9.6 0.64 0 7 13AB AC ADT T T P- - - + = (2) :k 9 7.2 0.48 0 17 13AB AC ADT T T+ - = (3) Making 60 NACT = in (1) and (3): 8 5 36 N 0 17 13AB ADT T- + + = (1¢) 9 7.2 28.8 N 0 17 13AB ADT T+ - = (3¢) Multiply (1¢) by 9, (3¢) by 8, and add: 12.6 554.4 N 0 572.0 N 13 AD ADT T- = = Substitute into (1¢) and solve for :ABT 17 5 36 572 544.0 N 8 13AB ABT T æ ö÷ç= + ´ ÷ =ç ÷ç ÷è ø Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) 0.64(60 N) (572 N) 17 13 844.8 N P = + + = Weight of plate 845 NP= = ◀
117.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 117 PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 0.6 0 17 13AB AC ADT T T- + + = (1) 12 9.6 0.64 0 17 13AB AC ADT T T P- + - + = (2) 9 7.2 0.48 0 17 13AB AC ADT T T+ - = (3) Making 520 NADT = in Eqs. (1) and (3): 8 0.6 200 N 0 17 AB ACT T- + + = (1¢) 9 0.48 288 N 0 17 AB ACT T+ - = (3¢) Multiply (1¢) by 9, (3¢) by 8, and add: 9.24 504 N 0 54.5455 NAC ACT T- = = Substitute into (1¢) and solve for :ABT 17 (0.6 54.5455 200) 494.545 N 8AB ABT T= ´ + = Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) 0.64(54.5455 N) (520 N) 17 13 768.00 N P = + + = Weight of plate 768 NP= = ◀ x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm
118.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 118 PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION 0: 0AB AC AD PS = + + + =F T T T j Free-Body Diagram at A: i j k i j k i j k 20 100 25 105 ft 60 100 18 118 ft 20 100 74 126 ft AB AB AC AC AD AD = - - + = = - + = = - - - = We write T λ i j k 4 20 5 21 21 21 AB AB AB AB AB AB T T AB T = = æ ö÷ç= - - + ÷ç ÷ç ÷è ø T λ i j k 30 50 9 59 59 59 AC AC AC AC AC AC T T AC T = = æ ö÷ç= - + ÷ç ÷ç ÷è ø T λ i j k 10 50 37 63 63 63 AD AD AD AD AD AD T T AD T = = æ ö÷ç= - - - ÷ç ÷ç ÷è ø Substituting into the Eq. 0S =F and factoring , , :i j k y A 100 ft B C O D 60 ft z x 74 ft 18 ft 20 ft 25 ft 20 ft
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 119 PROBLEM 2.111 (Continued) 4 30 10 21 59 63 20 50 50 21 59 63 5 9 37 0 21 59 63 AB AC AD AB AC AD AB AC AD T T T T T T P T T T æ ö÷ç- + - ÷ç ÷ç ÷è ø æ ö÷ç+ - - - + ÷ç ÷ç ÷è ø æ ö÷ç+ + - ÷ =ç ÷ç ÷è ø i j k Setting the coefficients of , , ,i j k equal to zero: :i 4 30 10 0 21 59 63AB AC ADT T T- + - = (1) :j 20 50 50 0 21 59 63AB AC ADT T T P- - - + = (2) :k 5 9 37 0 21 59 63AB AC ADT T T+ - = (3) Set 840 lbABT = in Eqs. (1) - (3): 30 10 160 lb 0 59 63AC ADT T- + - = (1¢) 50 50 800 lb 0 59 63AC ADT T P- - - + = (2¢) 9 37 200 lb 0 59 63AC ADT T+ - = (3¢) Solving, 458.12 lb 459.53 lb 1552.94 lbAC ADT T P= = = 1553 lbP = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 120 PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION 0: 0AB AC AD PS = + + + =F T T T j Free-Body Diagram at A: i j k i j k i j k 20 100 25 105 ft 60 100 18 118 ft 20 100 74 126 ft AB AB AC AC AD AD = - - + = = - + = = - - - = We write T λ i j k 4 20 5 21 21 21 AB AB AB AB AB AB T T AB T = = æ ö÷ç= - - + ÷ç ÷ç ÷è ø T λ i j k 30 50 9 59 59 59 AC AC AC AC AC AC T T AC T = = æ ö÷ç= - + ÷ç ÷ç ÷è ø T λ i j k 10 50 37 63 63 63 AD AD AD AD AD AD T T AD T = = æ ö÷ç= - - - ÷ç ÷ç ÷è ø Substituting into the Eq. 0S =F and factoring , , :i j k y A 100 ft B C O D 60 ft z x 74 ft 18 ft 20 ft 25 ft 20 ft
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 121 PROBLEM 2.112 (Continued) 4 30 10 21 59 63 20 50 50 21 59 63 5 9 37 0 21 59 63 AB AC AD AB AC AD AB AC AD T T T T T T P T T T æ ö÷ç- + - ÷ç ÷ç ÷è ø æ ö÷ç+ - - - + ÷ç ÷ç ÷è ø æ ö÷ç+ + - ÷ =ç ÷ç ÷è ø i j k Setting the coefficients of , , ,i j k equal to zero: :i 4 30 10 0 21 59 63AB AC ADT T T- + - = (1) :j 20 50 50 0 21 59 63AB AC ADT T T P- - - + = (2) :k 5 9 37 0 21 59 63AB AC ADT T T+ - = (3) Set 590 lbACT = in Eqs. (1) – (3): 4 10 300 lb 0 21 63AB ADT T- + - = (1¢) 20 50 500 lb 0 21 63AB ADT T P- - - + = (2¢) 5 37 90 lb 0 21 63AB ADT T+ - = (3¢) Solving, 1081.82 lb 591.82 lbAB ADT T= = 2000 lbP = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 122 PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Free-Body Diagram at A 16 30 34 34 N æ ö÷ç= + ÷ç ÷ç ÷è ø N i j and (175 lb)W= = -W j j i j k T λ i j k ( 30 ft) (20 ft) (12 ft) 38 ft 15 10 6 19 19 19 AC AC AC AC AC AC AC T T T AC T - + - = = = æ ö÷ç= - + - ÷ç ÷ç ÷è ø i j k T λ i j k ( 30 ft) (24 ft) (32 ft) 50 ft 15 12 16 25 25 25 AB AB AB AB AB AB AB T T T AB T - + + = = = æ ö÷ç= - + + ÷ç ÷ç ÷è ø Equilibrium condition: 0S =F 0AB AC+ + + =T T N W z 16 ft 8 ft B A C O x y 4 ft 30 ft 32 ft 12 ft
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 123 PROBLEM 2.113 (Continued) Substituting the expressions obtained for , , ,AB ACT T N and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 15 15 16 0 25 19 34AB ACT T N- - + = (1) From j: 12 10 30 (175 lb) 0 25 19 34AB ACT T N+ + - = (2) From k: 16 6 0 25 19AB ACT T- = (3) Solving the resulting set of equations gives: 30.8 lb; 62.5 lbAB ACT T= = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 124 PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P =-(45 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force ( 45 lb) .= -P k 15 15 16 0 25 19 34AB ACT T N- - + = (1) 12 10 30 (175 lb) 0 25 19 34AB ACT T N+ + - = (2) 16 6 (45 lb) 0 25 19AB ACT T- - = (3) Solving the resulting set of equations simultaneously gives: 81.3 lbABT = ◀ 22.2 lbACT = ◀ z 16 ft 8 ft B A C O x y 4 ft 30 ft 32 ft 12 ft
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 125 PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting 792 NP = gives: 8 5 0.6 0 17 13AB AC ADT T T- + + = (1) 12 9.6 0.64 792 N 0 17 13AB AC ADT T T- - - + = (2) 9 7.2 0.48 0 17 13AB AC ADT T T+ - = (3) Solving Equations (1), (2), and (3) by conventional algorithms gives 510.00 NABT = 510 NABT = ◀ 56.250 NACT = 56.2 NACT = ◀ 536.25 NADT = 536 NADT = ◀ x y z A B C DO 250 130 360 360 320 450 480 Dimensions in mm
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 126 PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that 2880 NP = and 0.Q = SOLUTION 0: 0A AB AC ADS = + + + + =F T T T P Q Where P=P i and Q=Q j i j k i j k i j k (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm (960 mm) (720 mm) (220 mm) 1220 mm AB AB AC AC AD AD =- - + = =- - - = =- + - = T λ i j k T λ i j k T λ i j k 48 12 19 53 53 53 12 3 4 13 13 13 48 36 11 61 61 61 AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD æ ö÷ç= = = - - + ÷ç ÷ç ÷è ø æ ö÷ç= = = - - - ÷ç ÷ç ÷è ø æ ö÷ç= = = - + - ÷ç ÷ç ÷è ø Substituting into 0,AS =F setting (2880 N)P = i and 0,Q = and setting the coefficients of , ,i j k equal to 0, we obtain the following three equilibrium equations: 48 12 48 : 2880 N 0 53 13 61AB AC ADT T T- - - + =i (1) 12 3 36 : 0 53 13 61AB AC ADT T T- - + =j (2) 19 4 11 : 0 53 13 61AB AC ADT T T- - =k (3) y x z 220 mm 240 mm 960 mm Q P A B C D O 380 mm 320 mm 960 mm
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 127 PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: 1340.14 N 1025.12 N 915.03 N AB AC AD T T T = = = 1340 NABT = ◀ 1025 NACT = ◀ 915 NADT = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 128 PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that 2880 NP = and 576 N.Q = SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 48 12 48 0 53 13 61AB AC ADT T T P- - - + = (1) 12 3 36 0 53 13 61AB AC ADT T T Q- - + + = (2) 19 4 11 0 53 13 61AB AC ADT T T- - = (3) Setting 2880 NP = and 576 NQ = gives: 48 12 48 2880 N 0 53 13 61AB AC ADT T T- - - + = (1¢) 12 3 36 576 N 0 53 13 61AB AC ADT T T- - + + = (2¢) 19 4 11 0 53 13 61AB AC ADT T T- - = (3¢) Solving the resulting set of equations using conventional algorithms gives: 1431.00 N 1560.00 N 183.010 N AB AC AD T T T = = = 1431 NABT = ◀ 1560 NACT = ◀ 183.0 NADT = ◀ y x z 220 mm 240 mm 960 mm Q P A B C D O 380 mm 320 mm 960 mm
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 129 PROBLEM 2.118 Three cables are connected at D, where an upward force of 30 kN is applied. Determine the tension in each cable. SOLUTION Free-Body Diagram of Point D: 0: 0D DA DB DCS = + + + =F T T T P Where P=P j i i j k i j k (1 m) 1 m (0.75 m) (3 m) (1 m) 3.25 m (1 m) (3 m) (1.5 m) 3.5 m DA DA DB DB DC DC =- = = + - - = = + - + = ( )T λ i T λ i j k T λ i j k 3 12 4 13 13 13 2 6 3 7 7 7 DA DA DA DA DA DB DB DB DB DB DC DC DC DC DC DA T T T DA DB T T T DB DC T T T DC = = = - æ ö÷ç= = = - - ÷ç ÷ç ÷è ø æ ö÷ç= = = - + ÷ç ÷ç ÷è ø Substituting into 0,DS =F setting (30 kN)P = j and setting the coefficients of , ,i j k equal to 0, we obtain the following three equilibrium equations: 3 2 : 0 13 7DA DB DCT T T- + + =i (1)
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 130 PROBLEM 2.118 (Continued) 12 6 : 30 kN 0 13 7DB DCT T- - + =j (2) 4 3 : 0 13 7DB DCT T- + =k (3) Solving the system of linear equations using conventional algorithms gives: 8.50 kNDAT = ◀ 19.50 kNDBT = ◀ 14.00 kNDCT = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 131 PROBLEM 2.119 For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1800 lb. PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: :i 4 30 10 0 21 59 63AB AC ADT T T- + - = (1) :j 20 50 50 0 21 59 63AB AC ADT T T P- - - + = (2) :k 5 9 37 0 21 59 63AB AC ADT T T+ - = (3) Substituting for 1800 lbP = in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 4 30 10 0 21 59 63AB AC ADT T T- + - = (1¢) 20 50 50 1800 lb 0 21 59 63AB AC ADT T T- - - + = (2¢) 5 9 37 0 21 59 63AB AC ADT T T+ - = (3¢) 973.64 lb 531.00 lb 532.64 lb AB AC AD T T T = = = 974 lbABT = ◀ 531 lbACT = ◀ 533 lbADT = ◀ y A 100 ft B C O D 60 ft z x 74 ft 18 ft 20 ft 25 ft 20 ft
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 132 PROBLEM 2.120 Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb cylinder is suspended from point D as shown. SOLUTION Free-Body Diagram of Point D: The forces applied at D are: , , andDA DB DCT T T W where 180.0 lb .= -W j To express the other forces in terms of the unit vectors i, j, k, we write j k i j k i j k (18 in.) (22 in.) 28.425 in. (24 in.) (18 in.) (16 in.) 34.0 in. (24 in.) (18 in.) (16 in.) 34.0 in. DA DA DB DB DC DC = + = =- + - = = + - = 180 lb D A B C 18 in. 16 in. 22 in. 24 in. 24 in.
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 133 PROBLEM 2.120 (Continued) and T λ j k T λ i j k T λ i j k (0.63324 0.77397 ) ( 0.70588 0.52941 0.47059 ) (0.70588 0.52941 0.47059 ) DA Da DA Da DA DB DB DB DB DB DC DC DC DC DC DA T T DA T DB T T DB T DC T T DC T = = = + = = = - + - = = = + - Equilibrium Condition with W= -W j 0: 0DA DB DCF WS = + + - =T T T j Substituting the expressions obtained for , , andDA DB DCT T T and factoring i, j, and k: ( 0.70588 0.70588 ) (0.63324 0.52941 0.52941 ) (0.77397 0.47059 0.47059 ) DB DC DA DB DC DA DB DC T T T T T W T T T - + + + - - - i j k Equating to zero the coefficients of i, j, k: 0.70588 0.70588 0DB DCT T- + = (1) 0.63324 0.52941 0.52941 0DA DB DCT T T W+ + - = (2) 0.77397 0.47059 0.47059 0DA DB DCT T T- - = (3) Substituting 180 lbW = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives, 119.7 lbDAT = ◀ 98.4 lbDBT = ◀ 98.4 lbDCT = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 134 PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that 1000 N,W = determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with i j k T λ i j k T i j k 2 2 2 (0.78 m) (1.6 m) (0 m) ( 0.78 m) (1.6 m) (0) 1.78 m [ (0.78 m) (1.6 m) (0 m) ] 1.78 m ( 0.4382 0.8989 0 ) AB AB AB AB AB AB AB AB AB T T AB T T =- + + = - + + = = = = - + + = - + + and i j k T λ i j k T j k 2 2 2 (0) (1.6 m) (1.2 m) (0 m) (1.6 m) (1.2 m) 2 m [(0) (1.6 m) (1.2 m) ] 2 m (0.8 0.6 ) AC AC AC AC AC AC AC AC TAC T T AC T = + + = + + = = = = + + = + and i j k T λ i j k T i j k 2 2 2 (1.3 m) (1.6 m) (0.4 m) (1.3 m) (1.6 m) (0.4 m) 2.1 m [(1.3 m) (1.6 m) (0.4 m) ] 2.1 m (0.6190 0.7619 0.1905 ) AD AD AD AD AD AD AD AD TAD T T AD T = + + = + + = = = = + + = + + y xz 0.78 m 0.40 m 0.40 m P O B F E C W A D 1.60 m 0.86 m 1.20 m 1.30 m
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 135 PROBLEM 2.121 (Continued) Finally, i j k T λ i j k T i j k 2 2 2 (0.4 m) (1.6 m) (0.86 m) ( 0.4 m) (1.6 m) ( 0.86 m) 1.86 m [ (0.4 m) (1.6 m) (0.86 m) ] 1.86 m ( 0.2151 0.8602 0.4624 ) AE AE AE AE AE AE AE AE AE T T AE T T = - + - = - + + - = = = = - + - = - + - With the weight of the container ,W= -W j at A we have: 0: 0AB AC AD WS = + + - =F T T T j Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: 0.4382 0.6190 0.2151 0AB AD AET T T- + - = (1) 0.8989 0.8 0.7619 0.8602 0AB AC AD AET T T T W+ + + - = (2) 0.6 0.1905 0.4624 0AC AD AET T T+ - = (3) Knowing that 1000 NW = and that because of the pulley system at B ,AB ADT T P= = where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. 378 NP = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 136 PROBLEM 2.122 Knowing that the tension in cable AC of the system described in Problem 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that 1000 N,W = determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Here, as in Problem 2.121, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition AB ADT T P= = and using the linear algebraic equations of Problem 2.131 with 150 N,ACT = we obtain (a) 454 NP = ◀ (b) 1202 NW = ◀ y xz 0.78 m 0.40 m 0.40 m P O B F E C W A D 1.60 m 0.86 m 1.20 m 1.30 m
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 137 PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 270 lb, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Free-Body Diagram of Point A: = = - + - = æ ö÷ç= - + - ÷ç ÷ç ÷è ø ( 48 in.) (72 in.) (16 in.) 88 in. 6 9 2 11 11 11 AB ABT AB T AB T T T λ i j k i j k = = + + - = æ ö÷ç= + - ÷ç ÷ç ÷è ø S = + + + + = (24 in.) (72 in.) ( 13 in.) 77 in. 24 72 13 77 77 77 0: 0 AC AC AB AC T AC T AC T T F T λ i j k i j k T T Q P W
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 138 PROBLEM 2.123 (continued) Setting coefficients of i, j, k equal to zero: æ ö÷ç- + + = - ÷ + =ç ÷ç ÷è ø 6 24 18 : 0 0 11 77 77 T T P T Pi (1) æ ö÷ç+ + - = ÷ - =ç ÷ç ÷è ø 9 72 135 : 0 0 11 77 77 T T W T Wj (2) æ ö÷ç- - + = - ÷ + =ç ÷ç ÷è ø 2 13 27 : 0 0 11 77 77 T T Q T Qk (3) Data: 77 270 lb 270 lb 154.0 lb 135 W T T æ ö÷ç= = ÷ =ç ÷ç ÷è ø Substituting for T in Eqn . (1) and (2): 36.0 lbP = ◀ 54.0 lbQ = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 139 PROBLEM 2.124 For the system of Prob. 2.123, determine W and P knowing that Q = 36 lb. PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 270 lb, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3). Setting 36 lbQ = we have: Eq. (3): 77 36 lb 27 T æ ö÷ç= ÷ç ÷ç ÷è ø 102.667 lbT = Eq. (1): 18 (102.667 lb) P 0 77 - + = 24.0 lbP = ◀ Eq. (2): 135 (102.667 lb) W 0 77 - = 180.0 lbW = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 140 PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force (341 N)=P j is applied to collar A, determine (a) the tension in the wire when y 155 mm,= (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION For both Problems 2.125 and 2.126: Free-Body Diagrams of Collars: 2 2 2 2 ( )A B x y z= + + Here 2 2 2 2 (0.525 m) (0.20 m) y z= + + or 2 2 2 0.23563 my z+ = Thus, when y given, z is determined, Now λ i j k i j k 1 (0.20 )m 0.525 m 0.38095 1.90476 1.90476 AB AB AB y z y z = = - + = - + Where y and z are in units of meters, m. From the F.B. Diagram of collar A: 0: 0x z AB ABN N P T lS = + + + =F i k j Setting the j coefficient to zero gives (1.90476 ) 0ABP y T- = With 341 N 341 N 1.90476AB P T y = = Now, from the free body diagram of collar B: 0: 0x y AB ABN N Q TS = + + - =F i j k λ Setting the k coefficient to zero gives (1.90476 ) 0ABQ T z- = And using the above result for ,ABT we have 341 N (341 N)( ) (1.90476 ) (1.90476)AB z Q T z z y y = = = 200 mm x y y z zB Q P A O
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 141 PROBLEM 2.125 (Continued) Then from the specifications of the problem, 155 mm 0.155 my = = 2 2 2 0.23563 m (0.155 m) 0.46 m z z = - = and (a) 341 N 0.155(1.90476) 1155.00 N ABT = = or 1155 NABT = ◀ and (b) 341 N(0.46 m)(0.866) (0.155 m) (1012.00 N) Q = = or 1012 NQ = ◀
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Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 142 PROBLEM 2.126 Solve Problem 2.125 assuming that 275 mm.y = PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force (341 N)=P j is applied to collar A, determine (a) the tension in the wire when y 155 mm,= (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION From the analysis of Problem 2.125, particularly the results: 2 2 2 0.23563 m 341 N 1.90476 341 N AB y z T y Q z y + = = = With 275 mm 0.275 m,y = = we obtain: 2 2 2 0.23563 m (0.275 m) 0.40 m z z = - = and (a) 341 N 651.00 (1.90476)(0.275 m) ABT = = or 651 NABT = ◀ and (b) 341 N(0.40 m) (0.275 m) Q = or 496 NQ = ◀ 200 mm x y y z zB Q P A O
143.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 143 PROBLEM 2.127 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have 180 (20 10 ) 150 g = - + = Then 2 2 2 (48 N) (60 N) 2(48 N)(60 N)cos150 104.366 N R R = + - = and 48 N 104.366 N sin sin150 sin 0.22996 13.2947 a a a = = = Hence: 180 80 180 13.2947 80 86.705 f a= - - = - - = 104.4 N=R 86.7° ◀ A 55° 25° 85° P Q
144.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 144 PROBLEM 2.128 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: (80 N)cos40xF = + 61.3 NxF = ◀ (80 N)sin 40yF = + 51.4 NyF = ◀ 120-N Force: (120 N)cos70xF = + 41.0 NxF = ◀ (120 N)sin 70yF = + 112.8 NyF = ◀ 150-N Force: (150 N)cos35xF = - 122.9 NxF = - ◀ (150 N)sin35yF = + 86.0 NyF = ◀ 80 N 120 N 150 N 30° 35° 40° y x
145.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 145 PROBLEM 2.129 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION xR = (200 lb)sin 40 (400 lb)cos 40xF PS = + - 177.860 lbxR P= - (1) yR = (200 lb)cos40 (400 lb)sin 40yFS = + 410.32 lbyR = (2) (a) For R to be vertical, we must have 0.xR = Set 0xR = in Eq. (1) 0 177.860 lb 177.860 lb P P = - = 177.9 lbP = ◀ (b) Since R is to be vertical: 410 lbyR R= = 410 lbR = ◀ α α 200 lb 400 lb P
146.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 146 PROBLEM 2.130 Knowing that 55a = and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 300 lb sin35 sin50 sin95 AC BCF T = = (a) 300 lb sin35 sin95 ACF = 172.7 lbA CF = ◀ (b) 300 lb sin50 sin95 BCT = 231 lbBCT = ◀ 30° 20° α 300 lb A B C
147.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 147 PROBLEM 2.131 Two cables are tied together at C and loaded as shown. Knowing that 360 N,P = determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free Body: C (a) 12 4 0: (360 N) 0 13 5x ACTS = - + =F 312 NACT = ◀ (b) 5 3 0: (312 N) (360 N) 480 N 0 13 5y BCTS = + + - =F 480 N 120 N 216 NBCT = - - 144.0 NBCT = ◀ A B P Q = 480 N C 3 4 600 mm 250 mm
148.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 148 PROBLEM 2.132 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of a. SOLUTION Free-Body Diagram: C Force Triangle Force triangle is isosceles with 2 180 85 47.5 b b = - = (a) 2(800 N)cos 47.5° 1081 NP = = Since 0,P > the solution is correct. 1081 NP = ◀ (b) 180 50 47.5 82.5a = - - = 82.5a = ◀ 35° A B C P 50° α
149.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 149 PROBLEM 2.133 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles qx, qy, and qz that the force forms with the coordinate axes. SOLUTION (a) (120 lb)cos 60 cos 20xF = 56.382 lbxF = 56.4 lbxF = + ◀ (120 lb)sin 60 103.923 lb y y F F =- =- 103.9 lbyF = - ◀ (120 lb)cos 60 sin 20 20.521 lb z z F F = - = - 20.5 lbzF = - ◀ (b) 56.382 lb cos 120 lb x x F F q = = 62.0xq = ◀ 103.923 lb cos 120 lb y y F F q - = = 150.0yq = ◀ 20.52 lb cos 120 lb z z F F q - = = 99.8zq = ◀ 36° 60° 48° 20° x y z A B C E D
150.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 150 PROBLEM 2.134 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION i j k T λ T i j k i j k 2 2 2 (900 mm) (600 mm) (920 mm) (900 mm) (600 mm) (920 mm) 1420 mm 2130 N [ (900 mm) (600 mm) (920 mm) ] 1420 mm (1350 N) (900 N) (1380 N) CA CA CA CA CA CA CA T CA T CA =- + - = + + = = = = - + - =- + - ( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T=- = = - ◀ x y z A B D C O 600 mm 920 mm 360 mm 900 mm
151.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 151 PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N. SOLUTION (600 N)[sin40 sin25 cos40 sin40 cos25 ] (162.992 N) (459.63 N) (349.54 N) (450 N)[cos55 cos30 sin55 cos55 sin30 ] (223.53 N) (368.62 N) (129.055 N) (386.52 N) (828.25 N) (220.49 N) (3R = + + = + + = + - = + - = + = + + = P i j k i j k Q i j k i j k R P Q i j k 2 2 2 86.52 N) (828.25 N) (220.49 N) 940.22 N + + = 940 NR = ◀ 386.52 N cos 940.22 N x x R R q = = 65.7xq = ◀ 828.25 N cos 940.22 N y y R R q = = 28.2yq = ◀ 220.49 N cos 940.22 N z z R R q = = 76.4zq = ◀ z xO y 30° 25° 40° 55° P Q
152.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 152 PROBLEM 2.136 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Free Body Diagram at A: Since tension inBA CT = cable BAC, it follows that A B AC BACT T T= = ( 17.5 in.) (60 in.) 17.5 60 62.5 in. 62.5 62.5 (60 in.) (25 in.) 60 25 65 in. 65 65 (80 in.) (60 in.) 4 3 100 in. 5 5 AB BAC AB BAC BAC AC BAC AC BAC BAC AD AD AD AD AD T T T T T T T T T - + æ ö- ÷ç= = = + ÷ç ÷ç ÷è ø + æ ö÷ç= = = + ÷ç ÷ç ÷è ø + æ ö÷ç= = = + ÷ç ÷çè ø i j T λ i j i k T λ j k i j T λ i j (60 in.) (45 in.) 4 3 75 in. 5 5AE AE AE AE AET T T ÷ - æ ö÷ç= = = - ÷ç ÷ç ÷è ø j k T λ j k D x E O B 25 in. 17.5 in. 45 in. 60 in. 80 in. y C A z P
153.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 153 PROBLEM 2.136 (Continued) Substituting into 0,AS =F setting ( 200 lb) ,= -P j and setting the coefficients of i, j, k equal to ,f we obtain the following three equilibrium equations: From 17.5 4 : 0 62.5 5BAC ADT T- + =i (1) From 60 60 3 4 : 200 lb 0 62.5 65 5 5BAC AD AET T T æ ö÷ç + ÷ + + - =ç ÷ç ÷è ø j (2) From 25 3 : 0 65 5BAC AET T- =k (3) Solving the system of linear equations using conventional algorithms gives: 76.7 lb; 26.9 lb; 49.2 lbBAC AD AET T T= = = ◀
154.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 154 PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when 9 in.,x = (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. SOLUTION Free-Body Diagrams of Collars: A: B: i j k λ (20 in.) 25 in.AB AB x z AB - - + = = Collar A: 0: 0y z AB ABP N N TS = + + + =F i j k λ Substitute for A Bλ and set coefficient of i equal to zero: 0 25 in. ABT x P - = (1) Collar B: 0: (60 lb) 0x y AB ABN N T¢ ¢S = + + - =F k i j λ Substitute for A Bλ and set coefficient of k equal to zero: 60 lb 0 25 in. ABT z - = (2) (a) 2 2 2 2 9 in. (9 in.) (20 in.) (25 in.) 12 in. x z z = + + = = From Eq. (2): 60 lb (12 in.) 25 in. ABT- 125.0 lbA BT = ◀ (b) From Eq. (1): (125.0 lb)(9 in.) 25 in. P = 45.0 lbP = ◀ 20 in. x x y z z B Q P A O
155.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 155 PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when 120 lbP = and 60 lb.Q = SOLUTION See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below: 0 25 in. ABT x P = = (1) 60 lb 0 25 in. ABT z - = (2) For 120 lb,P = Eq. (1) yields (25 in.)(20 lb)ABT x = (1¢) From Eq. (2): (25 in.)(60 lb)ABT z = (2¢) Dividing Eq. (1¢) by (2¢), 2 x z = (3) Now write 2 2 2 2 (20 in.) (25 in.)x z+ + = (4) Solving (3) and (4) simultaneously, 2 2 2 4 400 625 45 6.7082 in. z z z z + + = = = From Eq. (3): 2 2(6.7082 in.) 13.4164 in. x z= = = 13.42 in., 6.71 in.x z= = ◀ 20 in. x x y z z B Q P A O
156.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 156 PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC. SOLUTION Free-Body Diagram of Point C: 2 3 (1600 kg)(9.81 m/s ) 15.6960(10 ) N 15.696 kN W W W = = = A B C 1600 kg 960 mm 1100 mm 400 mm
157.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 157 PROBLEM 2.F2 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram needed to determine the magnitudes of the forces TC and TD. SOLUTION Free-Body Diagram of Point E: 40° TBTA TC TD
158.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 158 PROBLEM 2.F3 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. SOLUTION Free-Body Diagram of Point A: 65 lb 60 lb C A B h 15 in.
159.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 159 PROBLEM 2.F4 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F. SOLUTION Free-Body Diagram of Point B: 1 1 250 N 765 N 1015 N 8.25 tan 30.510 14 10 tan 22.620 24 E AB BC W q q - - = + = = = = = Use this free body to determine TAB and TBC. Free-Body Diagram of Point C: 1 1.1 tan 10.3889 6CDq - = = Use this free body to determine TCD and WF. Then weight of skier WS is found by 250 NS FW W= - ◀ 14 m 24 m 6 m 8.25 m 10 m 1.10 m A B C DF E
160.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 160 PROBLEM 2.F5 Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A. SOLUTION Free-Body Diagram of Point A: A B C D O 4.20 m 4.20 m 3.30 m 5.60 m 2.40 m x y z
161.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 161 PROBLEM 2.F6 A container of mass m = 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point A: 2 (120 kg)(9.81m/s ) 1177.2 N W = = x y z A B D C O 600 mm 320 mm 360 mm 500 mm 450 mm
162.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 162 PROBLEM 2.F7 A 150-lb cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable. SOLUTION Free-Body Diagram of Point C: O B C A y x z 15 ft 7.2 ft 3.6 ft 10.8 ft 10.8 ft P
163.
Copyright © McGraw-HillEducation. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 163 PROBLEM 2.F8 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A. SOLUTION Free-Body Diagram of point A: y A 90 ft 30 ft O B 30 ft 20 ft 45 ft z D C 60 ft 65 ft x |
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4320 | https://brightchamps.com/en-us/math/math-questions/6.35-as-a-fraction | Table Of Contents
Summarize this article:
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Last updated on August 5, 2025
6.35 as a Fraction
Numbers can be categorized into different types. A fraction is one of its kinds. It is always represented in the form of p/q, where p is the numerator and q is the denominator. A fraction represents a whole and a fractional part. Decimals represent the fractional part of numbers. For example, 1/2, the numbers in decimal are expressed with a decimal point (.), For example, 6.35, we are going to learn how to convert a decimal to a fraction.
What is 6.35 as a Fraction?
Answer
The answer for 6.35 as a fraction will be 127/20.
Explanation
Converting a decimal to a fraction is a task for students that can be done easily. You can follow the steps mentioned below to find the answer.
Step 1: Firstly, any decimal number should be converted to a fraction for easy calculation. Here, 6.35 is the number on the numerator and the base number 1 will be the denominator. Then, 6.35 becomes 6.35/1.
Step 2: To remove the decimal from a fraction, you need to multiply both the numerator and denominator by 100 (because there are 2 decimal places). 6.35/1 × 100/100 = 635/100
Step 3: Here 5 is the GCD of 635 and 100. Now, to make the fraction simpler, divide the numerator and denominator by 5. 635/100 = 127/20 Hence, 6.35 is in the form of the fraction 127/20.
Thus, 6.35 can be written as a fraction 127/20.
Important Glossaries for 6.35 as a Fraction
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4321 | https://faculty.valpo.edu/lpudwell/slides/parkingfunctions_FL2023.pdf | Permutations Parking Functions Recursion Polygons Trees Wrapping up Patterns in Parking Functions Lara Pudwell joint work with Ayomikun Adeniran (Colby College) 54th Southeastern International Conference on Combinatorics, Graph Theory and Computing March 10, 2023 Patterns in Parking Functions Lara Pudwell 1 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Definition A permutation is a list where order matters.
Sn is the set of all permutations of {1, 2, . . . , n}.
Examples: S1 = {1} S2 = {12, 21} S3 = {123, 132, 213, 231, 312, 321} |Sn| = n!
Patterns in Parking Functions Lara Pudwell 2 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Visualize π = π1π2 · · · πn ∈Sn by plotting the points (i, πi) in the xy-plane.
123 132 213 231 312 321 Patterns in Parking Functions Lara Pudwell 3 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up 5 6 2 7 1 9 3 4 8 π = 562719348 Patterns in Parking Functions Lara Pudwell 4 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Patterns in Parking Functions Lara Pudwell 5 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Patterns in Parking Functions Lara Pudwell 5 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up 562719348 contains the pattern 132 Patterns in Parking Functions Lara Pudwell 5 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up 562719348 contains the pattern 1234 Patterns in Parking Functions Lara Pudwell 5 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up 562719348 avoids the pattern 4321 Patterns in Parking Functions Lara Pudwell 5 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Big question How many permutations of length n contain the pattern ρ?
Or, alternatively...
Big question How many permutations of length n avoid the pattern ρ?
(depends on what ρ is!) Notation Sn(ρ) is the set of permutations of length n avoiding ρ.
Patterns in Parking Functions Lara Pudwell 6 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Definition A parking function is a sequence a1 · · · an ∈[n]n such that if b1 ≤b2 ≤· · · ≤bn is the increasing rearrangement of a1 · · · an then bi ≤i for all 1 ≤i ≤n.
Patterns in Parking Functions Lara Pudwell 7 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Definition A parking function is a sequence a1 · · · an ∈[n]n such that if b1 ≤b2 ≤· · · ≤bn is the increasing rearrangement of a1 · · · an then bi ≤i for all 1 ≤i ≤n.
Examples: 11111, 32123, 45312 11111, 12233, 12345 Patterns in Parking Functions Lara Pudwell 7 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Definition A parking function is a sequence a1 · · · an ∈[n]n such that if b1 ≤b2 ≤· · · ≤bn is the increasing rearrangement of a1 · · · an then bi ≤i for all 1 ≤i ≤n.
Examples: 11111, 32123, 45312 11111, 12233, 12345 Nonexamples: 22222, 51244, 15151 22222, 12445, 11155 Patterns in Parking Functions Lara Pudwell 7 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Definition A parking function is a sequence a1 · · · an ∈[n]n such that if b1 ≤b2 ≤· · · ≤bn is the increasing rearrangement of a1 · · · an then bi ≤i for all 1 ≤i ≤n.
Examples: 11111, 32123, 45312 11111, 12233, 12345 Nonexamples: 22222, 51244, 15151 22222, 12445, 11155 Observations There are (n + 1)n−1 parking functions of size n.
Every permutation of size n is a parking function of size n.
Patterns in Parking Functions Lara Pudwell 7 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up History Jel´ ınek and Mansour (2009) Consider parking functions as words on [n]n Determined all equivalence classes of patterns of length at most 5 Patterns in Parking Functions Lara Pudwell 8 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up History Jel´ ınek and Mansour (2009) Consider parking functions as words on [n]n Determined all equivalence classes of patterns of length at most 5 Remmel and Qiu (2016) Consider parking functions as labeled Dyck paths (bijection of Garsia and Haiman) Each Dyck path is associated with a permutation (many-to-one correspondence) Determined number of 123-avoiding parking functions Patterns in Parking Functions Lara Pudwell 8 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up History Jel´ ınek and Mansour (2009) Consider parking functions as words on [n]n Determined all equivalence classes of patterns of length at most 5 Remmel and Qiu (2016) Consider parking functions as labeled Dyck paths (bijection of Garsia and Haiman) Each Dyck path is associated with a permutation (many-to-one correspondence) Determined number of 123-avoiding parking functions Current project: Extend Remmel and Qiu’s work Count parking functions avoiding a subset of S3.
Patterns in Parking Functions Lara Pudwell 8 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Parking functions of size 2 Sequences: 11 12 21 Patterns in Parking Functions Lara Pudwell 9 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Parking functions of size 2 Sequences: 11 12 21 Blocks: {1, 2}, ∅ {1}, {2} {2}, {1} Patterns in Parking Functions Lara Pudwell 9 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Parking functions of size 2 Sequences: 11 12 21 Blocks: {1, 2}, ∅ {1}, {2} {2}, {1} Dyck paths: 1 2 1 2 2 1 Patterns in Parking Functions Lara Pudwell 9 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Parking functions of size 2 Sequences: 11 12 21 Blocks: {1, 2}, ∅ {1}, {2} {2}, {1} Dyck paths: 1 2 1 2 2 1 Associated permutations: 12 12 21 Patterns in Parking Functions Lara Pudwell 9 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Parking function: 6144231 Blocks: {2, 7}, {5}, {6}, {3, 4}, ∅, {1}, ∅ Dyck path: 2 7 5 6 3 4 1 Associated permutation: 2756341 Patterns in Parking Functions Lara Pudwell 10 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Warmup Notation Let pfn(ρ) be the number of parking functions of size n whose associated permutations avoid ρ.
Proposition pfn(21) = Cn (nth Catalan number) 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 Patterns in Parking Functions Lara Pudwell 11 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Warmup Proposition pfn(12) = 1 7 6 5 4 3 2 1 Patterns in Parking Functions Lara Pudwell 12 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Patterns P pfn(P), 1 ≤n ≤6 OEIS 123, 132, 231 1, 3, 5, 7, 9, 11 A005408 123, 132, 312 1, 3, 6, 10, 15, 21 A000217 123, 213, 231 123, 231, 312 123, 213, 312 1, 3, 7, 13, 21, 31 A002061 123, 132, 213 1, 3, 6, 17, 43, 123 A143363 132, 213, 231 1, 3, 8, 22, 64, 196 A014138 132, 231, 312 132, 213, 312 213, 231, 312 1, 3, 9, 28, 90, 297 A000245 132, 231, 321 1, 3, 9, 29, 98, 342 A077587 132, 213, 321 1, 3, 10, 35, 126, 462 A001700 132, 312, 321 213, 231, 321 213, 312, 321 1, 3, 11, 41, 154, 582 A076540 231, 312, 321 1, 3, 10, 38, 154, 654 A001002 Patterns in Parking Functions Lara Pudwell 13 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Patterns P pfn(P), 1 ≤n ≤6 OEIS 123, 231 1, 3, 8, 17, 31, 51 A105163 123, 312 1, 3, 9, 21, 41, 71 A064999 123, 132 1, 3, 8, 24, 75, 243 A000958 123, 213 1, 3, 9, 28, 90, 297 A000245 132, 231 1, 3, 10, 36, 137, 543 A002212 132, 213 1, 3, 11, 45, 197, 903 A001003 132, 312 213, 231 231, 312 132, 321 1, 3, 12, 52, 229, 1006 new 213, 321 1, 3, 13, 60, 275, 1238 new 213, 312 1, 3, 12, 54, 259, 1293 new 231, 321 1, 3, 12, 55, 273, 1428 A001764 312, 321 1, 3, 13, 63, 324, 1736 new Patterns in Parking Functions Lara Pudwell 14 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(132, 213, 312) = pfn(213, 231, 312) = 3(2n)!
(n + 2)!(n −1)! = Cn+1 −Cn Sn(132, 213, 312) Sn(213, 231, 312) Patterns in Parking Functions Lara Pudwell 15 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Sn(213, 231, 312) Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn Patterns in Parking Functions Lara Pudwell 16 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Sn(213, 231, 312) Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn Two cases: 1 Last block has one element 2 Last block is empty Patterns in Parking Functions Lara Pudwell 16 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Sn(213, 231, 312) Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn Two cases: 1 Last block has one element (a(n −1, k)) 2 Last block is empty Case 1? Deleting/reinserting last block (and standardizing) is bijection {1, 2}, ∅, {7}, {6}, {5}, {4}, {3} ↔{1, 2}, ∅, {6}, {5}, {4}, {3} Patterns in Parking Functions Lara Pudwell 16 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Sn(213, 231, 312) Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn Two cases: 1 Last block has one element (a(n −1, k)) 2 Last block is empty (a(n, k −1)) Case 2? Bijection via moving last element before decreasing run.
1 2 3 5 4 ↔ 1 2 5 4 3 1 2 3 5 4 ↔ 1 2 5 4 3 Patterns in Parking Functions Lara Pudwell 16 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Sn(213, 231, 312) Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn Two cases: 1 Last block has one element (a(n −1, k)) 2 Last block is empty (a(n, k −1)) In general: a(n, k) = a(n −1, k) + a(n, k −1).
Patterns in Parking Functions Lara Pudwell 16 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn In general: a(n, k) = a(n −1, k) + a(n, k −1).
Patterns in Parking Functions Lara Pudwell 17 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn In general: a(n, k) = a(n −1, k) + a(n, k −1).
a(n, k) gives triangle A030237, i.e. Catalan’s triangle with the rightmost diagonal removed.
Patterns in Parking Functions Lara Pudwell 17 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Let a(n, k) be the number of size n parking functions whose associated permutation begins with k −1 ascents.
a(n, 1) = 1 a(n, n) = Cn In general: a(n, k) = a(n −1, k) + a(n, k −1).
a(n, k) gives triangle A030237, i.e. Catalan’s triangle with the rightmost diagonal removed.
pfn(132, 213, 312) = pfn(213, 231, 312) = n X k=1 a(n, k) = Cn+1 −Cn Patterns in Parking Functions Lara Pudwell 17 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(231, 312, 321) = ⌊n 2 ⌋ X k=0 1 n + 1 2n −k n + k n + k k (OEIS A001002) (number of dissections of a convex (n + 2)-gon into triangles and quadrilaterals) Patterns in Parking Functions Lara Pudwell 18 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(231, 312, 321) = ⌊n 2 ⌋ X k=0 1 n + 1 2n −k n + k n + k k (OEIS A001002) (number of dissections of a convex (n + 2)-gon into triangles and quadrilaterals) Catalan object 1: dissections of a convex (n + 2)-gon into triangles Catalan object 2: 21-avoiding parking functions Patterns in Parking Functions Lara Pudwell 18 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Small cases: {1} {1} , {2} {1, 2} , ∅ Patterns in Parking Functions Lara Pudwell 19 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Small cases: {1} {1} , {2} {1, 2} , ∅ General cases: Patterns in Parking Functions Lara Pudwell 19 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Small cases: {1} {1} , {2} {1, 2} , ∅ General cases: {n −1} , {n} {n −1, n} , ∅ Patterns in Parking Functions Lara Pudwell 19 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Small cases: {1} {1} , {2} {1, 2} , ∅ General cases: {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} Patterns in Parking Functions Lara Pudwell 19 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up General case: {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} n = 3: Permutations Parking Functions Recursion Polygons Trees Wrapping up General case: {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} n = 3: {1, 2, 3} , ∅, ∅ {1, 2} , {3} , ∅ Permutations Parking Functions Recursion Polygons Trees Wrapping up General case: {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} n = 3: {1, 2, 3} , ∅, ∅ {1, 2} , {3} , ∅ {1} , {2, 3} , ∅ {1} , {2} , {3} Permutations Parking Functions Recursion Polygons Trees Wrapping up General case: {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} n = 3: {1, 2, 3} , ∅, ∅ {1, 2} , {3} , ∅ {1} , {2, 3} , ∅ {1} , {2} , {3} {1, 2} , ∅, {3} Patterns in Parking Functions Lara Pudwell 20 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {3} , {4} {1, 2} , ∅ Patterns in Parking Functions Lara Pudwell 21 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {3} , {4} {1, 2} , ∅ {1, 2, 3} , {4} , ∅, {5} , ∅ Patterns in Parking Functions Lara Pudwell 21 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(231, 312, 321) = ⌊n 2 ⌋ X k=0 1 n + 1 2n −k n + k n + k k (OEIS A001002) (number of dissections of a convex (n + 2)-gon into triangles and quadrilaterals) Avoiding {231, 312, 321} Patterns in Parking Functions Lara Pudwell 22 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(231, 312, 321) = ⌊n 2 ⌋ X k=0 1 n + 1 2n −k n + k n + k k (OEIS A001002) (number of dissections of a convex (n + 2)-gon into triangles and quadrilaterals) Avoiding {231, 312, 321} Patterns in Parking Functions Lara Pudwell 22 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} Patterns in Parking Functions Lara Pudwell 23 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} Patterns in Parking Functions Lara Pudwell 23 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} {5} , {4} , {6} {1, 3} , {2} , ∅ Patterns in Parking Functions Lara Pudwell 23 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} {5} , {4} , {6} {1, 3} , {2} , ∅ {1, 3} , {2, 5} , {4} , {6} , ∅, {7} , ∅ Patterns in Parking Functions Lara Pudwell 23 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Larger example {n −1} , {n} {n −1, n} , ∅ {n −1} , ∅, {n} {5} , {4} , {6} {1, 3} , {2} , ∅ {1, 3} , {2, 5} , {4} , {6} , ∅, {7} , ∅ {1, 3} , {2, 5} , {4} , {6} , ∅, {7, 8} , ∅, ∅ Patterns in Parking Functions Lara Pudwell 23 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Theorem pfn(231, 321) = 3n n 2n + 1 (OEIS A001764) Sn(231, 321) 3n n 2n + 1 counts ternary trees non-crossing trees Patterns in Parking Functions Lara Pudwell 24 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Strategy for pfn(231, 321) 1 bijection between Dyck paths and rooted ordered trees 2 bijection between parking functions and non-crossing trees via...
▶labeling Dyck paths ▶arranging tree vertices on circle Patterns in Parking Functions Lara Pudwell 25 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Strategy for pfn(231, 321) 1 bijection between Dyck paths and rooted ordered trees 2 bijection between parking functions and non-crossing trees via...
▶labeling Dyck paths ▶arranging tree vertices on circle Patterns in Parking Functions Lara Pudwell 25 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Reframing the Dyck path/tree bijection a b c d e f g Patterns in Parking Functions Lara Pudwell 26 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 ?
Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 2 ?
1 ?
2 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 ?
2 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 3 2 4 5 ?
1 4 2 3 5 ?
1 5 2 3 4 ?
1 ?
2 3 4 5 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 5 2 3 4 ?
Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 5 2 3 4 6 ?
1 5 2 3 4 ?
6 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 5 2 3 4 ?
6 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
1 5 2 3 4 7 6 Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Labelling the Dyck path to avoid {231, 321} Characterization of {231, 321}-avoiding permutations The digit d must be either first or second among the digits {d, d + 1, . . . , n}.
corresponds to 2 · 4 · 2 = 16 different {231, 321}-avoiding parking functions.
a b c d e f g Patterns in Parking Functions Lara Pudwell 27 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up corresponds to 2 · 3 = 6 different {231, 321}-avoiding parking functions.
a b c d e Patterns in Parking Functions Lara Pudwell 28 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up 1 2 3 4 5 1 2 4 3 5 1 2 5 3 4 1 3 2 4 5 1 4 2 3 5 1 5 2 3 4 Patterns in Parking Functions Lara Pudwell 29 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e Patterns in Parking Functions Lara Pudwell 30 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up b a c d f e g 1 2 4 3 5 7 6 {1, ?} a is left of 1 subtree, so ? is replaced with smallest remaining number.
{1, 2}, {?} c is left of 2 subtrees, so ? is replaced with 2nd smallest remaining number.
{1, 2}, {4}, {3, 5, ?} e is left of 0 subtrees, so ? remains.
{1, 2}, {4}, {3, 5, ?}, {6} Patterns in Parking Functions Lara Pudwell 31 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Recap pfn(132, 213, 312) = pfn(213, 231, 312) = Cn+1 −Cn pfn(231, 312, 321) = P⌊n 2 ⌋ k=0 1 n+1 2n−k n+k n+k k pfn(231, 321) = 3n n 2n + 1 Patterns in Parking Functions Lara Pudwell 32 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Recap pfn(132, 213, 312) = pfn(213, 231, 312) = Cn+1 −Cn pfn(231, 312, 321) = P⌊n 2 ⌋ k=0 1 n+1 2n−k n+k n+k k pfn(231, 321) = 3n n 2n + 1 Forthcoming: results for avoiding any set of 2 or more patterns in S3 Patterns in Parking Functions Lara Pudwell 32 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up Pattern P pfn(P), 1 ≤n ≤6 OEIS 123 1, 3, 11, 48, 232, 1207 new (Remmel & Qiu) 132 1, 3, 13, 69, 417, 2759 A243688 231 213 1, 3, 14, 81, 533, 3822 new 312 321 1, 3, 15, 97, 728, 6024 new “Number of Sylvester classes of 1-multiparking functions of length n.” Patterns in Parking Functions Lara Pudwell 33 / 34 Permutations Parking Functions Recursion Polygons Trees Wrapping up For further reading...
A. Adeniran and L. Pudwell, Pattern Avoidance in Parking Functions, arXiv:2209.04068 A.M. Garsia and M. Haiman, A Remarkable q, t-Catalan Sequence and q-Lagrange Inversion, J. Algebraic Combin. 5 (1996), 191–244.
V. Jel´ ınek and T. Mansour, Wilf-equivalence on k-ary words, compositions, and parking functions, Electron. J. Combin. 16 (2009), #R58, 9pp.
J. Remmel and D. Qiu, Patterns in ordered set partitions and parking functions, Permutation Patterns 2016 (slides), available electronically at
Richard Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 2001.
The On-Line Encyclopedia of Integer Sequences at oeis.org.
Thanks for listening!
slides at faculty.valpo.edu/lpudwell email: Lara.Pudwell@valpo.edu Patterns in Parking Functions Lara Pudwell 34 / 34 |
4322 | https://math.stackexchange.com/questions/4051411/number-of-functions-with-a-fixed-number-of-elements-in-the-range | combinatorics - Number of functions with a fixed number of elements in the range - Mathematics Stack Exchange
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Number of functions with a fixed number of elements in the range
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Find the number of functions f:A⟶B f:A⟶B, such that the range contains exactly 3 3 elements. Given that n(A)=4 n(A)=4 and n(B)=5 n(B)=5
Here's what I tried. Range has to have 3 3 elements. So the way to choose 3 3 elements out of 5 5 is (5 3)=10(5 3)=10
Now, I faced a problem while trying to order the choices. What I tried to do was, use the stars and bars method to find the solutions to
x 1+x 2+x 3=4 x 1+x 2+x 3=4
where,
x 1=x 1= Number of elements in the domain mapped to 1st element in range
x 2=x 2= Number of elements in the domain mapped to 2nd element in range
x 3=x 3= Number of elements in the domain mapped to 3rd element in range
And the Answer would have been 10×10×(The number of solutions obtained for the above equation)
But the answer is said to be 360 360 and my method doesn't work out, can someone point out where I went wrong?
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asked Mar 6, 2021 at 15:15
Techie5879Techie5879
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Your approach is correct; you forgot that the three elements in the range can be arranged in 3!3! ways as first, second, third.
Hence answer should be
(5 3)×3!×(4 2)=360(5 3)×3!×(4 2)=360
Note that you're not looking for number of solutions to x 1+x 2+x 3=4 x 1+x 2+x 3=4. Rather the positive integral solutions of this equation : (2,1,1)(2,1,1) and its permutations; indicate the type of mapping. (2,1,1)(2,1,1) means exactly one element in the range maps to 2 2 elements of the domain. Hence (4 2)(4 2) (onto) functions on any range.
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edited Mar 6, 2021 at 15:41
answered Mar 6, 2021 at 15:25
cosmo5cosmo5
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Yes I forgot to write positive integral solutions. In the equation x 1+x 2+x 3=4 x 1+x 2+x 3=4, we have x 1≥1 x 1≥1, x 2≥1 x 2≥1, x 3≥1 x 3≥1 Correct?Techie5879 –Techie5879 2021-03-06 15:44:53 +00:00 Commented Mar 6, 2021 at 15:44
Yes. But the point is we don't have to count the solutions. Please read the last paragraph.cosmo5 –cosmo5 2021-03-06 15:46:15 +00:00 Commented Mar 6, 2021 at 15:46
Yes I read it. Your (4 2)(4 2) comes as the number of positive integral solutions right? But I think thats the number of non-negative solutions, not positive, is it not?Techie5879 –Techie5879 2021-03-06 15:50:40 +00:00 Commented Mar 6, 2021 at 15:50
@Techie We understand that all mappings from A (say {1,2,3,4}) to B (say {a,b,c}) can be done in only one way namely, two elements in A map to single element in B while other two map to one each e.g., 1→a 1→a, 2→b 2→b, 3→c 3→c, 4→a 4→a. So number of maps is number of ways we can choose two elements of A to map to one element in B. ie, (4 2)(4 2).cosmo5 –cosmo5 2021-03-06 15:57:05 +00:00 Commented Mar 6, 2021 at 15:57
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As you said, we first choose any three elements, which can be done in (5 3)(5 3) ways.
Now, you need to apply inclusion-exclusion principle, for mapping a set of 4 4 elements to a set of 3 3 elements. Basically, you only need to count the number of onto functions from a set of 4 4 elements to a set of 3 3 elements, which can be given by
∑k=0 3(−1)k(3 k)(3−k)4∑k=0 3(−1)k(3 k)(3−k)4
Which evaluates to 36 36, now multiplying this by 10 10 gives 360 360.
For calculating the number of onto functions from a set of m m elements to a set of n n elements, you may see this nice post.
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answered Mar 6, 2021 at 15:26
V.GV.G
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Why do we need the principle of inclusion exclusion here?Techie5879 –Techie5879 2021-03-06 15:29:09 +00:00 Commented Mar 6, 2021 at 15:29
Because, you have first fixed the three elements in the range. Now, you need to map all the four elements in the first set to all the three elements in the second set. Think about it, if you just said the number of ways would be 3 4 3 4, would it be correct? No, because this includes the case when a particular element in the second set (from the choosen 3 3) wasn't mapped. @Techie5879 V.G –V.G 2021-03-06 15:31:06 +00:00 Commented Mar 6, 2021 at 15:31
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The solutions to the equation (with x 1,x 2,x 3≥1 x 1,x 2,x 3≥1) are (2,1,1)(2,1,1), (1,2,1)(1,2,1), (1,1,2)(1,1,2), i.e. exactly one range element has two preimages.
choose range: (5 3)(5 3) choices
choose range element that has 2 2 preimages: 3 3 choices
choose two preimages for that element: (4 2)(4 2) choices
choose one preimage for second range element: (2 1)(2 1) choices
(5 3)⋅3⋅(4 2)⋅(2 1)=360(5 3)⋅3⋅(4 2)⋅(2 1)=360
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edited Mar 6, 2021 at 15:35
answered Mar 6, 2021 at 15:29
hgmathhgmath
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You cannot use stars and bars here, because the objects to be placed are assumed there to be indistinguishable. You should use the Stirling number of the second kind instead:
N=(5 3){4 3}3!=360,N=(5 3){4 3}3!=360,
where the factor 3!3! accounts for the fact that our "boxes" are also distinguishable.
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edited Mar 6, 2021 at 15:40
answered Mar 6, 2021 at 15:35
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4323 | https://www.youtube.com/watch?v=kZEQj4OGMTA | Simplifying Exponential Expressions | Exponents and Scientific Notation | College Algebra
Minute Math
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A math video lesson on the Simplifying Exponential Expressions. This lesson is based on the open education resource College Algebra by OpenStax. This lesson covers Simplifying Exponential Expressions which is under the section Exponents and Scientific Notation in the College Algebra textbook. #oer #exponentialexpressions #collegealgebra
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Transcript:
hi I'm Shaun Gannon and this is minute math and today we're learning about simplifying exponential expression minute math when you need healthy newsman head back okay we've been practicing this so let's go through some examples and see what we can to help solve these or learn to solve these or simplify these so let's get busy first one let's say we have one zoom in a little here we have 6 M Squared and to the negative 1 power all to the third power okay well my first step here is what I want to do is I'm going to multiply or really distribute this exponent use the power of a product rule to all parts so we have 6 to the third power times M squared to the third power and n to the negative 1 to the third power from there we're going to use the power rule all right and we're going to kind of multiply some these things out or we can okay so 6 to the third is just well 6 to the third M to the 2 to the third power multiply 2 times 3 which is 6 and n to the negative 1 and then take the negative 1 times the 3 and we have a negative 3 and 2 negative 3 from there we can simplify a little bit more okay remember or number 6 to the third power is 216 M to the sixth really can't simplify any more but we have n to the negative third power bring that bad boy down to the denominator and we have N to the positive third power and then we have our answer not too bad let's get another one let's go with B okay so we have 17 to the fifth power times the 17 to the negative fourth power times 17 to the negative third power okay this one's actually not too bad notice we have the same base across the board okay since we've seen base being multiplied we just really have to and have the exponents so he has 17 ^ 5 plus and negative 4 plus a negative 3 well what is that simplified it be well 5 plus a negative 4 is a 1 plus a negative 3 is a negative 2 so we have 17 to negative 2 power which is 1 over 17 squared don't forget that and 17 squared is 1 over or this 289 so if 289 and there we have it one over 289 let's go with another one can fit in here using these U to the negative 1 V over V to the negative 1 power all squared okay we can distribute that square to the numerator and the denominator so we have the U to the negative 1 V squared over V to the negative 1 all squared okay well let's distribute that squared by multiplying it to each exponent so negative 1 times 2 is the negative 2 so we have U to the negative 2 and V to the first power 1 times 2 is 2 maybe square all over so now that our same thing negative 1 times 2 negative 2 so in V to the negative 2 power we don't really like negative exponents so if it's the numerator bring to the denominator make a positive and vice versa so the U squared is going to go down to the denominator and be a positive you or U to the negative 2 is positive u squared in the denominator V squared comes up so we have V squared times a V squared here we add our exponents right in multiplication 2 plus 2 is a 4 and so we have V to the positive 4 power over you square and there's our answer all right let's keep going let's go with break it down here a little bit on D D we have negative 2 a to the third B to the negative 1 times 5 a to negative 2 B to the second power ok so let's simplify this well sentence all being multiplied we can multiplication really across the board right every single one of these little bits here is being multiplied let's rearrange and put our like terms next to each other negative 2 times 5 right times the A's a to the third times a to the negative 2 let's put a little dots under ones that we've dealt with so don't forget times B to the negative 1 and B Square and I think I've gotten everything there okay so let's simplify that not too bad I combine our like terms negative 2 times 5 is negative 10 a to the third times a to the negative 2 remember multiplication add the exponents 3 plus a negative 2 positive 1 so we have a to the positive 1 power and we don't really need to write that positive 1 power but you can I guess and that same thing same base with it B add the exponents negative 1 plus 2 is the positive 1 so if B to the first power and that's a negative 10 times a times B let's keep this train going we got here okay e x squared square root of 2 also the fourth power times x squared square root of 2 all to the negative 4 power okay so this one a little tricky okay one thing we notice is that we have the same base here write x squared square root of 2 so what I'm going to do is remember a rule we add the exponents for multiplying that so we have x squared square root of 2 all to the exponent of 4 plus a negative 4 right here well 4 plus a negative 4 is 0 x squared square root of 2 all to 0 power and then this is pretty easy right anything to the 0 power is just 1 and your answer just came out to be easy one what it saves you time doing it this way because if you had try simplify all that and done it yes you got to one but it it take you a while all right and lastly let's get with F here 3w squared o to the fifth power over 6w negative 2 power square okay well that's 5th power here ok we remember can add to each one of our bases with the exponents so we have 3 to the fifth power times W squared to the fifth power over same thing with the denominator 6 squared times W or W to the negative 2 power to the second power remember that little X birthday over to let's simplify this where we can all right 3 to the fifth power is 243 and then we add the multiply the exponents 2 times 5 is 10 so we have W to the tenth power over 6 squared is 36 multiply the exponents negative 2 times 2 is a negative 4 so we have W to the negative 4 okay now we have to bring the numerator to the denominator to the numerator with this negative exponent so let's do that 243 W to the 10 times W to the positive fourth power over 36 okay and now let's simplify where we can 243 over 36 that does simplify to be 27 over 4 and then 10 we multiplication with the same base right we have the exponents 10 plus 4 is 14 we have W to the 14th power and there's our answer 27w to the 14th power over for so now we have our solutions we got went through a lot of examples but now you should know how to simplify exponential expressions hit the like button and comment down below if you learned something and subscribe this helps us make more videos thanks for watching mad minute math when you need help use minute math in it math minute half when you need help you use minute math minute math tutor calm |
4324 | https://artofproblemsolving.com/community/c776104h1742531_some_lemmas_on_cyclic_quadrilaterals?srsltid=AfmBOop77UtE4bCXXtXoN5XnsGF8bn6kQSpJcPh6vYXcN5GCgC13OSOm | Magic of Hogwarts : Some Lemmas on Cyclic Quadrilaterals
Community » Blogs » Magic of Hogwarts » Some Lemmas on Cyclic Quadrilaterals
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Magic of Hogwarts =================
Some Lemmas on Cyclic Quadrilaterals
by AlastorMoody, Nov 23, 2018, 7:33 PM
We will first check out some basic properties of a cyclic quadrilateral,
If points lie on circle , then is a cyclic quadrilateral
For proving a quadrilateral cyclic, proving any of the following will prove the desired quadrilateral cyclic,
Criteria 1#
In Quadrilateral if,
1)
2)
Then it is a cyclic quadrilateral
Criteria 2#
In Quadrilateral if,
1)
2)
3)
4)
Then it is a cyclic quadrilateral
Lemma 1:
If side , then, is to in , or
proof
Trivial. Angle chasing
Lemma 2:
Power of Point Theorem:If side , then,
proof
Using the ratio of sides of the similar triangle stated in Lemma 1, we can prove Lemma 2 or Power of Point Theorem
Some Advantages:
Lemma 3:
Power of point (in disguise):If is tangent to at , then,
proof
See what happens when
Lemma 4:
is such a cyclic quadrilateral, that, is a diameter, and if is foot of the altitude from to ,then,
proof(1)
Since, is cyclic
proof(2)
Easy to spot isogonal lines and and Then the result follows
Lemma 5:
If a cyclic quadrilateral is a trapezium, it has to be an isosceles trapezium
proof
Let the Quadrilateral be , where ,Let ,
For not equal to , this is not possible
Lemma 6:
If is the orthocenter of , then is the incenter of the orthic triangle of
proof
There are six cyclic quadrilaterals, they are anyways sufficient to prove the equal angles
Lemma 7:
If is the orthocenter of , then reflection of over any side of lies on circumcircle of
proof
Let be the reflection of over side and let the foot of perpendicular from , and be , and , Let , Therefore,
Lemma 8:
If be the reflection of orthocenter of over mid-point of side , then, lies on and is the diameter
proof
Coming soon!!
Lemma 9:
Let be inscribed in a circle with center and Let line be tangent at , and let two points on either side of be and that lie on line , such, that is closer to than and is closer to than , Then,
and
proof
Let and By simple angle chasgin, we have,
Hence, which implies the desired result
Lemma 10:
Let be inscribed in circle , Let angle bisector , such , Prove,
If is the perpendicular bisector of , then lies on
proof
simple angle chasing for the first part and then phantom points for the second.......
Lemma 11:
Let with on the midpoint of and is tangent to ,such that , then is tangent to .
proof
Since, , Let's assume that does intersect at
Therefore, by Power of point theorem, implies the desired result!!
Problems:
1# JBMO ShortList 2015 G1 Around the triangle the circle is circumscribed, and at the vertex tangent to this circle is drawn. The line , which is parallel to this tangent intersects the lines and at the points and , respectively. Prove that the points belong to the same circle.
2# (own ) The point is outside the circle . Two tangent lines, passing from the point touch the circle at the points and . The media n intersects the circle at the point , Prove,
3# JBMO ShortList 2015 G2 The point is outside the circle . Two tangent lines, passing from the point touch the circle at the points and . The media n intersects the circle at the point and the line intersects again the circle at the point . Prove that the lines and are parallel.
4# Indian RMO 1992 P4 Let be a quadrilateral inscribed in circle , such that and , If is radius of , then prove,
5# USAMO 2003 P4 Let be a triangle. A circle passing through and intersects segments and at and , respectively. Lines and intersect at , while lines and intersect at . Prove that if and only if .
6# ISL 2010 G1 Let be an acute triangle with the feet of the altitudes lying on respectively. One of the intersection points of the line and the circumcircle is The lines and meet at point Prove that
7# Evan Chen's EGMO Let be a cyclic quadrilateral, Let be incenter of and Let be the incenter of , prove, that is also cyclic quadrilateral
8# USAJMO 2011 P5 Points lie on a circle and point lies outside the circle. The given points are such that
(i) lines and are tangent to ,
(ii) are collinear, and
(iii) .
Prove that bisects .
9# Source: Unknown A circle touches the sides of a quadrilateral at respectively. intersects at . The line passing through and perpendicular to intersects at respectively. Prove that if then .
10# Indian RMO 1999 P3 Let be a square and points on sides respectively such that . If is the midpoint of show that where are points of intersection of with the lines .
11# Indian RMO 2000 P5 The internal bisector of angle in a triangle with meets the circumcircle of the triangle in . Joi n to the center of the circle and suppose that meets in , possibly when extended. Given that is perpendicular to , show that is parallel to .
12# Source:Unknown Let is , such . Let is point on such that . The circumcircle of cut at , show that, .
13# JBMO Shortlist 2011 G2 Let and be the altitudes of . A line passing through and parallel to i ntersects the line a t the point . If is the orthocenter of , find the angle .
14# IMO 1990 P1/ ISL 1990 P11 Chords and of a circle intersect at a point inside the circle. Let be an interior point of the segment . The tangent line at to the circle through ,, and intersects the lines and at and , respectively. If
find in terms of .
15# Canada MO 1999 P3 Let be a point inside the circle . Consider the set of chords of that contains . Prove that their midpoints all lie on a circle.
16# IMO 2006 P1 Let be triangle with incenter . A point in the interior of the triangle satisfies Show that , and that equality holds if and only if .
17# USAMO 1999 P6 Let be an isosceles trapezoid with . The inscribed circle of triangle meets at . Let be a point on the (internal) angle bisector of such that . Let the circumscribed circle of triangle meet line at and . Prove that the triangle is isosceles.
I guess , these problems are enough for warm-up/practice, others can be added in the comments section!
This post has been edited 45 times. Last edited by AlastorMoody, Jan 15, 2019, 5:45 PM
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Don't forget IMO 2006/1
by Kagebaka, Dec 23, 2018, 8:25 PM
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4325 | https://www.geeksforgeeks.org/python/python-ordered-set/ | Ordered Set - Python
Last Updated :
23 Jul, 2025
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An ordered set is a collection that combines the properties of both a set and a list. Like a set, it only keeps unique elements, meaning no duplicates are allowed. Like a list, it keeps the elements in the order they were added.
In Python, the built-in set does not maintain the order of elements, while a list allows duplicates. An ordered set solves this problem by keeping the uniqueness of a set while preserving the order of insertion.
input_dataSet = {"Prince", "Aditya", "Praveer", "Shiv"}
Output in case of unordered set: {"Aditya, "Prince", "Shiv", "Praveer"}, It can be random position on your side
Output in case of ordered set: {"Prince", "Aditya", "Praveer", "Shiv"}
Explanation: As you know in Python if you print this set more one time than, every time you will getting the random position of the items for the same dataset. But in case of ordered set you will getting the same dataset every time in same order you had inserted items.
Using the Ordered Set Module (or class)
In default, you have an unordered set in Python but for creating the ordered set you will have to install the module named ordered-set by pip package installer as mentioned below:
How to Install the ordered set module
By using the pip package installer download the ordered-set module as mentioned below
pip install ordered_set
Syntax of ordered Set:
orderedSet(Listname)
Now, for more clarification let's iterate the ordered set because the set cannot be iterated as mentioned below:
Python
````
from ordered_set import OrderedSet
createOrderedSet = OrderedSet(
['GFG', 'is', 'an', 'Excellent', 'Excellent', 'platform'])
print(createOrderedSet)
for item in createOrderedSet:
print(item, end=" ")
````
from ordered_set import OrderedSet from ordered_set import OrderedSet
createOrderedSet = OrderedSet(createOrderedSet = OrderedSet
['GFG', 'is', 'an', 'Excellent', 'Excellent', 'platform']) 'GFG' 'is' 'an' 'Excellent' 'Excellent' 'platform'
print(createOrderedSet) print createOrderedSet
for item in createOrderedSet: for item in createOrderedSet
print(item, end=" ") print item end = " "
Output:
OrderedSet(['GFG', 'is', 'an', 'Excellent', 'platform']) GFG is an Excellent platform
Table of Content
Ordered set using the dictionary data structure
Ordered set using the list data structure
Using the Dictionary Data Structure
We can use the dictionary data structure to create the ordered set because the dictionary is itself the ordered data structure in which we will use set items as keys because keys are unique in the dictionary and at the place of value we can create the empty string. Let's take a look at implementation as explained below:
Python
````
d = {"Prince": "", "Aditya": "",
"Praveer": "", "Prince": "", "Shiv": ""}
print(d)
for key in d.keys():
print(key, end=" ")
````
d = {"Prince": "", "Aditya": "", d = "Prince" "" "Aditya" ""
"Praveer": "", "Prince": "", "Shiv": ""} "Praveer" "" "Prince" "" "Shiv" ""
print(d) print d
for key in d.keys(): for key in d keys
print(key, end=" ") print key end = " "
Output
{'Prince': '', 'Aditya': '', 'Praveer': '', 'Shiv': ''}
Prince Aditya Praveer Shiv
Using the list Data Structure
A list is a built-in data structure that stores an ordered collection of items. However, unlike a set, lists can contain duplicate values. To create an ordered set using a list, you need to manually remove duplicates while keeping the order of insertion intact. Let's take a look at implementation as explained below:
Python
````
def removeduplicate(data):
countdict = {}
for element in data:
if element in countdict.keys():
countdict[element] += 1
else:
countdict[element] = 1
data.clear()
for key in countdict.keys():
data.append(key)
dataItem = ["Prince", "Aditya", "Praveer", "Prince", "Aditya", "Shiv"]
removeduplicate(dataItem)
print(dataItem)
````
def removeduplicate(data): def removeduplicate data
countdict = {} countdict =
for element in data: for element in data
if element in countdict.keys(): if element in countdict keys
countdict[element] += 1 countdict element += 1
else: else
countdict[element] = 1 countdict element = 1
data.clear() data clear
for key in countdict.keys(): for key in countdict keys
data.append(key) data append key
dataItem = ["Prince", "Aditya", "Praveer", "Prince", "Aditya", "Shiv"] dataItem = "Prince" "Aditya" "Praveer" "Prince" "Aditya" "Shiv"
removeduplicate(dataItem) removeduplicate dataItem
print(dataItem) print dataItem
Output
```
['Prince', 'Aditya', 'Praveer', 'Shiv']
```
P
princekumaras
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4326 | https://www.scribd.com/document/317102348/field-and-wave-electromagnetics-cheng-pdf | Field and Wave Electromagnetics Cheng PDF | PDF
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4327 | https://proofwiki.org/wiki/Sum_of_Even_Integers_is_Even | Sum of Even Integers is Even
From ProofWiki
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1 Theorem
2 Proof 1
2.1 Basis for the Induction
2.2 Induction Hypothesis
2.3 Induction Step
3 Proof 2
4 Also see
Theorem
The sum of any finite number of even integers is itself even.
Proof 1
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
: The sum of $n$ even integers is an even integer.
$\map P 1$ is trivially true, as this just says:
: The sum of $1$ even integer is an even integer.
The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even.
So $\map P 0$ is also true.
Basis for the Induction
$\map P 2$ is the case:
: The sum of any two even integers is itself even.
Consider two even integers $x$ and $y$.
Since they are even, they can be written as $x = 2 a$ and $y = 2 b$ respectively for integers $a$ and $b$.
Therefore, the sum is:
: $x + y = 2 a + 2 b = 2 \paren {a + b}$
Thus $x + y$ has $2$ as a divisor and is even by definition.
This is our basis for the induction.
$\Box$
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
: The sum of any $k$ even integers is itself even.
Then we need to show:
: The sum of any $k+1$ even integers is itself even.
Induction Step
This is our induction step:
Consider the sum of any $k + 1$ even integers.
This is the sum of:
: $k$ even integers (which is even by the induction hypothesis)
and:
: another even integer.
That is, it is the sum of two even integers.
By the basis for the induction, this is also even.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
That is:
: The sum of any finite number of even integers is itself even.
$\blacksquare$
Proof 2
Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ even numbers.
By definition of even number, this can be expressed as:
: $S = \set {2 s_1, 2 s_2, \ldots, 2 s_n}$
where:
: $\forall k \in \closedint 1 n: r_k = 2 s_k$
Then:
| | | | | | | | | | | | | | | |
--- --- --- --- --- --- ---
| | | | | | (\ds \sum_{k \mathop = 1}^n r_k) | (=) | | | | (\ds \sum_{k \mathop = 1}^n 2 s_k) | | | | |
| | | | | | (\ds ) | (=) | | | | (\ds 2 \sum_{k \mathop = 1}^n s_k) | | | | |
Thus, by definition, $\ds \sum_{k \mathop = 1}^n r_k$ is even.
$\blacksquare$
Also see
Sum of Even Number of Odd Integers is Even
Retrieved from "
Categories:
Proven Results
Even Integers
Euclidean Number Theory
Sum of Even Integers is Even
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Book Title
Combinatorial Geometry
ISBN
9780471588900
Subject Area
Mathematics
Publication Name
Combinatorial Geometry
Publisher
Wiley & Sons, Incorporated, John
Item Length
9.4 in
Subject
Geometry / General
Publication Year
1995
Series
Wiley Series in Discrete Mathematics and Optimization Ser.
Type
Textbook
Format
Hardcover
Language
English
Item Height
1.2 in
Author
János Pach, Pankaj K. Agarwal
Item Weight
26.6 Oz
Item Width
6.4 in
Number of Pages
384 Pages
Category
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About this product
Product Identifiers
Publisher
Wiley & Sons, Incorporated, John
ISBN-10
0471588903
ISBN-13
9780471588900
eBay Product ID (ePID)
834559
Product Key Features
Number of Pages
384 Pages
Publication Name
Combinatorial Geometry
Language
English
Publication Year
1995
Subject
Geometry / General
Type
Textbook
Author
János Pach, Pankaj K. Agarwal
Subject Area
Mathematics
Series
Wiley Series in Discrete Mathematics and Optimization Ser.
Format
Hardcover
Dimensions
Item Height
1.2 in
Item Weight
26.6 Oz
Item Length
9.4 in
Item Width
6.4 in
Additional Product Features
Edition Number
1
Intended Audience
Scholarly & Professional
LCCN
94-048203
Dewey Edition
20
Series Volume Number
37
Illustrated
Yes
Dewey Decimal
516/.13
Table Of Content
ARRANGEMENTS OF CONVEX SETS. Geometry of Numbers. Approximation of a Convex Set by Polygons. Packing and Covering with Congruent Convex Discs. Lattice Packing and Lattice Covering. The Method of Cell Decomposition. Methods of Blichfeldt and Rogers. Efficient Random Arrangements. Circle Packings and Planar Graphs. ARRANGEMENTS OF POINTS AND LINES. Extremal Graph Theory. Repeated Distances in Space. Arrangement of Lines. Applications of the Bounds on Incidences. More on Repeated Distances. Geometric Graphs. Epsilon Nets and Transversals of Hypergraphs. Geometric Discrepancy. Hints to Exercises. Bibliography. Indexes.
Synopsis
A complete, self-contained introduction to a powerful and resurging mathematical discipline Combinatorial Geometry presents and explains with complete proofs some of the most important results and methods of this relatively young mathematical discipline, started by Minkowski, Fejes Tth, Rogers, and Erd's. Nearly half the results presented in this book were discovered over the past twenty years, and most have never before appeared in any monograph. Combinatorial Geometry will be of particular interest to mathematicians, computer scientists, physicists, and materials scientists interested in computational geometry, robotics, scene analysis, and computer-aided design. It is also a superb textbook, complete with end-of-chapter problems and hints to their solutions that help students clarify their understanding and test their mastery of the material. Topics covered include: Geometric number theory Packing and covering with congruent convex disks Extremal graph and hypergraph theory Distribution of distances among finitely many points Epsilon-nets and Vapnik Chervonenkis dimension Geometric graph theory Geometric discrepancy theory And much more, A complete, self-contained introduction to a powerful and resurgingmathematical discipline . Combinatorial Geometry presents andexplains with complete proofs some of the most important resultsand methods of this relatively young mathematical discipline, started by Minkowski, Fejes Toth, Rogers, and Erd s. Nearly halfthe results presented in this book were discovered over the pasttwenty years, and most have never before appeared in any monograph.Combinatorial Geometry will be of particular interest tomathematicians, computer scientists, physicists, and materialsscientists interested in computational geometry, robotics, sceneanalysis, and computer-aided design. It is also a superb textbook, complete with end-of-chapter problems and hints to their solutionsthat help students clarify their understanding and test theirmastery of the material. Topics covered include: Geometric number theory Packing and covering with congruent convex disks Extremal graph and hypergraph theory Distribution of distances among finitely many points Epsilon-nets and Vapnik--Chervonenkis dimension Geometric graph theory Geometric discrepancy theory And much more, A complete, self-contained introduction to a powerful and resurging mathematical discipline Combinatorial Geometry presents and explains with complete proofs some of the most important results and methods of this relatively young mathematical discipline, started by Minkowski, Fejes Tóth, Rogers, and Erd's. Nearly half the results presented in this book were discovered over the past twenty years, and most have never before appeared in any monograph. Combinatorial Geometry will be of particular interest to mathematicians, computer scientists, physicists, and materials scientists interested in computational geometry, robotics, scene analysis, and computer-aided design. It is also a superb textbook, complete with end-of-chapter problems and hints to their solutions that help students clarify their understanding and test their mastery of the material. Topics covered include: Geometric number theory Packing and covering with congruent convex disks Extremal graph and hypergraph theory Distribution of distances among finitely many points Epsilon-nets and Vapnik--Chervonenkis dimension Geometric graph theory Geometric discrepancy theory And much more, A complete, self-contained introduction to a powerful and resurging mathematical discipline. Combinatorial Geometry presents and explains with complete proofs some of the most important results and methods of this relatively young mathematical discipline, started by Minkowski, Fejes Toth, Rogers, and Erd's.
LC Classification Number
QA167.P33 1995
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Limits
1 chapter
Limits are a fundamental concept in mathematics that describe the value a function or sequence approaches as its variable approaches a certain point. They are used to understand the behavior of functions at extreme values or near points of discontinuity.
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Limit of a Function
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What is a Limit?
INTRODUCTION AND DEFINITION OF LIMITS
Limits are a basic concept in mathematics that describe the behavior of functions as their input values approach a specific point. This concept is crucial for understanding infinitesimally small changes and forms the basis for differential and integral calculus. Limits allow students to formalize and analyze concepts such as infinity, continuity, and derivatives.
BASIC DEFINITION: The limit of a function at a point describes the value that the function approaches as its argument (or input value) approaches that specific point. Formally, this is expressed as the limit of the function f(x) as x approaches the value c.
UNDERSTANDING LIMITS
Limits are used to analyze various mathematical situations:
CONTINUITY OF FUNCTIONS: The concept of a limit allows for the formal definition of continuous functions, meaning that the function has no jumps, holes, or breaks in its course.
DERIVATIVES AND TANGENTS: The limit of the difference quotient of a function, as the interval between two points approaches zero, describes the concept of a derivative, which represents the slope of the tangent to the graph of the function.
BEHAVIOR AT INFINITY: The limit of a function as x approaches infinity describes the behavior of the function at the "edge" of its domain or when its values increase significantly.
METHODS FOR FINDING LIMITS
There are various methods for finding limits, including:
DIRECT SUBSTITUTION: If the function is continuous at point c, then the limit of the function is equal to the value of the function at that point.
FACTORIZATION AND CANCELLATION: For rational functions where division by 0 occurs, factorization and cancellation can help reveal the limit value.
Understanding and using limits is of fundamental importance in many areas of mathematics and applied sciences:
DIFFERENTIAL AND INTEGRAL CALCULUS: Limits are the foundation for calculating derivatives and integrals, which are central to the analysis and modeling of dynamic systems.
ANALYSIS AND TOPOLOGY: Limits allow for the study of continuity, convergence, and other properties of functions and sequences in more abstract mathematical contexts.
CONCLUSION
Limits are a key concept in mathematics that enable precise analysis of changes and the behavior of functions. Their ubiquity in differential and integral calculus, as well as more broadly in mathematical analysis, demonstrates their central place in the foundations of mathematics. Understanding limits is essential for anyone involved in mathematical analysis, engineering, and other sciences that focus on a quantitative understanding of the world.
Limits in Calculus - Definition and Finding Methods |
4330 | https://www.cdc.gov/std/treatment-guidelines/hpv.htm | Human Papillomavirus (HPV) Infection - STI Treatment Guidelines
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Sexually Transmitted Infections Treatment Guidelines, 2021
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Human Papillomavirus (HPV) Infection
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Anogenital Warts
HPV-Associated Cancers and Precancers
Approximately 150 types of HPV have been identified, at least 40 of which infect the genital area (1194). The majority of HPV infections are self-limited and are asymptomatic or unrecognized. Sexually active persons are usually exposed to HPV during their lifetime (838,1195,1196). Oncogenic, high-risk HPV infection (e.g., HPV types 16 and 18) causes the majority of cervical, penile, vulvar, vaginal, anal, and oropharyngeal cancers and precancers (1197), whereas other HPV infection (e.g., HPV types 6 and 11) causes genital warts and recurrent respiratory papillomatosis. Persistent oncogenic HPV infection is the strongest risk factor for development of HPV-attributable precancers and cancers. A substantial proportion of cancers and anogenital warts are attributable to HPV in the United States. An estimated 34,800 new HPV-attributable cancers occurred every year during 2012–2016 (1198). Before HPV vaccines were introduced, approximately 355,000 new cases of anogenital warts occurred every year (1199).
Prevention
HPV Vaccines
Three HPV vaccines are licensed in the United States: Ceravrix, a 2-valent vaccine (2vHPV) that targets HPV types 16 and 18; Gardasil, a 4-valent vaccine (4vHPV) that targets HPV types 6, 11, 16, and 18; and Gardasil 9, a 9-valent vaccine (9vHPV) that targets HPV types 6, 11, 16, 18, 31, 33, 45, 52, and 58. Types 16 and 18 account for 66% of all cervical cancers, whereas the five additional types targeted by the 9-valent vaccine account for 15%. Types 6 and 11 cause >90% of genital warts. Only 9vHPV vaccine is available in the United States.
ACIP recommendations for HPV vaccination ( include the following:
Routine HPV vaccination for all adolescents at age 11 or 12 years.
Administering vaccine starting at age 9 years.
Catch-up vaccination through age 26 years for those not vaccinated previously.
Not using HPV vaccination for all adults aged >26 years. Instead, shared clinical decision-making between a patient and a provider regarding HPV vaccination is recommended for certain adults aged 27–45 years not vaccinated previously.
A 2-dose vaccine schedule (at 0- and 6–12-month intervals) is recommended for persons who initiate vaccination before their 15th birthday.
A 3-dose vaccine schedule (at 0-, 1–2-, and 6-month intervals) for immunocompromised persons regardless of age of initiation.
HPV vaccines are not recommended for use in pregnant women. HPV vaccines can be administered regardless of history of anogenital warts, abnormal Pap test or HPV test, or anogenital precancer. Women who have received HPV vaccine should continue routine cervical cancer screening (see Cervical Cancer). HPV vaccine is available for eligible children and adolescents aged <19 years through the Vaccines for Children (VFC) program (additional information is available at or by calling CDC INFO 800-232-4636). For uninsured persons aged <19 years, patient assistance programs are available from the vaccine manufacturers. Prelicensure and postlicensure safety evaluations have determined that the vaccine is well tolerated. With >120 million doses of HPV vaccines distributed in the United States, robust data demonstrate that HPV vaccines are safe ( Impact-monitoring studies in the United States have demonstrated reductions of genital warts as well as the HPV types contained within the quadrivalent vaccine (1200–1203). Settings that provide STI services should either administer the vaccine to eligible clients within the routine and catch-up age groups through age 26 years who have not started or completed the vaccine series, or link these persons to another facility equipped to provide the vaccine. Clinicians providing services to children, adolescents, and young adults should be knowledgeable about HPV and the vaccine ( HPV vaccination has not been associated with initiation of sexual activity or sexual risk behaviors (1204,1205).
Abstaining from sexual activity is the most reliable method for preventing genital HPV infection. Persons can decrease their chances of infection by practicing consistent and correct condom use and limiting their number of sex partners. Although these interventions might not fully protect against HPV, they can decrease the chances of HPV acquisition and transmission.
Diagnostic Considerations
HPV tests are available for detecting oncogenic types of HPV infection and are used in the context of cervical cancer screening and management or follow-up of abnormal cervical cytology or histology (see Cervical Cancer). These tests should not be used for male partners of women with HPV or women aged <25 years, for diagnosis of genital warts, or as a general STI test.
Application of 3%–5% acetic acid, which might cause affected areas to turn white, has been used by certain providers to detect genital mucosa infected with HPV. The routine use of this procedure to detect mucosal changes attributed to HPV infection is not recommended because the results do not influence clinical management.
Treatment
Treatment is directed to the macroscopic (e.g., genital warts) or pathologic precancerous lesions caused by HPV. Subclinical genital HPV infection typically clears spontaneously; therefore, specific antiviral therapy is not recommended to eradicate HPV infection. Precancerous lesions are detected through cervical cancer screening; HPV-related precancer should be managed on the basis of existing guidance (see Cervical Cancer).
Counseling
Key Messages for Persons with Human Papillomavirus Infection
When counseling persons with anogenital HPV infection, the provider should discuss the following:
Anogenital HPV infection is common. It usually infects the anogenital area but can infect other areas, including the mouth and throat. The majority of sexually active persons get HPV at some time during their lifetime, although most never know it.
Partners tend to share HPV, and it is not possible to determine which partner transmitted the original infection. Having HPV does not mean that a person or his or her partner is having sex outside the relationship.
Persons who acquire HPV usually clear the infection spontaneously, meaning that HPV becomes undetectable with no associated health problems.
If HPV infection persists, genital warts, precancers, and cancers of the cervix, anus, penis, vulva, vagina, head, or neck might develop.
Discussion of tobacco use, and provision of cessation counseling, is important because of its contribution to the progression of precancer and cancer.
The types of HPV that cause genital warts are different from the types that can cause cancer.
Many types of HPV are sexually transmitted through anogenital contact, mainly during vaginal and anal sex. HPV also might be transmitted during oral sex and genital-to-genital contact without penetration. In rare cases, a pregnant woman can transmit HPV to an infant during delivery.
Treatments are available for the conditions caused by HPV but not for the virus itself.
Having HPV does not make it harder for a woman to get pregnant or carry a pregnancy to term. However, certain precancers or cancers that HPV can cause, and the surgical procedures needed to treat them, can affect a woman’s ability to get pregnant or carry a pregnancy to term.
No HPV test can determine which HPV infection will become undetectable and which will persist or progress to disease. However, in certain circumstances, HPV tests can determine whether a woman is at increased risk for cervical cancer. These tests are not for detecting other HPV-related problems, nor are they useful for women aged <25 years or men of any age.
Prevention
Three HPV vaccines can prevent diseases and cancers caused by HPV. The 2vHPV, 4vHPV, and 9vHPV vaccines protect against the majority of cervical cancer cases, although the 4vHPV and 9vHPV vaccines also protect against the majority of genital warts. Only 9vHPV vaccine is available in the United States. HPV vaccines are safe and effective and are recommended routinely for adolescents aged 11–12 years. Catch-up vaccination is also recommended for older adolescents and young adults through age 26 years ( Shared clinical decision-making is recommended regarding HPV vaccination for certain adults aged 27–45 years who are not adequately vaccinated per guidance (
Condoms used consistently and correctly can lower the chances of acquiring and transmitting HPV and developing HPV-related diseases (e.g., genital warts or cervical cancer). However, because HPV can infect areas not covered by a condom, condoms might not fully protect against HPV.
Limiting the number of sex partners can reduce the risk for HPV. However, even persons with only one lifetime sex partner can get HPV.
Abstaining from sexual activity is the most reliable method for preventing genital HPV infection.
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4331 | https://www.youtube.com/watch?v=8wgyYsslim0 | Finding Trig Values Given a Circle with Radius "r"
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This video goes through 1 example of how to find trig values when given a circle with radius "r".
trigonometry #unitcircle #mathematics
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now I am real quickly up here in the corner going to write the definitions for each of the trig functions that you should have gotten from the video that you watched last week all right it's sin Theta = r y over R it was cine Theta isal X over R tangent Theta is equal to y overx and then the other three trig functions which is cosecant Theta which was R over y secant Theta which was given to you as R overx and coent Theta which was X over y all right now you were supposed to watch a video last week which talked about that introduced it all that kind of things all right now what I want to make a connection here is yes you can use those formulas all right regardless of whether you use right triangle definitions or whether you use are definitions that are based on a unit circle but in this case with a different radius all right you're still going to get the same answers and it just depends on how you are going to want to visualize this okay so in a question where they say okay here's a point -125 all right it's on a circle with a radius with something other than r r does not equal one okay so in other words R does not equal one find six trig values now I'm just going to do that they are a little more wordy in your directions but we know what we're mean there we're trying to find all six trick values all right so what this is this is a point it's on a circle but the radius is not R okay so I'm going to attempt to draw that over here okay okay so -125 if I were to plot that point I would go -2 I would go up five okay -12 up five that's putting this this point right here -125 is in that second quadrant all right now it is a circle okay I kind of big on that but it's a circle it's on a circle all right and the radius which would be from here to here all right I don't know what that radius is right now okay now it's not one watch what I do the bow tie triangle that we talked about all right the bow tie triangle in my second quadrant is this so there's my right triangle in that second quadrant all right and from here to here is also the radius from here to here's the radius all right it's halfway across Circle so it doesn't matter where I draw it all right now if this point point is -125 what's that mean that means I started at the origin and I went to the left 12 or -12 and if I use that negative then the signs of my trig function in the second coordinate are going to be correct second second quadrant is going to be correct because this is a -2 I went -12 to get there and then I went up five okay now the only value I don't know is that radius all right if I'm going to use these definitions I need to know that measurement because this would be y over you know this is X this is y so for S it would be five over whatever R is y over R okay or I could also use my right triangle definitions so it doesn't make any difference all right but basically what this boils down to is okay I'm not on a circle with R equals 1 so I have to find the radius that that's got to be my step one okay so step one find R and you're going to do that with just a good old fashioned Pagan theorem okay so -122 + 5^ 2 = R 2 all right 144 again this is basic arithmetic you guys can do this I don't really need to be working all this part out let's see that's 169 okay so then my R is 13 okay so my R is 13 so that value right there this side is 13 okay so now I've got all my sides I could use these defs I can also use my right triangle trig definitions all right now if I'm going to then find each of my trig values let's go back to black all right I need to find sign I need to find cosine I need to find tangent I need to find cosecant I need to find secant and I need to find cose or cotangent sorry okay so let's do right triangle definition right triangle definitions here's my Theta right triangle definition for sign is opposite over hypotenuse right so I could do opposite over hypotenuse which would give me a 5 over 13 okay so I could do it with right triangle trick all right or I could do it with Y over R all right I'll write that here y over R well what's my y value it's five and what's my R value that I found it's 133 so it doesn't make any difference this is using these definitions the red right there is using your right triangle definitions but because I know what quadrant it is all right I'm in this the second quadrant all right sign is positive in the second quadrant okay sign is positive in the second quadrant so I got a positive answer okay all right let's go ahead and do the right triangle definition first okay so cosine is adjacent over hypotenuse so adjacent over hypotenuse which would be a -12 over hypotenuse which is 13 all right so I could do it that way or I could come up here and I could say okay I'm using X over R well my x value is a --12 and my r value is 13 all right so I'm just trying to get you to see that you can do these questions more than one way all right and then if we go back and we look at you know that chart about what's the values all right well cosine in the second quadrant is supposed to be negative and my answer is negative all right the only two that's positive is s and cosecant which doesn't that make sense since these are reciprocal and if this is negative then this one will be negative all right depending on what I come up with here which better be negative it'll be negative over here as well okay so let's use our right triangle definition first tangent is opposite over adjacent so opposite over adjacent opposite is five and adjacent is -12 which makes it 512 put that negative in the top so it's not tacky all right but again could I come up here and use these definitions as well that's y overx my yv value is five my x value is -12 again moving that negative to the top so that it looks nice okay would be the way to go all right and again tangent should be negative in the second quadrant it is all right now for these again you memorize these definitions you memorize your right triangle definitions or you know that these two are reciprocals so this is 135 this is put the negative in the top -3 over 12 and again put the negative in the top-12 over 5 okay so um here let's in our let's do this okay this is right triangle trig okay using those definitions all right and the blue is the quote unit based on a unit circle okay so we're except not a unit circle here but our unit circle are not equal to one definitions okay so you same problem I can work in a multitude of ways all right |
4332 | https://pubmed.ncbi.nlm.nih.gov/2463205/ | Sudan stain of fecal fat: new insight into an old test - PubMed
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. 1989 Feb;96(2 Pt 1):421-7.
doi: 10.1016/0016-5085(89)91566-7.
Sudan stain of fecal fat: new insight into an old test
M R Khouri1,G Huang,Y F Shiau
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1 Department of Medicine, Veterans Administration Medical Center, Philadelphia, Pennsylvania.
PMID: 2463205
DOI: 10.1016/0016-5085(89)91566-7
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Sudan stain of fecal fat: new insight into an old test
M R Khouri et al. Gastroenterology.1989 Feb.
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. 1989 Feb;96(2 Pt 1):421-7.
doi: 10.1016/0016-5085(89)91566-7.
Authors
M R Khouri1,G Huang,Y F Shiau
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1 Department of Medicine, Veterans Administration Medical Center, Philadelphia, Pennsylvania.
PMID: 2463205
DOI: 10.1016/0016-5085(89)91566-7
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The 72-h fecal fat determination is used as the gold standard to document the presence of steatorrhea. Although the Sudan stain for fecal fat is advocated as a sensitive screening test, a quantitative correlation between the 72-h fecal fat quantitation and the fecal Sudan stain is lacking. This study was designed to examine the staining properties of different classes of purified lipids in an experimentally defined artificial matrix, and to elucidate the reasons for the lack of quantitative correlation between these two tests. Our results indicate that the "neutral fat" stain without acidification or heating identifies triglyceride; and at an appropriate pH, the "neutral stain" also identifies fatty acid. The "split fat" stain with acidification and heating identifies both triglyceride and fatty acid. After acidification, fatty acid soaps are converted to the nonionized fatty acid. Thus, fatty acid soaps can be identified indirectly as fat droplets that are stained by the split fat stain. Although cholesterol is stained with Sudan stain after heating, upon cooling, cholesterol forms crystals of anhydrous cholesterol, making its staining pattern distinct. Neither the neutral fat nor the split fat stain can detect phospholipid or cholesteryl ester. The 72-h fecal fat determination is a measure of the total fatty acid content after a specimen is saponified. The resulting fatty acids are derived from a variety of endogenous and exogenous sources, including free fatty acids, soaps of fatty acids, triglycerides, cholesterol esters, and phospholipids. Therefore, the 72-h fecal fat quantitation does not differentiate between the primary sources of the measured fatty acid. It is concluded that the 72-h fecal fat determination is not specific for documenting triglyceride (fat) malabsorption. Until new methods are developed that specifically measure fecal triglyceride and fatty acid, the Sudan stain of fecal fat appears to be a more specific method for detecting the presence of triglyceride and fatty acid in a matrix.
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Sudan stain and quantitative fecal fat.Simko V.Simko V.Gastroenterology. 1990 Jun;98(6):1722-3. doi: 10.1016/0016-5085(90)91126-q.Gastroenterology. 1990.PMID: 1692555 No abstract available.
Dietary fat intake, 72-hour excretion, and Sudan stain for fecal fat.[No authors listed][No authors listed]Gastroenterology. 1989 Aug;97(2):550-1. doi: 10.1016/0016-5085(89)90107-8.Gastroenterology. 1989.PMID: 2473001 No abstract available.
Old insight into a new insight into an old test.Bernstein LH.Bernstein LH.Gastroenterology. 1989 Aug;97(2):552-3. doi: 10.1016/0016-5085(89)90113-3.Gastroenterology. 1989.PMID: 2473002 No abstract available.
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Sudan stain and quantitative fecal fat.Simko V.Simko V.Gastroenterology. 1990 Jun;98(6):1722-3. doi: 10.1016/0016-5085(90)91126-q.Gastroenterology. 1990.PMID: 1692555 No abstract available.
Fecal triglyceride excretion is not excessive in pancreatic insufficiency.Khouri MR, Ng SN, Huang G, Shiau YF.Khouri MR, et al.Gastroenterology. 1989 Mar;96(3):848-52.Gastroenterology. 1989.PMID: 2464525
Diagnostic significance of the sudan III staining for fecal fat.Masamune O, Takahashi T, Nagasaki A, Iwabuchi J, Ishikawa M.Masamune O, et al.Tohoku J Exp Med. 1977 Aug;122(4):397-402. doi: 10.1620/tjem.122.397.Tohoku J Exp Med. 1977.PMID: 72437
Effect of dietary fat composition on rat colon plasma membranes and fecal lipids.Awad AB, Chattopadhyay JP, Danahy ME.Awad AB, et al.J Nutr. 1989 Oct;119(10):1376-82. doi: 10.1093/jn/119.10.1376.J Nutr. 1989.PMID: 2685200 Review.
Plasma lipid transport.Spector AA.Spector AA.Clin Physiol Biochem. 1984;2(2-3):123-34.Clin Physiol Biochem. 1984.PMID: 6386279 Review.
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The importance of stool tests in diagnosis and follow-up of gastrointestinal disorders in children.Kasırga E.Kasırga E.Turk Pediatri Ars. 2019 Sep 25;54(3):141-148. doi: 10.14744/TurkPediatriArs.2018.00483. eCollection 2019.Turk Pediatri Ars. 2019.PMID: 31619925 Free PMC article.Review.
Determination of fecal fat concentration by near infrared spectrometry for the screening of pancreatic steatorrhea.Ventrucci M, Cipolla A, Di Stefano M, Ubalducci GM, Middonno M, Ligabue A, Roda E.Ventrucci M, et al.Int J Pancreatol. 1998 Feb;23(1):17-23. doi: 10.1007/BF02787499.Int J Pancreatol. 1998.PMID: 9520087
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4333 | https://study.com/academy/lesson/work-definition-characteristics-and-examples.html | Work | Definition, Formula & Examples - Lesson | Study.com
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Science/Physics/Energy
Work | Definition, Formula & Examples
Work (Physics) Lessons
Practice Applying Energy Formulas to Systems Work and Energy Lab Practice Applying Work & Kinetic Energy Formulas Work & Power Lesson Plan
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Contributors: Maram Ghadban, Elizabeth Friedl
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Elizabeth FriedlView bio
What is work and what is its formula? Learn the definition of work in physics and how to calculate the value of work done by a force using a formula with some examples. Updated: 11/21/2023
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The Definition of Work ----------------------
Work is defined as the product of an external force acting on an object and the displacement the force has caused. It is considered as a measure of energy that's transferring in or out of an object, and it can either be positive, negative, or even zero.
(i) If the external force and the object's displacement are in the same direction, then the work will be positive. Examples of such are the following:
Pushing an object on a smooth horizontal surface
Riding a bicycle
Kicking a stationary ball
Throwing a stone forward into a pond
Work is positive: the external force and the displacement of the object are at the same direction.
(ii) If the external force and the object's displacement are in the opposite direction, then the work will be negative. Examples of such are the following:
Throwing an object in the air. The displacement is upward and the force (gravity) is in the downward direction.
Fluid resistance to flow, viscosity, is an example of negative work.
Pushing an object on a rough surface. The frictional force and the displacement are in opposite directions.
Kicking a ball in motion.
Negative work: the external force and the displacement of the object are at opposite directions.
(iii) If the external force or the object's displacement is equal to zero, or if they are perpendicular to another, then the work is equal to zero. Examples of such are the following:
Carrying a box.
Pushing a sturdy wall, there is no work done because there was no displacement.
Sitting on a chair.
Zero Work: the external force and the displacement of the object are perpendicular to one another.
What is Work?
The word "work" is used in diverse contexts; doing house chores and completing homework is considered work, while taking tests and preparing meals is also considered work. Any general activity that involves physical or/and mental strain is called work. Well, that's at least how people define it in day-to-day life. Work in physics is quantitative and has a precise definition; it determines the amount of energy transferred when an object is displaced by an external force. It has a value with an actual meaning, unlike the "work" that is used generally. What meaning could be derived from labor and toil when neither of them is quantifiable? To summarize:
Work in physics is quantitative; it gives information about the energy of the displacement of an object that is subjected to an external force.
Work in general is non-quantitative; it has no numerical values nor mathematical formula that could be used to make objective conclusions. It is defined as any activity that requires physical and/or mental strain.
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0:01 What Is Work?
1:58 Calculating Work
3:38 Lesson Summary
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Work Formula ------------
Work is a scalar quantity; it is the dot product of two vectors: force and displacement. As shown:
W=→F⋅→d
W is work measured in Joules (J), which is equal to Newtons-meter (Nm).
→F is the force vector, measured in Newtons (N).
→d is the displacement vector, measured in meters (m).
The dot product of two vectors yields a scalar quantity. That is the general rule. It is strange, however, that work, which is a scalar quantity, can be either positive or negative depending on the direction of the vectors. If it does depend on the direction, how can it still be considered a scalar quantity? Well, that is because work communicates the magnitude of the energy required for an object's displacement. It doesn't genuinely depend on the direction in the coordinate system that is being used; it would still remain the same value if the coordinate system was rotated and flipped. However, that is not the case for the vectors. Their components change as the coordinate system is rotated. Their inner product, work, remains constant in spite of all that flipping. Overall, work is purely magnitude and has no direction.
Work also depends on the angle between the direction of motion and the external force:
If the angle between the vectors is 0∘<Θ<90∘, then the work is positive.
Positive work: angle between vectors is between 0 and 90 degrees.
If the angle between the vectors is 90∘<Θ<180∘, then the work is negative.
Negative work: angle between vectors is between 90 and 180 degrees.
If the angle between the vectors is Θ=90∘, then the work is equal to zero.
Zero Work: angle between the two vectors is equal to 90 degrees.
The work formula can be updated to include the angle between the vectors:
W=→F⋅→d c o s Θ
Cosine Θ is used in dot products, whereas sine Θ is used in cross products.
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How to Calculate Work ---------------------
The work formula can be used in calculating the amount of energy of an object moving as a result of being subjected to an external force. The formula is simple enough to use, and it is quite straightforward. Firstly, the directions of the vectors must be determined. It is always a good strategy to draw a free-body diagram. It will help in identifying the directions of all forces acting on an object (external force, gravitational, friction, etc..). Next, the displacement needs to be determined. This is an easy enough task. In most cases, it is already a given and it can be deduced easily from the problem statement. The following step is determining the force that is acting on the object. This can be done by recalling Newton's second law, which states that an object remains in the state it is in unless a force acts on it and causes it to accelerate. It confirms that an object's acceleration is dependent on its mass and on the force applied, as shown:
F=m∗a
The following example shows how this formula can be used:
A 5kg ball accelerated from rest to 15 m/s in 3.5 seconds on a smooth surface. Find the force that acted on the ball.
The acceleration is equal to the difference in final and initial velocities divided by the change in time:
a=Δ v/Δ t=(v 2−v 1)(t 2−t 1)=(15−0)(3.5−0)=4.3 m/s 2
Newton's second law can be used to find the force:
F=m∗a=5 k g∗4.3 m/s 2=21.5 k g m/s 2=21.5 N
After finding the values and the directions of the vectors, determine the angle between the vectors, then simply plug in all the variables in the work equation.
Examples of Positive, Negative, Or Zero Work
Practice Problem Positive Work
A 2 kg ball is suspended in the air and it drops to the ground from a height of 1.5 m. Compute the work.
The direction of both the force and displacement vectors is the same, and as the picture shows they are parallel. The angle between the vectors is equal to zero degrees.
Positive work: falling ball
The force that is acting on the ball is gravitational. It is falling due to gravity and its mass:
F=m∗g=2 k g∗9.81 m/s 2=19.62 N
The displacement of the ball is the height from which it fell. The work can now be calculated:
W=19.62∗1.5∗c o s(0)=29.43 N m=29.43 J
Note: cos (0) is equal to positive one.
Practice Problem Negative Work
Find the work done by friction if the frictional force acting on a box displaced at 3 m is 12 N.
W=12∗3∗c o s(180)=−36 J
Negative work: frictional force
Note: cos(180) is equal to negative one.
Practice Problem Zero Work
A chocolate box is tightly secured in a trolley that's traveling at 12 m/s. The weight of the chocolate box is 0.5 kg. What is the work done by the chocolate box?
The force (weight) of the chocolate box is acting vertically and its motion is in the horizontal direction (force is perpendicular on the displacement). The angle between the vectors is 90 degrees. Using the work formula:
W=→F⋅→d c o s(90)=0 J
Note: cos(90) is equal to zero.
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Lesson Summary --------------
Work is a term that is used in a wide range of contexts, it is usually associated with any activity that involves physical and mental toll. The day-to-day definition of work is quite vague. It is non-quantifiable and it cannot be derived from a formula, unlike the work that physics defines. Work in physics is quantitative. It defines the magnitude of energy that is associated with an object's displacement after subjecting it to an external force. It is the dot product of two vectors, force and displacement, as shown:
W=→F⋅→d c o s Θ
Work has the units of Joules (J), which is equal to Newtons-meter (Nm). Work is a scalar quantity, meaning it doesn't depend on the direction. Work can be positive, negative, and even zero depending on the direction of the vectors and the angle between them. If the force and the displacement are in the same direction, then the work is positive. An example of such is a ball falling mid-air. If the vectors are in opposite directions, then the resultant work is negative. The work caused by frictional forces is an example. The work is zero if either there is no force applied or there is no displacement. It can also be zero if the directions of the vectors are perpendicular to one another. It is best to draw a free body diagram when attempting to compute work. This would help in knowing the direction of each vector and in assessing the angle between the vectors.
Work is positive if 0∘<Θ<90∘.
Work is negative if 90∘<Θ<180∘.
Work is zero if Θ=90∘.
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Video Transcript
What Is Work?
On a typical day, you probably wake up, get dressed, eat breakfast, and head off to work. After you spend all day at your job, you go home, eat dinner, walk the dog, maybe watch some TV, and then go to bed. In this sense, work can be just about anything - construction, typing on a keyboard, driving a bus, teaching a class, cooking food, treating patients, and so much more.
But in physics, work is more specific. This is the displacement of an object due to force. How much work is done depends on the distance the object is moved. This makes it easy to put work into a solvable equation: work = force distance. While this equation is fairly straightforward, there are three important things to note. First, the object must move over some distance in order for work to be done. Second, the force and the distance of movement must be in the same direction. And finally, the force must be constant.
The units we use for work are joules (J), named for James Prescott Joule. Though he is now known for his work in science, he actually preferred brewing beer… until he realized how science could help him be a better brewer!
The joule is a combination of both of the components on the right side of our work equation: force and distance. Quite simply, a joule is a Newton-meter (Nm), and one joule of work is done when a force of 1 N is exerted over a distance of 1 m.
This amount of work is on par with lifting an apple over your head. I bet you didn't realize that counted as work, but in the world of physics, it does! You can probably see that 1 J isn't really practical to use for larger amounts of work, so instead we use kilojoule (kJ), which is 1000 J, or megajoule (MJ), which is 1,000,000 J.
Regardless of the amount of work done, it involves three key components: the amount of force, the distance displaced, and the cause of the displacement, which is the force itself.
Calculating Work Done
Work is an interesting concept because the same amount of work can be done in different situations. For example, if you lift a 5 N load 10 m in the air, the amount of work done is: 5 N 10 m, or 50 J.
But say you lift twice the weight over half the distance. In this case, the amount of work done is: 10 N 5 m…also 50 J! The work done is the same because even though the displacement distance is less, the weight of the object is more, so it takes more force to displace it.
But if you were to lift the 10 N object the original distance of 10 m, the amount of work done in this case is twice as much because now 10 N 10 m = 100 J.
Can you see how the amount of work done increases if you increase either the force (the weight), the distance, or both? Lifting the same amount of weight twice as high means twice the amount of work is done. Likewise, lifting twice the weight over the same distance also doubles the amount of work.
However, simply holding an object in the air doesn't count as work. The object must move over some distance in order for work to be done, the force and the distance of movement must be in the same direction, and that force must be constant. So if you move that apple up in the air, you're doing work on the apple, but holding it in place is not doing any work because the apple isn't moving. Likewise, you can push on a wall all day, but if that wall doesn't move, no work is done on it!
Lesson Summary
Our everyday work is different for each individual. But in physics, work is only one thing: the displacement of an object due to force. Anytime work is done there are three components involved: the amount of force (in Newtons), the distance of the displacement (in meters), and the cause of the displacement (the force).
In multiplying force and distance for work, we end up with the unit of joules. 1 J of work is equal to a force of 1 N exerted over a distance of 1 m, about the same as lifting an apple over your head. Because of this, we often work with kJ or MJ for very large amounts of work.
The amount of work done can be increased by increasing the amount of force, the distance of displacement, or both. But in order for work to be done, there must be a displacement of the object. Pushing on a wall may be difficult and tiring, but if you don't move that wall, no work is done on it.
Learning Outcomes
Review this lesson so that you can:
Assess the components of 'work'
Define work in physics
Write the formula that measures the amount of work that has been done
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FAQ
What is the formula for work?
Work is the dot product of two vectors, force and displacement. As shown:
W = F d
Where W is work, F is the force exerted on an object, and d is the object's displacement.
What is work and its unit?
Work is the dot product of force and displacement. The unit of force is Newtons (N) and the unit of displacement is (m).
W = F d
Nm = Nm
The unit of work is Newton-meters, which is equal to 1 Joule (J).
Nm = J
What is the definition of work in physics?
Work is a quantitative scalar property that is equal to the dot product of the force acting on an object and the displacement of an object. It has the units of Joules, which is equivalent to Newton-meters. Work can be positive, negative, or zero depending on the direction of the vectors.
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Ch 1. AP Physics 1: Systems
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7:10 Next Lesson Work Done by a Variable Force
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4334 | https://suncatcherstudio.com/printables/math/fraction-strips/ | Skip to content
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4335 | https://stackoverflow.com/questions/875377/which-floor-is-redundant-in-floorsqrtfloorx | Stack Overflow
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Which floor is redundant in floor(sqrt(floor(x)))?
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I have floor(sqrt(floor(x))). Which is true:
The inner floor is redundant.
The outer floor is redundant.
math
floor
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edited Sep 16, 2012 at 15:32
Bill the Lizard
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asked May 17, 2009 at 20:06
unj2unj2
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4
Why call it a "Floor and Ceiling Question"?
Renze de Waal
– Renze de Waal
2009-05-17 21:01:03 +00:00
Commented May 17, 2009 at 21:01
6
Actually, the following very general result can be proved by extending Elazar Leibovich's idea below: If f(x) is any continuous, monotonically increasing function such that: f(x) is an integer ==> x is an integer, then floor(f(x)) = floor(f(floor(x))) and similarly, ceiling(f(x)) = ceiling(f(ceiling(x))). (Reference: Concrete Mathematics by Graham, Knuth, Patashnik; Pg 71, Eq. 3.10).
Ashutosh Mehra
– Ashutosh Mehra
2009-05-18 01:42:42 +00:00
Commented May 18, 2009 at 1:42
My next question was to develop a general framework. Thanks for the reference. I will definitely take a look at the findings.
unj2
– unj2
2009-05-18 02:13:27 +00:00
Commented May 18, 2009 at 2:13
@Ashutosh: Btw your blog looks cool.
unj2
– unj2
2009-05-18 02:16:09 +00:00
Commented May 18, 2009 at 2:16
Add a comment |
9 Answers 9
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38
Obviously the outer floor is not redundant, since for example, sqrt(2) is not an integer, and thus floor(sqrt(2)) sqrt(2).
It is also easy to see that sqrt(floor(x)) sqrt(x) for non-integer x. Since sqrt is a monotone function.
We need to find out whether or not floor(sqrt(floor(x)))==floor(sqrt(x)) for all rationals (or reals).
Let us prove that if sqrt(n) then sqrt(n+1), for integers m,n. It is easy to see that
n
Therefor by the fact that sqrt is montone we have that
sqrt(n) < m -> sqrt(n+1) < m+1 -> sqrt(n+eps)
Therefor floor(sqrt(n))=floor(sqrt(n+eps)) for all 0 and integer n. Assume otherwise that floor(sqrt(n))=m and floor(sqrt(n+eps))=m+1, and you've got a case where sqrt(n) however sqrt(n+eps)>=m+1.
So, assuming the outer floor is needed, the inner floor is redundant.
To put it otherwise it is always true that
floor(sqrt(n)) == floor(sqrt(floor(n)))
What about inner ceil?
It is easy to see that floor(sqrt(n)) floor(sqrt(ceil(n))). For example
floor(sqrt(0.001))=0, while floor(sqrt(1))=1
However you can prove in similar way that
ceil(sqrt(n)) == ceil(sqrt(ceil(n)))
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edited May 15, 2012 at 8:54
mikebloch
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answered May 17, 2009 at 20:39
Elazar LeibovichElazar Leibovich
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2 Comments
unj2
unj2
Does your solution imply that an inner ceiling would also be redundant?
unj2
unj2
I cant see any problem here. Unless somebody proves otherwise, this is correct.
18
The inner one is redundant, the outer one of course not.
The outer one is not redundant, because the square root of a number x only results in an integer if x is a square number.
The inner one is redundant, because the square root for any number in the interval [x,x+1[ (where x is an integer) always lies within the interval [floor(sqrt(x)),ceil(sqrt(x))[ and therefore you don't need to floor a number before taking the square root of it, if you are only interested the integer part of the result.
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edited May 17, 2009 at 21:12
answered May 17, 2009 at 20:40
Simon LehmannSimon Lehmann
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5 Comments
unj2
unj2
There were a lot of good proofs but this one is the sweetest. Thanks.
Elazar Leibovich
Elazar Leibovich
Who said there's no whole integer in [sqrt(x),sqrt(x+1)]? If for example sqrt(x)==0.9 and sqrt(x+1)=1.2, then the result of floor(sqrt(floor(x))) might be different than floor(sqrt(x))
Simon Lehmann
Simon Lehmann
I'm glad you like it. After reading all the different and well done actual proofs I thought my explanation might be too informal for some tastes.
unj2
unj2
Wow. Is there a hole in the solution?
Simon Lehmann
Simon Lehmann
@Elazar: I wrote [sqrt(x),sqrt(x)+1[ and not [sqrt(x),sqrt(x+1)[, but even then you are right (though, your example is wrong). I've corrected the interval to what I actually meant. The sqrt(x) and sqrt(x+1) of any integer x lies within the same integer interval.
5
Intuitively I believe the inner one is redundant, but I can't prove it.
You're not allowed to vote me down unless you can provide a value of x that proves me wrong. 8-)
Edit: See v3's comment on this answer for proof - thanks, v3!
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edited May 17, 2009 at 20:26
answered May 17, 2009 at 20:14
RichieHindleRichieHindle
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3 Comments
unj2
unj2
I too tried out examples and feel it in my gut that the inner one is redundant but its damn hard to prove that i am correct.
GSerg
GSerg
Indeed. There was an answer saying the inner one is redundant, but before I could upvote it it got downvoted and deleted. Still, I believe that's the case.
v3.
v3.
Can't you just say that floor(sqrt(x)) only changes value on integer values of x? If x is not an integer, then flooring it won't change the integer part of the square root.
4
The inner floor is redundant
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answered May 17, 2009 at 20:09
ichibanichiban
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2 Comments
Peter PerháÄ
Peter PerháÄ
there, i'll give ya the ten points lost :) I was surprised Richie got that many upvotes for an answer like that...
ichiban
ichiban
Anybody who downvotes me, needs to downvote Richie. I give my answer before him.
4
The inner floor is redundant. A proof by contradiction:
Assume the inner floor is not redundant. That would mean that:
floor(sqrt(x)) != floor(sqrt(x+d))
for some x and d where floor(x) = floor(x+d). Then we have three numbers to consider: a = sqrt(x), b = floor(sqrt(x+d)), c = sqrt(x+d). b is an integer, and a < b < c. That means that a^2 < b^2 < c^2, or x < b^2 < x+d. But if b is an integer, then b^2 is an integer. Therefore floor(x) < b^2, and b^2 <= floor(x+d), and then floor(x) < floor(x+d). But we started by assuming floor(x) = floor(x+d). We've reached a contradiction, so our assumption is false, and the inner floor is redundant.
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answered May 17, 2009 at 20:40
Ned BatchelderNed Batchelder
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3
If x is an integer then the inner floor is redundant.
If x is not an integer then neither are redundant.
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answered May 17, 2009 at 20:12
MackerMacker
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6 Comments
unj2
unj2
the question really is why for the second part.
majkinetor
majkinetor
So much bla-bla around, while Macker only got it right and concise.
unj2
unj2
the bla bla was trying to convince ourselves that it is true.
Macker
Macker
Strange that you say it's wrong but have agreed with the same conclusions in your choice of answer. I'm confused. You didn't ask for a proof in your question btw.
unj2
unj2
I am sorry the question is still open. Could you edit it with your proof?
|
3
The outer floor is not redundant. Counterexample: x = 2.
floor(sqrt(floor(2))) = floor(sqrt(2)) = floor(1.41...)
Without the outer floor the result would be 1.41...
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answered May 17, 2009 at 20:19
bbmudbbmud
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2
If the inner floor were not redundant, then we would expect that floor(sqrt(n)) != floor(sqrt(m)), where m = floor(n)
note that n - 1 < m <= n. m is always less than or equal to n
floor(sqrt(n)) != floor(sqrt(m)) requires that the values of sqrt(n) and sqrt(m) differ by at least 1.0
however, there are no values n for which the sqrt(n) differs by at least 1.0 from sqrt(n + 1), since for all values between 0 and 1 the sqrt of that value is < 1 by definition.
thus, for all values n, the floor(sqrt(n)) == floor(sqrt(n + 1)). This is in contradiction to the original assumption.
Thus the inner floor is redundant.
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edited May 17, 2009 at 23:44
answered May 17, 2009 at 20:53
DemiDemi
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2 Comments
Elazar Leibovich
Elazar Leibovich
"floor(sqrt(n)) != floor(sqrt(m)) requires that the values of sqrt(n) and sqrt(m) differ by at least 1.0" doesn't seem to be true, if sqrt(n) = 0.999 and sqrt(m) = 1, their different by far less than 1, however floor(sqrt(n))!=floor(sqrt(m))
melfar
melfar
"note that n <= m < n + 1" That doesn't seem to be true, since floor(n) can be < n
0
If n^2 <= x < (n+1)^2 where n is an integer, then
n <= sqrt(x) < n+1, so floor(sqrt(x)) = n;
n^2 <= floor(x) < (n+1)^2, so n <= sqrt(floor(x)) < n+1, so floor(sqrt(floor(x))) = n.
Therefore, floor(sqrt(floor(x))) = floor(sqrt(x)), which implies the inner floor is redundant.
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answered Apr 4, 2019 at 15:28
SnzFor16MinSnzFor16Min
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4336 | https://www.ck12.org/flexi/geometry/quadrilaterals/what-are-the-four-properties-of-a-rhombus/ | What are the four properties of a rhombus?
A rhombus is a quadrilateral with four congruent sides. All rhombuses are parallelograms. Rhombus and its PropertiesThe Properties of a Rhombus: 1. Opposite sides are parallel, i.e., @$\begin{align}AB \parallel CD\end{align}@$ and @$\begin{align}BC \parallel DA\end{align}@$. 2. All four sides are congruent, i.e., @$\begin{align}AB = BC = CD = DA\end{align}@$. 3. Opposite angles are congruent, i.e., @$\begin{align}\angle{ABC} = \angle{CDA}\end{align}@$ and @$\begin{align}\angle{DAB} = \angle{BCD}\end{align}@$. 4. Diagonals are the interior angle bisectors, i.e., @$\begin{align}\angle BAC = \angle DAC\end{align}@$, @$\begin{align}\angle BDC = \angle BDA\end{align}@$, @$\begin{align}\angle{DBC} = \angle{DBA}\end{align}@$ and @$\begin{align}\angle BCA = \angle DCA\end{align}@$. 5. Diagonals intersect each other at right angles, i.e., @$\begin{align}\angle AOB\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle BOC\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle COD\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle DOA\end{align}@$@$\begin{align}= 90^\circ\end{align}@$.
A rhombus is a quadrilateral with four congruent sides. All rhombuses are parallelograms.
The Properties of a Rhombus:
Opposite sides are parallel, i.e., @$\begin{align}AB \parallel CD\end{align}@$ and @$\begin{align}BC \parallel DA\end{align}@$.
All four sides are congruent, i.e., @$\begin{align}AB = BC = CD = DA\end{align}@$.
Opposite angles are congruent, i.e., @$\begin{align}\angle{ABC} = \angle{CDA}\end{align}@$ and @$\begin{align}\angle{DAB} = \angle{BCD}\end{align}@$.
Diagonals are the interior angle bisectors, i.e., @$\begin{align}\angle BAC = \angle DAC\end{align}@$, @$\begin{align}\angle BDC = \angle BDA\end{align}@$, @$\begin{align}\angle{DBC} = \angle{DBA}\end{align}@$ and @$\begin{align}\angle BCA = \angle DCA\end{align}@$.
Diagonals intersect each other at right angles, i.e., @$\begin{align}\angle AOB\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle BOC\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle COD\end{align}@$@$\begin{align}=\end{align}@$@$\begin{align}\angle DOA\end{align}@$@$\begin{align}= 90^\circ\end{align}@$.
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Home>Grade 4>Dividing Two-Digit Numbers (Elementary)
Dividing Two-Digit Numbers (Elementary)
Directions: Using the digits 1 to 9 at most one time each, fill in the boxes to make the smallest (or largest) quotient.
Hint
What does the number on the left represent? What does the number on the right represent?
Answer
Answers will vary depending on whether decimal answers are allowed. 98÷ 12 is one answer for the largest quotient. 12 ÷ 98 is the smallest quotient.
Source: Robert Kaplinsky
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Tags 4.NBT.65.NBT.6DOK 3: Strategic ThinkingRobert Kaplinsky
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7 comments
Brad ShonkOctober 21, 2016 at 9:18 am
I changed it a little bit. Make the largest quotient by filling in the boxes using whole numbers 1 – 9 no more than one time each. No remainders. I know using 98 as the dividend will be what I want to use. I can’t use 12 or 13 as the divisor. Wait a minute I can use 14 as the divisor. That’s the answer 98 divided by 14 equals 7.
Any thoughts. Reply
Ellen MetzgerNovember 3, 2016 at 10:20 am
I like this version, too. You could also make the smallest quotient possible, not including quotients less than one. The smallest whole number quotient possible will be 2, given that you can’t repeat any of the digits, and there are actually many ways (I counted 21) of getting 2 as the quotient of two 2-digit numbers using the digits 1-9 no more than one time each. Students could try to find them all, or you could limit the digits to 1 through 6, and then there are a lot fewer solutions (I counted 4). Reply
Miss JacksonJanuary 3, 2017 at 6:15 am
my students discovered that 96 divided by 12 is 8 therefore making it larger than your answer of 7. GO WAMPUS CATS! Reply
Robert KaplinskyFebruary 2, 2017 at 11:14 am
Yeah, lots of answers are possible depending on what assumptions you make. For example, Miss Jackson’s 96 divided by 12 appears to be the largest whole number quotient. If you allow for decimals, that does change things. Well done everyone! Reply
Arthur L.January 29, 2019 at 8:09 am
98 / 11 = 8.90909091 is bigger than 8. P.S. Chinstrap Penguins are the best. Reply
Miller MathJanuary 30, 2019 at 11:14 am
However, you used two of the same digit but I would definitely commend a student for making that attempt. Reply
KianahJanuary 10, 2020 at 5:55 am
98 ÷ 12 is one answer for the largest quotient. 12 ÷ 98 is the smallest quotient. I think Reply
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3 Consecutive Integers Problem
GRE Quant Problem Solving
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Hi all –
I don’t understand the solution to this question. Couldn’t the answer technically be anything? For example:
24 x 32 x 40 = 30, 720.
This product is a multiple of 8. You could go on like this with even bigger multiples of 8. So why is 3072, specifically, the answer?
Screenshot (79)1920×778 82.7 KB
The solution cites the reason to be because “the answer can be written in terms of 8 x 16 x 24”, but I’m not really sure what that means.
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Archi
Jan 18
This is an interesting question that I prefer to solve using algebra. Let’s start with the phrase “consecutive multiples” of 8, which can be represented as 8y, 8y + 8, and 8y + 16. This ensures that any value substituted for y yields consecutive multiples of 8.
Now, we need to find the largest number that the product (8y)(8y + 8)(8y + 16) is a multiple of. We can express this as:
Here assume m to be the divisor that we have to find.
(8y)(8(y + 1))(8(y + 2)) / m
Combining the 8s gives us (512y(y + 1)(y + 2))/m. Since we need at least three 8s for it to be an integer, m must contain at least three 8s. Now next thing is, finding the number which is always a divisor of y(y + 1)(y + 2)
We know that, the product of three consecutive integers y, y + 1, and y + 2 will always be a multiple of both 2 and 3. Thus, it is also a multiple of 6. The product 512 6 equals 3072. We multiplied by 6 and not 3 or 2 because we are looking for the largest number.Therefore, m must have at least 512 and 6 as factors.
While higher numbers may yield larger multiples, to satisfy the requirement across all values of y, 3072 is the largest guaranteed divisor. We cannot assume that consecutive integers will always be divisible by numbers beyond 6. For instance, while pairs of consecutive integers with one odd in the middle will always include two even integers and thus be a multiple of 8, 512 8 (4096) but this does not ensure that it the largest number divisor or for that matter divisor at first place for all values of y for eg what if we get 3 consecutive multiples of 8 where the middle one isn’t odd? Hence, we conclude that in “must” scenario, 3072 must be our largest divisor. If we go beyond this, we can’t ensure it works for every value of y.
Does this make sense?
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Manipulating circuit into two-port network for feedback analysis
I am trying to apply two-port feedback analysis to a circuit similar to the one shown below. I can't represent the forward amplifier as a two-port network because the input signal and feedback signal go to different inputs.
Is it a valid operation to refer the feedback signal to the non-inverting input of the forward amplifier to form a two-port network as shown in the second figure?
Then I can calculate the forward amplifier gain and the feedback quantity, and the summing node is clear. Referring the feedback signal to the non-inverting input can be done using equations for the ideal inverting and non-inverting gains of the forward amplifier, because the non-ideality (i.e. finite loop gain) is the same for both gains.
simulate this circuit – Schematic created using CircuitLab
simulate this circuit
1 Answer 1
If all R values are equal, the use of Vin- vs Vin+ for feedback is irrelevant in this 1st circuit.
The Vin- current is relative to Vout , while the Vin+ is relative to 0V. So the gain of Vo/Vin is a huge difference due to Aol.
So the real problem is the feedback is not differential. The 2nd problem is choosing a stable gain for a differential feedback.
To make this easier to understand...
The non-inverting signal input with negative feedback causes both O.A. inputs to follow the input with a "virtual" null difference. So if all R's are equal the gain= +2 and the use of a Vout/2 buffered then matches and follows the input.
But using the Vin+ referenced to 0V using Vin- with a gain of -1 will not produce a Vfb of 0V to match the grounded reference of Vin+. Now both inputs are 0V but Vfb is still Vout/2 changing it's polarity won't help.
Using 741's here
The label nodes are connected.
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4340 | https://brainly.com/question/15747999 | [FREE] Given the three side lengths, how can you tell if a triangle is a right triangle? Answer: Square all side - brainly.com
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Given the three side lengths, how can you tell if a triangle is a right triangle?
Answer:
Square all side lengths. If the longest side length squared is equal to the sum of the squares of the other two side lengths, then it is a right triangle.
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Square all side lengths. If the longest side length squared is equal to the sum of the squares of the other two side lengths, Then it is a right triangle.
Given that,
The three side length should be given.
Now in order to prove the triangle is a right triangle. The following information should be considered.
The side length should be squared.
The longest length side should be squared and equivalent to the sum of the square of the other two side length.
Also we called as the pythagoras theorem.
Learn more: brainly.com/question/17429689
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To determine if a triangle is a right triangle, identify the longest side, square all side lengths, and use the Pythagorean theorem to check if the sum of the squares of the two shorter sides equals the square of the longest side. If this condition is met, the triangle is a right triangle. Otherwise, it is not.
Explanation
To determine if a triangle is a right triangle based on its side lengths, you can use the Pythagorean theorem. Here’s how to do it step-by-step:
Identify the Side Lengths: Label the three sides of the triangle as a, b, and c, where c is the longest side.
Square Each Side Length: Calculate the square of each side. This gives you a², b², and c².
Apply the Pythagorean Theorem: According to the theorem, for a triangle to be classified as a right triangle, the following equation must hold true:
a 2+b 2=c 2
This means that the sum of the squares of the two shorter sides (a and b) should equal the square of the longest side (c).
Check the Condition: If the equation is satisfied, then the triangle is a right triangle. If not, it is not a right triangle.
Example:
Consider a triangle with sides of lengths 3, 4, and 5:
Alert that 5 is the longest side, so we set c = 5, a = 3, and b = 4.
Calculate each side squared: 3 2=9, 4 2=16, and 5 2=25.
Check the equation: 9+16=25. Since this equation holds true, the triangle with sides 3, 4, and 5 is indeed a right triangle.
Examples & Evidence
For example, for a triangle with sides 6, 8, and 10, you would check: 6 2+8 2=36+64=100, which is equal to 1 0 2=100. Thus, it is a right triangle.
The Pythagorean theorem is a well-established principle in mathematics, stating that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is used widely in various fields including construction, physics, and engineering.
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Square all side lengths. If the longest side length squared is equal to the sum of the squares of the other two side lengths, Then it is a right triangle.
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### Free Mathematics solutions and answers
Community Answer 5.0 1 If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then what type of triangle was formed? right acute obtuse scalene
Community Answer 1 How does this model demonstrate the Pythagorean Theorem? a. The sum of the lengths of the shortest and the longest sides is equal to twice the length of the middle side. So double the length of the longer leg of any right triangle is equal to the sum of the shorter leg and the hypotenuse. b. The sum of the area of the two smaller squares is equal to the area of the larger square. So the sum of the lengths of the two legs of any right triangle squared is equal to the length of the hypotenuse squared. c. The sum of the area of the smallest and the largest squares is equal to the area of the middle square. So the sum of the lengths of the shorter leg and the hypotenuse of any right triangle squared is equal to the length of the middle leg squared. d. The length of the longest side minus two equals the length of the middle side. The length of the middle side minus two equals the length of the shortest side. So the length of the short leg of any right triangle is equal to the length of the middle leg minus 2, and the length of the hypotenuse is equal to the length of the middle leg plus 2.
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
New questions in Mathematics
If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them?
Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216
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How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1
Is this equation correct? 6 3⋅7 3=4 2 3
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4341 | https://janetaylor.net/thriving-another-word-for-success/ | Skip to content
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Thriving – Another Word for Success?
Jane Taylor Personal Growth
Is thriving another word for success in life? Another year has passed and I have been reflecting on my focus for this year. I love the time to sit and contemplate. During this process, one of the words that come to mind is thriving, so I went to investigate some more.
What Does Thrive Mean?
When I was researching, I found many definitions on thrive. However, as a person, I was more interested in thriving, which is an adjective (i.e. describes a person, place or thing). As a person, thriving is what many of my clients want and means –
“prosperous and growing; flourishing” ~ Cambridge Dictionary and Lexico/Oxford
“characterized by success or prosperity” ~ Merriam-Webster Dictionary
“Something that’s thriving is doing very well.” ~ Vocabulary.com
“Improving, growing, or succeedingsteadily” ~ the Free Dictionary
“continuing to be successful, strong, healthy, etc” ~ Oxford Learner’s Dictionary
“[Thriving] appears to come down to an individual experiencing a sense of development, of getting better at something, and succeeding at mastering something. In the simplest terms, what underpins it is feeling good about life and yourself and being good at something.” ~ Brown, Arnold, Fletcher and Standage
Any other definitions on thriving that resonate with you? If so, feel free to share them in the comments below 🙂
7 Inspirational Quotes on Thriving…
Following are some inspirational quotes I found that help with defining thriving –
“My mission in life is not merely to survive, but to thrive; and to do so with some passion, some compassion, some humor, and some style.” ~ Maya Angelou
“When women thrive, all of society benefits, and succeeding generations are given a better start in life.” – Kofi Annan
“To fully thrive, we must not only eliminate the stressors but also actively seek joyful, loving, fulfilling lives that stimulate growth processes.” ~ Bruce Lipton
“We can only flourish when we are truly grounded in self. keep that in mind when journeying.” ~ Alexandra Elle
“We all make mistakes, but the people who thrive from their mistakes are the successful ones.” ~ Henry Cloud
“We are not held back by the love we didn’t receive in the past, but by the love we’re not extending in the present.” ~ Marianne Williamson
Do you have another inspirational quote on thriving? If so, feel free to share it below!
Questions for Reflection – What Does Thrive Mean to You?
Following are some questions for reflection on thrive / thriving. After readying the above information,
what does thrive and/or thriving mean to you?
can you remember a time in your life when you were thriving? What were you doing? Who were you being? Who was around you at the time?
is thriving a better word for success in your life?
Time to Thrive in Your Sport and Life?
Are you ready to thrive in your work and life? If so. what small step can you take today to continue to fulfil your potential?
If you are ready to reclaim your courage and take thenext step towards freedom and opening your heart,why not join our Toolkit?
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What Are the Differences Between Rectangular Waveguides and Circular Waveguides?
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What Are the Differences Between Rectangular Waveguides and Circular Waveguides?
06 Mar March 6, 2025
What Are the Differences Between Rectangular Waveguides and Circular Waveguides?
SEO Gelsey2025-03-25T14:00:01+08:00
Fundamental Design Characteristics and Mode Propagation
The basic structure of a waveguidegreatly shapes its electromagnetic traits and mode behavior.
Shape and Mode Support
Rectangular and circular waveguides vary widely in their physical forms. This leads to unique ways of guiding electromagnetic waves. Here’s a comparison table:
ApplicationRectangular WaveguidesCircular Waveguides
Radar TechnologyPreferred for high power, low loss (e.g., X-band, 8-12 GHz)Used in systems needing circular motion
AerospaceLess common, lacks flexibility under stress Ideal for mobile radar due to durability, flexibility
TelecommunicationsFavored for satellite links (cost-effective, low attenuation)Suits circular polarization (e.g., broadcasting, radar joints)
The waveguide’s form determines which electromagnetic waves it can carry.
Cut-off Frequency
The cut-off frequency marks the point below which a mode stops traveling. It differs notably between rectangular and circular waveguides due to their distinct cross-sectional designs. This trait heavily affects their use based on frequency needs. Here’s a comparison:
AspectRectangular Waveguide (WR-90)Circular Waveguide (3.1 cm Diameter)
Dimensions2.286 cm x 1.016 cm 3.1 cm diameter
Dominant ModeTE10 TE11
Cut-off Frequency6.56 GHz 5.17 GHz
Typical ApplicationsX-band radar, satellite communication Radar, telecommunications
Operating Range8.2-12.4 GHz Similar range, broader lower frequencies
Cost ImpactHigher for narrower bands Lower for wider, softer frequencies
For rectangular waveguides, each mode ties to a specific cut-off frequency based on its width and height. This suits X-band radar and satellite links. Circular waveguides of similar size show a gentler cut-off for their main mode. Choosing them can save money for satellite links needing wider spans and lower frequencies.
The cut-off frequency is a vital factor in picking waveguides. It sways both size and expense of the setup.
Power Handling and Loss
Power handling and loss are key elements in judging waveguide fit for various tasks.
Power Handling
Rectangular waveguidesshine with their superb power-handling skills. Their broader surfaces allow heat to spread out efficiently. They often serve in high-power radar setups, like air traffic control systems at 10 GHz. Here, they handle about 1 megawatt peak pulse power. Such systems need strong power handling to avoid signal twists or breakdowns.
Power Loss
In rectangular waveguides, power loss drops thanks to their flat sides. These shorten current travel time, cutting conductor losses and resistance. Circular waveguides, however, tend to lose more power. Their longer conduction paths are to blame. Though the loss gap might be tiny—around 0.1 dB/m—it matters in systems like microwave relay networks.
Rectangular waveguides typically outshine circular ones in power handling and energy retention.
Bandwidth and Frequency Response
Rectangular straight waveguides deliver top-notch results in keeping single-mode action over broad frequency ranges. Their boxy shape creates a larger gap between the base mode and higher ones. This offers a wider usable bandwidth than circular options. This quality is precious in systems craving precise frequency control and low mode mixing. Surface treatments can boost frequency response further. Special designs can also tweak bandwidth for specific needs.
Application-Specific Considerations
Choosing between rectangular and circular waveguides often rests on the task’s unique needs. Here’s a comparison table:
ApplicationRectangular WaveguidesCircular Waveguides
Radar TechnologyPreferred for high power, low loss (e.g., X-band, 8-12 GHz)Used in systems needing circular motion
AerospaceLess common, lacks flexibility under stress Ideal for mobile radar due to durability, flexibility
TelecommunicationsFavored for satellite links (cost-effective, low attenuation)Suits circular polarization (e.g., broadcasting, radar joints)
Radar Technology
Rectangular waveguides pop up often in radar tech. They send high power with scant signal loss. They’re especially apt for X-band radar systems, like air traffic control, spanning 8 to 12 GHz. The TE10 mode in rectangular waveguides focuses microwave energy tightly. This is crucial for accurately heating radar targets.
Aerospace Applications
Circular waveguides lead in aerospace uses. Their adaptability and toughness shine under shifting pressures and temperatures. The round shape resists shakes, twists, and bends well. It keeps signals steady. This makes them perfect for mobile radar in vehicles and planes where strength and flex are key.
Telecommunications
In telecom, rectangular waveguides win for satellite links. They’re budget-friendly and lose less signal. Crafting them is simpler and cheaper, yielding big savings. Circular waveguides, though, fit systems needing circularly polarized waves. Think broadcasting antennas or spinning radar joints.
Manufacturing and Handling Considerations
How waveguides are made and managed also shapes their fit for a job.
Cost
Rectangular waveguides usually cost less to craft than circular ones. They’re cut from flat metal sheets, needing simpler tools. For instance, a rectangular waveguide for an X-band radar might be 30% cheaper than a circular one. This price edge makes them a smart pick when bulk matters.
Structural Strength and Flexibility
Circular waveguides deliver steady material quality and greater durability. Their even cross-section stands out. They’re ideal where external forces hit, like in aerospace. The round form shrugs off vibrations and bends, keeping signals true. Plus, they skip exact inner alignment, resisting jams better.
Antenna Parameters
Several antenna traits tie closely to waveguide performance. These help outline how an antenna works.
Gain: Gauges energy an antenna can beam, factoring in feed losses. Higher gain sends more power one way.
Half Power Beam Width (HPBW): The angle between half-power points of the main lobe. It shows an antenna’s focus.
Return Loss/VSWR: The share of energy bounced back to the input. It reveals system match quality.
Efficiency: The ratio of radiated energy to supplied energy, counting all losses.
Polarization: The path electrical energy swings in—linear, circular, or elliptical.
AEM Antennas can offer more details on these traits.
RFecho: Your Reliable Waveguide Supplier
RFechoexcels in crafting high-performing antennas and microwave parts. Their range includes standard gain horns, reflector antennas, and various waveguide pieces. RFecho pledges top-quality service and aids customers in hitting their targets.
Conclusion
To wrap up, both rectangular and circular waveguides are vital cogs in RF and microwave systems. Yet, they bring unique perks and drawbacks. Rectangular waveguides shine in tasks needing high power, single-mode runs, and thriftiness. They’re perfect for radar and satellite links. Circular waveguides, with their bendability and mode variety, suit aerospace and circular polarization needs better. Grasping these contrasts is essential for engineers and designers. It helps them pick the best waveguide setup for their aims.
References:
Advanced Microwave Technologies co., Ltd. 2025. _How does a Rectangular Straight Waveguide compare to a Circular Waveguide?_
Dolph Microwave. 2024. _5 differences between rectangular and circular waveguide._
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介词的固定搭配其实是有一定规律的，对于每个介词的含义以及细微的区别还是先理解之后，学习来就很容易了，硬背肯定是背不完的。
先放张图，有助于加深理解：
介词虽然容易混淆，但是也算英语学习中比较基础的部分了。底下的内容很长，要是觉得看着太累的话，推荐给大家一个不小心发现的免费的英语直播课，比较适合英语基础相对薄弱的人，会有老师很耐心地讲英语词汇、语法等基础知识，还有同步训练和配套的学习资源，点这个链接就可以听课了：
授课老师全都颜值又高又厉害，呜呜呜。。
安利完毕，进入正题，长文预警！
常用的介词搭配，大致可以总结如下：
时间介词
A. 时间点的表达：at, on
在具体某一时间点前，用at：
at 5 o'clock 在五点钟
at midnight 在午夜
at sunset 在日落时分
at the moment 现在
at present 当下
at the same time 同时
在具体某一天或日期前，用on：
on Friday 在星期五
on 12 March 1999 在1999年3月12日
on Christmas Day 在圣诞节这天
注意，at Christmas / at Easter表示在圣诞节／在复活节时期，而不是在具体节日那天。
在某一天的早上/下午/晚上也要用介词on
on the evening of May the first
在五月一日的晚上
注意，last / next / this / every 修饰的时间前不使用任何介词：
I'll see you next Friday.
下周五见。
They got married last March.
他们去年三月结婚了。
B. 时间段的表达: in, during, over, throughout
一、在较大的时间单位前（月、年、季节、世纪、年代等），用in表示某事或某活动发生在该时间段内：
in October 在十月
in 1968 在1968年
in the 1980s 在1980年代
in the 18th century 在18世纪
in winter 在冬天
in the Middle Ages 在中世纪
in a few minutes / in six months类似的表达说明未来某一时间：
The train will be leaving in a few minutes.
火车几分钟后就要开了（火车还没开，从现在开始的几分钟后就会开了）。
Jack has gone away. He'll be back in a week.
杰克走了。他一周后回来（人还没回，从现在开始的一周后就会回来了）。
during用于表示某事在某个时期发生过，或者持续了一段时间，后面可以加时间，也可以加表示事件的名词（illness, holiday, visit等）：
She sneezed during the performance.
她在表演过程中打了个喷嚏。（在某个时间段内发生过）
Weather conditions have been improving during the past few days.
天气状况在过去几天里有所改善。（在某个时间段内持续发生）
over用来表示某件事在某段时间内持续了部分或全部时间：
Weather conditions have been improving over the past few days.
天气状况在过去几天里有所改善。
They chatted over a cup of coffee.
他们一边喝咖啡一边聊天。
throughout用来强调某件事在某个时间段内不间断发生，持续全部时间，相当于during the whole...:
She sneezed throughout the performance. (= during the whole performance).
演出过程中她一直在打喷嚏。
C. 表达到某个时间点：until, by
until用来表示某事会一直持续到某个特定时间，意为“直到...”：
We have to be at home until 2:30.
我们得在家里待到两点半。
not...until表示“直到...才...”
We can’t leave home until 2:30.
我们两点半才能离开家。
by用来表示某事会在某个特定时间之前或最近的某个时间发生，意为“到...为止”：
We have to be at home by 2:30.
我们必须在2:30之前到家。（在两点半或两点半之前我们必须回到家）
地点介词
A. in
in表示“在某一空间范围内”：
There's no one in the room / in the building / in the garden.
房间里/大楼里/花园里一个人也没有。
What have you got in your hand / in your mouth?
你手里/嘴里拿着什么?
in表示“在某一地理范围内”,如国家、城镇、山区、河海等：
When we were in Italy, we spent a few days in Venice.
我们在意大利的时候，在威尼斯呆了几天。
I have a friend who lives in a small village in the mountains.
我有一个朋友住在山里的一个小村庄里。
Look at those people swimming in the sea / in the river.
看那些在海里/河里游泳的人。
B. at
at表示“在某一个具体的点”：
Who is the man standing at the bus stop / at the door / at the window?
站在公共汽车站/门口/窗口的那个人是谁?
Turn left at the traffic lights / at the church / at the roundabout.
在红绿灯处/在教堂/在环形交叉路口向左转。
Angela's house is the white one at the end of the street.
安琪拉的房子是在街的尽头的那座白色的房子。
When you leave the hotel, please leave your key at reception.
当你离开旅馆时，请把钥匙留在接待处。
2.at表示“在顶端／底端”：
Write your name at the top / at the bottom of the page.
把你的名字写在这页的顶部/底部。
C. on
on表示“在......上”。例如：
I sat on the floor / on the ground / on the grass / on a chair / on the beach.
我坐在地板上/地上/草地上/椅子上/海滩上。
There's a dirty mark on the wall / on the ceiling / on your nose / on your shirt.
墙上/天花板上/你的鼻子上/你的衬衫上有一块污迹。
Have you seen the notice on the notice board / on the door?
你看到布告栏/门上的通知了吗?
其他带有on的常见表达
(1) 在左边/右边 on the left / on the right
(2) 在某个楼层 on the ground floor 在一楼（英国的一楼叫做ground floor, first floor是二楼）
(3) 在第几页 on page seven
(4) 在农场、操场等开阔空间里 on the farm, on the playground
D. 易混点辨析
in vs. at
（1）in强调在“内，里”；at强调在“点”。试比较：
There were a lot of people in the shop. It was very crowded.
商店里有很多人。那里非常拥挤（强调商店里面这个空间）。
Go along this road, then turn left at the shop.
沿着这条路走，然后在商店左转（强调在商店这个位置所采取的行动，商店看做一个整体）。
（2）in后面加地点，一般是面积比较大的地理单位，如城镇、国家，at后面的地点一般都是建筑、机构：
Two years later, I met Chris again in Edinburgh.
两年后，我和克里斯在爱丁堡再次见面。
We met at the post office this morning.
今天早上我们在邮局见面了。
at vs. on
at强调在“点”； on强调在“上”，可以是表面附着的。试比较：
There is somebody at the door. Shall I go and see who it is?
有人在敲门。要我去看看是谁吗（强调在门那个位置）？
There is a notice on the door. It says "Do not disturb".
门上有个通知，上面写着“请勿打扰”（强调在门上附着）。
方式介词
A. by
by表示用某种方法、手段、途径等:
They met force by force.
他们以暴力对付暴力。
I defeated him by knocking down his argument.
我通过驳斥他的论点挫败了他。
The stone rolled down the mountain by gravity.
这块石头由于重力作用而滚下山。
by后面加表示交通方式的名词，意为乘坐（交通工具），注意后接的名词前面通不用冠词:
to travel by boat/bus/car/plane 乘船╱公共汽车╱轿车╱飞机
to travel by air/land/sea 坐飞机；经陆路╱海路
B. in
表示用某种媒介或语言：
You can do this in a different way.
你可以用不同的方法做这件事。
What’s this in French?
这个用法语怎么说？
C. with
with表示“用...工具“，后接具体的工具、手段或其他行为方式：
He plays table tennis with his left hand.
他用左手打乒乓球。
She cloaked her sorrow with laughter.
她用笑来掩饰她的悲痛。
D. through
through表示“通过（方法、手段）”、“经由”:
We learn a second language through listening, speaking, reading and writing.
我们通过听、说、读、写来学习第二语言。
These plants take in water through their roots.
这些植物通过根部吸收水分。
表趋向与方位
A. 表趋向：across & over
across和over用来表示在（或到达）桥、路、边境或河流等的另一边：
Mike lives in the house across/over the road from ours.
迈克住在我们家马路对面的房子里。
The truck came towards them across/over the bridge.
卡车从桥上向他们驶来。
表示到达某个比较高的或者高度比宽度长的物体的对面去时，用over：
He jumped over the fence into the garden.
他跳过篱笆进入花园。
表示某个我们认为是平面的事物，或表示国家或海洋等区域时，用across，不用over：
The programme was broadcast across Australia.
该节目在澳大利亚全国播出。
The figures moved rapidly across the screen.
这些数字在屏幕上迅速移动。
B. 表趋向：along & through
表示沿着某种呈线状的事物（例如道路、河流）时，用along：
They walked along the footpath until they came to a small bridge.
他们沿着人行道一直走到一座小桥跟前。
through用来强调谈论的是发生在三维空间里的运动，周围被很多事物包围，而不是发生在一个二维空间或平面区域中：
He pushed his way through the crowd of people to get to her.
他从人群中挤过去找她。
through经常表示从空间的一边或一头运动到另一边或另一头：
She walked through the forest to get to her grandmother's house.
她穿过森林去奶奶家。
C. 表方位：above& over，below& under
表示一物的位置比另一物高时，可以用above或over:
Above/Over the door was a sign saying, ‘Mind your head’.
门的上方有一块牌子，上面写着“小心碰头”。
当某物不是垂直于另一物之上时，常用above:
They lived in a village in the mountains above the lake.
他们住在湖上群山中的一个村庄里。（山在湖的海拔位置之上，并不垂直悬空于湖面）
说明某物覆盖在另一物之上并与之接触时，用over而不用above：
She put a quilt over the bed.
她把被子盖在床上。
below是above的反义词，under是over的反义词，它们之间的区别与above和over之间的区别类似：
It’s hard to believe that there is a railway line below/under the building.
很难相信在大楼下面有一条铁路。
(表示一物的位置比另一物低时，可以用below或under)
Her head was below the lever of the table so nobody noticed her.
她的头低于桌子的高度，所以没有人注意到她。
（头并不是在桌子的下方，而是相对于桌子的高度较低，用below）
She hid the present under a blanket.
她把礼物藏在了一张毯子下面。
(在某物下方且有接触，用under)
D. 易混点辨析：between vs. among
当有两个人或事物时，用between；当有三个或超过三个以上的人或事物时，用among：
The treaty was signed between Great Britain and France.
这项条约是英国和法国之间签订的。（两国之间，用between）
He stood among all his friends in the room and felt very happy.
他站在房间里所有的朋友中间，感到非常高兴。（all说明朋友数量大于三个）
注意：between说明后面所跟内容的独立性（Great Britain和France都指的不同对象）；而among表示后面的内容所代表的一个整体／整体的一部分（all his friends看做一个整体）。如果between后面跟了三个对象，则表示“这三者，彼此之间都....”：
There is a disagreement between Neil, John, and Margaret.
尼尔、约翰和玛格丽特之间有分歧（三个人彼此之间谁也不同意谁。）
表排除
A. except & except for
except和except for后的内容不包括在前面提及的整体内：
The price of the holiday includes all meals except (for) lunch.
假期的费用包括午餐以外的所有餐费。
Everyone has been invited except (for) me.
除了我之外的每个人都被邀请了。
except与except for的区别
(1) 表示主句的描述不完全准确时，用except for：
The car was undamaged in the accident, except for a broken headlight.
除了一个坏了的前灯，汽车在事故中没有损坏。
（主句描述的“汽车在事故中没有被损坏”不完全准确，因为车灯坏了）
Except for the weather, the holiday couldn't have been better.
（主句描述的“假期真的太棒了”不完全准确，因为天气不太好）
(2) except for可以置于句首，但except不可：
Except for the service, I enjoy everything about this restaurant.
除了服务以外，我觉得这家餐厅还不错。
(3) 当后面接介词短语、动词的不定式和that引导的从句时，用except：
There is likely to be rain everywhere today except in Scotland.
除在苏格兰之外的任何地区今天都可能有雨。（后面接介词短语）
I rarely need to go into the city center except to do some shopping.
我几乎不去市中心，除非我要购物的时候。（后面接带to的不定式）
The boy didn't do anything except cry last night.
这个男孩昨晚除了哭以外什么也没做。
（注：当句子中使用的谓语是单词do的任何形式时，那么except的后面接不带to的不定式，即动词原形）
They look just like the real thing, except that they are made of plastic.
它们看起来就像真的东西，除了他们是由塑料做的。
B. besides& apart from
besides后的内容是包括在范围内的：
Besides cricket, I enjoy watching football and basketball.
除了板球，我还喜欢看足球和篮球。
I've got no family besides my parents.
除了父母，我没有其他亲人了。
注意：besides还可以作为副词，表示“况且，再说”，表示补充：
I don't really want to go. Besides, it's too late now.
我并不真的想去。况且现在太晚了。
apart from作介词时，用法和besides相似，表示“除了...还...”，后面的内容是也是包括在主句范围内的：
Apart from cricket, I enjoy watching football and basketball.
除了板球，我还喜欢看足球和篮球。
C. but& but for
but的意义与except（for）相似，但but常用在表否定意义的词汇后：
There was no way but upwards, towards the light.
没有别的路，只有向上，向着光明。
Immediately after the operation he could see nothing but vague shadows.
手术一结束，他除了模糊的影子什么也看不见。
but for表示要不是因为...，一些另外的事情就可能（不）发生：
But for his broken leg he would probably have been picked for the national team by now.
要不是他的腿摔断了，他现在可能已经入选国家队了。
（注：would have been为虚拟语气，用于表达对过去的情况的假设）。
The country would now be self-sufficient in food but for the drought last year.
要不是去年的干旱，这个国家现在在粮食方面是自给自足的。
介词短语：名词+介词
A. 名词+for
当前面的名词表示要求/需要/原因等，且前后存在因果关系时，后面一般用介词for：
The company closed down because there wasn't enough demand for its product.
这家公司由于产品供不应求而倒闭。
There's no excuse for behavior like that. There's no need for it.
没有任何借口可以为那样的行为辩解，也无需辩解。
The train was late, but nobody knew the reason for the delay.
火车晚点了，但是没有人知道晚点的原因。
B. 名词+of
当前面的名词与介词后面的名词存在所属关系时，后面一般用介词of：
Rachel showed me a picture of her family.
瑞秋向我展示了她一张全家福。
I had a map of the town, so I was able to find my way around.
我有这个小镇的地图，因此我可以找到路。
C. 名词+to
当表示“针对，对于...”意思时，名词后面一般用介词to：
I hope we find a solution to the problem.
我希望我们可以找到针对这个问题的解决方法。
I was surprised at her reaction to my suggestion.
他对我提出的建议所做出的反应，令我很惊讶。
His attitude to his job is very positive.
他对工作的态度积极向上。
D. 名词+with/between
当前面的名词表示“关系，联系”时，后面一般用介词with/between，其中between是表示两者间的关系：
Do you have a good relationship with your parents?
你跟你父母关系良好吗？
There are some differences between British and American English.
英式英语和美式英语之间存在差别。
E. 名词+in
当前面的名词表示变化时，后面的介词用in：
There has been an increase in the number of road accidents recently.
最近交通事故的数量有所增加。
Last year there was a big fall in sales.
去年销售额大幅下降。
介词短语：形容词+介词
A. 表示人的品质、能力、知觉的形容词
当形容词表示人的品质时，一般后面用介词of，介词后面加人称代词或人名：
Thank you. It was very kind of you to help me.
谢谢你，你人真好，乐意帮助我。
It was stupid of me to go out without a coat in such cold weather.
我真傻，这么冷的天竟然没穿外套就出门。
表示有能力/无能力做某事，用capable/incapable of:
I’m sure you are capable of passing the examination.
我确定你有能力通过考试。
表示意识到、察觉到，用aware/ conscious of:
She slipped away without him being aware of it.
她悄悄离开，没有让他发觉。
He became acutely conscious of having failed his parents.
他深深感到自己辜负了父母的期望。
B. 与表达情绪的形容词搭配的介词
当形容词表达烦恼/沮丧/兴奋/抱歉等情绪时，一般后面用介词about：
Lisa is upset about not being invited to the party.
丽萨因为没有获邀参加派对而感到沮丧
I am so excited about watching the football game.
观看足球赛时我特别兴奋。
当形容词表示对收到的东西或某事的结果具有某种感情或态度时，一般后面的介词用with:
They were satisfied with the present I gave them.
他们对我送的礼物很满意。
Were you happy with the exam result?
你对考试结果还满意吗？
当形容词表达惊喜和震惊等意思时，一般搭配at：
I was surprised at the gifts my mother gave me on my birthday.
妈妈送我的生日礼物，令我感到十分惊喜。
I hope you weren't shocked at what I said.
我希望你不会为我所说的话而感到震惊。
She was deeply ashamed of her behaviour at the party.
她对自己在聚会上的行为深感羞愧。
He didn’t trust me. He was suspicious of my motives.
他不信任我。他对我的动机表示怀疑。
表示害怕/羞耻/嫉妒/怀疑/容忍等情绪的形容词，常和of搭配：
Are you afraid of spiders?
你害蜘蛛吗？
C. 其他常见的形容词+介词的搭配
good/bad/ at 擅长，不擅长
similar to 与...相似
different from 与...不同
interested in 感兴趣
keen on 热衷于，喜欢
proud of 对...感到骄傲
fond of 喜欢
typical of 典型
full/ short of 充满/缺乏
dependent on 依赖于
independent of不依赖于
famous for 因...出名
responsible for 对...负责
介词短语：动词+介词
A. 常见的动词+介词组合
一些动词后面常跟特定的介词，同一动词与不同介词搭配有不同的意义。下表列出了一些常见的动词+介词组合：
B. 介词对词组意义的影响
about通常表示“关于某件事”：
They began to learn about nutrition when they were at primary school.
他们从小学就开始学习营养知识。
We talked about a lot of things in the meeting.
我们在会议上谈论了许多事情。
Lucy asked about you yesterday when I met her.
我昨天遇见露西时，她问起你。
for和ask, argue连用表示人们想通过该动作得到什么：
He finished the drink quickly and asked for another.
他很快喝完了饮料，又要了一杯。
The British Sports Minister is reported to be ready to argue for an experimental lifting of the ban.
据报道，英国体育大臣准备争取试行取消那条禁令。
of和talk, how，learn连用表示讨论、拥有或得到信息，比about更为正式：
Some talk of him as future presidential material.
一些人谈到他是当未来总统的材料。（talk about有同样意义，但是不够正式）
The whole country knew of Churchill’s love of cigars.
全国都知道丘吉尔热爱雪茄。（know about有同样意义，但是不够正式）
I have just learned of the death of Dr. Jones.
我刚听说琼斯博士的去世的消息。（learn about有同样意义，但是不够正式）
on和talk或argue连用意为“关于”，“就......而言”：
I was asked to talk on my research.
我被邀请谈谈我的研究。
We agreed on a time to meet.
我们约定了见面的时间。（就...达成一致）
with和动词连用引出所涉及的人：
I used to argue with Leo for hours.
我过去常常和利奥争论几个小时。
Adam thinks we should accept the offer, and I agree with him.
亚当认为我们应该接受这个提议，我同意他的看法。
注意，agree with还表示赞同某个想法或行为：
I agree with letting children choose the clothes they wear.
我同意让孩子们选择他们穿的衣服。
码字好辛苦~点个赞鼓励一下我呗~
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JimmyGs
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JimmyGs来啦~
介词这个东西，看着很玄，理解起来很玄，不过做起题来...还是很玄
不过，要弄好介词，你首先要知道的是，英语的本质是图形文字。不像中文的象形文字，文字本身就是象形。
就拿on为例，你以为它的意思是“在...上面”？
No No No...
它的本义其实是“接触”这个画面。就是这个
它的反义词就是off，“掉落”
所以用最直观的图像来看：
turn on 打开电源（就是开关接触上了）
turn off 关闭电源（开关不接触了，脱落了）
到这里还是很好理解的，那我们再来推导一下，看看这些词组：
put on the table 虽然意思是“放在桌面上”，不过你想象一下它的画面，是不是就是“把它跟桌面接触”？
keep on 坚持，是不是就是“保持接触，不要断”？
pay attention on sth. 注意某事，用了“on”，是不是就是“把注意力跟sth接触”的画面？
这样一想，画面一出来，这个短语根本不用背呀，一想就知道了呀~
再来一个“about”，你以为它的意思是“关于”？
No No No...
它的画面是“环绕”
看到这个 图，就是“about”的画面了，而且你也学到了一个新单词——“roundabout”，“环岛”。其实就是简单的“round 圆形”➕“about 环绕”。
这是最直观的，那我们再来看看这些短语 ：
talk about 谈论关于... ，它的画面其实是“围绕...来谈论”，神奇吧！！英语就是这样的图形文字
learn about 它不是“学习关于...”，而是“围绕...来学习”
这就是为什么 care about 除了有“关心...”，它还有“对...感兴趣”的意思，“围绕着...来关心”嘛~
辣么~最后，来猜猜 walk about 是啥子意思？
更多英语学习方法、英语思维探讨，w➕farmerjim 咯~下次再见
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Fernando阿肥子
英语教学那些事儿
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虽然但是，介词这种东西只有部分是通过中文可以理解的，有一部分介词是通过中文解释中国人也没办法理解的。
英语介词非常发达，原因是英语是静态语言，名词发达，介词通常用在名词前面，所以介词发达，这是作为动态语言的中文所不擅长的，以中文为母语的人往往不能完全理解英语特有的介词系统的全部用法，即便通过中文详细解释，都难以掌握，就好像很多英语学习者一直无法准确感受到虚拟语气的用法一样（这也是中文没有的）。
再比如中文量词发达，这也是外国人学中文的时候很懵的地方，我们学他们的介词也有这种问题：一条狗，一本书，一辆车，一栋楼，一匹马，一滩水，一架飞机，一朵花，一头狼，一尊佛，一扇窗...
试问你是否推荐外国人学中文的时候通过理解的方式去学习中文的量词？我们都是通过积累的，而非通过推敲，因为推敲和强制记忆反而行不通。
结论：介词学习多阅读，从实践中积累才是最正确的方式。
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4344 | https://www.semanticscholar.org/paper/Period-Three-Implies-Chaos-Li-Li/144be7a4c12d40f45aca578292d353d74d00e3ab | DOI:10.1007/978-0-387-21830-4_6
Corpus ID: 197457607
Period Three Implies Chaos
@article{Li1975PeriodTI,
title={Period Three Implies Chaos},
author={Tien-Yien Li and James A. Yorke},
journal={American Mathematical Monthly},
year={1975},
volume={82},
pages={985-992},
url={
}
Tien-Yien Li, J. Yorke
Published 1 December 1975
Mathematics
American Mathematical Monthly
The way phenomena or processes evolve or change in time is often described by differential equations or difference equations. One of the simplest mathematical situations occurs when the phenomenon can be described by a single number as, for example, when the number of children susceptible to some disease at the beginning of a school year can be estimated purely as a function of the number for the previous year. That is, when the number x n+1, at the beginning of the n + 1st year (or time period…
View via Publisher
3,539 Citations
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4345 | https://edu.rsc.org/resources/aromatic-chemistry-16-18/4010285.article | Aromatic chemistry 16–18 | Resource | RSC Education
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Starters for 10: Advanced level 2 (16–18)
6 Aromatic chemistry
1 Introduction
2 Kinetics
3 Equilibria
4 Acids and bases
5 Carbonyl chemistry
6 Aromatic chemistry
7 Compounds with amine groups
8 Polymers
9 Structure determination
10 Organic synthesis
11 Thermodynamics
12 Periodicity
13 Redox equilibria
14 Transition metal chemistry
15 Inorganics in aqueous solution
Aromatic chemistry
No comments
Test learners knowledge of aromatic chemistry
The topics covered in this Starter for ten activity are: Naming aromatic compounds, industrially important molecules. structure of benzene, electrophilic substitution, and synthestic routes with benzene.
Example questions
Name the following aromatic derivatives using IUPAC nomenclature
Name the benzene derivative and outline why it is industrially important.
The benzene ring is present in many important compounds and therefore determining its structure was very important. The German chemist Kekulé (1865) had suggested that benzene existed as a hexagonal carbon ring with alternate single and double bonds. Professor Ingold of London University prepared some crystals of hexamethylbenzene and sent them to the crystallographers at the University of Leeds for analysis. There Kathleen Lonsdale showed by mathematical analysis of her X-ray diffraction pattern that Kekulé’s proposed structure was incorrect. Kathleen later became a Professor at University College London and only the second woman fellow of the Royal Society.
Draw the structure Kekulé proposed
Why could the crystallographers not work on benzene rather than the hexamethyl derivative?
If Kekulé’s structure were true, what would the X-ray of benzene show? (you can use pictures to illustrate your answer where appropriate)
What features of benzene were shown up by the X-ray analysis that helped to disprove the Kekulé structure?
If benzene had the structure proposed by Kekulé, what kind of reaction mechanisms would it show?
The enthalpy of hydrogenation of cyclohexene is -120 kJ.mol-1. Using this information what would you expect the enthalpy of hydrogenation of benzene to be if the Kekulé structure holds true? The actual enthalpy of hydrogenation of benzene is -208 kJ.mol-1. What does this infer about the actual structure of benzene?
Notes
The full question and answer sheet is available from the ‘Downloads’ section below. An editable version is also available.
Downloads
Aromatic
PDF, Size 0.32 mb
#### Aromatic - editable
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Starters for 10: Advanced level 2 (16–18)
1### Introduction
2### Kinetics
3### Equilibria
4### Acids and bases
5### Carbonyl chemistry
6Currently reading
Aromatic chemistry
7### Compounds with amine groups
8### Polymers
9### Structure determination
10### Organic synthesis
11### Thermodynamics
12### Periodicity
13### Redox equilibria
14### Transition metal chemistry
15### Inorganics in aqueous solution
Level
16-18 years
Use
Handout
Starter
Download
Editable
Category
Structure and bonding
Organic chemistry
Equations, formulas and nomenclature
Reactions and synthesis
Specification
England
A/AS level
AQA Chemistry
Organic chemistry
Introduction to organic chemistry
Nomenclature
Apply IUPAC rules for nomenclature to name organic compounds limited to chains and rings with up to six carbon atoms each.
Aromatic chemistry
Bonding
The nature of the bonding in a benzene ring, limited to planar structure and bond length intermediate between single and double.
Delocalisation of p electrons makes benzene more stable than the theoretical molecule cyclohexa-1,3,5-triene.
Students should be able to: use thermochemical evidence from enthalpies of hydrogenation to account for this extra stability.
Electrophilic substitution
Electrophilic attack on benzene rings results in substitution, limited to monosubstitutions.
Nitration is an important step in synthesis, including the manufacture of explosives and formation of amines.
Friedel–Crafts acylation reactions are also important steps in synthesis.
Students should be able to outline the electrophilic substitution mechanism of: nitration (including the generation of the nitronium ion).
Students should be able to outline the electrophilic substitution mechanism of: acylation (using AlCl₃ as a catalyst).
Amines
Preparation
Aromatic amines, prepared by the reduction of nitro compounds, are used in the manufacture of dyes.
Edexcel Chemistry
Topic 6: Organic Chemistry I
Topic 6A: Introduction to organic chemistry
4. be able to name compounds relevant to this specification using the rules of International Union of Pure and Applied Chemistry (IUPAC) nomenclature Students will be expected to know prefixes for compounds up to C₁₀
Topic 18: Organic Chemistry III
Topic 18A: Arenes - benzene
1. understand that the bonding in benzene has been represented using the Kekulé and the delocalised model, the latter in terms of overlap of p-orbitals to form π-bonds
2. understand that evidence for the delocalised model of the bonding in benzene is provided by data from enthalpy changes of hydrogenation and carbon-carbon bond lengths
4 iii. a mixture of concentrated nitric and sulfuric acids
4 iv. halogenoalkanes and acyl chlorides with aluminium chloride as catalyst (Friedel-Crafts reaction)
5. understand the mechanism of the electrophilic substitution reactions of benzene (halogenation, nitration and Friedel-Crafts reactions), including the generation of the electrophile
Topic18B: Amines, amides, amino acids and proteins
12. know that aromatic nitro-compounds can be reduced, using tin and concentrated hydrochloric acid, to form amines
OCR Chemistry A
Module 6: Organic chemistry and analysis
6.1 Aromatic compounds, carbonyls and acids
6.1.1 Aromatic compounds
a) the comparison of the Kekulé model of benzene with the subsequent delocalised models for benzene in terms of p-orbital overlap forming a delocalised π-system
b) the experimental evidence for a delocalised, rather than Kekulé, model for benzene in terms of bond lengths, enthalpy change of hydrogenation and resistance to reaction
c) use of IUPAC rules of nomenclature for systematically naming substituted aromatic compounds
di) the electrophilic substitution of aromatic compounds with: i) concentrated nitric acid in the presence of concentrated sulfuric acid
dii) the electrophilic substitution of aromatic compounds with: ii) a halogen in the presence of a halogen carrier
diii) the electrophilic substitution of aromatic compounds with: iii) a haloalkane or acyl chloride in the presence of a halogen carrier (Friedel–Crafts reaction) and its importance to synthesis by formation of a C–C bond to an aromatic ring
e) the mechanism of electrophilic substitution in arenes for nitration and halogenation
Wales
A/AS level
WJEC Chemistry
Unit 4: ORGANIC CHEMISTRY AND ANALYSIS
4.2 Aromaticity
(a) structure of and bonding in benzene and other arenes
(c) mechanism of electrophilic substitution, such as in the nitration, halogenation and Friedel-Crafts alkylation of benzene, as the characteristic reaction of arenes
Northern Ireland
A/AS level
CCEA Chemistry
Unit A2 2: Analytical, Transition Metals, Electrochemistry and Organic Nirtrogen Chemistry
5.7 Amines
5.7.4 explain the formation of phenylamine by reduction of nitrobenzene using tin and concentrated hydrochloric acid, to form the phenylammonium salt, followed by liberation of the free amine by addition of alkali;
5.7.8 explain the formation of benzenediazonium chloride from phenylamine and recall the coupling of diazonium ions with phenol and phenylamine;
Unit A2 1: Further Physical and Organic Chemistry
4.10 Aromatic chemistry
4.10.1 explain the structure of the benzene molecule with reference to delocalised π electrons; and
4.10.3 explain the mechanisms of the monobromination, mononitration, monoalkylation and monoacylation of benzene, including equations for the formation of the electrophile;
4.10.4 recall the names of the electrophiles for the bromination and nitration of benzene; and
Republic of Ireland
Leaving Certificate
Chemistry
5. Fuels and heats of reactions
5.3 Aromatic Hydrocarbons
Depth of treatment
Structure of benzene, methylbenzene and ethylbenzene as examples of aromatic compounds.
7. Organic chemistry
7.2 Planar Carbon
Activities
Solubility of propanone in (i) cyclohexane and (ii) water.
Depth of treatment
Simple explanation of the use of the circle to represent the identical bonds in benzene, intermediate between double and single.
Sigma and pi bonding in benzene.
Aromatic compouds.
An indication of the range and scope of aromatic chemistry (structures not required).
7.3 Organic Chemical Reaction Types
Depth of treatment
Unreactivity of benzene with regard to addition reactions, relative to ethene.
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4346 | https://www.oit.edu/sites/default/files/document/chapter-2-2.pdf | 2 2.2 Naming of Alkanes Frequently we will be looking at large, complicated organic molecules and we will want to refer to a small portion of that molecule by name without referring to the whole molecule. To take a simple example, let’s look at the branched isomer of butane: We can regard this molecule as being a chain propane molecule with a CH3 group being substituted on the middle carbon atom and providing a branch point. Note that this CH3 group is almost but not quite a methane molecule. There is one less H atom so that the CH3 group can be bonded by its fourth bond to the propane chain. We call this CH3 group a methyl group. Similarly in the molecule: We have a -CH2CH3 group being branching off of the middle C atom of a pentane molecule. Again the -CH2CH3 group is almost an ethane molecule except that one hydrogen has been removed so that the bond can be used to attach the group to the main chain. The -CH2CH3 group is called an ethyl group. These are names for alkyl groups for any given number of carbons, but we will be concerned only with the names for chains ranging from 1-10 carbon atoms. Number of C Atoms Name 1 methyl 2 ethyl 3 propyl 4 butyl 5 pentyl 3 6 hexyl 7 heptyl 8 octyl 9 nonyl 10 decyl The first four names (which are by far the most common) can be remembered by the mnemonic device: Methyl, ethyl Propyl, butyl After that, It’s all futile! There are 2 isomers with the formula C4H10: The unbranched chain isomer is called butane as indicated by the rules given above. The branched chain isomer was originally called isobutane, because it was an isomer of butane. For the formula C5H12 there are 3 isomers as shown below. The first molecule is the unbranched chain pentane; the second molecule has a methyl(-CH3) branch off of a 4-C chain, and the third has 2 methyl(CH3) groups coming off the middle C atom of the 3 carbon chain: The first molecule is called pentane as before; the second isomer was called isopentane, and the third isomer was called neopentane because it was a new isomer (neo-is the Latin for new). 4 If we look at the table below however, we see that naming these isomers by simply putting prefixes in front of the name of the straight chain isomer will not be a very easy task for organic molecules containing a large number of C atoms. We will simply run out of names. Chemical Formula Number of Isomers C6H14 5 C7H16 9 C10H22 75 C15H32 4347 C40H82 62,500,000,000,000 Clearly a more systematic way of naming molecule is necessary and organic chemists have developed such a system, called IUPAC nomenclature (International Union of Pure and Applied Chemistry). We will present the rules necessary to name alkanes in this unit, and will present additional rules for naming other functional groups in subsequent units. Although systematic names are used by professional chemists, it should be noted that most drug and food molecules are referred to by common non-systematic names. These names must simply be memorized. Rules of Systematic Nomenclature 1. Name the longest (straight) unbranched carbon chain in the structure. 2. Preceding the name of the longest chain, write down in alphabetical order the names of each of the alkyl groups which are attached to the main chain. 3. If there are several groups of the same kind, list the group only once, using appropriate prefix: di- for 2, tri- for 3, tetra- for 4, penta- for 5, hexa- for 6, hepta- for 7, octa- for 8 to indicate how many of the groups there are. 4. Assign a number, as a prefix, to each of the alkyl groups in the name to indicate the position of the group on the main chain. Start numbering from whichever end of the main chain results in the lowest sum of numbers. .Examples: Draw and name 2 isomers of butane (C4H10) 5 butane (as before) 2-methylpropane 3 isomers with the formula (C5H12) pentane (as before) 2-methylbutane longest chain is 4 C’s, so name compound as a derivative of butane with a methyl group hanging on 2nd C from the end We might draw as still another isomer and name it 3-methylbutane. Closer inspection shows that if we simply flip the molecule over from left to right we will have 2-methylbutane, which is the preferred name, since it has a smaller number. 2,2-dimethylpropane longest “straight” chain is 3C’s so name it propane; two methyl groups hanging off second C so we have 2,2-dimethylpropane Note that IUPAC rules require that the number 2 be repeated to make it absolutely clearly that both methyl groups are on #2 C of the unbranched chain. Draw 4 isomers with formula C6H14 6 Solution: 1) hexane 2) 2-methyl pentane. (We want to start numbering from left to right to keep the numbering as small as possible. If we move the branch down one C on the main chain we have: The name of this molecule is 3-methyl pentane and it does not matter from which end of the molecule we start the numbering. If we move the methyl group down another C on the main chain we have You might initially think this is a new isomer, 4-methylpentane, but closer inspection should make you realize this molecule is the same as 2-methylpentane, just flipped from left to right. If we move the methyl branch 1 C further down the chain we have . Careful inspection of this conformation should reveal that it is NOT a branched isomer at all. It is just a different conformation of an unbranched 6 C hexane chain. It may be easier to visualize this if you take a pencil (or pen) and start from left to right. You can cover all 6 C atoms in one continuous stroke of the pencil. 4) There are however two more isomers of C6H14 which have two methyl groups branching off a 4 C unbranched chain. 2,2 dimethylbutane 2,3 dimethylbutane Additional naming practice: 7 a) The longest straight chain is 6 C’s so name it hexane; add the branching chains in alphabetical order: ethyl and methyl. We still need to indicate where the ethyl and methyl groups are hooked on and that’s where the numbers come in. Numbering from left to right we obtain: 3-ethyl-2-methylhexane Sum of numbers is 5 Numbering from right to left: 4-ethyl-5-methylhexane. The sum of the numbers is 9, so numbering from left to right is preferred. Draw the line-bond notation for the above molecule. b)Name: Longest straight chain, ignoring branches, is 7 carbons. We have 3 methyl groups and one ethyl group branching off the main chain so our in front we have ethyl trimethyl Now add the numbers, try starting from either end (then add the sum of the numbers). Numbering from left to right we get 4-ethyl-2,2,5-trimethylheptane. Sum of #’s is 13 Numbering from the right to left we have 4-ethyl-3,6,6-trimethylheptane.Sum of #’s is 19 First answer is correct name because the sum of numbers is smaller. c)Name At first glance it might appear that the longest chain(reading straight across) has 6 C, but look again and notice there is a 4 C branch on C #4 (from the left). This “branch” is longer than the 2 C chain continuing horizontally. Since free rotation allows putting the chain in whichever direction we wish, the longest unbranched chain contains 8 C atoms Numbering from left to right we have: 4-ethyl-2,2,3-trimethyloctane. Sum of numbers is 11 Number from right to left we have 5-ethyl-6,7,7 trimethyloctane. Sum of numbers is 25 First name is the correct one. 8 Naming alkanes with halogens In naming a molecule which contain halogens (F,Cl,Br, and I) we treat the halogen just like an alkyl group and indicate its presence with the following names: fluoro for F chloro for C1 bromo for Br iodo for I Older common names for simple halogenated compounds often use the ionic form of the halogen name i.e. fluoride, chloride, bromide, and iodide even though no ionic bond is present. CH3Cl is named chloromethane. The older name is methyl chloride. CH3Br is named bromomethane. The older name is methyl bromide . Let’s look at some examples: a)Two molecules with which we introduced the concept of structural isomers: 1,1-dichloroethane 1,2-dichloroethane The longest carbon chain in both cases is ethane and in both cases there are two chlorine atoms so name them both dichloroethane. However in the first case both C1 atoms are on the same C atom so call it 1,1-dichloroethane. In the second molecule the C1 atoms are on different carbon atoms so we indicate that as 1,2-dichloroethane. b)Two isomers with the formula C2H3C13 1,1,1-trichloroethane 1,1,2-trichloroethane (not 2,2,2-trichloroethane (not 1,2,2-trichloroethane) and not 1-trichloroethane) c) A molecule with both alkyl and halogen groups 9 Numbering from left to right: 5-chloro-3,3,4 trimethylhexane Sum of numbers is 15 Numbering from right to left: 2-chloro-3,4,4-trimethylhexane Sum of number is 13. Numbering from right to left gives the correct name. d) 2-chloro-3-ethylpentane 1,3-dichloro-3-ethyl-4-methylpentane What would the numbering be if we numbered from right to left? Examples with practical uses: Freons Dichlorodifluoromethane chlorotrifluoromethane trichlorofluoromethane (Freon12) (Freon13) (Freon 11) Freon 12 and other Freons have been used as the gases in refrigerators and air conditioners, but when they escape to the upper atmosphere, UV radiation can break the C-Cl bond and produce Cl free radicals (molecules with unpaired electrons) which can then destroy ozone in the upper atmosphere. Chlorofluorocarbons(CFC’s) were phased out in developed countries for spray cans in 1978, for refrigerators and air conditioners in 1995, and for medical inhalers (metered dose inhalers or MDI’s for inhaled drugs such as albuterol) as of 2008. They have been replaced with hydrofluoroalkanes (HFAs) 134a and 227 which are less damaging to the ozone layer. (The hydrofluoroalkanes are less damaging to the ozone layer because the C-F bond is not as susceptible breakage by UV radiation as the C-Cl bond.) 10 HFA 134a has the structure and the systematic name 1,1,1,2-tetrafluoroethane HFA 227 has the wedge and dash structure and the systematic name 1,1,1,2,3,3,3-heptafluoropropane. Notice that both of these molecules contain only F halogen atoms and no Cl atoms. The C-F bond is much more resistant to UV light than the C-Cl bond. The substitution of HFAs for CFCs is more complicated than you might think and has required extensive research in the last two decades. The HFAs are more expensive to produce and large scale testing of efficacy and toxicity of the newly reformulated inhalers was required by the FDA. The cost of metered dose inhalers (MDIs) using HFAs are substantially more expensive than the old inhalers using CFCs. Two albuterol inhalers reformulated with HFAs. Bromomethane (Methyl bromide) IUPAC: bromomethane Common: methylbromide Bromomethane (methyl bromide) has been used as a specialized soil fumigant, especially for growing strawberries. It is extremely toxic and great care must be taken in applying it. Methyl bromide is also a very potent ozone destroyer and there have been calls for taking it off the market for many years. 11 Trichloromethane(chloroform) Trichloromethane (chloroform) was one of the earliest general anesthetics ( a molecule which induces sleep as well as analgesia). It was administered by breathing the vapors of liquid chloroform and was first used during childbirth in 1847. It competed strongly with ether for the next 60 years as the general surgical anesthetic of choice. It has a sweet smell and is much less flammable and less irritating than ether, but the risk of putting the patient to sleep permanently is higher than for ether. Lethal overdose with chloroform was all too frequent and it could cause cardiac arrhythmias. Accumulated data published in 1934 showed that deaths from chloroform anesthesia were over four times as frequent as those with ether anesthesia (about 1 in 3000 for chloroform vs 1 in 14,000 for ether anesthesia). It is no longer use as an anesthetic. It continued to be used in some cough suppressants and toothpastes and other cosmetics until such use was banned in 1976 based on data from rats that it could cause birth defects and liver cancer. Current exposure in the US is very small, primarily from chlorinated water reacting with organic compounds in the water to form trace amounts of chloroform in drinking water. Tetrachloromethane(carbon tetrachloride) Tetrachloromethane(carbon tetrachloride) was commonly used as a drycleaning solvent in the first half of the twentieth century to remove grease and dirty from clothes.. It was also used in the synthesis of the Freon gases for refrigerators and air conditioners. Exposure to CCl4 was associated with neurological, liver, and kidney damage and it was replaced as a dry cleaning solvent. Like the Freon molecules, CCl4 can cause ozone depletion and its use has dropped dramatically. 12 IUPAC: 2-bromo-2-chloro 1,1,1-trifluoroethane Common: halothane This molecule has the common name of halothane(suggest why!). It came on the market n 1956 and replaced ether as the anesthetic of choice because it was less irritating to the airway and it was not flammable. It did however have some adverse effects including cardiac depression (slowing of the heart rate) and rare (1 in 35,000) cases of hepatitis. It gradually lost “market share” to safer anesthetics. Current inhalation anesthetics: The current most commonly used inhalation anesthetics are halogenated ethers shown below which eliminate the flammability problem of ether and are generally less irritating to the airway than ether. Isoflurane Desflurane Sevoflurane The mechanism of action of general anesthetics is poorly understood, is very complex, and probably involves many different factors. They appear to inhibit movement of the action potential along neurons as well as activating or inactivating a wide variety of neuronal receptors. Topical anesthetics that work by chilling the skin surface Name the following three molecules: a) b) c) 1 chloroethane 1,2-dichloro-1,1,2,2-tetrafluoroethane 1,1,1,3,3-pentafluoropropane ethyl chloride 13 a)The first molecule has the common name ethyl chloride and has been used as a mild topical anesthetic while doing minor surgeries (e.g. removing a deep splinter). It has a low boiling point (12 oC) and boils off the skin (at 37oC), lowering the skin temperature in the process and decreasing pain sensation. It is occasionally used as a recreational inhalant. Its use is decreasing.. b)The second molecule, 1,2-dichloro-1,1,2,2-tetrafluroethane, is marketed under the trade name, Frigiderm, works by a similar mechanism. It has a boiling point of 4oC. c) The third molecule, 1,1,1,3,3-pentafluoropropane, is marketed under the name Gebauer’s Spray and Stretch. It is marketed for reducing pain of sprains. (Boiling point = 14oC) Perfluoroalkanes (completely fluorinated C chains) are liquids which dissolve large amounts of oxygen while remaining biologically inert. Such fluids were the basis for the oxygenated liquid in the movie The Abyss. Perfluoroalkanes have been investigated for filling the lungs of premature infants with respiratory distress syndrome. Clinical trials of one such liquid (Liquivent) were disappointing and it was not approved for market. Perfluorochemicals are also accumulating in the environment due to their invertness, albeit in small concentrations. They can accumulate in the fat of animals (including humans). Although their short-term toxicity appears to be low, long term data is minimal and there is increasing concern about the long term biological effects of perfluorochemicals accumulating in the body. 14 Perfluorochemicals have also been investigated for their use as artificial blood which could be used instead of real blood for blood transfusions. Perfluorochemicals would have the advantage of not having to match blood types and would not require testing for viral contamination with HIV and hepatitis which are necessary for blood transfusions. More recent research has been done on an emulsion of perfluorochemical, water, and lecithin) with the trade name Oxygent TM. There is definitely a demand for a blood substitute, but that product has not yet been widely marketed. |
4347 | https://www.reddit.com/r/learnmath/comments/j8hclr/are_multiplication_and_division_really_inverse/ | Are multiplication and division really inverse operations? : r/learnmath
Skip to main contentAre multiplication and division really inverse operations? : r/learnmath
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r/learnmath•5 yr. ago
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Are multiplication and division really inverse operations?
1)Counterexample:
20=0
I can't reverse this multiplication by using division. Once we have 0 we can't get our 2 back. For this reason I conclude that division and multiplication are not inverse operations.
UPDATE:
2) Another argument. If they were inverse, then division would get as input one number and output two numbers. But it's not as real division work. So, they are not inverse.
Am I right? If not, why? If yes, then why are we taught that they are inverse?
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4348 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5079093/ | Common pitfalls in statistical analysis: The use of correlation techniques - PMC
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Perspect Clin Res
. 2016 Oct-Dec;7(4):187–190. doi: 10.4103/2229-3485.192046
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Common pitfalls in statistical analysis: The use of correlation techniques
Rakesh Aggarwal
Rakesh Aggarwal
1 Department of Gastroenterology, Sanjay Gandhi Postgraduate Institute of Medical Sciences, Lucknow, Uttar Pradesh, India
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1, Priya Ranganathan
Priya Ranganathan
1 Department of Anaesthesiology, Tata Memorial Centre, Mumbai, Maharashtra, India
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1 Department of Gastroenterology, Sanjay Gandhi Postgraduate Institute of Medical Sciences, Lucknow, Uttar Pradesh, India
1 Department of Anaesthesiology, Tata Memorial Centre, Mumbai, Maharashtra, India
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Address for correspondence: Dr. Priya Ranganathan, Department of Anaesthesiology, Tata Memorial Centre, Ernest Borges Road, Parel, Mumbai - 400 012, Maharashtra, India. E-mail: drpriyaranganathan@gmail.com
Copyright: © Perspectives in Clinical Research
This is an open access article distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as the author is credited and the new creations are licensed under the identical terms.
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PMCID: PMC5079093 PMID: 27843795
Abstract
Correlation is a statistical technique which shows whether and how strongly two continuous variables are related. In this article, which is the eighth part in a series on ‘Common pitfalls in Statistical Analysis’, we look at the interpretation of the correlation coefficient and examine various situations in which the use of technique of correlation may be inappropriate.
Keywords: Biostatistics; correlation; “data interpretation, statistical”
INTRODUCTION
We often have information on two numeric characteristics for each member of a group and are interested in finding the degree of association between these characteristics. For instance, an obstetrician may decide to look up the records of women who delivered in her hospital in the previous year to find out whether there is a relationship between their family incomes and the birth weights of their babies. The relationship here means whether the two variables fluctuate together, i.e., does the birth weight increase (or decrease) as the income increases.
“Correlation” is a statistical tool used to assess the degree of association of two quantitative variables measured in each member of a group. Although it is a very commonly used tool in medical literature, it is also often misunderstood. This piece describes what “correlation” implies and the situations in which it may be used, as also its pitfalls and the situations where it should not be used. To illustrate various concepts, we use scatter plots, a graphical method of showing values of two variables for each individual in a group.
MEASUREMENT OF CORRELATION: CORRELATION COEFFICIENT
The degree of correlation between any two variables on a continuous scale is mathematically expressed as the correlation coefficient (also known as Pearson's correlation coefficient or “r”), a number whose values can vary between −1.0 and +1.0. Thus, it has a sign (+ or −) and a magnitude.
Direction
Two variables are said to be “positively” correlated [Figure 1a-c] when their values change in tandem, i.e., increasing values of one are associated with increasing values of the other. By contrast, a “negative” correlation [Figure 1d-f] exists when increasing values of one variable are associated with a decrease in the values of the other. Variables with no or little discernible relationship [Figure 1g] are said to have “no correlation.”
Figure 1.
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Scatter plots of relationship between values of two quantitative variables and their corresponding correlation coefficient (r) values. “r” can vary between − 1.0 and + 1.0. If as the values of one variable (say on X-axis) increase, those of the other variable (on Y-axis) increase, “r” is positive (a-c); however, if the latter decrease, “r” is negative (d-f). When the values of two variables have no clear relation, “r” is zero (g). The absolute values of “r” are higher when the individual data points are closer to a line showing the linear trend (a > b > c; d > e > f)
Magnitude
The absolute value of r represents the strength of association. A value of 1.0 implies a perfect linear relationship between the two variables, i.e., all observations lie on a straight line [Figure 1a and d], whereas 0 indicates the absence of any linear relationship [Figure 1g]. Higher values (closer to 1.0) imply that individual observations lie close to an imaginary line describing the relationship between the two variables [Figure 1b and e], and lower values imply that the observations are more spread out [Figure 1c and f].
INTERPRETATION OF VALUE OF CORRELATION COEFFICIENT
Square of correlation coefficient (r 2), known as coefficient of determination, represents the proportion of variation in one variable that is accounted for by the variation in the other variable. For example, if height and weight of a group of persons have a correlation coefficient of 0.80, one can estimate that 64% (0.80 × 0.80 = 0.64) of variation in their weights is accounted for by the variation in their heights.
It is possible to calculate P value for an observed correlation coefficient to determine whether a significant linear relationship exists between the two variables of interest or not. However, with medium- to large-sized samples, these methods show even small correlation coefficients to be highly significant and hence their use is generally eschewed.
WHEN SHOULD CORRELATION NOT BE USED?
The correlation coefficient looks for a linear relationship. Hence, it can be fallacious in situations where two variables do have a relationship, but it is nonlinear. For instance, hand-grip strength initially increases with age (through childhood and adolescence) and then declines (e.g., Figure 2a). In such cases, “r” could be low (r = 0 for the data in Figure 2a), even though there is a clear relationship.
Correlation analysis assumes that all the observations are independent of each other. Thus, it should not be used if the data include more than one observation on any individual. For instance, in the above example, if hand-grip strength had been measured twice in some subjects that would be an additional reason not to use correlation analysis.
If one (or a few) individual observation in the sample is an outlier, i.e., located far away from the others, it may introduce a false sense of relationship [Figure 2b]. Please note that the data points in this figure are identical to those in Figure 1g, except for the addition of one outlier. On excluding this outlier, the value of r would drop from 0.71 to 0!
If the dataset has two subgroups of individuals whose values for one or both variables differ from each other [Figure 2c], this can lead to a false sense of relationship overall, even when none exists within each subgroup. For instance, let us consider a group of 20 men and 20 women. If one plots their heights (on X-axis) and hemoglobin levels (on Y-axis), most women may end up in the left lower corner (shorter and lower hemoglobin) and most men in the right upper corner (taller and higher hemoglobin), suggesting a false relationship (a positive “r” value) between height and hemoglobin levels.
With very small sample size (say 3–6 observations), a relationship may appear to be present even though none exists.
Linear correlation analysis applies only to data on a continuous scale. It should not be used when one or both variables have been measured using an ordinal scale, for example, patients’ assessment of pain severity on a scale of 0–10, where higher number means worse pain but similar differences (say from 1 to 3 and from 6 to 8) do not necessarily imply similar change in pain. In these cases, a Spearman's rank correlation method should be used.
Relationship between a variable and one of its components (e.g., aggregate marks vs. marks in one subject). For instance, it would be fallacious to use correlation to assess the relationship of height of a group of persons with the lengths of their body's lower segments since the lower segment forms a part of the overall height.
Heteroscedasticity or a situation in which the one variable has unequal variability across the range of values of a second variable. For instance, if one looks at the relationship of annual health expenditure versus the annual income of a family, the former is likely to vary more for richer persons than for poor persons [Figure 2d].
Figure 2.
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Situations in which linear correlation should not be used: (a) two variables have a relationship which is nonlinear (analysis of data points in this figure shows r = 0, thus failing to detect the relationship), (b) the data have one or a few outliers (one outlier at right upper end resulted in a false relationship with r = 0.71; exclusion of this point reduces r to near zero), (c) when the data have two subgroups, within each of which there is no correlation, and (d) when variability in values on Y-axis changes with values on X-axis. Each situation is described further in the text
Many of the above pitfalls are easily avoided if one first makes a scatter plot for the data and visually inspects it for nonlinear relationships, outliers, or presence of obvious subgroups.
In addition, correlation analysis is also often inappropriately used to measure agreement between two methods of measuring the same thing (e.g., tumor volume measured using ultrasound and computed tomography). This will be discussed in the next article in this series.
A FINAL CAUTION: CORRELATION DOES NOT MEAN CAUSATION
A relationship between two variables is sometimes taken as evidence that one causes the other. This is, however, often not true, and hence the popular statistical adage: “Correlation does not imply causation.” You may wish to visit for some interesting insights into how correlation can arise without any causative link.
Examples of such noncausative correlation include (i) countries’ annual per capita chocolate consumption and the number of Nobel laureates per 10 million population; (ii) weekly ice-cream consumption and a number of drowning incidents in swimming pools. These are due to the association of both the variables being studied to national income and hot weather, respectively.
ENDPIECE
Correlation analysis is a very powerful tool to explore relationships in data. However, one must be careful to use it only when it is applicable. Many of these problems can be avoided by a careful thought about the data, plotting the raw data (to look for nonlinear relationships, outliers, and heteroscedasticity of data), and by thinking in terms of coefficient of determination in preference to the correlation coefficient.
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Nil.
Conflicts of interest
There are no conflicts of interest.
REFERENCES
1.Messerli FH. Chocolate consumption, cognitive function, and Nobel laureates. N Engl J Med. 2012;367:1562–4. doi: 10.1056/NEJMon1211064. [DOI] [PubMed] [Google Scholar]
2.Maurage P, Heeren A, Pesenti M. Does chocolate consumption really boost Nobel Award chances? The peril of over-interpreting correlations in health studies. J Nutr. 2013;143:931–3. doi: 10.3945/jn.113.174813. [DOI] [PubMed] [Google Scholar]
Articles from Perspectives in Clinical Research are provided here courtesy of Wolters Kluwer -- Medknow Publications
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Abstract
INTRODUCTION
MEASUREMENT OF CORRELATION: CORRELATION COEFFICIENT
INTERPRETATION OF VALUE OF CORRELATION COEFFICIENT
WHEN SHOULD CORRELATION NOT BE USED?
A FINAL CAUTION: CORRELATION DOES NOT MEAN CAUSATION
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4349 | https://www.youtube.com/watch?v=KoVv8fNU48U | How To Convert (Change) Base Three(3) To Base Ten(10)@fischermath271
Fischer Math
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This video describes how to convert base 3 (three) to base 10 (ten). Place 0, 1, 2, 3, and 4 on the numbers to be converted, this shows the powers of 3. Then start to write out like this (1x3⁴)+ (0x3³)+(1x3²)+(2x3¹)+(1x3^0)=81+0+9+6+1=97 base 10 @fischermath271
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Transcript:
let's solve this let's convert this um Base to Base 10 solution we have 1 0 1 2 1 B three this can be written as assume this is zero 1 2 3 and four this can be WR as 1 3^ 4 + 0 3 power of 3 + 1 3 ra to power of 2 plus 2 3^ of 1 + 1 3^ of 0 so we begin to solve 3^ 4 will give us 81 plus this will give us Z so 0 3^ 3 27 plus 1 3 power 2 9+ 2 3 itself plus 1 1 anything power power 0 is 1 this is in base three in base 10 sorry base 10 |
4350 | https://math.jhu.edu/~lindblad/201/l8.pdf | Lecture 8: 3.1 Image and Kernel of a Linear Transformation If T: X →Y is a transformation then the set X is called the domain of T. The set Im(T) of all images T(x) when x varies over all points in the domain is called the image of T, or sometimes the range. Note that the image need not be all of the target space Y .
T is said to be onto if each y ∈Y is the image T(x) of at least one x ∈X.
T is said to be one-to-one if each y ∈Y is the image of at most one x ∈X.
T is called invertible if its one-to-one and onto.
Ex 7 Define T :R2 →R3 by T(x) = Ax, where A= 1 0 2 1 0 1 . Is T onto? What is the image?
Sol The image of T is all combinations of the column vectors of A T(x) = 1 0 2 1 0 1 x1 x2 = 1 2 0 x1 + 0 1 1 x2 for any x1 and x2. The image is the plane ’spanned’ by the two column vectors.
Given vectors v1, . . . , vk and scalars λ1, . . . , λk, the vector w = λ1v1 + · · · + λkvk is called a linear combination of the vectors v1, . . . , vk, with weights λ1, . . . λk.
The set of all linear combinations of a v1, ..., vn is called the span of v1, ..., vn and is denoted by Span(v1, ..., vn). The set {v1, ..., vn} span (is a spanning set for) V if every vector in V can be written as a linear combination of v1, ..., vn.
Th The image of a linear transformation T(x) = Ax is the span of the column vectors of A.
Pf T(x) = Ax = " | | v1 · · · vn | | # "x1 .
.
.
xn # = x1v1 + · · · + xnvn The image of a linear transformation T(x)=Ax is also called the column space of A, Col(A).
A subspace W of Rn is a subset which is closed under addition and scalar multiplication: (a) 0 ∈W, (b) u∈W and v∈W then u+v∈W, (c) w∈W and k is a scalar then kw∈W.
Ex A plane ax1 + bx2 + cx3 = 0 going through the origin in space is a subspace of R3.
Th The image of a linear transformation T(x) = Ax, from Rn →Rm is a subspace of Rm.
Pf For a proof see the proof of Theorem 3.1.4 in the textbook.
Alternatively it follows from the previous theorem and the following theorem: Th If v1, ..., vn ∈Rm then Span(v1, ..., vn) is a subspace of Rm.
Pf (b) follows from that sums of linear combinations are linear combination. In fact let W = Span(v1, ..., vn). Then if u = c1v1 + · · · + cnvn ∈W and w = d1v1 + · · · + dnvn ∈W it follows that u+w = (c1 +d1)v1 +· · ·+(cn +dn)vn ∈W since it is also a linear combination.
1 2 The kernel The kernel, Ker(T), of a linear transformation T : Rn →Rm is the set of all x in the domain such that T(x) = 0. It is a proper subset of the domain Rn unless T is the zero map.
Ex Let T:R3 →R2 by T(x)=Ax, where A= 1 2 0 0 1 1 . Is T one-to-one? What is the kernel?
Sol Ax = 0 has nontrivial solutions since there are more variables than equations. Hence there are infinitely many points such that T(x) = 0 so T is not one-to-one. Explicitly 1 2 0 0 0 1 1 0 ∼ 1 0 −2 0 0 1 1 0 , ⇔ x1 = 2x3 x2 = −x3 x3 = free The kernel is hence the subspace spanned by the line x = 2 −1 1 t, for any parameter t.
The kernel of a linear transformation T(x)=Ax is also called the null space of A, Nul(A) Th The kernel of a linear transformation T(x) = Ax, from Rn →Rm is a subspace of Rn.
Pf We must verify the three properties (a), (b), (c) in the definition of subspace.
(a) 0 ∈Nul A since A0 = 0.
(b) If u, v ∈Nul A, show that u + v ∈Nul A. A(u + v) = Au + Av = 0 + 0 = 0.
(c) If u ∈Nul A, show that λu ∈Nul A. A(λu) = λAu = λ0 = 0.
Ex 1 Find an explicit description of Nul A where A = 3 6 6 3 9 6 12 13 0 3 .
Sol Row reduction to solve Ax = 0; 3 6 6 3 9 0 6 12 13 0 3 0 ∼(1)/3 1 2 2 1 3 0 6 12 13 0 3 0 ∼ (2)−6(1) 1 2 2 1 3 0 0 0 1−6 −15 0 ∼(1)−2(2) 1 2 0 13 33 0 0 0 1−6 −15 0 Hence Ax=0 ⇔ x1 + 2x2 + 13x4 + 33x5 = 0 x3 −6x4 −15x5 = 0 . x2, x4, x5 are free so the sol. is x1 x2 x3 x4 x5 = −2x2 −13x4 −33x5 x2 6x4 + 15x5 x4 x5 = x2 −2 1 0 0 0 + x4 −13 0 6 1 0 + x5 −33 0 15 0 1 Hence Nul A = Span{u, v, w}, is the span of the three vectors u, v, w above.
3 We always have that 0 ∈Ker(A). When is Ker(A) = {0}?
Th (a) If A is m × n then Ker(A) = {0} if and only if rank(A) = n.
(b) If A is m × n and Ker(A) = {0} then m ≤n.
(c) If A is n × n then Ker(A) = {0} if and only if A is invertible.
Th For an n × n matrix A the following statements are equivalent: (i) A is invertible (ii) Ax = b has a unique solution x for all b.
(iii) Rref(A) = I.
(iv) rank(A) = n.
(v) Im(A) = Rn.
(vi) Ker(A) = 0 4 Summary and Questions If T: X →Y is a transformation then the set X is called the domain of T. The set Im(T) of all images T(x) when x varies over all points in the domain is called the image of T, or sometimes the range. Note that the image need not be all of the target space Y .
T is said to be onto if each y ∈Y is the image T(x) of at least one x ∈X.
T is said to be one-to-one if each y ∈Y is the image of at most one x ∈X.
T is called invertible if its one-to-one and onto.
Given vectors v1, . . . , vk and scalars λ1, . . . , λk, the vector w = λ1v1 + · · · + λkvk is called a linear combination of the vectors v1, . . . , vk (with weights λ1, . . . λk).
The set of all linear combinations of a v1, ..., vn is called the span of v1, ..., vn and is denoted by Span(v1, ..., vn). The set {v1, ..., vn} is said to span W if W = Span(v1, ..., vn).
Th The image of a linear transformation T(x) = Ax is the span of the column vectors of A.
Pf T(x) = Ax = " | | v1 · · · vn | | # "x1 .
.
.
xn # = x1v1 + · · · + xnvn The kernel, Ker(T), of a linear transformation T : Rn →Rm is the set of all x in the domain such that T(x) = 0. It is a proper subset of the domain Rn unless T is the zero map.
A subspace W of Rn is a subset which is closed under addition and scalar multiplication: (a) 0 ∈W, (b) u∈W and v∈W then u+v∈W, (c) w∈W and k is a scalar then kw∈W.
Ex A plane ax1 + bx2 + cx3 = 0 going through the origin in space is a subspace of R3.
Th If v1, ..., vn ∈Rm then Span(v1, ..., vn) is a subspace of Rm.
Th The image of a linear transformation T(x) = Ax, from Rn →Rm is a subspace of Rm.
Th The kernel of a linear transformation T(x) = Ax, from Rn →Rm is a subspace of Rn.
Question Which of the following are subspaces: (a) the plane x1 + 2x2 −4x3 = 1, (b) The span of the vectors (1, 2, 4) and (2, 4, 8)? (c) The Kernel of the matrix corresponding to rota-tion by 90 degrees counterclockwise? (d) The circle x2 1+x2 2 = 1. (e) The ball x2+y2+z2 ≤1.
Question Which of the following transformations have a nontrivial kernel (i.e. containing more than just 0)? (a) Rotation by π/2 counterclockwise, (b) Projection of the plane onto the x axis. (c) Reflection in the x axis. |
4351 | https://grayepc.weebly.com/uploads/2/3/6/2/23621820/sol_-_day_4.pdf | 1 Solutions Preparation ‐ A solution that has a known, accurate concentration is a standard solution.
‐ Good quality glassware and procedures are needed to prepare a standard solution ‐ There are 2 ways to create a standard solution o By dissolving a solid o By diluting a more concentrated solution Preparing a standard solution from a solid Step 1: Calculate what mass of solid is needed to make the required volume and concentration of solution n = cv m = nM Step 2: Measure out the mass of solid using an electronic scale Step 3: Dissolve the solid in half of the required volume of water Step 4: Transfer dissolved solid into a volumetric flask (make sure the volumetric flask is the size needed) Step 5: Fill the volumetric flask to the marking for the volume of the solution needed Step 6: Place top on volumetric flask and invert to mix. Example What are the steps for preparing 100 mL of a 0.750 mol/L sodium hydroxide solution?
.
2 Practice Sheet 6 1.
To test the hardness of water (Figure 3), an industrial chemist performs an analysis using 100.0 mL of a 0.250 mol/L standard solution of ammonium oxalate. What mass of ammonium oxalate, (NH4)2C2O4(s), is needed to make the standard solution?
2.
Calculate the mass of solid lye (sodium hydroxide) (Figure 4) needed to make 500 mL of a 10.0 mol/L strong cleaning solution.
3.
List several examples of solutions that you prepared from solids in the last week.
4.
You have been asked to prepare 2.00 L of a 0.100 mol/L aqueous solution of cobalt (II) chloride for an experiment. What mass of CoCl2 will you use? Write out the steps necessary to produce the solution of cobalt (II) chloride.
5. A technician prepares 500.0 mL of a 0.0750 mol/L solution of potassium permanganate as part of a quality‐control analysis in the manufacture of hydrogen peroxide. Calculate the mass of potassium permanganate required to prepare the solution.
3 Preparing a standard solution by dilution ‐ A stock solution is an initial, usually concentrated, solution from which samples are taken for a dilution.
‐ We can prepare solutions by diluting an existing solution to a desired concentration. Formula: C1V1 = C2V2 C1 = initial concentration C2 = final concentration V1 = initial volume V2 = final volume To make a standard solution using dilution follow the following steps.
Step 1: Calculate the volume of the stock solution needed to make the solution. C1V1 = C2V2 Step 2: Measure the volume of the stock solution using a pipette Step 3: Add the volume of stock solution to the appropriate sized volumetric flask Step 4: Fill volumetric flask to line Step 5: Put on cap and invert to mix Example: What are the steps to make 250mL of 1.0 mol/L hydrochloric acid from a concentrated (12 mol/L) solution of hydrochloric acid?
.
4 Practice Sheet 7 1.
12 mL of concentrated NH3(aq) (14 mol/L) is diluted to make 250mL of solution for cleaning windows. What is the concentration of the solution made?
2.
Radiator antifreeze (ethylene glycol) is diluted with an appropriate quantity of water to prevent freezing of the mixture in the radiator. A 4.00 L container of 94% V/V antifreeze is diluted to 9.00 L. Calculate the concentration of the final solution.
3. Many solutions are prepared in the laboratory from purchased concentrated solutions. Calculate the volume of concentrated 17.8 mol/L stock solution of sulphuric acid a laboratory technician would need to make 2.00 L of 0.200 mol/L solution by dilution of the original concentrated solution.
4. In a study of reaction rates, you need to dilute the copper(II) sulfate solution prepared in Investigation 5.3. You take 5.00 mL of 0.005000 mol/L CuSO4(aq) and dilute it to a final volume of 100.0 mL.
(a) Determine the final concentration of the dilute solution.
(b) What mass of CuSO4(s) is present in 10.0 mL of the final dilute solution?
(c) Can this final dilute solution be prepared directly using the pure solid? Defend your answer .
5 1. (a) Briefly describe two different ways of making a soluon. (b) When should you use each method?
2.
In an analysis for sulfate ions in a water treatment plant, a technician needs 100 mL of 0.125 mol/L barium nitrate soluon. What mass of pure barium nitrate is required?
3. A 1.00 L bole of purchased acec acid is labelled with a concentraon of 17.4 mol/L. A technician dilutes this enre bole of concentrated acid to prepare a 0.400 mol/L soluon. Calculate the volume of diluted soluon prepared.
4. A 10.00 mL sample of a test soluon is diluted in a laboratory to a final volume of 250.0 mL. The concentraon of the diluted soluon is 0.274 g/L. Determine the concentraon of the original test soluon.
. |
4352 | https://sites.science.oregonstate.edu/~gablek/CH334/Chapter5/fischer.htm | OREGON STATE UNIVERSITY
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4353 | https://math.stackexchange.com/questions/1017999/writing-a-trigonometric-function-in-terms-of-another-for-theta-in-a-given-quadra | algebra precalculus - Writing a trigonometric function in terms of another for theta in a given quadrant - Mathematics Stack Exchange
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Writing a trigonometric function in terms of another for theta in a given quadrant
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Asked 10 years, 10 months ago
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tan θ tanθ in terms of cos θ cosθ in quadrant III
I suppose that using the second pythagorean identity: tan 2 θ+1=sec 2 θ tan 2θ+1=sec 2θ
I have a general idea of how to manipulate this equation... but I am stuck in terms of arriving at the correct answer. This is as far as I have gotten:
tan 2 θ=sec 2 θ−1 tan 2θ=sec 2θ−1
tan 2 θ=1 cos 2 θ−1 tan 2θ=1 cos 2θ−1
tan 2 θ=1 cos 2 θ−(sin 2 θ+cos 2 θ)tan 2θ=1 cos 2θ−(sin 2θ+cos 2θ)
I feel very lost going from here. I probably didn't even head in the right direction to begin with. I would prefer a hint to set me on the right path rather than a direct answer.
algebra-precalculus
trigonometry
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asked Nov 12, 2014 at 4:28
Cherry_DeveloperCherry_Developer
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In general
tan 2 θ=sin 2 θ cos 2 θ=1−cos 2 θ cos 2 θ tan 2θ=sin 2θ cos 2θ=1−cos 2θ cos 2θ
Now in the I I I I I I Quadrant tan θ≥0 tanθ≥0
So,
tan θ=+1−cos 2 θ cos 2 θ−−−−−−−−−√tanθ=+1−cos 2θ cos 2θ
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answered Nov 12, 2014 at 4:29
lab bhattacharjeelab bhattacharjee
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Why is the answer negative and yours is positive?Cherry_Developer –Cherry_Developer 2014-11-12 05:06:40 +00:00 Commented Nov 12, 2014 at 5:06
@Cherry_Developer, The answer may not be right always, as math-only-math.com/all-sin-tan-cos-rule.htmllab bhattacharjee –lab bhattacharjee 2014-11-12 05:10:38 +00:00 Commented Nov 12, 2014 at 5:10
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4354 | https://www.engineeringtoolbox.com/nitrogen-d_977.html | Engineering ToolBox
- Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Nitrogen Gas - Specific Heat vs. Temperature
Specific heat of Nitrogen Gas - N2 - at temperatures ranging 175 - 6000 K
Specific heat (C) is the amount of heat required to change the temperature of a mass unit of a substance by one degree.
Isobaric specific heat (Cp) is used for substances in a constant pressure (ΔP = 0) system.
I sochoric specific heat (Cv) is used for substances in a constant-volume , (= isovolumetric or isometric ) closed system.
The specific heat - CP and CV - will vary with temperature. When calculating mass and volume flow of a substance in heated or cooled systems with high accuracy - the specific heat (= heat capacity) should be corrected according values in the table below.
Nitrogen accounts for 78 % of the atmospheric air volume. Nitrogen is an inert, neutral and colorless gas.
Specific heat of Nitrogen Gas - N2 - at temperatures ranging 175 - 6000 K :
Nitrogen Gas - N2- Specific Heat vs. Temperature
| Temperature - T - (K) | Specific Heat - cp - (kJ/kgK) |
| 175 | 1.039 |
| 200 | 1.039 |
| 225 | 1.039 |
| 250 | 1.039 |
| 275 | 1.039 |
| 300 | 1.040 |
| 325 | 1.040 |
| 350 | 1.041 |
| 375 | 1.042 |
| 400 | 1.044 |
| 450 | 1.049 |
| 500 | 1.056 |
| 550 | 1.065 |
| 600 | 1.075 |
| 650 | 1.086 |
| 700 | 1.098 |
| 750 | 1.110 |
| 800 | 1.122 |
| 850 | 1.134 |
| 900 | 1.146 |
| 950 | 1.157 |
| 1000 | 1.167 |
| 1050 | 1.177 |
| 1100 | 1.187 |
| 1150 | 1.196 |
| 1200 | 1.204 |
| 1250 | 1.212 |
| 1300 | 1.219 |
| 1350 | 1.226 |
| 1400 | 1.232 |
| 1500 | 1.244 |
| 1600 | 1.254 |
| 1700 | 1.263 |
| 1800 | 1.271 |
| 1900 | 1.278 |
| 2000 | 1.284 |
| 2100 | 1.290 |
| 2200 | 1.295 |
| 2300 | 1.300 |
| 2400 | 1.304 |
| 2500 | 1.307 |
| 2600 | 1.311 |
| 2700 | 1.314 |
| 2800 | 1.317 |
| 2900 | 1.320 |
| 3000 | 1.323 |
| 3500 | 1.333 |
| 4000 | 1.342 |
| 4500 | 1.349 |
| 5000 | 1.355 |
| 5500 | 1.362 |
| 6000 | 1.369 |
The values above apply to undissociated states. At high temperatures above 1500 K dissociation becomes appreciable and pressure is a significant variable.
See also other properties of Nitrogen at varying temperature and pressure : Density and specific weight, Dynamic and kinematic viscosity, Prandtl number, Specific heat (Heat capacity), Thermal conductivity and thermal diffusivity, and thermophysical properties at standard conditions,
as well as Specific heat of Air - at Constant Pressure and Varying Temperature, Air - at Constant Temperature and Varying Pressure, Ammonia, Butane, Carbon dioxide, Carbon monoxide, Ethane, Ethanol, Ethylene, Hydrogen, Methane, Methanol, Oxygen, Propane and Water.
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4355 | https://llego.dev/posts/generating-valid-parentheses-combinations-python/ | Generating Valid Parentheses Combinations in Python - llego.dev
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Generating Valid Parentheses Combinations in Python
Updated:Aug 23, 2023 | at 03:12 AM
Parentheses are a fundamental concept in computer programming and mathematics to control order of operations and group expressions. Generating all possible valid combinations of parentheses is a common technical interview coding challenge. This task tests a developer’s understanding of recursion, stacks, tree traversal, and dynamic programming.
In this comprehensive guide, we will examine different methods to generate valid parentheses combinations in Python.
Table of Contents
Open Table of Contents
Overview
Method 1: Recursive Backtracking
Method 2: Iterative Using Stack
Method 3: Recursive with Memoization
Method 4: Iterative Dynamic Programming
Method 5: BFS Tree Traversal
Summary and Recommendations
Overview
The problem statement is:
Given an integer n, generate all possible valid combinations of n pairs of parentheses.
For example, if n = 3, the valid combinations are:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
Some key points:
Parentheses must be balanced - every opening bracket must have a matching closing bracket.
The number of left and right parentheses must be equal to n.
Order matters - ”(()))” is invalid while ”((()))” is valid.
Cannot have more than n pairs of parentheses.
This type of combinatorial generation problem lends itself well to recursion and dynamic programming solutions. We will examine 5 different methods:
Recursive Backtracking
Iterative Using Stack
Recursive with Memoization
Iterative Dynamic Programming
BFS Tree Traversal
For each approach, we will break down the key steps with example code and explanations. Pros and cons of each method will also be analyzed.
Method 1: Recursive Backtracking
Recursive backtracking is a straightforward technique for this problem. We start with an empty string, then recursively add either a left or right parenthesis at each step. Backtracking allows exploring all possible combinations.
Python Implementation
```
def generate_parens(n):
combinations = []
generate(combinations, "", 0, 0, n)
return combinations
def generate(combinations, current, openN, closeN, max):
if len(current) == max2:
combinations.append(current)
return
if openN < max:
generate(combinations, current+"(", openN+1, closeN, max)
if closeN < openN:
generate(combinations, current+")", openN, closeN+1, max)
```
How it Works
Base cases are when we reach the maximum number of parentheses, we add current combination to output list.
Recursively call generate, increasing open or close based on constraints.
Key is tracking number of open and close parens added so far.
Try adding left parenthesis if number of open parens < n.
Try right parenthesis if close parens < open parens (to keep valid).
Analysis
Simple recursive backtracking is easy to implement.
However, repeatedly exploring the same subpaths leads to exponential time complexity O(2^n). Not efficient for larger n.
Recursive calls with string copying at each step also requires O(n) space complexity.
Method 2: Iterative Using Stack
We can optimize the backtracking approach by using a stack to iteratively build the combinations.
Python Implementation
```
def generate_parens(n):
combinations = []
stack = [('(', 1, 0)]
while stack:
current, openN, closeN = stack.pop()
if len(current) == n2:
combinations.append(current)
else:
if openN < n:
stack.append((current+'(', openN+1, closeN))
if closeN < openN:
stack.append((current+')', openN, closeN+1))
return combinations
```
How it Works
Use stack to track current combination string along with open/close counts.
Push next combinations by appending ( if open < n, ) if close < open.
Pop finished combinations and continue building.
Avoid repeated recursive calls.
Analysis
Iterative stack-based approach improves time complexity to O(n x 2^n) by removing function call overhead.
Space complexity reduced to O(n) by iteratively updating a single combination string rather than recursing on copies.
Method 3: Recursive with Memoization
We can optimize the recursive backtracking solution using memoization - storing results of solved subproblems to avoid recomputing them.
Python Implementation
```
from functools import lru_cache
@lru_cache(maxsize=None)
def generate_parens(n):
if n == 0:
return ['']
combinations = []
for c in generate_parens(n-1):
combinations.append('('+c+')')
combinations.append('()'+c)
combinations.append(c+'()')
return combinations
```
How it Works
Decorating the recursive function with lru_cache memoizes results.
For a given n, recursively call on n-1 and add ’()’ around previous results.
Avoids re-exploring same subpaths by caching prior results.
Analysis
Memoization reduces exponential time to polynomial O(n x 4^n) by caching subproblems.
But still O(n) space complexity to store recursive call stack.
Method 4: Iterative Dynamic Programming
We can further optimize using iterative bottom-up dynamic programming.
Python Implementation
```
def generate_parens(n):
combinations =
for i in range(1,n+1):
temp = []
for c in combinations[i-1]:
temp.append('('+c+')')
temp.append('()'+c)
temp.append(c+'()')
combinations.append(temp)
return combinations[-1]
```
How it Works
Build up combinations for n by extending combinations for n-1.
Avoid recursion using bottom-up iterative dynamic programming.
Analysis
Runs in O(n x 4^n) time by iteratively combining prior results.
Constant O(n) space needed for current/prior combination arrays.
Most optimal method overall.
Method 5: BFS Tree Traversal
We can model this problem as traversing a tree breadth-first and collect valid leaf nodes.
Python Implementation
```
from collections import deque
def generate_parens(n):
queue = deque([('(', 1, 0)])
combinations = []
while queue:
current, openN, closeN = queue.popleft()
if len(current) == n2:
combinations.append(current)
else:
if openN < n:
queue.append((current+'(', openN+1, closeN))
if closeN < openN:
queue.append((current+')', openN, closeN+1))
return combinations
```
How it Works
Use BFS queue to traverse tree level by level.
Append ( to current if open < n, ) if close < open.
Leaf nodes give valid combinations.
Analysis
Generates combinations in lexicographic order.
Time complexity O(n x C(2n, n)) as we build full tree.
Space O(n) for queue.
Slower and more memory intensive than dynamic programming.
Summary and Recommendations
Backtracking recursion is simple to implement but inefficient without optimization.
Stack-based iterative backtracking improves time complexity.
Memoization reduces repeated work by caching prior results.
Iterative bottom-up dynamic programming is the most optimal approach.
BFS traversal generates combinations in lexicographic order.
For interview coding questions, I recommend either the iterative stack-based or dynamic programming solutions, which offer the best time/space complexity. Make sure to explain the algorithm and analysis to the interviewer.
Some follow-up opportunities:
Return combinations in lexicographic sorted order.
Handle invalid parentheses sequences.
Generate expressions with minimum removals to make valid.
In summary, generating valid parentheses combinations evaluates fundamental coding concepts like recursion, trees, dynamic programming, and backtracking search. Mastering this question demonstrates strong technical and analytical skills for programming interviews.
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Contents
(Top)
1 Definition
1.1 Unit symbol
1.2 Dimensional analysis
2 Conversions
2.1 Between degrees
2.2 Between gradians
3 Usage
3.1 Mathematics
3.2 Physics
3.3 Prefixes and variants
4 History
4.1 Pre-20th century
4.2 As an SI unit
5 See also
6 Notes
7 References
8 External links
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From Wikipedia, the free encyclopedia
SI derived unit of angle
"㎭" redirects here; not to be confused with Rad (radiation unit).
For other uses, see Radian (disambiguation).
| Radian |
| An arc of a circle with the same length as the radius of that circle subtends an angle of 1 radian. The circumference subtends an angle of 2π radians. |
| General information |
| Unit system | SI |
| Unit of | angle |
| Symbol | rad |
| Conversions |
| 1 rad in ... | ... is equal to ... |
| |
| milliradians | 1000 mrad |
| turns | 1/2π turn ≈ 0.159154 turn |
| degrees | 180/π° ≈ 57.295779513° |
| gradians | 200/π grad ≈ 63.661977g |
The radian, denoted by the symbol rad, is the unit of angle in the International System of Units (SI) and is the standard unit of angular measure used in many areas of mathematics. It is defined such that one radian is the angle subtended at the center of a plane circle by an arc that is equal in length to the radius. The unit is defined in the SI as the coherent unit for plane angle, as well as for phase angle. Angles without explicitly specified units are generally assumed to be measured in radians, especially in mathematical writing.
Definition
One radian is defined as the angle at the center of a circle in a plane that is subtended by an arc whose length equals the radius of the circle. More generally, the magnitude in radians of a subtended angle is equal to the ratio of the arc length to the radius of the circle; that is, θ = s/r, where θ is the magnitude in radians of the subtended angle, s is arc length, and r is radius. A right angle is exactly π/2 radians.
One complete revolution, expressed as an angle in radians, is the length of the circumference divided by the radius, which is 2πr/r, or 2π. Thus, 2π radians is equal to 360 degrees. The relation 2π rad = 360° can be derived using the formula for arc length, . Since radian is the measure of an angle that is subtended by an arc of a length equal to the radius of the circle, . This can be further simplified to . Multiplying both sides by 360° gives 360° = 2π rad.
Unit symbol
The International Bureau of Weights and Measures and International Organization for Standardization specify rad as the symbol for the radian. Alternative symbols that were in use in 1909 are c (the superscript letter c, for "circular measure"), the letter r, or a superscript R, but these variants are infrequently used, as they may be mistaken for a degree symbol (°) or a radius (r). Hence an angle of 1.2 radians would be written today as 1.2 rad; archaic notations include 1.2 r, 1.2rad, 1.2c, or 1.2R.
In mathematical writing, the symbol "rad" is often omitted. When quantifying an angle in the absence of any symbol, radians are assumed, and when degrees are meant, the degree sign ° is used.
Dimensional analysis
See also: § As an SI unit
Plane angle may be defined as θ = s/r, where θ is the magnitude in radians of the subtended angle, s is circular arc length, and r is radius. One radian corresponds to the angle for which s = r, hence 1 radian = 1 m/m = 1. However, rad is only to be used to express angles, not to express ratios of lengths in general. A similar calculation using the area of a circular sector θ = 2A/r2 gives 1 radian as 1 m2/m2 = 1. The key fact is that the radian is a dimensionless unit equal to 1. In SI 2019, the SI radian is defined accordingly as 1 rad = 1. It is a long-established practice in mathematics and across all areas of science to make use of rad = 1.
Giacomo Prando writes "the current state of affairs leads inevitably to ghostly appearances and disappearances of the radian in the dimensional analysis of physical equations". For example, an object hanging by a string from a pulley will rise or drop by y = rθ centimetres, where r is the magnitude of the radius of the pulley in centimetres and θ is the magnitude of the angle through which the pulley turns in radians. When multiplying r by θ, the unit radian does not appear in the product, nor does the unit centimetre—because both factors are magnitudes (numbers). Similarly in the formula for the angular velocity of a rolling wheel, ω = v/r, radians appear in the units of ω but not on the right hand side. Anthony French calls this phenomenon "a perennial problem in the teaching of mechanics". Oberhofer says that the typical advice of ignoring radians during dimensional analysis and adding or removing radians in units according to convention and contextual knowledge is "pedagogically unsatisfying".
In 1993 the American Association of Physics Teachers Metric Committee specified that the radian should explicitly appear in quantities only when different numerical values would be obtained when other angle measures were used, such as in the quantities of angle measure (rad), angular speed (rad/s), angular acceleration (rad/s2), and torsional stiffness (N⋅m/rad), and not in the quantities of torque (N⋅m) and angular momentum (kg⋅m2/s).
At least a dozen scientists between 1936 and 2022 have made proposals to treat the radian as a base unit of measurement for a base quantity (and dimension) of "plane angle". Quincey's review of proposals outlines two classes of proposal. The first option changes the unit of a radius to meters per radian, but this is incompatible with dimensional analysis for the area of a circle, πr2. The other option is to introduce a dimensional constant. According to Quincey this approach is "logically rigorous" compared to SI, but requires "the modification of many familiar mathematical and physical equations". A dimensional constant for angle is "rather strange" and the difficulty of modifying equations to add the dimensional constant is likely to preclude widespread use.
In particular, Quincey identifies Torrens' proposal to introduce a constant η equal to 1 inverse radian (1 rad−1) in a fashion similar to the introduction of the constant ε0.[a] With this change the formula for the angle subtended at the center of a circle, s = rθ, is modified to become s = ηrθ, and the Taylor series for the sine of an angle θ becomes: where is the angle in radians. The capitalized function Sin is the "complete" function that takes an argument with a dimension of angle and is independent of the units expressed, while sin is the traditional function on pure numbers which assumes its argument is a dimensionless number in radians. The capitalised symbol can be denoted if it is clear that the complete form is meant.
Current SI can be considered relative to this framework as a natural unit system where the equation η = 1 is assumed to hold, or similarly, 1 rad = 1. This radian convention allows the omission of η in mathematical formulas.
Defining radian as a base unit may be useful for software, where the disadvantage of longer equations is minimal. For example, the Boost units library defines angle units with a plane_angle dimension, and Mathematica's unit system similarly considers angles to have an angle dimension.
Conversions
Conversion of common angles
| Turns | Radians | Degrees | Gradians |
| 0 turn | 0 rad | 0° | 0g |
| 1/72 turn | π/36 or 𝜏/72 rad | 5° | 5+5/9g |
| 1/24 turn | π/12 or 𝜏/24 rad | 15° | 16+2/3g |
| 1/16 turn | π/8 or 𝜏/16 rad | 22.5° | 25g |
| 1/12 turn | π/6 or 𝜏/12 rad | 30° | 33+1/3g |
| 1/10 turn | π/5 or 𝜏/10 rad | 36° | 40g |
| 1/8 turn | π/4 or 𝜏/8 rad | 45° | 50g |
| 1/2π or 𝜏 turn | 1 rad | approx. 57.3° | approx. 63.7g |
| 1/6 turn | π/3 or 𝜏/6 rad | 60° | 66+2/3g |
| 1/5 turn | 2π or 𝜏/5 rad | 72° | 80g |
| 1/4 turn | π/2 or 𝜏/4 rad | 90° | 100g |
| 1/3 turn | 2π or 𝜏/3 rad | 120° | 133+1/3g |
| 2/5 turn | 4π or 2𝜏 or α/5 rad | 144° | 160g |
| 1/2 turn | π or 𝜏/2 rad | 180° | 200g |
| 3/4 turn | 3π or ρ/2 or 3𝜏/4 rad | 270° | 300g |
| 1 turn | 𝜏 or 2π rad | 360° | 400g |
Between degrees
As stated, one radian is equal to . Thus, to convert from radians to degrees, multiply by .
For example:
Conversely, to convert from degrees to radians, multiply by .
For example:
Radians can be converted to turns (one turn is the angle corresponding to a revolution) by dividing the number of radians by 2π.
Between gradians
One revolution corresponds to an angle of radians, which equals one turn, and to 400 gradians (400 gons or 400g). To convert from radians to gradians multiply by , and to convert from gradians to radians multiply by . For example,
Usage
Mathematics
In calculus and most other branches of mathematics beyond practical geometry, angles are measured in radians. This is because radians have a mathematical naturalness that leads to a more elegant formulation of some important results.
Results in analysis involving trigonometric functions can be elegantly stated when the functions' arguments are expressed in radians. For example, the use of radians leads to the simple limit formula
which is the basis of many other identities in mathematics, including
Because of these and other properties, the trigonometric functions appear in solutions to mathematical problems that are not obviously related to the functions' geometrical meanings (for example, the solutions to the differential equation , the evaluation of the integral , and so on). In all such cases, it is appropriate that the arguments of the functions are treated as (dimensionless) numbers—without any reference to angles.
The trigonometric functions of angles also have simple and elegant series expansions when radians are used. For example, when x is the angle expressed in radians, the Taylor series for sin x becomes:
If y were the angle x but expressed in degrees, i.e. y = πx / 180, then the series would contain messy factors involving powers of π/180:
In a similar spirit, if angles are involved, mathematically important relationships between the sine and cosine functions and the exponential function (see, for example, Euler's formula) can be elegantly stated when the functions' arguments are angles expressed in radians (and messy otherwise). More generally, in complex-number theory, the arguments of these functions are (dimensionless, possibly complex) numbers—without any reference to physical angles at all.
Physics
The radian is widely used in physics when angular measurements are required. For example, angular velocity is typically expressed in the unit radian per second (rad/s). One revolution per second corresponds to 2π radians per second.
Similarly, the unit used for angular acceleration is often radian per second per second (rad/s2).
For the purpose of dimensional analysis, the units of angular velocity and angular acceleration are s−1 and s−2 respectively.
Likewise, the phase angle difference of two waves can also be expressed using the radian as the unit. For example, if the phase angle difference of two waves is (n × 2π) radians, where n is an integer, they are considered to be in phase, whilst if the phase angle difference of two waves is (n × 2π + π) radians, with n an integer, they are considered to be in antiphase.
A unit of reciprocal radian or inverse radian (rad−1) is involved in derived units such as meter per radian (for angular wavelength) or newton-metre per radian (for torsional stiffness).
Prefixes and variants
Metric prefixes for submultiples are used with radians. A milliradian (mrad) is a thousandth of a radian (0.001 rad), i.e. 1 rad = 103 mrad. There are 2π × 1000 milliradians (≈ 6283.185 mrad) in a circle. So a milliradian is just under 1/6283 of the angle subtended by a full circle. This unit of angular measurement of a circle is in common use by telescopic sight manufacturers using (stadiametric) rangefinding in reticles. The divergence of laser beams is also usually measured in milliradians.
The angular mil is an approximation of the milliradian used by NATO and other military organizations in gunnery and targeting. Each angular mil represents 1/6400 of a circle and is 15/8% or 1.875% smaller than the milliradian. For the small angles typically found in targeting work, the convenience of using the number 6400 in calculation outweighs the small mathematical errors it introduces. In the past, other gunnery systems have used different approximations to 1/2000π; for example Sweden used the 1/6300 streck and the USSR used 1/6000. Being based on the milliradian, the NATO mil subtends roughly 1 m at a range of 1000 m (at such small angles, the curvature is negligible).
Prefixes smaller than milli- are useful in measuring extremely small angles. Microradians (μrad, 10−6 rad) and nanoradians (nrad, 10−9 rad) are used in astronomy, and can also be used to measure the beam quality of lasers with ultra-low divergence. More common is the arc second, which is π/648000 rad (around 4.8481 microradians).
SI multiples of radian (rad)
| Submultiples | Multiples |
| Value | SI symbol | Name | Value | SI symbol | Name |
| 10−1 rad | drad | deciradian | 101 rad | darad | decaradian |
| 10−2 rad | crad | centiradian | 102 rad | hrad | hectoradian |
| 10−3 rad | mrad | milliradian | 103 rad | krad | kiloradian |
| 10−6 rad | μrad | microradian | 106 rad | Mrad | megaradian |
| 10−9 rad | nrad | nanoradian | 109 rad | Grad | gigaradian |
| 10−12 rad | prad | picoradian | 1012 rad | Trad | teraradian |
| 10−15 rad | frad | femtoradian | 1015 rad | Prad | petaradian |
| 10−18 rad | arad | attoradian | 1018 rad | Erad | exaradian |
| 10−21 rad | zrad | zeptoradian | 1021 rad | Zrad | zettaradian |
| 10−24 rad | yrad | yoctoradian | 1024 rad | Yrad | yottaradian |
| 10−27 rad | rrad | rontoradian | 1027 rad | Rrad | ronnaradian |
| 10−30 rad | qrad | quectoradian | 1030 rad | Qrad | quettaradian |
History
Pre-20th century
The idea of measuring angles by the length of the arc was in use by mathematicians quite early. For example, al-Kashi (c. 1400) used so-called diameter parts as units, where one diameter part was 1/60 radian. They also used sexagesimal subunits of the diameter part. Newton in 1672 spoke of "the angular quantity of a body's circular motion", but used it only as a relative measure to develop an astronomical algorithm.
The concept of the radian measure is normally credited to Roger Cotes, who died in 1716. By 1722, his cousin Robert Smith had collected and published Cotes' mathematical writings in a book, Harmonia mensurarum. In a chapter of editorial comments, Smith gave what is probably the first published calculation of one radian in degrees, citing a note of Cotes that has not survived. Smith described the radian in everything but name – "Now this number is equal to 180 degrees as the radius of a circle to the semicircumference, this is as 1 to 3.141592653589" –, and recognized its naturalness as a unit of angular measure.
In 1765, Leonhard Euler implicitly adopted the radian as a unit of angle. Specifically, Euler defined angular velocity as "The angular speed in rotational motion is the speed of that point, the distance of which from the axis of gyration is expressed by one." Euler was probably the first to adopt this convention, referred to as the radian convention, which gives the simple formula for angular velocity ω = v/r. As discussed in § Dimensional analysis, the radian convention has been widely adopted, while dimensionally consistent formulations require the insertion of a dimensional constant, for example ω = v/(ηr).
Prior to the term radian becoming widespread, the unit was commonly called circular measure of an angle. The term radian first appeared in print on 5 June 1873, in examination questions set by James Thomson (brother of Lord Kelvin) at Queen's College, Belfast. He had used the term as early as 1871, while in 1869, Thomas Muir, then of the University of St Andrews, vacillated between the terms rad, radial, and radian. In 1874, after a consultation with James Thomson, Muir adopted radian. The name radian was not universally adopted for some time after this. Longmans' School Trigonometry still called the radian circular measure when published in 1890.
In 1893 Alexander Macfarlane wrote "the true analytical argument for the circular ratios is not the ratio of the arc to the radius, but the ratio of twice the area of a sector to the square on the radius." However, the paper was withdrawn from the published proceedings of mathematical congress held in connection with World's Columbian Exposition in Chicago (acknowledged at page 167), and privately published in his Papers on Space Analysis (1894). Macfarlane reached this idea or ratios of areas while considering the basis for hyperbolic angle which is analogously defined.
As an SI unit
See also: § Dimensional analysis
As Paul Quincey et al. write, "The status of angles within the International System of Units (SI) has long been a source of controversy and confusion." In 1960, the General Conference on Weights and Measures (CGPM) established the SI and the radian was classified as a "supplementary unit" along with the steradian. This special class was officially regarded "either as base units or as derived units", as the CGPM could not reach a decision on whether the radian was a base unit or a derived unit. Richard Nelson writes "This ambiguity [in the classification of the supplemental units] prompted a spirited discussion over their proper interpretation." In May 1980 the Consultative Committee for Units (CCU) considered a proposal for making radians an SI base unit, using a constant α0 = 1 rad, but turned it down to avoid an upheaval to current practice.
In October 1980 the CGPM decided that supplementary units were dimensionless derived units for which the CGPM allowed the freedom of using them or not using them in expressions for SI derived units, on the basis that "[no formalism] exists which is at the same time coherent and convenient and in which the quantities plane angle and solid angle might be considered as base quantities" and that "[the possibility of treating the radian and steradian as SI base units] compromises the internal coherence of the SI based on only seven base units". In 1995 the CGPM eliminated the class of supplementary units and defined the radian and the steradian as "dimensionless derived units, the names and symbols of which may, but need not, be used in expressions for other SI derived units, as is convenient". Mikhail Kalinin writing in 2019 has criticized the 1980 CGPM decision as "unfounded" and says that the 1995 CGPM decision used inconsistent arguments and introduced "numerous discrepancies, inconsistencies, and contradictions in the wordings of the SI".
At the 2013 meeting of the CCU, Peter Mohr gave a presentation on alleged inconsistencies arising from defining the radian as a dimensionless unit rather than a base unit. CCU President Ian M. Mills declared this to be a "formidable problem" and the CCU Working Group on Angles and Dimensionless Quantities in the SI was established. The CCU met in 2021, but did not reach a consensus. A small number of members argued strongly that the radian should be a base unit, but the majority felt the status quo was acceptable or that the change would cause more problems than it would solve. A task group was established to "review the historical use of SI supplementary units and consider whether reintroduction would be of benefit", among other activities.
See also
Angular frequency
Minute and second of arc
Steradian, a higher-dimensional analog of the radian which measures solid angle
Trigonometry
Notes
^ Other proposals include the abbreviation "rad" (Brinsmade 1936), the notation (Romain 1962), and the constants ם (Brownstein 1997), ◁ (Lévy-Leblond 1998), k (Foster 2010), θC (Quincey 2021), and (Mohr et al. 2022).
^ The International System of Units (PDF), V3.01 (9th ed.), International Bureau of Weights and Measures, Aug 2024, p. 138, ISBN 978-92-822-2272-0: "One radian is the angle subtended at the centre of a circle by an arc that is equal in length to the radius."
^ The International System of Units (PDF), V3.01 (9th ed.), International Bureau of Weights and Measures, Aug 2024, p. 138, ISBN 978-92-822-2272-0: "The radian is the coherent unit for plane angle. ... The radian is also the coherent unit for phase angle."
^ Ocean Optics Protocols for Satellite Ocean Color Sensor Validation, Revision 3. National Aeronautics and Space Administration, Goddard Space Flight Center. 2002. p. 12.
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4357 | https://www.sciencedirect.com/science/article/abs/pii/S1047847708000750 | Intracellular precipitation of hydroxyapatite mineral and implications for pathologic calcification - ScienceDirect
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Journal of Structural Biology
Volume 162, Issue 3, June 2008, Pages 468-479
Intracellular precipitation of hydroxyapatite mineral and implications for pathologic calcification
Author links open overlay panel Fereshteh Azari a b, Hojatollah Vali a c d, Jean-Luc Guerquin-Kern e f, Ting-Di Wu e f, Alain Croisy e f, S. Kelly Sears d, Maryam Tabrizian b c, Marc D.McKee a c
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Abstract
In contrast to physiologic biomineralization occurring in bones, teeth and otoconia in vertebrates, calcification of soft tissues occurs in many pathologic conditions. Although similarities have been noted between the two processes, and despite the important clinical consequences of ectopic calcification, the molecular mechanisms regulating ectopic calcification are poorly understood. Although calcification is mainly extracellular, intracellular calcification has been reported and might indeed contribute to pathologic calcification of soft tissues. To better understand the process of intracellular calcification as a potential origin for pathologic calcification, and to examine the role of proteoglycans in this process, we investigated a pattern of intracellular nucleation and growth of hydroxyapatite in Madin–Darby Canine Kidney (MDCK) epithelial cells using electron microscopy, secondary ion mass spectroscopy (NanoSIMS), cytochemical staining, immunolabeling and biochemical analysis. We report here that under mineralizing cell culture conditions where β-glycerophosphate (βGP) was added as an exogenous organic source of phosphate, βGP-cleaving alkaline phosphatase activity increased and hydroxyapatite crystals subsequently nucleated within intracellular, membrane-bounded compartments. The small, leucine-rich proteoglycan decorin was also upregulated and associated with mineral in these cultures. Such information provides insight into the mechanisms leading to pathologic calcification and describes a process whereby hydroxyapatite deposition in cells might lead to ectopic calcification.
Introduction
Unlike the physiologic mineralization that occurs in vertebrate bones (including cartilage), teeth and otoconia, ectopic calcification occurs when mineral precipitates pathologically in soft tissues. Debilitating conditions having pathologic calcification include urolithiasis, cardiovascular disease and atherosclerosis, medial arterial calcification (elastocalcinosis), end-stage renal disease, arthritis and calciphylaxis (Giachelli, 2004). Moreover, many implantable prosthetic devices are prone to calcification and thus require replacement—perhaps the best known being the revision surgeries required for bioprosthetic heart valves (Schoen and Levy, 2005). Physiologic precipitation of calcium-phosphate crystals, deposited mainly as a highly substituted hydroxyapatite, occurs predominantly in the extracellular matrices of bone, cartilage and teeth. The mineralization process is controlled by resident cells in these tissues via the secretion of mineral-binding extracellular matrix proteins and proteoglycans, the production and transport of inhibitory pyrophosphate to the extracellular compartment, and by enzymes of the alkaline phosphatase family that can degrade mineralization inhibitors (Addison et al., 2007). Another layer of control over skeletal and dental mineralization is exerted by systemically circulating, mineral-inhibiting proteins such as fetuin-A (Jahnen-Dechent et al., 2008). While still a matter of much discussion, resident cells might also regulate mineralization by releasing membranous vesicles into the extracellular matrix (matrix vesicles) that subsequently act as a nidus for mineralization (Anderson, 2003).
In contrast to physiologic mineralization, calcification of soft tissues occurs ectopically in pathologic conditions—often with devastating clinical consequences. Although some similarities have been noted between the two processes (Donley and Fitzpatrick, 1998, Mody et al., 2003), and despite the important clinical significance of soft tissue calcification, the molecular mechanisms regulating ectopic calcification are poorly understood relative to those occurring normally in bones and teeth. Indeed, ectopic calcification has been sub-classified into metastatic, dystrophic and metabolic forms, but the mineralization mechanisms discriminating between them are likewise unknown (Bonewald et al., 2003). Until fairly recently, pathologic calcification has generally been considered as a passive process involving the spontaneous precipitation of hydroxyapatite in response to tissue injury or necrosis (Steitz et al., 2001). However, several lines of evidence have recently emerged suggesting some ectopic calcification likewise is a cell-regulated process, similar to that occurring elsewhere, where mineral precipitation depends on a balance between pro-calcific and anti-calcific regulatory molecules (Giachelli, 2005).
Although calcification is mainly an extracellular phenomenon occurring in extracellular matrices (McKee et al., 2005), intracellular calcification has been reported. For example, Purkinje cells were reported to contain electron-dense vesicular structures, mitochondria and lysosomes that all contained abundant calcium-phosphate mineral deposits (Ando et al., 2004). In malakoplakia, granulomatous lesions contain abundant macrophages with calcified lysosomes called Michaelis–Gutmann bodies (Ghadially, 2001). Also, mitochondrial calcification has been reported in heart transplant recipients treated with cyclosporine (Millane et al., 1994), and in myocardial calcification that can occur pathologically in the perinatal period (Drut et al., 1998). In bone-forming osteoblasts, intracellular calcification has been observed in the cytoplasm and in vesicular/vacuolar structures (Rohde and Mayer, 2007, Stanford et al., 1995).
Proteins that regulate hydroxyapatite crystal growth include extracellular matrix proteins and proteoglycans such as, for example, osteopontin, osteocalcin, bone sialoprotein, dentin matrix protein-1, amelogenin, enamelin, biglycan and decorin (Boskey et al., 1997a, Giachelli, 1999). In addition to calcium and regulatory protein levels, the phosphate/pyrophosphate balance is also thought to be a key molecular determinant of vertebrate mineralization (McKee et al., 2005). Proteoglycans—proteins with glycosaminoglycan side chains—are ubiquitously expressed and have a wide range of functions (Lander and Selleck, 2000). With regard to mineralization, most studies have focused on the function of proteoglycans in the extracellular matrix, with some studies reporting that proteoglycans control the size and shape of hydroxyapatite by binding to crystal surfaces (Boskey et al., 1997a, Boskey et al., 1997b, Campo and Betz, 1987, Moriguchi et al., 2004, Rees et al., 2001, Waddington and Langley, 1998). Whereas substantial characterization has been done, and functional roles determined, for the large proteoglycans of cartilage and intervertebral disc, much less is known about the small leucine-rich proteoglycans (SLRPs) in mineralized tissues (Young et al., 2006), whose family members include decorin, biglycan, fibromodulin and lumican.
While the focus of proteoglycan research initially was on secreted, large extracellular matrix macromolecules, other reports have proposed intracellular functions for smaller proteoglycans based on their associations with zymogen storage granules in pancreas, mast-cell granules, internalization of growth factors and polyamines, nuclear components, and other intracellular organelles (Kolset and Gallagher, 1990, Kolset et al., 2004). To better understand the process of intracellular calcification as a potential origin for pathologic calcification, and to examine the role of proteoglycans in this process, we investigated a pattern of intracellular nucleation and growth of hydroxyapatite in Madin–Darby Canine Kidney (MDCK) cells using electron microscopy, secondary ion mass spectroscopy (NanoSIMS) having detection abilities at the nano-scale, cytochemical staining, immunolabeling and biochemical analysis. Our selection of MDCK epithelial cells for this study was based on our interest in the pathologic calcification occurring during nephrolithiasis (both calcium oxalate and hydroxyapatite formation), and on the fact that the epithelial tubular cells in the kidney have been implicated in this process (Chau et al., 2003, Lieske and Toback, 2000). We report here in vitro that hydroxyapatite is nucleated within intracellular, membrane-bound compartments of MDCK epithelial cells, and that proteoglycans are associated with calcification sites in these cultures. Such information provides insight into the mechanisms leading to pathologic calcification, and describes a process whereby hydroxyapatite crystal deposition in cells might lead to the extensive calcification associated with many diseases.
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Cell culture and induction of calcification
Madin–Darby Canine Kidney (MDCK) cells (provided by Dr. I.R. Nabi, Department of Cellular and Physiological Sciences, University of British Columbia, Canada) were cultured in Dulbecco’s Modified Eagle’s Medium (DMEM, Sigma, St. Louis, MO) containing 4.5 g/l glucose, 10% fetal bovine serum (Invitrogen, Burlington, ON) supplemented with 10 mM Hepes, 100 U/ml of penicillin, 100 μg/ml streptomycin and non-essential amino acids. The cells were incubated under conventional cell culture conditions at 37°C
βGP-induced calcium-phosphate precipitation in MDCK epithelial cell cultures
In mineralization studies using cultured bone and tooth cells, βGP is commonly added to the media to provide a source of phosphate ions—after cleavage by alkaline phosphatase—for mineralization. To determine whether such inorganic phosphate facilitates mineralization of renal epithelial cells in cell culture, MDCK cells were grown for 28 days in the presence of 10 mM βGP. As shown in Fig. 1A, mineral deposits were observed by von Kossa staining in a time-dependent manner after addition of this
Discussion
Ectopic precipitation of hydroxyapatite and other minerals in hard and soft tissues alike can lead to debilitating loss of tissue function. Pathologic calcification is particularly prevalent in heart, blood vessels, lung, kidney, and articular cartilage (Giachelli, 2004). Although the mechanisms driving this undesirable calcification are not well understood, recent studies suggest that in some cases it is a regulated process controlled or induced by local cellular activity, and many
Acknowledgments
This research was funded by the Canadian Institutes of Health Research (MDM), the Natural Sciences and Engineering Council of Canada (HV), and the Fonds Quebecois de la Recherche sur la Nature et les Technologies (MT) supporting the Centre for Biorecognition and Biosensors (CBB) based at McGill University. FA, HV, MT and MDM are members of CBB. MDM is also a member of the Jamson T.N. Wong Laboratories for calcified tissue research of the McGill Centre for Bone and Periodontal Research.
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4358 | https://math.stackexchange.com/questions/2632396/obtaining-probabilities-given-cumulative-probability-distribution-function | Obtaining probabilities given cumulative probability distribution function - Mathematics Stack Exchange
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Obtaining probabilities given cumulative probability distribution function
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Asked 7 years, 8 months ago
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Let the cumulative distribution function of random variable X X is given as:
F(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪0 1 5(1+x 3)1 5[3+(x−1)2]1 if x<0 if 0≤x<1 if 1≤x<2 if x≥2 F(x)={0 if x<0 1 5(1+x 3)if 0≤x<1 1 5[3+(x−1)2]if 1≤x<2 1 if x≥2
It is required to obtain the following probabilities:
P(0<X<2)P(0<X<2)
P(0≤X≤1)P(0≤X≤1)
P(0.5≤X≤1.5)P(0.5≤X≤1.5).
Although I know how to solve this problem but most of the time I get wrong result. Whenever a cumulative distribution function has a discontinuity, I get scare. This function has discounting ity at countable points but I think that should not affect the results. How, I obtained the above probabilities are as follows:
My approach:
P(0<X<2)=P(X<2)−P(X≤0)=3 5 P(0<X<2)=P(X<2)−P(X≤0)=3 5
P(0≤X≤1)=P(X≤1)−P(X≤0)=3 5 P(0≤X≤1)=P(X≤1)−P(X≤0)=3 5
This should not have any problem since function is continuous at both points. Hence, P(0.5≤X≤1.5)=1 5[3+(1.5−1)2]−1 5[1+0.5 3]P(0.5≤X≤1.5)=1 5[3+(1.5−1)2]−1 5[1+0.5 3]
I think my steps should be correct. Please take a moment to review my steps and my rationale for solving these types of problems. Thanks
probability
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edited Feb 2, 2018 at 15:00
user173262
asked Feb 2, 2018 at 6:48
userNoOneuserNoOne
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At x=1 x=1 your cdf has two values...zoli –zoli 2018-02-02 06:50:05 +00:00 Commented Feb 2, 2018 at 6:50
I guess it has a typo. Let me correct it userNoOne –userNoOne 2018-02-02 06:50:56 +00:00 Commented Feb 2, 2018 at 6:50
Then correct it at 2 2 as well. Which corrections point at the root of your problem. At an x x the cdf is well defined.zoli –zoli 2018-02-02 06:52:40 +00:00 Commented Feb 2, 2018 at 6:52
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your results are correct. You can compute such probabilities using the left limits of F F. Define F(x−):=lim n→∞F(x−1 n)F(x−):=lim n→∞F(x−1 n). Then
P(a<X<b)P(a≤X≤b)P(a<X≤b)P(a≤X<b)=F(b−)−F(a),=F(b)−F(a−),=F(b)−F(a),=F(b−)−F(a−).P(a<X<b)=F(b−)−F(a),P(a≤X≤b)=F(b)−F(a−),P(a<X≤b)=F(b)−F(a),P(a≤X<b)=F(b−)−F(a−).
So in your example you get that
P(0<X<2)P(0≤X≤1)P(0.5≤X≤1.5)=F(2−)−F(0)=4 5−1 5=3 5,=F(1)−F(0−)=3 5−0=3 5,=F(1.5)−F(0.5−)=…...P(0<X<2)=F(2−)−F(0)=4 5−1 5=3 5,P(0≤X≤1)=F(1)−F(0−)=3 5−0=3 5,P(0.5≤X≤1.5)=F(1.5)−F(0.5−)=…...
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answered Feb 2, 2018 at 7:58
CetttCettt
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Thanks Cett, it was very helpful.userNoOne –userNoOne 2018-02-02 08:08:54 +00:00 Commented Feb 2, 2018 at 8:08
Thumb's up for you and congrats for your 1500 reputation.userNoOne –userNoOne 2018-02-02 08:09:29 +00:00 Commented Feb 2, 2018 at 8:09
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Power is the amount of energy consumed or produced over time.
Power is a measurement of the energy used to move vehicles (automobiles, trucks, tractors, trains, boats and airplanes).
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The Ohm's Law Calculator suite has automatic unit conversions for the input and output of the equations. The equations are as follows. Click on the formula for a pop-up calculator with the formula, or click on the description link for the page dedicated to that formula.
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Power (P) = I • V: Computes the power (watts) as a function of current (amps) and potential (volts).
Power (P) = R • I²: Computes the power (watts) as a function of current (amps) and resistance (ohms).
Power (P) = V² / R: Computes the power (watts) as a function of potential (volts) and resistance (ohms).
Potential (volts, millivolts)
Volts (V) = I • R: Computes the potential (volts) as a function of current (amps) and resistance (ohms).
Volts (V) = P / I: Computes the potential (volts) as a function of power (watts) and current (amps).
Volts(V) = √(P • R): Computes the potential (volts) as a function of power (watts) and resistance (ohms).
Resistance (ohms, milliohms, kiloohms)
Resistance (R) = V²/P: Computes the resistance (ohms) as a function of potential (volts) and power (watts).
Resistance (R) = P / I²: Computes the resistance (ohms) as a function of power (watts) and current (amps).
Resistance (R) = V / I: Computes the resistance (ohms) as a function of potential (volts) and current (amps).
Current (amps, milliamps, microamps, gilberts)
Current (I) = V / R: Computes the current (amps) as a function of potential (volts) and resistance (ohms).
Current (I) = P / V: Computes the current (amps) as a function of the power (watts) and the potential (volts).
Current (I) = √(P/R): Computes the current (amps) as a function of the power (watts) and the resistance (ohms). |
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Annotating Exemplars
Providing Strong Models for Student Writers
One highly effective strategy for helping students develop and transfer writing knowledge is to increase their awareness about the genres they must write in.
In How Learning Works: Seven Research-Based Principles for Smart Teaching, Ambrose et al. claim that making explicit the ways experts organize and present knowledge in their fields strongly improves student learning. Without this assistance, novice learners will often “come up with knowledge organizations that are superficial and/or do not lend themselves to abstraction or problem-solving”(58).
Annotation = Being Meta
Annotating a text – labeling the strong rhetorical and cognitive features – helps students become more aware of how to think and write in disciplines. Annotations demystify the process, make explicit what is often implicit for experts, and provide a valuable model for students to strive toward. In short, they provide "sign posts" for students navigating a new landscape. Because you are an expert in the genres of writing you assign (lab reports, white papers, memos, manuals, analytical essays, etc.), you are in a powerful position to assist students with acquiring rhetorical and cognitive knowledge specific to your field. Providing rich annotations helps students undergo a “cognitive apprenticeship” (Bazerman) and become familiar with the types of discourses (Gee) and genres that disciplines use.
Guidelines to Help Students Learn and Analyze Writing Genres
Places that are familiar and important to us may not appear intelligible or hospitable to students we try to bring into our worlds.
—Charles Bazerman, "Where is the Classroom?"
(Adapted from Scenes of Writing: Strategies for Composing with Genres(Devitt, Reiff, Bawarshi 2004)
1. Collect samples of the genre. Your course reading list can serve as an initial collection. Choose one or two exemplars from your reading list and provide annotations.
2. Identify the scene and the situation in which the genre is used.
Setting: Where does the genre appear? With what other genres does this genre interact?
Subject: What topics, issues, ideas, questions, etc. does the genre address? When people use this genre, what is it they are interacting about?
Purposes: Why do writers write this genre and why do readers read it? What purposes does the genre fulfill for people who use it?
3. Identify and Describe Patterns in the Genre’s Features. What recurrent features do the samples share? (Or, how does one sample share features with others in its kind?) What content is typically included? What excluded? How is the content treated? What sorts of examples are used? What counts as evidence (personal testimony, facts, etc.?) What rhetorical appealsare used? What appeals to logos (logic), pathos (feeling), and ethos (authority) appear? How are texts in the genre structured? What are their parts, and how are they organized? In what format are texts in this genre formatted? What layout or appearance is common? How long is a typical text in this genre? What types of sentences do texts in this genre typically use? How long are they? Are they simple or complex, passive or active? What diction (types of words) is most common? Is a type of jargon used? How would you describe the writer’s voice?
4. Analyze what These Patterns Reveal about the Situation and the Scene.
What do these rhetorical patterns reveal about the genre, its situation, and the scene in which it is used? Why are these patterns significant? What can you learn about the actions being performed through the genre by observing language patterns? What do participants have to know or believe to understand or appreciate the genre? Who is invited to the genre and who is excluded? What roles for writers or readers does it encourage or discourage?
Ways to Annotate
You can provide annotations in several formats.
Inserted comments in a word processing file
Sticky notes attached to a pdf file
Written comments at the end of the text or on an accompanying handout with samples cited from the text
Online annotations via a social bookmarking resource, such as Diigo.
An oral commentary provided on a screencast of the text.
Choose a format that works best for your course and learning community.
What to Annotate
Research on the use of models consistently shows the benefits of working with strong (exemplar) texts. Exemplar texts can be professional pieces from your field, and/or exemplars of student writing. A good practice is to identify and annotate at least one sample of excellent student work from your course each term. This will allow you to build your archive, and it will allow you to show future students what A-level writing entails in your course.
Getting Started: Sample Annotations
A strong annotation will identify the “writing moves” (Graff) and thinking patterns that you would like your students to emulate. The following phrases can be adapted to fit your own annotation style and needs:
The writer effectively introduces the topic/ question at issue/ argument/ problem by . . .
Here, the writer organizes complex ideas, concepts, and information (Provide examples) so that each new element builds on . . .
In this sentence/ paragraph/ section, the writer develops the topic/ argument/ problem thoroughly by selecting the most significant and relevant facts, (provide examples, such as extended definitions, concrete details, quotations, or other information) appropriate to the audience’s knowledge of the topic. . . .
Notice here how the writer uses appropriate and varied transitions and syntax to link the major sections of the text, create cohesion, and clarify the relationships among complex ideas and concepts. . . .
The writer uses precise language and domain-specific vocabulary (provide examples) to manage the complexity of the topic. . . .
The writer establishes and maintains a ________ style and ________tone while attending to the norms and conventions of the (identify specific discipline) in which the writer is writing. . . .
The writer provides a concluding section that follows from and supports the information or explanation . . .
The writer demonstrates good command of the conventions of standard written English . . .
The CTLT can assist instructors and staff with developing annotations. Drop by or make an appointment!
Useful Sources
Ambrose, Susan, et al. How Learning Works: Seven Research-Based Principles forSmart Teaching. San Francisco: Jossey-Bass, 2010.
Bazerman, Charles. “Where is the Classroom?” English Basics, Winter 1992.Reprinted in Learning and Teaching Genre, ed. Freedman and Medway. Boynton-Cook, 1994.
Devitt, Amy, et al. Scenes of Writing: Strategies for Composing with Genres. Harlow, England: Longman, 2004.
Gee, James Paul.Social Linguistics and Literacies. New York: Routledge, 1996.
Graff, Gerald and Cathy Birkenstein. They Say, I Say: The Moves that Matter inAcademic Writing.New York: Norton, 2007.
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View Edit To edit this page, request access to the workspace. Already have an account? Log in! Perfect Square ============== Page history last edited by Andrea Grieser;)15 years, 2 months ago A perfect square is a rational number whose square root is also a rational number. In other words, there exists a number that when multiplied by itself gives you the perfect square. An example of a perfect square is 9. There exists a number, 3, that when squared gives us 9. The perfect squares between 1 and 100 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 You can create this list by squaring the first 10 numbers: 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8 2, 9 2, 10 2 We can use our knowledge of perfect squares to help usestimate the square root of numbers that are not perfect squares. Perfect Square ============== #### Page Tools ### Insert links Insert links to other pages or uploaded files. PagesImages and files Insert a link to a new page 1. Loading... 1. No images or files uploaded yet. Insert image from URL Tip: To turn text into a link, highlight the text, then click on a page or file from the list above. ### Comments (0) You don't have permission to comment on this page. Printable version PBworks / Help Terms of use / Privacy policy / GDPR About this workspace Contact the owner / RSS feed / This workspace is public### Join this workspace To join this workspace, request access. Already have an account? Log in! ### Navigator Algebra Relations Right Triangles Scale Drawings Sets Sigma Notation (Summation Notation) Simplifying Radicals Slide and Divide Method for Factoring Polynomials Slope Slope-Intercept Form of a Line Solving Equations Solving Equations for a Variable Solving Inequalities Solving Linear Systems of Equations Solving Linear Systems of Inequalities Solving Quadratic Equations Solving Quadratic Equations by Completing the Square Spheres Standard Deviation Standard Form of a Line Statistical Data Representation Statistical Deviation Statistical Variation Statistics Subtracting Polynomials Surface Area Synthetic Division Systems of Equations and Inequalities Three-Dimensional Figures Transformations Triangles Trinomials Volume Wiki rules X-intercept Y-intercept Zero Product Property optionsPagesFiles ### SideBar Mrs. G's Algebra Wiki. See rules for more information. Virginia Standards of Learning (SOLs) 2009 Index ### Recent Activity Show 0 new item s Logicedited by Andrea Grieser Deductive Reasoningedited by Andrea Grieser Inductive Reasoningedited by Andrea Grieser Logicedited by Andrea Grieser Deductive Reasoningedited by Andrea Grieser Deductive Reasoningedited by Andrea Grieser Deductive Reasoningadded by Andrea Grieser More activity... |
4363 | https://www.doubtnut.com/qna/277385680 | Prove the following Identities sin ( x + y ) - 2 sin x + sin ( x - y ) cos ( x + y ) - 2 cos x + cos ( x - y ) = tan x
To prove the identity sin(x+y)−2sinx+sin(x−y)cos(x+y)−2cosx+cos(x−y)=tanx, we will simplify the left-hand side step by step. Step 1: Rewrite the numerator using sine addition formulas We know the sine addition formulas: sin(a+b)=sinacosb+cosasinb, sin(a−b)=sinacosb−cosasinb. Using these, we can rewrite sin(x+y) and sin(x−y): sin(x+y)=sinxcosy+cosxsiny, sin(x−y)=sinxcosy−cosxsiny. Substituting these into the numerator gives: sin(x+y)−2sinx+sin(x−y)=(sinxcosy+cosxsiny)−2sinx+(sinxcosy−cosxsiny). Step 2: Combine terms in the numerator Combining the terms in the numerator: =sinxcosy+cosxsiny−2sinx+sinxcosy−cosxsiny, =2sinxcosy−2sinx. Factoring out 2sinx: =2sinx(cosy−1). Step 3: Rewrite the denominator using cosine addition formulas Now, we apply the cosine addition formulas: cos(x+y)=cosxcosy−sinxsiny, cos(x−y)=cosxcosy+sinxsiny. Substituting these into the denominator gives: cos(x+y)−2cosx+cos(x−y)=(cosxcosy−sinxsiny)−2cosx+(cosxcosy+sinxsiny). Step 4: Combine terms in the denominator Combining the terms in the denominator: =cosxcosy−sinxsiny−2cosx+cosxcosy+sinxsiny, =2cosxcosy−2cosx. Factoring out 2cosx: =2cosx(cosy−1). Step 5: Simplify the fraction Now we have: 2sinx(cosy−1)2cosx(cosy−1). We can cancel 2(cosy−1) from the numerator and the denominator (assuming cosy≠1): =sinxcosx=tanx. Conclusion Thus, we have shown that: sin(x+y)−2sinx+sin(x−y)cos(x+y)−2cosx+cos(x−y)=tanx. This proves the identity. ---
To prove the identity
sin(x+y)−2sinx+sin(x−y)cos(x+y)−2cosx+cos(x−y)=tanx,
we will simplify the left-hand side step by step.
Step 1: Rewrite the numerator using sine addition formulas
We know the sine addition formulas:
sin(a+b)=sinacosb+cosasinb,
sin(a−b)=sinacosb−cosasinb.
Using these, we can rewrite sin(x+y) and sin(x−y):
sin(x+y)=sinxcosy+cosxsiny,
sin(x−y)=sinxcosy−cosxsiny.
Substituting these into the numerator gives:
sin(x+y)−2sinx+sin(x−y)=(sinxcosy+cosxsiny)−2sinx+(sinxcosy−cosxsiny).
Step 2: Combine terms in the numerator
Combining the terms in the numerator:
=sinxcosy+cosxsiny−2sinx+sinxcosy−cosxsiny,
=2sinxcosy−2sinx.
Factoring out 2sinx:
=2sinx(cosy−1).
Step 3: Rewrite the denominator using cosine addition formulas
Now, we apply the cosine addition formulas:
cos(x+y)=cosxcosy−sinxsiny,
cos(x−y)=cosxcosy+sinxsiny.
Substituting these into the denominator gives:
cos(x+y)−2cosx+cos(x−y)=(cosxcosy−sinxsiny)−2cosx+(cosxcosy+sinxsiny).
Step 4: Combine terms in the denominator
Combining the terms in the denominator:
=cosxcosy−sinxsiny−2cosx+cosxcosy+sinxsiny,
=2cosxcosy−2cosx.
Factoring out 2cosx:
=2cosx(cosy−1).
Step 5: Simplify the fraction
Now we have:
2sinx(cosy−1)2cosx(cosy−1).
We can cancel 2(cosy−1) from the numerator and the denominator (assuming cosy≠1):
=sinxcosx=tanx.
Conclusion
Thus, we have shown that:
sin(x+y)−2sinx+sin(x−y)cos(x+y)−2cosx+cos(x−y)=tanx.
This proves the identity.
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4364 | https://people.csail.mit.edu/siracusa/doc/hyptest-faq.pdf | Hypothesis Testing 1 Introduction This document is a simple tutorial on hypothesis testing. It presents the basic concepts and definitions as well as some frequently asked questions associated with hypothesis testing. Most of the material presented has been taken directly from either Chapter 4 of Scharf or Chapter 10 of Wasserman .
2 Hypotheses Let X be a random vector with the range χ and distribution Fθ(x). The parameter θ is belongs to the parameter space Θ. Let Θ = Θ0 S Θ1 S ... S ΘM−1 be a disjoint covering of the parameter space. We define Hi as the hypothesis is that θ ∈Θi.
An M-ary hypothesis test chooses which of the M disjoint subsets contain the unknown parameter θ.
When M = 2 we have a binary hypothesis test. For the remainder of this document we will only discuss binary hypothesis tests (H0 : θ ∈Θ0 versus H1 : θ ∈Θ1).
2.1 Null and Alternative Hypotheses In many binary hypothesis tests H0 is referred to as the null hypothesis and H1 the alternative hypothesis.
This is due to the fact that in most binary tests the hypothesis H0 is set up to be refuted in order to support an alternative hypothesis H1. That is, H0 usually represents the absence of some effect/factor/condtion. For example when testing if a new drug is better than a placebo for relieving a set of symptoms the null hypothesis H0 says the new drug has the same effect as the placebo. With tests of this form it is common to talk about a hypothesis test in terms accepting or rejecting the null hypothesis.
2.2 Simple vs. Composite When Θi contains a single element θi hypothesis Hi is said to be simple. Otherwise it is composite. A binary hypothesis test can be simple vs simple, simple vs composite, composite vs simple, or composite vs composite. Here are some simple examples: H0 : θ = θ0 versus H1 : θ = θ1 (simple vs simple) H0 : θ = 0 versus H1 : θ ̸= 0 (simple vs composite) H0 : θ < 0 versus H1 : θ > 0 (composite vs composite) 2.3 One-Sided and Two-Sided Tests A hypothesis test is considered to be two-sided if the is of the form: H0 : θ = θ0 versus H1 : θ ̸= θ0 where the alternative hypothesis H1 “lies on both sides of H0”. A test of the form H0 : θ ≤θ0 versus H1 : θ > θ0 1 or H0 : θ ≥θ0 versus H1 : θ < θ0 is called a one-sided test. Note that one-sided and two-sided tests are only defined for scalar parameter spaces and at least one hypothesis must be composite.
2.4 Frequentist vs. Bayesian View of Hypotheses Thus far we discussed hypothesis testing in terms of determining which subset of a parameter space an unknown θ lies. The classic/frequentist approach to hypothesis testing treats θ as deterministic but unknown.
A Bayesian approach treats θ as a random variable and assumes there is a distribution on the possible θ in the parameter space. That is, one can define a prior on each hypothesis being true. Discussion of advantageous and disadvantageous of each of these will be spread throughout the following sections.
3 Binary Hypothesis Testing Give a sample x of a random vector X whose range is χ and has the distribution Fθ(x) a binary hypothesis test (H0 : θ ∈Θ0 versus H1 : θ ∈Θ1) takes the form φ(x) = ( 0 ∼H0, x ∈A 1 ∼H1, x ∈Ac, (1) This equation is read as, “the test function φ(x) equals 0, and the hypothesis H0 is accepted, if the measure-ment x lies in the acceptance region A (where A ⊂χ). If the measurement lies outside this region then test function equals 1 and hypothesis H0 is rejected and H1 is accepted.” Usually the region A is of the form: A = {x : T(x) < c} (2) where T is a test statistic and c is a critical value. The trick is to find the appropriate test statistic T and an appropriate critical value c. We will be more explicit about what “appropriate” means in the following sections.
3.1 Type I and Type II Errors There are two types of errors a binary hypothesis test can make. A type I error or false alarm is when H0 is true, but x ∈Ac. That is, the test chooses H1 when H0 is true. A type II error or miss is when H1 is true, but x ∈A. That is, the test chooses H0 when H1 is true.
3.2 Size and Power If H0 is simple (Θ0 = {θ0}), the size or probability of false alarm is α = Pθ0(φ(x) = 1) = Eθ0[φ(x)] = PF A.
(3) where Eθ0[φ(x)] indicates that φ(x) is averaged under the density function fθ0(x).
If H0 is composite, the size is defined to be α = sup θ∈Θ0 Eθ[φ(x)].
(4) The size is the worst-case probability of choosing H1 when H0 is true. A test is said to have level α if its size is less than or equal to α.
A hit or detection is when H1 is true, and x ∈Ac. If H1 is simple, the power or probability of detec-tion is β = Pθ1(φ(x) = 1) = Eθ1[φ(x)] = PD.
(5) 2 If H1 is composite, the power is defined for each θ ∈Θ1 as β(θ). In fact everything can be defined in terms of this power function: β(θ) = Pθ(φ(x) = 1) = Pθ(x ∈Ac) (6) That is the power of a composite test is defined for each θ ∈Θ1 and the size can be written as: α = sup θ∈Θ0 β(θ) (7) A receiver operating characteristic (ROC) curve is a plot of β versus α (for a simple vs simple hypothesis test). Usually multiple β, α pairs are obtained by adjusting the threshold / critical value c in Equation 2.
3.3 Bias A test φ(x) is said to be unbiased if its power is never smaller than its size. That is, β(θ) ≥α ∀θ ∈Θ1 (8) 3.4 Best and Uniformly Most Powerful Test For a simple versus simple binary hypothesis test, φ(x) is the best test of size α if it has the most power among all tests of size α. That is, if φ(x) and φ′(x) are two competing tests each of which has size α, then β ≥β′. The best test maximizes the probability of detection (power) for a fixed probability of false alarm (size). Neyman-Pearson will show us the form of the best test for a fixed α in the next section.
A test φ(x) is said to be uniformly most powerful (UMP) of size α if it has size α and its power is uniformly (for all θ) greater than the power of any other test φ′(x) whose size is less than or equal to α. That is: sup θ∈Θ0 Eθ[φ(x)] = α sup θ∈Θ0 Eθ[φ′(x)] ≤α Eθ[φ(x)] ≥Eθ[φ′(x)] ∀θ ∈Θ1 In general a UMP test may be difficult to find or may not exist. One strategy to proving a test is UMP is to find the best test for a particular θ and then show the test does not depend on θ. The Karlin-Rubin theorem shows how to obtain the UMP test for certain one-sided hypothesis tests (See ).
4 Neyman-Pearson Lemma The Neyman-Pearson Lemma shows how to find the most powerful or best test of size α when H0 and H1 are both simple. The lemma tells us the “appropriate” test statistic T to maximize the power given a fixed α.
The test is a slight generalization of the test defined in 1. The lemma states that: φ(x) = 1, fθ1(x) > kfθ0(x) γ, fθ1(x) = kfθ0(x) 0, fθ1(x) < kfθ0(x), (9) or alternatively φ(x) = 1, T(x) > k γ, T(x) = k 0, T(x) < k (10) for some k ≥0, 0 ≤γ ≤1, is the most powerful test of size α > 0 for testing H0 versus H1. T(x) = fθ1(x)/fθ0(x) = L(x) and is called the likelihood ratio. When φ(x) is 1 or 0 it is the same as in Equation 1. However, when φ(x) = γ we “flip a γ coin” to select H1 with probability γ (when the coin comes up heads).
3 Proof. Consider any test φ′(x) such that α′ ≤α. We have Z [φ(x) −φ′(x)][fθ1(x) −kfθ0(x)] ≥0 (11) β −β′ ≥k(α −α′) (12) ≥0.
(13) 4.1 Setting the size to α The question remains of how to set k to produce a test of size α. The size for this test is: α = Eθ0[φ(x)] = Pθ0[L(x) > k] + γPθ0[L(x) = k] (14) If there exists a k0 such that Pθ0[L(x) > k0] = α (15) then we set γ = 0 and pick k = k0. Otherwise there exists a k′ 0 such that Pθ0[L(x) > k′ 0] ≤α < Pθ0[L(x) ≥k′ 0] (16) We can then use k = k′ 0 and choose the γ that solves: γPθ0[L(x) = k′ 0] = α −Pθ0[L(x) > k′ 0] (17) 4.2 Interpretation So the Neyman-Pearson tells us that the best / most powerful test for a fixed size alpha is one that uses that makes decisions by thesholding the likelihood ratio L(x). We refer to such tests as likelihood ration tests (LRT) . Note that the test statistic is the likelihood ratio L(x) and is a random variable. If L(x) = k with probability zero (which is most likely the case for continuous x) then the threshold k is found as: α = Pθ0[L(x)] > k] = Z ∞ k fθ0(L)dL (18) where fθ0(L) is the density function for L(x) under H0.
5 Bayesian Hypothesis Testing In the previous sections we discussed simple binary hypothesis testing in the following framework. Given a measurement x drawn from the distribution Fθ(x), how do we choose whether θ = θ0 or θ = θ1. We defined hypothesis H0 : θ = θ0 and H1 : θ = θ1 and look for a test of H1 versus H0 that is “optimal”. We talked about optimality in terms of maximizing the power (β) of such a test for a fixed size α. The parameter θ (and the hypothesi) are treated at deterministic but unknown quantities. That is either H1 or H0 is true and we don’t know which one. We don’t have any prior knowledge of how likely H1 or H0 is to occur or how likely any parameter choice is. Note that the power and size are both defined in terms of one of the hypothesis being true.
The Bayesian approach to hypothesis testing treats θ and the hypothesis H as unknown random variables.
Here we are introducing the random variable H to represent the hypothesis. If H = i it means hypothesis Hi is true. We can think of the test function φ as an estimator for H. The conceptual framework is as follows.
Mother Nature selects H and the parameter θ from a joint distribution f(θ, H) = p(H)f(θ|H). She does this by first choosing H according to p(H) and then picks a θ according to f(θ|H). Note that f(θ|H = i) = 0 for all θ / ∈Θi. Her selection determines from which distribution Fθ(x) Father Nature draws his measurement.
This measurement is given to the experimenter and he or she must decide between estimate the value of H via a decision function φ. Each time the experiment is run a parameter θ is chosen by Mother nature, and the 4 experimenter outputs a decision ˆ H = φ(x). The goal in Bayesian hypothesis testing to design a test φ that is “optimal” / gives the best performance. Here “optimality” is described in terms of the Bayes risk which is described below.
To be more concrete let us consider the simple versus simple binary hypothesis test we have been dis-cussing so far.
Let p(H = H0) = p0 and p(H = H1) = 1 −p0.
Since the hypothesis is simple f(θ|H = Hi) = δ(θ −θi). Mother nature selects H according to p(H). In the simple binary case we are considering this is equivalent to picking θ = θ0 with probability p0 and θ = θ1 with probability p1 = 1 −p0.
Depending on her choice Father Nature then draws a a measurement x from either Fθ0(x) or Fθ1(x). Our goal will be to obtain ˆ H = φ(x) that minimizes the Bayes risk.
5.1 Cost of Decisions A cost or loss function is defined for each possible pairing of the true hypothesis H and decision ˆ H = φ(x).
That is for the pair (H, φ(x)) = (i, j) we assign a nonnegative cost C[H = i, φ(x) = j] = cij. We say cij the loss incurred when Mother Nature selects the hypothesis Hi and the experimenter decides to choose Hj. That is for a simple binary hypothesis test we give values for c00, c11, c01 and c10. Normal we do not associate a loss/cost for making a correct decision, i.e. c00 = c11 = 0.
5.2 Risk We define the risk R(Hi, φ(x)) for each Hi as the expected loss given H = Hi for particular test function φ: R(H, φ) = Ex[C[H, φ(x)]] = ( c00P00 + c01P01, θ = θ0 c10P10 + c11P11, θ = θ1 (19) where Pij = p(φ(x) = 1|H = Hi). This is equivalent to pθi(φ(x) = j) in the simple binary hypothesis case.
5.3 Bayes Risk The Bayes risk is the average risk over the distribution of H that Mother Nature used (for the binary case, only P(H = 0) = p0 is needed).
R(p0, φ) = EH[R(H, φ)] = p0R(H = 0, φ) + (1 −p0)R(H = 1, φ) (20) Given that we known the prior p0, the optimal test φ is defined to be the one that minimizes the Bayes Risk: φ = arg min φ′ R(p0, φ′) (21) It turns out that the solution to this equation/optimization has the form (see 6.432 notes or Scharf Chapter 5): φ(x) = ( 1, L(x) > η 0, L(x) < η (22) where L(x) = f(x|H = 1) f(x|H = 0) = R θ f(x|H = 1, θ)f(θ|H = 1)dθ R θ f(x|H = 0, θ)f(θ|H = 0)dθ (23) is the likelihood ratio and for the simple versus simple case is L(x) = f(x|θ1) f(x|θ0) = fθ1(x) fθ0(x) (24) and the threshold η = p0(c10 −c00) (1 −p0)(c01 −c11) (25) 5 So once again we see that the “optimal” test is a likelihood ratio test. Here “optimal” is in terms of minimizing Bayes risk. The test statistic is the likelihood ratio and the acceptance region depends on the threshold η which is based on the priors and decision costs.
Minimizing risk sounds like the right way to think about these problems. However this approach requires us to have some knowledge about the prior on θ and be able to assign a cost to each possible outcome. This may be difficult in some applications. For example, what is the prior probability of a missile coming toward your home? We will quickly discuss what can be done when we can assign costs to decision but don’t know the a priori probabilities of each hypothesis in the Minmax section below. When we don’t know the prior AND cannot think of a meaningful costs we go back to Neyman-Pearson testing.
6 Test Statistics and Sufficiency When we talk about the hypothesis H being a random variable we can consider the following Markov chain H →X →T(X) where T(X) is some test statistic. Any test statistic is said to be sufficient for H if p(H|X) = p(H|T(X)) That is, if T(X) is sufficient it tells us everything we need to know about the observation x in order to estimate H and the Markov chain can be written as H →T(X) →X.
Note that the likelihood ratio L(X) was to optimal test statistic for our hypothesis test. It takes our K dimension observation x and maps it to a scalar value. It can be shown that L(X) is a sufficient statistic for H.
This was just a quick note. More details can be found in the 6.432 notes.
7 Minimax Tests As we eluded to before we may be able to associated a cost with each of the possible outcomes of a hypothesis test but have no idea what the prior on H is (or even worse the full f(θ, H)). In such cases we can play a game in which we assume Mother Nature is really mean and will choose a prior that makes whatever test we choose to look bad. That is for a simple versus simple binary hypothesis test: maxpominφR(p0, φ) (26) To combat this, we will try to find a φ that minimizes the worst she can do: minφmaxpoR(p0, φ) (27) Seciont 5.3 of Scharf shows how to find such a minmax detector / test function φ. It is also shown that maxpominφR(p0, φ) = minφmaxpoR(p0, φ) (28) This topic is also discussed in the 6.432 notes.
8 Generalized Likelihood Ratio Test A discussed before when dealing with composite hypotheses in Neyman-Pearson framework we wish to find the UMP test for a fixed size α (put your Bayes hat away for a bit). However, it is typically the case that such a test does not exist. A generalized likelihood ratio test is a way to deal with general composite hypothesis test. Again we will focus on the binary case with H0 : θ ∈Θ0 versus H1 : θ ∈Θ1. The generalized likelihood ratio is a test statistic with the following form: LG(x) = supθ∈Θ1 f(x|θ) supθ∈Θ0 f(x|θ) (29) We see that for a simple versus simple hypothesis test LG(x) = L(x). This test statistic is rather intuitive.
If top part of the fraction in Equation 29 is greater than the bottom part then the data is best explained when 6 θ ∈Θ1. If the opposite is true then the data is best explained by θ ∈Θ0. Instead of using Equation 29 as the test statistic one generally uses: λ(x) = supθ∈Θ f(x|θ) supθ∈Θ0 f(x|θ) (30) instead, where Θ = Θ0 S Θ1. It can be shown in general that λ(x) = max(LG(x), 1). This may seem a little strange in that Θ and Θ0 are nested (they are not disjoint). Much more can be said about this, but for now I will simply paraphrase Wasserman’s tutorial on this subject and say that in practice using λ(x) instead of LG(X) has little effect in practice and theoretically properties of λ(x) are usually much easier to obtain.
So the generalized likelihood ratio test gives us a test statistic that makes some intuitive sense. We can threshold this statistic and calculate it’s size and power if we can derive it’s distribution under each hypothesis.
Now let’s put our Bayes hat back on. We showed before that the likelihood ratio test minimizes the Bayes risk and that for a composite test the likelihood ratio is: L(x) = R θ f(x|H = 1, θ)f(θ|H = 1)dθ R θ f(x|H = 0, θ)f(θ|H = 0)dθ (31) That is we should integrate over all possible values of theta for each hypothesis in each hypothesis. However, typically our parameter space is extremely large. If we assume that f(x|H = 1, θ)f(θ|H = 1) contains a large peak (looks like a delta function) at ˆ θ1 = arg maxθ∈Θ1 f(x|H = 1, θ)f(θ|H = 1) and f(x|H = 1, θ)f(θ|H = 0) peaks at ˆ θ0 we can approximate the likelihood ratio as: ˆ L(x) = f(x| ˆ θ1)f( ˆ θ1|H = 1) f(x| ˆ θ0)f( ˆ θ1|H = 0) = maxθ∈Θ1 f(x|H = 1, θ)f(θ|H = 1) maxθ∈Θ0 f(x|H = 0, θ)f(θ|H = 0) (32) which is one possible interpretation of the generalized likelihood ratio ˆ LG(x) (ignore the details involved with max and sup).
9 Test of Significance In the Neyman-Pearson testing framework one fixes the size of the α of the test. However, different people may have different criteria for choosing an appropriate size. One experimenter may be happy with setting the size to α = 0.05 while another demands α is set to 0.01. In such cases it is possible that one experimenter accepts H0 while the other rejects it when given the same data x. Only reporting if H0 was accepted or rejected may not be very informative in such cases. If the experimenters both use the same test statistic (i.e.
both do a likelihood ratio test) it may be more useful for them to report the outcome of their experiment in terms of the significance probability or p-value of the test (also referred to as the observed size). We give a formal definition for the p-value below.
9.1 P-value If a test rejects H0 at a level α it will also reject at a level α′ > α. Remember that when a test rejects H0 at level α that means it’s size is less than or equal to α. (i.e. if we say test rejects at level .05 that means it’s size α ≤.05). The p-value is the smallest α at which a test rejects H0.
Suppose that for every α ∈(0, 1) we have a size α test with a rejection Ac α, then p-value = inf{α : T(x) ∈Ac α} (33) That is the p-value is the smallest level α at which an experimenter using the test statistic T would reject H0 on the basis of the observation x.
Ok, that definition requires a lot of thought to work through. It may be easier to understand what a p-value is if we explain how to calculate one: p-value = sup θ∈Θ0 Pθ(T(X) ≥T(x)) (34) 7 where x is the observed value of X. If H0 is a simple hypothesis and Θ0 = {θ0} then p-value = Pθ(T(X) ≥T(x)) = P(T(X) ≥T(x)|H0) (35) It is important remember that T(X) is a random variable with a particular distribution under H0 and that T(x) is a number, the value of the test statistic for the observed x. In the case of a simple H0 the p-value is the probability of obtaining a test statistic value greater than the one you observed when H0 is true. Another way to look at it is that the p-value is the size of a test using your observed T(x) as the threshold for rejecting H0.
If the test statistic T(X) has a continuous distributen then under a simple H0 : θ = θ0, the p-value has a uniform distribution between 0 and 1. If we reject H0 when the p-value is less than α then the probability of false alarm (or size of the test) is α. That is we can set up a test to have size α by making the test function be: φ(x) = ( 1, p-value ≥α 0, p-value < α, (36) So a small p-value is strong evidence against H0. However, note that a large p-value is NOT strong evidence in favor of H0. A large p-value can mean H0 is true OR H0 is false but the test has low power β. It also important to note that the p-value is not the probability that the null hypothesis is true . That is, in almost every case the p-value ̸= p(H0|x). We will shown one exception in the next section.
9.2 P-values for One-Sided Tests Let’s look at a one-sided tests where Θ ⊂R and θ0 is entirely to one side of θ1. In this case, p-values will sometimes have a Bayesian justification. For example if X ∼N(θ, σ2) and p(θ) = 1, then p(θ|x) is N(x, σ2). We test H0 : θ ≤θ0 against H1 : θ > θ0, p(H0|x) = p(θ ≤θ0|x) = Φ θ0 −x σ .
(37) The p-value is p-value = p(X ≥x) = 1 −Φ x −θ0 σ = p(H0|x) (38) because Φ is symmetric.
10 Permutation Tests We showed how to calculate a p-value or significance in the previous section. This calculation requires knowing the distribution (the cdf) of the test statistic T(X) under H0. However, in many cases it may be difficult to obtain this distribution (i.e. distribution of X may be unknown). A permutation tests are tests based on non-parametric estimates of significance. It does not rely on the distribution of the test statistic. The basic procedure is as follows: 1. Compute the observed value of the test statistic tobs = T(x) 2. Obtain a new sample xs that obeys the null hypothesis H0 via a resampling function π.
That is xs = π(x).
3. Compute ts = T(xs) 4. Repeat Steps 2 and 3 B times and let t1, ..., tB denote the resulting values.
5. Calculate an approximate ˆ p-value = 1 B PB j=1 I(tj > tobs) where I(true) = 1 and I(false) = 0.
6. Reject H0 (choose H1) if ˆ p-value > α 8 where step 5 uses the empirical cumulative distribution obtained from samples in step 2 to estimate the p-value. The question remains on what we mean by obeys the null hypothesis in step 2. The resampling function π obeys H0 if each new sample xs is equality likely when H0 is true.
Take for example an observation x = (y1, y2, ..., yN, z1, ..., zM) which is N observations of some random variable Y and M observations of the random variable Z. If H0 was that both Z and Y have the same mean one possible test statistic would be T(x) = | 1 N PN i=1 xi − 1 M PN+M j=N+1 xj|. We let π(x) produces a new N+M sample that is a random permutation on the order of the elements in xs. There are (N+M)! possible permutations each of which is equally likely under H0.
A simple introduction to permutation tests can be found in . In Joeseph P. Romano shows that for any finite set of transformations π ∈Π that are a mapping of X onto itself and for which π(X) and X have the same distribution under H0 the testing procedure described above produces a test of size α.
References P. Good. Permutation Tests: A Practical Guide to Resampling Methods for Testing Hypotheses. Springer-Verlag, 1994.
J. P. Romano. On the behavoir of randomization tests without a group invariance assumption. Journal of the American Statistical Association, 85(411):686–692, 1990.
L. Scharf. Statistical Signal Processing: Detection, Estimation, and Time Series Analysis. Addison-Wesley Publishing Company, 1991.
L. Wasserman. All of Statistics : A Concise Course in Statistical Inference. Springer-Verlag New York, Inc., 2004.
9 |
4365 | https://www.navarra.es/NR/rdonlyres/435E1C2C-DFA9-4612-916C-8FC30D8EB9B1/235272/mayusculasdefinitivoweb.pdf | USO DE LAS MAYÚSCULAS CUESTIONES GENERALES: Las mayúsculas deben llevar tilde y diéresis si les corresponde: Ángel, SANGÜESA. Solo se exceptúan las siglas puras, que nunca llevan tilde: CIA, no CÍA. Los símbolos internacionales de medidas (m, cm, kW), son de forma fija, sin presentar variaciones nunca entre mayúscula y minúscula. 1. MAYÚSCULA CONDICIONADA POR LA PUNTUACIÓN Se escriben en mayúscula las palabras que van detrás de punto o son la primera palabra de un escrito. Cuando se usan puntos suspensivos, si lo que viene tras ellos es un periodo sintáctica-mente independiente, comienza con mayúscula: Compramos de todo: mariscos, carne, vino… La comida fue un éxito. Pero si lo que sigue a los puntos es continuación sintáctica de lo anterior, no: Me pareció bastante… tonto. Tras dos puntos, lo más habitual es la minúscula, pero si hay corte sintáctico claro, de-be ir en mayúscula, por ejemplo tras los dos puntos del encabezamiento de una carta (Estimado señor:), o cuando los dos puntos introducen una cita literal, o cuando lo que antecede a los dos puntos tiene función de epígrafe, como en una enciclopedia o diccio-nario (Rebozuelo: Se trata de una seta muy común…). También va en mayúscula lo que sigue a expresiones administrativas como EXPONE: o CERTIFICA: y similares. Los puntos del signo de interrogación y de exclamación, valen por puntos normales y, por tanto, lo que sigue va en mayúscula. Cuando varias exclamaciones o interrogaciones van seguidas, pueden no llevar nada entre ellas, en cuyo caso comienzan todas por ma-yúscula; pero si se consideran coordinadas, debe añadirse entre ellas coma, y la frase que sigue a esta coma comienza en minúscula: ¿Cómo te llamas?, ¿dónde trabajas?, ¿dónde vives? Si la interrogación o exclamación solo afecta a una parte de la frase, va minúscula tras el signo ¿ o ¡: A este chico ¿qué le pasa? 2. MAYÚSCULAS EN NOMBRES PROPIOS O EXPRESIONES EQUIVALEN-TES Se escriben con mayúscula inicial todos los nombres propios y también los nombres comunes o expresiones pluriverbales que por alguna razón (generalmente la antonoma-sia) tienen función de nombres propios. Seres personales Los nombres de divinidades y demonios se escriben con mayúscula inicial: Jehová, Alá, Júpiter… Algunas religiones usan el nombre común dios, por antonomasia como nombre propio de su divinidad; en este caso debe escribirse en mayúscula, Dios. No así los pronombres que se refieren a él, como solía hacerse antes: Señor, yo te suplico. Dios en su infinita bondad. Los nombres comunes que por antonomasia funcionan como nombre de un dios o de-monio van también con mayúscula inicial: el Maligno, el Todopoderoso, el Mesías, el Señor. En los apellidos con partícula, esta se escribe con minúscula cuando va precedida del nombre de pila: Luis de Pedro. Pero si el apellido va solo, la partícula lleva mayúscula inicial: El señor De Pedro; De Cospedal dijo; el señor Del Burgo (pero J. I. del Burgo). En los apellidos con artículo, este se escribe siempre con mayúscula, en todos los casos: Antonio La Torre, el señor La Torre. Los apellidos en plural usados para nombrar una familia, van en mayúscula: los Pérez, los Soprano, los Borbones. Pero cuando para nombrar una dinastía o linaje se usan patronímicos (adjetivos deriva-dos del nombre propio de un antepasado fundador), van en minúscula: los abasíes, los nazaríes, los pelópidas. Sobrenombres, apodos y seudónimos se escriben en mayúscula, aunque si llevan artí-culo, este va en minúscula: el Greco, el Cordobés, el Pescadilla. Los tratamientos, títulos nobiliarios u otros, cargos, empleos y dignidades son nom-bres comunes y deben ir en minúscula: rey, papa, presidente, ministro, duque, director general, jefe de sección, minis- tro. Sin embargo, la abreviatura de los tratamientos va con mayúscula: Sr., D. (don), D.ª (doña), S. (san), etc. A veces coincide el cargo con el nombre de la institución; entonces debe distinguirse con la mayúscula cuándo se trata de uno u otro: El defensor del pueblo dijo en la cámara… (cargo). Las oficinas del Defensor del Pueblo… (institución). Estos títulos pueden usarse a veces por antonomasia, es decir, se puede usar uno de estos nombres comunes como equivalente de un nombre propio: El Papa (el actual papa Benedicto XVI). El Rey, (Juan Carlos I, rey de España). Incluso en estos casos, vale la minúscula: Su santidad visitará Navarra. Nótese que Su Santidad Pablo I visitará… sería incorrecto. En general, cuando hay antonomasia, es decir, cuando se usa un nombre común como equivalente del nombre personal, este se escribe en mayúscula: el Mantuano (Virgilio), el Magnánimo (Alfonso V), el Todopoderoso (Dios), el Mesías (Jesucristo). Nombres de lugar Los nombres propios geográficos (continentes, países, ciudades, pueblos, comarcas, mares, ríos, etc.) van con mayúscula inicial. El artículo que frecuentemente llevan, se escribe casi siempre en minúscula (Nos bañamos en el Mediterráneo). Solo cuando, por ejemplo, un nombre de comarca (la Rioja, la Mancha), se convierte en nombre oficial de una entidad política, el artículo pasa a escribirse en mayúscula: La Rioja, La Mancha (Comunidad de La Rioja, Comunidad de Castilla-La Man-cha). El mismo tratamiento reciben las áreas geopolíticas: América Latina, Europa Occidental, Oriente, Occidente. En algunos nombres de países y ciudades, el artículo ha quedado soldado al nombre el artículo va en mayúscula: El Salvador, La Paz, Las Palmas. La diferencia se ve comparando por ejemplo la India con La Paz. En la primera, se pue-de introducir adjetivos entre el artículo y el nombre o sustituirlo por un demostrativo: La misteriosa India; aquella India que conocimos. En la segunda, no: La acogedora La Paz que conocimos. Esta Las Palmas de hoy no es la que conocimos. Muchos nombres de lugar llevan un genérico, un nombre común clasificador, que los precede (ciudad, río, mar, sierra, montaña, etc.); en general no se considera parte del nombre y va en minúscula: el río Ebro, la sierra de Gredos, el río Araxes, el cabo de Hornos, el mar Rojo, el río Grande. Pero a veces el genérico sí forma parte del nombre y debe ir entonces en mayúscula: Sierra Nevada, Ciudad Real, Cabo Verde, Ciudad del Cabo, el Río de la Plata, las Montañas Rocosas, los Picos de Europa. El Camino de Santiago lleva c mayúscula porque no es un camino sino una entidad tu-rístico-espiritual. Si el adjetivo que acompaña al genérico geográfico es de nacionalidad, o derivado de un nombre de lugar (Italia, Iberia, Balcanes, Grecia, etc.), toda la expresión va en minúscu-las: península ibérica, península itálica, arábiga, balcánica, islas británicas, meseta castellana. Estos genéricos geográficos se usan a veces solos, por antonomasia, para designar un lugar que dentro del contexto se identifica inequívocamente. En estos casos el genérico va en mayúscula: La Cordillera, sin más, los sudamericanos para hablar de los Andes. La Península para la península ibérica (para los españoles). Otra antonomasia frecuente es utilizar una expresión denominativa para designar como variante estilística un lugar. También estos casos se escriben en mayúscula: La Ciudad Condal, el Nuevo Mundo, la Ciudad Eterna. En las vías y espacios urbanos (calles, plazas, etc.), también hay muchas veces un nombre común genérico que las precede. Normalmente no forma parte del nombre y va en minúscula: calle Mayor, plaza de Merindades, paseo de Valencia. Pero si el adjetivo que acompaña al genérico lo precede, se escribe en mayúscula: Quinta Avenida. Si son de otro idioma, se respeta el uso original: Oxford Street. Estrellas, constelaciones planetas y satélites van en mayúscula: Osa Mayor, Estrella Polar, Venus. Sol y Luna se escriben con mayúsculas en contextos astronómicos, cuando claramente se habla de cuerpos celestes, especialmente si se mencionan junto con otros planetas o meteoros. Pero en otros contextos se escriben en minúscula: El sol brillaba espléndido aquella mañana. Lo vimos a la luz de la luna. Me gusta la luna llena. Tierra, con mayúscula, es el nombre del planeta; en otros casos se escribe en minúscula: El avión tomó tierra. Los puntos cardinales se escriben siempre en minúscula, en todos los casos (salvo que forme parte de un nombre propio: Corea del Norte). Las líneas imaginarias de la cor-teza terrestre van en minúscula: ecuador, trópicos, paralelo. Entidades Los nombres de instituciones, entidades, organismos, departamentos, divisiones administrativas, partidos políticos, establecimientos públicos, etc. llevan todas sus palabras significativas con mayúscula inicial: Departamento de Salud, Sección de Publicaciones, Biblioteca General, el Teatro Gayarre, la Casa Blanca, la Casa Rosada, el Instituto Cervantes, Archivo Gene-ral de Navarra, Universidad Pública de Navarra, Facultad de Medicina, la Puerta de Alcalá. En los nombres de edificios y monumentos se escribe el genérico en minúscula: El arco del Triunfo, la catedral de Barcelona, la iglesia de San Pedro, la torre de Pisa. Esta mayúscula sirve para distinguir cuándo hablamos de instituciones y cuándo de otras realidades: La Universidad española se enfrenta a un gran cambio. / Se decidió crear una universidad en Pamplona. La Iglesia no admite el aborto. / Las primeras iglesias cristianas se construyeron. El Ejército está descontento. / El ejército francés fue masacrado. Educación primaria, educación secundaria son nombres comunes y por tanto deben ir en minúscula. Sin embargo, cuando se refieren a divisiones administrativas, van en mayús-cula: El tercer ciclo de Educación Primaria. Los nombres de órdenes religiosas se escriben con mayúscula inicial: El Carmelo, la Compañía de Jesús. Si les acompaña la palabra orden, esta también va con mayúscula inicial: La Orden del Temple. Obras de creación Los títulos de libros, películas, cuadros, esculturas, piezas musicales, programas, etc. llevan mayúscula inicial solo en su primera palabra: El jinete polaco, Las cuatro estaciones, la Celestina, el Pensador. Igual los títulos de ponencias, discursos y conferencias. Y las oraciones: El Yo pecador, el Señor mío Jesucristo. Pero los libros sagrados la llevan en todas sus palabras significativas: Libro de los Reyes, Hechos de los Apóstoles. También los nombres de revistas y colecciones llevan todas sus palabras significativas con mayúscula inicial: Cuadernos de Etnología, Biblioteca de Autores Españoles, Colección Barco de Vapor. Documentos y textos legales Los documentos históricos, documentos oficiales, leyes, decretos, etc. llevan todas las palabras significativas de su nombre oficial con mayúscula inicial, excepto si el nombres es excesivamente largo, entonces solo la primera palabra lleva mayúscula: Edicto de Nantes, Tratado de Blois, Declaración Universal de los Derechos Humanos, Ley para la Ordenación del Sistema Educativo… Los nombres no oficiales con los que se conocen informalmente algunas leyes van en minúscula: ley de extranjería, ley de igualdad, ley de educación, ley de la dependencia. También los planes, proyectos y programas llevan mayúscula inicial en todas sus pa-labras significativas, salvo que el nombre resulte muy largo, en cuyo caso solo va con mayúscula inicial la primera palabra. Los premios, distinciones, certámenes y acontecimientos culturales y deportivos llevan mayúscula inicial en todas las palabras significativas de su nombre oficial: Premio Cervantes, Premios Goya, Bienal de Venecia, Juegos Olímpicos, Feria del Libro, Premio Príncipe de Viana de la Cultura. Disciplinas científicas Se escriben en minúscula. Solo cuando se refieren a asignaturas o a estudios reglados concretos van con mayúscula inicial: La medicina europea avanzó mucho en el s. XIX./ Estoy estudiando Medicina. Me faltan dos asignaturas de Derecho. / El derecho civil es una creación romana. Los nombres de asignaturas que no coinciden exactamente con el nombre de una disci-plina científica se recomienda poner todas las palabras significativas con mayúscula inicial salvo si el nombre es muy largo. Los nombres de cursos, congresos, seminarios, etc. llevan todas sus palabras significa-tivas con mayúscula inicial: III Congreso Mundial de Neonatología. Animales Los nombres científicos en latín de plantas y animales, que van siempre en cursiva, solo llevan mayúscula inicial en la primera palabra: Felis silvestris catus (gato común), Felis leo, Homo sapiens (no Homo Sapiens). Los nombres vernáculos van en minúscula: águila ratonera, vencejo común. Referencias temporales Los nombres de las dos grandes épocas de la historia humana, se escriben con minúscu-la inicial: historia y prehistoria, en todos los contextos, salvo que sean asignaturas. Las demás épocas y edades históricas, los periodos geológicos y los acontecimientos de la historia se escriben con mayúscula inicial: el Cuaternario, el Jurásico, el Paleolítico, la Edad del Hierro, la Edad Media, la Antigüedad, la Reforma, la Contrarreforma, la Reconquista, la Primera Guerra Mundial, la Gran Depresión, el Renacimiento, la Primavera de Praga, el Cisma de Occidente, Mayo del 68, el Viernes Negro. No llevan mayúscula inicial, en cambio, los adjetivos que siguen al nombre de un im-perio: El Imperio romano, bizantino, persa, etc.; las revoluciones van con mayúscula inicial en todas su palabras significativas (Revolución Industrial, Revolución Cultural), salvo si el adjetivo es de nacionalidad: Revolución francesa. Pero si el adjetivo precede al sustantivo, entonces va en mayúscula: La Edad Media tardía / la Baja Edad Media, el Bajo Imperio, el Celeste Imperio. En el caso de las guerras, no llevan mayúscula las palabras guerra o batalla: La guerra de los Cien Años, la guerra de Secesión, la guerra de la Independen-cia, la batalla de las Termópilas. Si la expresión es totalmente descriptiva, es mejor escribirla toda en minúscula: La guerra civil española. Pero si hay antonomasia, todo en mayúsculas: La Guerra Civil (se entiende por antonomasia que la española). Los nombres de festividades, religiosas o no, llevan todas sus palabras significativas con mayúscula inicial: Semana Santa, San Fermín (pero el nombre del santo es san Fermín), Año Nue-vo, Navidad, Primero de Mayo, Día de la Madre, etc. Los nombres de los días de la semana, meses y estaciones se escriben con minúscula inicial: martes, julio, verano, otoño. |
4366 | https://www.splashlearn.com/math-vocabulary/open-interval-and-closed-interval | Published Time: 2023-06-27T09:21:04+00:00
Open Interval and Closed Interval: Definition, Examples, Facts
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Open Interval and Closed Interval – Definition, Examples, FAQs
Home » Math Vocabulary » Open Interval and Closed Interval – Definition, Examples, FAQs
What Is Open Interval and Closed Interval?
What Is the Difference between Open Interval and Closed Interval?
Operations on Open Intervals and Closed Intervals
Solved Examples on Open Interval and Closed Interval
Practice Problems on Open Interval and Closed Interval
Frequently Asked Questions on Open Interval and Closed Interval
What Is Open Interval and Closed Interval?
There are two types of intervals known as open interval and closed interval. In an open interval (a, b), the endpoints are not included. In a closed interval [a, b], the endpoints are included.
Before we dive into the details of open intervals and closed intervals, let us first understand the definition of interval in math.
Interval: An interval is the collection of all real numbers in the given range. It contains all the real numbers between two given numbers, called endpoints of the interval. The endpoints are included or excluded based on the type of the interval.
Examples: (0,1),[−5,5],(1,9]
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What Is an Open Interval?
An interval in which the endpoints are not included is called an open interval. An open interval (a, b) represents the set of all real numbers x such that a <x<b.
We get an open interval if an inequality has the signs
less than <
greater than >
Example: (-3, 5) is an open interval that includes all the real numbers greater than -3 but less than 5. We can write is with the help of compound inequality −3<x<5. In the set notation, we can express the open interval (−3,5) as {x|−3<x<5}.
Open Interval Notation
A round bracket or parentheses ( ) is used to denote an open interval. An open interval is written as an ordered pair within round brackets as (a, b).
Open Interval on a Number Line
To represent an open interval on a number line, we use two hollow circles to denote the endpoints. These hollow circles show that the points are not included.
Open interval (-3, 5) on a number line:
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What Is a Closed Interval?
An interval in which the endpoints are included is called a closed interval. A closed interval [a, b] represents the set of all real numbers x such that a≤x≤b.
We get closed interval if an inequality has the signs
less than or equal to ≤
greater than or equal to ≥
Example: [2,7] is a closed interval which includes 2, 7 and all the real numbers between 2 and 7. We can write it with the help of compound inequality 2≤x≤7. In the set notation, we can express the open interval [2,7] as {x|2≤x≤7}.
Closed Interval Notation
Square brackets [] are used to denote a closed interval. We write a closed interval as an ordered pair within square brackets as [a,b].
Closed Interval on a Number Line
To represent a closed interval on a number line, we use two solid circles to denote the endpoints. Solid circles convey that these points are to be included.
Closed interval [2, 7] on a number line:
What Is the Difference between Open Interval and Closed Interval?
Open Interval vs. Closed IntervalOpen IntervalClosed intervalOpen intervals are denoted by an ordered pair (a, b) with round brackets or parentheses ( ).Closed intervals are denoted by an ordered pair [a, b] with square brackets [ ].Endpoints a and b are not included.Endpoints a and b are also included.Open interval (a, b) can be defined by an inequality a<x>b.Closed interval [a, b] can be defined by an inequality a≤x≥b.Endpoints are denoted by hollow circles on a number line.Endpoints are denoted by solid circles on a number line.
Operations on Open Intervals and Closed Intervals
Let’s discuss two operations on intervals that are similar to the operations we perform on sets.
Union of Two Intervals
The union of two intervals A and B included all the elements that are in A, in B, or both in A and B.
Examples:
(1,5)(2,7)\=(1,7)
4,9\=4,15)
Intersection of Two Intervals
The intersection of two intervals is the collection of all the elements that are common to both the intervals.
Examples:
(1,5)(2,7)\=(2,5)
[4,9\=(5,9)
Complement of an Interval
The complement of an interval with respect to the set of real numbers is the set of all real numbers that are not in the given interval.
Example: A\=(4,8)
Complement of A\=A′\=(−,4][8,)
Facts about Open Interval and Closed Interval
Open intervals do not include the endpoints but closed intervals always include the endpoints.
A half-open interval is an interval wherein one endpoint is included but the other is not.
For example, in the interval [a, b), the endpoint a is included, but b is not included.
Half-open interval (a,b] includes the endpoint b but not the endpoint a.
Conclusion
In this article, we learned about open and closed intervals, definitions, notations along with operations on them. Let’s solve a few examples and practice problems based on these concepts to understand the concept better.
Solved Examples on Open Interval and Closed Interval
1. Look at the two intervals [3,7] and (2,6). Which interval includes the number 3? Which interval includes 6?
Solution:
The closed interval [3,7] includes all the real numbers between 3 and 7, including both 3 and 7. The open interval (2, 6) includes all the real numbers greater than 2 but less than 6.
Both intervals include the number 3.
The closed interval [3,7] includes the number 6.
The open interval (2, 6) does not include 6.
2. Write the given inequalities as intervals.
i) 0<x<99
i) −5≤x≤8
Solution:
i) 0<x<99
Endpoints are not included. So, we use an open interval.
The inequality can be written as (0,99).
ii) −5≤x≤8
Endpoints are included. So, we use a closed interval.
The inequality can be written as [−5,8].
3. What is the union and intersection of the intervals (0,99) and [1,100]?
Solution:
(0,99) represents the inequality 0<x<99.
[1,100] represents the inequality 1≤x≤100.
The union of intervals includes the elements present in either of the intervals.
(0,99)∪[1,100]\=(0,100]
The intersection includes the elements common to both intervals.
(0,99)∩[1,100]\=[1,99)
4. Observe at the inequality −4<x<3 and state if it is an open interval or a closed interval. Represent it on a number line.
Solution:
The inequality is −4<x<3 is an open interval. This means x is less than 3 but greater than -4. Here, as -4 and 3 are not included, it is an open interval.
Practice Problems on Open Interval and Closed Interval
Open Interval and Closed Interval - Definition, Examples, FAQs
Attend this quiz & Test your knowledge.
1
Which of the following intervals includes the number 5?
[2,5)
(2,5)
(5,8]
[3,7]
CorrectIncorrect
Correct answer is: [3,7]
The interval that includes the number 5 is [3,7], as it includes all real numbers x such that 3≤x≤7.
2
Which statement is true about open intervals?
An open interval includes its endpoints.
An open interval does not include its endpoints.
An open interval can only include one endpoint.
An open interval is denoted by (a,b].
CorrectIncorrect
Correct answer is: An open interval does not include its endpoints.
The only correct statement is “An open interval does not include its endpoints.”
3
Which of the following intervals does not include the number 2?
(−1,2)
[−7,2]
[2,13)
(−10,13)
CorrectIncorrect
Correct answer is: (−1,2)
The interval that does not include 2 is (−1,2), which is an open interval and does not include 2.
4
Which of the following is a closed interval?
(0,5)
[2,6)
(3,8]
[4,9]
CorrectIncorrect
Correct answer is: [4,9]
The interval [4,9] is a closed interval because it includes both 4 and 9.
5
Represent 15≤y≤45 using an interval.
(15,45]
[15,45)
[15,45]
(15,45)
CorrectIncorrect
Correct answer is: [15,45]
15≤y≤45 can be written as [15,45].
Frequently Asked Questions on Open Interval and Closed Interval
What is a half-open interval?
A half-open interval simply means that one of the endpoints is included, and the other is not. Half-open intervals are written as (a,b] or [a,b). Here, only one interval is open and the other is closed.
How can I identify an open interval and a closed interval?
The easiest way to identify an open interval is by looking at the bracket used. Round brackets represent an open interval. On the other hand, a closed interval is denoted by the square brackets like [a, b].
How do we define real numbers as an interval?
Real numbers can be represented by an open interval (−,∞,∞).
Is (−,∞∞) an open interval or closed interval?
It is always an open interval.
What are examples of open and closed intervals in real life?
Suppose a courier company only accepts packages less than 30 kg. So, 30 kg would not be an acceptable weight. Thus, we can define the range between 0 and 30, both 0 and 30 not included. This can be represented by an open interval (0, 30).
Suppose a company is looking to hire people within the age group of 25 to 45 such that people who are aged 25 and 45 are also included. The endpoints of this example, 25 and 45, are also included, Hence, this would be a closed interval.
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4367 | https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Introductory_Statistics_(Shafer_and_Zhang)/03%3A_Basic_Concepts_of_Probability/3.01%3A_Sample_Spaces_Events_and_Their_Probabilities | 3.1.5
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3.1: Sample Spaces, Events, and Their Probabilities
Last updated
: Mar 26, 2023
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3: Basic Concepts of Probability
3.2: Complements, Intersections, and Unions
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Learning Objectives
To learn the concept of the sample space associated with a random experiment.
To learn the concept of an event associated with a random experiment.
To learn the concept of the probability of an event.
Sample Spaces and Events
Rolling an ordinary six-sided die is a familiar example of a random experiment, an action for which all possible outcomes can be listed, but for which the actual outcome on any given trial of the experiment cannot be predicted with certainty. In such a situation we wish to assign to each outcome, such as rolling a two, a number, called the probability of the outcome, that indicates how likely it is that the outcome will occur. Similarly, we would like to assign a probability to any event, or collection of outcomes, such as rolling an even number, which indicates how likely it is that the event will occur if the experiment is performed. This section provides a framework for discussing probability problems, using the terms just mentioned.
Definition: random experiment
A random experimentis a mechanism that produces a definite outcome that cannot be predicted with certainty. The sample space associated with a random experiment is the set of all possible outcomes. An event is a subset of the sample space.
Definition: Element and Occurrence
An event EE is said to occur on a particular trial of the experiment if the outcome observed is an element of the set EE.
Example 3.1.13.1.1: Sample Space for a single coin
Construct a sample space for the experiment that consists of tossing a single coin.
Solution
The outcomes could be labeled hh for heads andttfor tails. Then the sample space is the set: S={h,t}S={h,t}
Example 3.1.23.1.2: Sample Space for a single die
Construct a sample space for the experiment that consists of rolling a single die. Find the events that correspond to the phrases “an even number is rolled” and “a number greater than two is rolled.”
Solution
The outcomes could be labeled according to the number of dots on the top face of the die. Then the sample space is the set S={1,2,3,4,5,6}S={1,2,3,4,5,6}
The outcomes that are even are 2,4,and62,4,and6, so the event that corresponds to the phrase “an even number is rolled” is the set {2,4,6}{2,4,6}, which it is natural to denote by the letter EE. We write E={2,4,6}E={2,4,6}.
Similarly the event that corresponds to the phrase “a number greater than two is rolled” is the set T={3,4,5,6}T={3,4,5,6}, which we have denoted TT.
A graphical representation of a sample space and events is a Venn diagram, as shown in Figure 3.1.13.1.1. In general the sample space SS is represented by a rectangle, outcomes by points within the rectangle, and events by ovals that enclose the outcomes that compose them.
Example 3.1.33.1.3: Sample Spaces for two coines
A random experiment consists of tossing two coins.
Construct a sample space for the situation that the coins are indistinguishable, such as two brand new pennies.
Construct a sample space for the situation that the coins are distinguishable, such as one a penny and the other a nickel.
Solution
After the coins are tossed one sees either two heads, which could be labeled 2h2h, two tails, which could be labeled 2t2t, or coins that differ, which could be labeled dd Thus a sample space is S={2h,2t,d}S={2h,2t,d}.
Since we can tell the coins apart, there are now two ways for the coins to differ: the penny heads and the nickel tails, or the penny tails and the nickel heads. We can label each outcome as a pair of letters, the first of which indicates how the penny landed and the second of which indicates how the nickel landed. A sample space is then S′={hh,ht,th,tt}S′={hh,ht,th,tt}.
A device that can be helpful in identifying all possible outcomes of a random experiment, particularly one that can be viewed as proceeding in stages, is what is called a tree diagram. It is described in the following example.
Example 3.1.43.1.4: Tree diagram
Construct a sample space that describes all three-child families according to the genders of the children with respect to birth order.
Solution
Two of the outcomes are “two boys then a girl,” which we might denote bbgbbg, and “a girl then two boys,” which we would denote gbbgbb.
Clearly there are many outcomes, and when we try to list all of them it could be difficult to be sure that we have found them all unless we proceed systematically. The tree diagram shown in Figure 3.1.23.1.2, gives a systematic approach.
The diagram was constructed as follows. There are two possibilities for the first child, boy or girl, so we draw two line segments coming out of a starting point, one ending in a bb for “boy” and the other ending in a gg for “girl.” For each of these two possibilities for the first child there are two possibilities for the second child, “boy” or “girl,” so from each of the bb and gg we draw two line segments, one segment ending in a bb and one in a gg. For each of the four ending points now in the diagram there are two possibilities for the third child, so we repeat the process once more.
The line segments are called branches of the tree. The right ending point of each branch is called a node. The nodes on the extreme right are the final nodes; to each one there corresponds an outcome, as shown in the figure.
From the tree it is easy to read off the eight outcomes of the experiment, so the sample space is, reading from the top to the bottom of the final nodes in the tree,
S={bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg}
S={bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg}
Probability
Definition: probability
The probability of an outcome ee in a sample space SS is a number PP between 11 and 00 that measures the likelihood that ee will occur on a single trial of the corresponding random experiment. The value P=0P=0 corresponds to the outcome ee being impossible and the value P=1P=1 corresponds to the outcome ee being certain.
Definition: probability of an event
The probability of an event AA is the sum of the probabilities of the individual outcomes of which it is composed. It is denoted P(A)P(A).
The following formula expresses the content of the definition of the probability of an event:
If an event EE is E={e1,e2,...,ek}E={e1,e2,...,ek}, then
P(E)=P(e1)+P(e2)+...+P(ek)
P(E)=P(e1)+P(e2)+...+P(ek)
The following figure expresses the content of the definition of the probability of an event:
Since the whole sample space SS is an event that is certain to occur, the sum of the probabilities of all the outcomes must be the number 11.
In ordinary language probabilities are frequently expressed as percentages. For example, we would say that there is a 70%70% chance of rain tomorrow, meaning that the probability of rain is 0.700.70. We will use this practice here, but in all the computational formulas that follow we will use the form 0.700.70 and not 70%70%.
Example 3.1.53.1.5
A coin is called “balanced” or “fair” if each side is equally likely to land up. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair coin.
Solution
With the outcomes labeled hh for heads and tt for tails, the sample space is the set
S={h,t}
S={h,t}
Since the outcomes have the same probabilities, which must add up to 11, each outcome is assigned probability 1/21/2.
Example 3.1.63.1.6
A die is called “balanced” or “fair” if each side is equally likely to land on top. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair die. Find the probabilities of the events EE: “an even number is rolled” and TT: “a number greater than two is rolled.”
Solution
With outcomes labeled according to the number of dots on the top face of the die, the sample space is the set
S={1,2,3,4,5,6}
S={1,2,3,4,5,6}
Since there are six equally likely outcomes, which must add up to 11, each is assigned probability 1/61/6.
Since E={2,4,6}E={2,4,6},
P(E)=16+16+16=36=12
P(E)=16+16+16=36=12
Since T={3,4,5,6}T={3,4,5,6},
P(T)=46=23
P(T)=46=23
Example 3.1.73.1.7
Two fair coins are tossed. Find the probability that the coins match, i.e., either both land heads or both land tails.
Solution
In Example 3.1.33.1.3 we constructed the sample space S={2h,2t,d}S={2h,2t,d} for the situation in which the coins are identical and the sample space S′={hh,ht,th,tt}S'={hh,ht,th,tt} for the situation in which the two coins can be told apart.
The theory of probability does not tell us how to assign probabilities to the outcomes, only what to do with them once they are assigned. Specifically, using sample space S, matching coins is the event M={2h,2t} which has probability P(2h)+P(2t). Using sample space S′, matching coins is the event M′={hh,tt}, which has probability P(hh)+P(tt). In the physical world it should make no difference whether the coins are identical or not, and so we would like to assign probabilities to the outcomes so that the numbers P(M) and P(M′) are the same and best match what we observe when actual physical experiments are performed with coins that seem to be fair. Actual experience suggests that the outcomes in S' are equally likely, so we assign to each probability 14, and then...
P(M′)=P(hh)+P(tt)=14+14=12
Similarly, from experience appropriate choices for the outcomes in Sare:
P(2h)=14
P(2t)=14
P(d)=12
The previous three examples illustrate how probabilities can be computed simply by counting when the sample space consists of a finite number of equally likely outcomes. In some situations the individual outcomes of any sample space that represents the experiment are unavoidably unequally likely, in which case probabilities cannot be computed merely by counting, but the computational formula given in the definition of the probability of an event must be used.
Example 3.1.8
The breakdown of the student body in a local high school according to race and ethnicity is 51% white, 27% black, 11% Hispanic, 6% Asian, and 5% for all others. A student is randomly selected from this high school. (To select “randomly” means that every student has the same chance of being selected.) Find the probabilities of the following events:
B: the student is black,
M: the student is minority (that is, not white),
N: the student is not black.
Solution
The experiment is the action of randomly selecting a student from the student population of the high school. An obvious sample space is S={w,b,h,a,o}. Since 51% of the students are white and all students have the same chance of being selected, P(w)=0.51, and similarly for the other outcomes. This information is summarized in the following table:
OutcomewbhaoProbability0.510.270.110.060.05
Since B={b},P(B)=P(b)=0.27
Since M={b,h,a,o},P(M)=P(b)+P(h)+P(a)+P(o)=0.27+0.11+0.06+0.05=0.49
Since N={w,h,a,o},P(N)=P(w)+P(h)+P(a)+P(o)=0.51+0.11+0.06+0.05=0.73
Example 3.1.9
The student body in the high school considered in the last example may be broken down into ten categories as follows: 25% white male, 26% white female, 12% black male, 15% black female, 6% Hispanic male, 5% Hispanic female, 3% Asian male, 3% Asian female, 1% male of other minorities combined, and 4% female of other minorities combined. A student is randomly selected from this high school. Find the probabilities of the following events:
B: the student is black
MF: the student is a non-white female
FN: the student is female and is not black
Solution
Now the sample space is S={wm,bm,hm,am,om,wf,bf,hf,af,of}. The information given in the example can be summarized in the following table, called a two-way contingency table:
| Gender | Race / Ethnicity | | | | |
--- --- --- |
| White | Black | Hispanic | Asian | Others |
| Male | 0.25 | 0.12 | 0.06 | 0.03 | 0.01 |
| Female | 0.26 | 0.15 | 0.05 | 0.03 | 0.04 |
Since B={bm,bf},P(B)=P(bm)+P(bf)=0.12+0.15=0.27
Since MF={bf,hf,af,of},P(M)=P(bf)+P(hf)+P(af)+P(of)=0.15+0.05+0.03+0.04=0.27
Since FN={wf,hf,af,of},P(FN)=P(wf)+P(hf)+P(af)+P(of)=0.26+0.05+0.03+0.04=0.38
Key Takeaway
The sample space of a random experiment is the collection of all possible outcomes.
An event associated with a random experiment is a subset of the sample space.
The probability of any outcome is a number between 0 and 1. The probabilities of all the outcomes add up to 1.
The probability of any event A is the sum of the probabilities of the outcomes in A.
3: Basic Concepts of Probability
3.2: Complements, Intersections, and Unions |
4368 | http://www.geo.cornell.edu/geology/classes/Geo656/656notes09/656_09Lecture06.pdf | Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 56 February 4, 2009 THE U-TH-PB SYSTEM U and Th are, strictly speaking, rare earth elements, although they belong to the actinide series in-stead of the lanthanide series. The other rare earths we have met so far, Nd and Sm, are lanthanides. As in the lanthanide rare earths, an inner electron shell is being filled as atomic number increases in the actinides. Both U and Th generally have a valence of +4, but under oxidizing conditions, such as at the surface of the Earth, U has a valence of +6. In six-fold coordination, U4+ has an ionic radius of 89 pm1 (100 pico meters = 1 Å); U6+ has an ionic radius of 73 pm in 6-fold and 86 pm in 8-fold coordination. Th4+ has an ionic radius of 94 pm. These radii are not particularly large, but the combination of some-what large radius and high-charge is not readily accommodated in crystal lattices of most common rock-forming minerals, so both U and Th are highly incompatible elements. Th is relatively immobile under most circumstances. In its reduced form, U4+ is insoluble and therefore fairly immobile, but in the U6+ form, which is stable under a wide range of conditions at the surface of the Earth, U forms the soluble oxyanion complex, UO 4 2– . As a result, U can be quite mobile. U and Th can form their own phases in sedimentary rocks, uranite and thorite, but they are quite rare. In igneous and metamorphic rocks, U and Th are either dispersed as trace elements in major phases, or concentrated in accessory minerals (when they are present) such as zircon (ZrSiO4), which concentrates U more than Th, and monazite ([La,Ce,Th]PO4) which concentrations Th more than U. These elements may be also concen-trated in other accessory phases such as apatite (Ca5(PO4)3(OH)) and sphene (CaTi(SiO4)OH). However, zircon is by far and away the most important from a geochronological perspective. U and Th are refractory elements, and we can therefore expect the Th/U ratio of the Earth to be the same as chondrites or nearly so. There is, however, some debate about the exact terrestrial Th/U ratio, and we can be no more precise than to say it is 4±0.22. This ratio is 3.8 in the CI chondrite Orgueil, but may be low due to mobility of U in hydrous fluid in the CI parent body. The geochemical behavior of Pb is more complex than that of the elements we have discussed so far and consequently, less well understood. It is a relatively volatile element, so its concentration in the Earth is certainly much lower than in chondrites. It is also a chalcophile element. If the core contains, as some believe, S as the light element, it is possible that some of the Earth's Pb is in the core (it is, how-ever, difficult to distinguish loss of Pb from the Earth due to its volatility from loss of Pb from the sili-cate portion of the Earth due to extraction into the core). Pb can exist in two valence states, Pb2+ and Pb4+. Pb2+ is by far the most common state; the Pb4+ state is rare and restricted to highly alkaline or oxi-dizing solutions. The ionic radius of Pb2+ is 119 pm in 6-fold coordination and 129 pm in 8-fold coordi-nation. As a result of its large ionic size, Pb is an incompatible element, though not as incompatible as U and Th (incompatibility seems to be comparable to the light rare earths). The most common Pb min-eral is galena (PbS). In silicates, Pb substitutes readily for K (ionic radius 133 pm) particularly in potas-sium feldspar, but less so in other K minerals such as biotite. Most naturally occurring compounds of Pb are highly insoluble under most conditions. As a result, Pb is usually reasonably immobile. How-ever, under conditions of low pH and high temperature, Pb forms stable and somewhat soluble chlo-ride and sulfide complexes, so that Pb can sometimes be readily transported in hydrothermal solutions. 1In eight-fold coordination, the effective ionic radius of U4+ is 1.00Å. In zircon, a mineral which highly concentrations U, U is in 8-fold coordination. This is probably a pretty good indication that 8-fold coordination is the preferred con-figuration. The figure for 6-fold coordination is given for comparison to other radii, which have been for 6-fold coor-dination. Th has a radius of 1.05Å in 8-fold coordination. 2 The uncertainty results from the mobility of U. The CI carbonaceous chondrites experienced mild alteration in hy-drous conditions on the parent body. U was mobilized under these conditions and thus the U/Th ratio varies in these meteorites. For this reason, they cannot be used to precisely determine the U/Th ratio of the Solar System and the Earth. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 57 February 4, 2009 Although Pb is clearly less incompatible that U and Th, these 3 elements have been extracted from the mantle and concentrated in the crust to approximately the same degree. The reason for this is not yet completely understood, and we will discuss the problem later in the course. The U-Th-Pb system is certainly the most powerful tool in the geochronologist's tool chest. Table 6.1 summarizes this decay system. The reason for the power is simply that there are three parents decay-ing to 3 isotopes of Pb, and in particular, there are two isotopes of U which decay to Pb with very dif-ferent half lives. This is important because chemical processes will not change the ratio of the two U isotopes to each other and will not change the ratio of the two Pb daughter isotopes to each other. The point is best illustrated as follows. First we write the decay equation for each of the two U decay sys-tems: 207Pb = 235U(e !5t "1) 6.01 206Pb = 238U(e !8t "1) 6.02 where the asterisk designates radiogenic 206Pb and 207Pb, and λ5 and λ8 are the decay constants for 235U and 238U respectively. If we divide 6.01 by 6.02, we have: 207Pb 206Pb = 235U(e !5t "1) 238U(e !8t "1) 6.03 Now if the ratio of the U isotopes is everywhere the same (as it is at the present day), 6.03 can be writ-ten as 207Pb 206Pb = (e !5t "1) 137.88(e !8t "1) 6.04 The nice thing about equation 6.04 is that the only variable on the right hand side is time; in other words the 207Pb/206Pb is a function only of time. TABLE 6.1. Parameters of the U-Th-Pb System Parent Decay Mode λ Half-life Daughter Ratio 232Th α,β 4.948 x 10-11y-1 1.4 x 1010y 208Pb, 8 4He 208Pb/204Pb, 3He/4He 235U α,β 9.849 x 10-10y-1 7.07 x 108y 207Pb, 7 4He 207Pb/204Pb, 3He/4He 238U α,β 1.551 x 10-10y-1 4.47 x 109y 206Pb, 6 4He 206Pb/204Pb, 3He/4He In practice what this means is that the age is independent of the parent/daughter ratio; i.e., we not need to measure the parent/daughter ratio. We shall see that this property actually allows us to somewhat relax our requirement that the system remain closed. We can also see that although we could write an equation similar to 80.3 using 232Th and 208Pb instead of 235U and 207Pb, there would be lit-tle advantage to doing so because Th and U are different elements and could well be lost or gained in different proportions. If Madison Avenue were given the task of selling the U-Th-Pb system, they would probably say that you get 4 dating methods for the price of one: 238U-206Pb, 235U-207Pb, 232Th-208Pb, and 207Pb-206Pb. In a cer-tain sense, this is true. However, if you bought the package, you would probably quickly discover that the first three above, applied independently, were not particularly powerful, at least in comparison to either the Pb-Pb technique or simultaneous use of a combination of several techniques (an exception might be the 232Th-208Pb system, which might prove useful on separated Th-bearing minerals). The Pb-Pb method, as it is called, can be quite useful when applied independently, particularly where there is reason to believe that there has been some recent change in the parent/daughter ratio. We have men-tioned in an earlier lecture that the slope on a plot of 207Pb/204Pb vs. 206Pb/204Pb is proportional to age since: Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 58 February 4, 2009 !( 207Pb / 204Pb) !( 206Pb / 204Pb) = (e "5t #1) 137.88(e "8t #1) 6.05 Equation 6.05 is very similar to equation 6.04. We would use 6.04 when either there is no significant initial Pb, or the amount of initial Pb is sufficiently small that we can make a reasonable estimate of its isotopic composition and make a correction for it. We would use 6.05 when initial Pb is present in sig-nificant quantities and has an unknown composition. Figure 6.1 shows an example of a Pb-Pb isochron that yielded a reasonably precise age. Unlike a conventional isochron, the intercept in the Pb-Pb iso-chron has no significance, and the initial isotopic composition cannot be determined without some ad-ditional information about parent/daughter ratios. There are a couple of reasons why we might suspect parent/daughter ratios have changed, and hence might prefer the Pb-Pb approach to one involving parent-daughter ratios. First, the solubility of U under oxidizing conditions often leads to mobility (open-system behavior) in the zone of weathering. It has often been found that U-Pb ages are spurious, yet Pb-Pb ages seem correct. This circumstance appears to result from recent U mobility as erosion brings a rock into the weathering zone. A second situation where parent/daughter ratios would have experienced recent change is in magma generation. When melting occurs, the U and Pb isotope ratios in the magma will be identical to those in the source (because the isotopes of an element are chemically identical), but the U/Pb ratio (and Th/Pb) ratio will change, as the chemical behaviors of U and Pb differ. So conventional dating schemes cannot generally provide useful geochronological information about sources of magmas. However, the Pb-Pb dating method can, at least in principle, provide useful information, because the Pb isotope ratios of a magma are representative of the source and the method does not depend on parent/daughter ratios. Essen-tially, what we are doing is allowing volcanism to ‘sample’ the source, generally the mantle, but some-times the lower continental crust. The sample is representative of the isotopic composition of the source, but not representative of the elemental chemistry of the source. The relationship between Pb isotope ratios in mantle-derived magmas has lead to the conclusion that heterogeneities in the mantle must have existed for times on the order of 1-2 Ga. This is an extremely important constraint not only on the chemical evolution of the mantle, but also on its dynamics. The U-Pb system achieves it greatest power when we use the 238U-206Pb, 235U-207Pb, and 207Pb-206Pb methods in combination. In many instances, it can ac-tually be used to ‘see through’ open system be-havior and obtain an age of initial crystallization. We shall examine this in the next lecture. Th/U Ratios Provided Th/U ratios are constant and known in a set of samples we wish to date, we can calculate ages from 208Pb/204Pb–206Pb/204Pb iso-chrons just as we can using 207Pb and 206Pb. However, although U and Th are geo-10 20 30 40 50 60 Northwest Rhyolite 2.692 ± 0.011 Ga MSWD = 0.94 Noranda sulfides 206Pb/204Pb 14 16 18 20 22 24 G 204Pb 207Pb Figure 6.1. A Pb-Pb isochron obtained on volcanic rocks hosting the Noranda (Quebec) Cu-Zn sulfide deposit. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 59 February 4, 2009 chemically similar and the Th/U ratio is not likely to vary much, it would not be prudent to assume the ratio is actually constant for geochronological purposes. Furthermore, there is little reason to do so, since we can already compute the age using 207Pb and 206Pb. But it may be useful in some circumstances to turn the problem around and compute the Th/U ratio from the age and the slope of the data on a plot of 208Pb/204Pb vs. 206Pb/204Pb. The basis of this is as follows. We write the usual growth equations for 206Pb and 208Pb: 206Pb / 204Pb = ( 206Pb / 204Pb)0 + 238U / 204Pb(e !8t "1) 6.06 208Pb / 204Pb = ( 208Pb / 204Pb)0 + 232Th / 204Pb(e !2t "1) 6.07 Subtracting the initial ratio from each side of each equation and dividing 6.7 by 6.6 we have !( 208Pb / 204Pb) !( 206Pb / 204Pb) = 232Th / 204Pb(e !2t "1) 238U / 204Pb(e !8t "1) 6.08 or !( 208Pb / 204Pb) !( 206Pb / 204Pb) = !(e "2t #1) (e "8t #1) 6.09 where κ is used to designate the 232Th/238U ratio. Using µ to designate the 238U/204Pb ratio, the parent-daughter ratio of the Th-Pb system is the product µκ. Equation 6.09 tells us that the slope of a line on a plot of 208Pb/204Pb vs. 206Pb/204Pb is proportional to time and κ, provided that κ does not vary. If we can calculate t from the corresponding 207Pb/204Pb– 206Pb/204Pb slope, we can solve 6.09 for κ. If, however, κ varies linearly with µ, a straight line will still result on the 208Pb/204Pb vs. 206Pb/204Pb plot and our estimate of κ will be incorrect. THE U-TH-PB SYSTEM: ZIRCON DATING Zircon (ZrSiO4) is a mineral with a number of properties that make it extremely useful for geochro-nologists (Figure 6.2 ). First of all, it is very hard (hardness 71/2), which means it extremely resistant to mechanical weathering. Second, it is extremely resistant to chemical weathering and metamorphism. For geochronological purposes, these properties mean it is likely to remain a closed system. Third, it concentrates U (and Th to a lesser extent) and excludes Pb, resulting in typically very high 238U/204Pb ra-tios. It is quite possibly nature's best clock. Finally, it is reasonably common as an accessory phase in a variety of igneous and metamorphic rocks. The very high 238U/204Pb ratios in zircon (and similar high µ minerals such as sphere and apatite) pro-vide some special geochronological opportunities and a special diagram, the concordia diagram, has been developed to take advantage of them. The discussion that follows can be applied to any other system with extremely high 238U/204Pb ratios, but in practice, zircons constitute the principle target for Pb geo-chronologists. A concordia diagram is simply a plot of 206Pb/238U vs. 207Pb/235U. You should satisfy yourself that both of these ratios are proportional to time. In essence, the concordia diagram is a plot of the 238U– 206Pb age against the 235U–207Pb age. The ‘concordia’ curve on such a diagram that is the locus of points where the 238U–206Pb age equals the 235U–207Pb age. Such ages are said to be concordant. Figure 6.3 is an example of a concordia diagram. The best way to think about evolution of Pb/U ratios is to imagine that the diagram itself evolves with time, along with its axes, while the actual data point stays fixed. Let’s take a 4.0 Ga old zircon as an example. When it first formed, or “closed”, it would have plotted at the origin, because had anyone been around to analyze it, they would have found the 207Pb/235U and 206Pb/238U ratios to be 0. Ini-tially, 207Pb/235U would have increased rapidly, while the 206Pb/238U would have been increasing only slowly. This is because 4.0 Ga ago there was a lot of 235U around (recall that 235U has a short half-life). As time passed, the increase in 207Pb/235U would have slowed as the 235U was ‘used up’. So imagine Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 60 February 4, 2009 that the diagram initially 'grows' or 'expands' to the left, expanding downward only slowly. Had someone been around 3.0 Ga ago to deter-mine 'zircon' ages, he would have drawn it as it appears in Figure 6.4 (of course, he would have labeled the 3.0 Ga point as 0, the 4.0 Ga point as 1.0, etc.). Any zircon that has re-mained as a completely closed system since its crystallization must plot on the concordia line. What happens when a zircon gains or looses U or Pb? Let’s take the case of Pb loss, since that is the most common type of open-system behavior in zircons. The zircon must lose 207Pb and 206Pb in exactly the proportions they exist in the zircon because the two are chemically identical. In other words, a zircon will not lose 206Pb in preference to 207Pb or visa versa. Let’s take the specific case of a 4.0 Ga zircon that experi-enced some Pb loss during a metamorphic event at 3.0 Ga. If the loss was complete, the zircon would have been reset and would have plotted at the origin in Figure 6.4. We could not distinguish it from one that formed 3.0 Ga. Suppose now that that zircon had lost only half its Pb at 3.0 Ga. During the Pb loss, the 206Pb/238U and 207Pb/235U would have both de-creased by half. Consequently, the point would have migrated half way to the origin. At 3.0 Ga, therefore, it would have plotted on a ‘cord’, i.e., a straight line, between its initial position on the concordia curve, the 4.0 Ga point, and the origin (Figure 6.5a) at 3.0 Ga. Had it lost some other amount of Pb, say 30% or 80%, it would have plot-ted on the same cord, but further or nearer the origin. The line is straight because the loss of 207Pb is al-ways directly proportional to the loss of 206Pb. The origin in Figure 9.3a corresponds to the 3.0 Ga point on the concordia in Figure 6.5b. So, in Figure 6.5b, the zircon would lie on a cord between the 4.0 Ga and the 3.0 Ga point. We would say this is a 'discordant' zircon. Figure 6.2 Upper. Separated Zircon crystals. Notice the zoning. Lower. Strongly zoned zircon showing differing ages of spots ana-lyzed by ion probe. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 61 February 4, 2009 The intercepts of this cord with the concordia give the ages of initial crystallization (4.0 Ga) and meta-morphism (3.0 Ga). So if we can determine the cord on which this discordant zircon lies, we can de-termine the ages of both events from the intercepts of that cord with the concordia. Unfortunately, if our only data point is this single zircon, we can draw an infinite number of cords passing through this point, so the ages of crystallization and metamorphism are indeterminate. However, we can draw only 1 line through 2 points. So by measuring two zircons (or populations of zircons) that have the same crystallization ages and metamorphism ages, but have lost different amounts of Pb, and hence plot on different points on the same cord, the cord can be determined. The closure age and partial resetting ages can then be determined from the interecepts. (as usual in geochronology, however, we are reluc-tant to draw a line through only two points since any two points define some line; so at least three measurements are generally made). In practice, different zircon populations are selected based on size, appearance, magnetic properties, color, etc. While zircon is generally a trace mineral, only very small quantities, a few milligrams, are needed for a measurement. Indeed, it is possible to analyze single zir-cons and even parts of zir-cons. U gain would affect the position of zircons on the concordia diagram in the same manner as Pb loss; the two processes are es-sentially indistinguishable on the concordia diagram. U loss, on the other hand, moves the points away from the origin at the time of the loss (Figure 6.6). In this case, the zircons lie on an extension of a cord above the concordia. As is the case for Pb loss, the upper intercept of the cord 0 10 20 30 40 50 60 70 3.0 Ga 4.0 Ga 4.55 Ga 207Pb/235U 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 238U 206Pb Figure 6.4. A concordia diagram as it would have been drawn at 3.0 Ga. 0 0 0.2 0.4 0.6 0.8 1 1.2 20 40 60 80 100 4.5 Ga 4.0 Ga 3.5 Ga 3.0 Ga 2.5 Ga 2.0 Ga 1.5 Ga 1.0 Ga 0.5 Ga 207Pb/235U 238U 206Pb Figure 6.3. The concordia diagram. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 62 February 4, 2009 gives the initial age and the lower intercept gives the age of U loss. U loss in less common than Pb loss. This is true for two reasons. First, U is happy in the zircon, Pb is not. Second, Pb will occupy a site damaged by the alpha decay, making diffusion out of this site easier. Radiation damage is a significant problem in zircon geochronology, and one of the main reasons ages can be imprecise. U-rich zircons are particularly subject to radiation damage. Heavily damaged crystals are easily recognized under the microscope and are termed metamict. Pb gain in zircons is not predictable because the isotopic composition of the Pb gained need not be the same as the composition of the Pb in the zircon. Thus Pb gain would destroy any age relationships. However, Pb gain is much less likely than other open system behaviors. Zircons that have suffered multiple episodes of open system behavior will have U-Pb systematics that are difficult to interpret and could be incorrectly interpreted. For example, zircons lying on a cord be-tween 4.0 and 3.0 Ga that subsequently lose Pb and move on a second cord toward the 2.0 Ga could be interpreted as having a metamorphic age of 2.0 Ga and a crystallization age of between 4.0 and 3.0 Ga. Continuous Pb loss from zircons can also complicate the task of interpretation. The reason is that in continuous Pb loss, zircons do not define a straight line cord, but rather a slightly curved one. Again imagining that the concordia diagram grows with time, a zircon loosing Pb will always move toward the origin. However, the position of the origin rela-tive to the position of the zircon moves with time in a non-linear fashion. The result is a non-linear evo-lution of the isotopic composition of the zircon. Given the mechanical and chemical stability of zircon, it should not be surprising that the oldest terrestrial material yet identified is zircon. Until a decade ago, the oldest dated terrestrial rocks were the Isua gneisses in Greenland. These are roughly 3850 Ma old. Work published in 1989, revealed that the Acasta gneisses of the Slave Province (Northwest Territories, Canada) are 3.96 Ga old. These ages were determined using an ion probe to date the cores of zircon crystals extracted from 0 0.2 0.4 0.6 0.8 1 1.2 206Pb/238U 3.0 Ga 4.0 Ga 0 10 20 30 40 50 70 80 90 60 4.55 Ga 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0 10 20 30 40 50 60 70 207Pb/235U 3.0 Ga 4.0 Ga 4.55 Ga 207Pb/235U 206Pb/238U Crystallization Age Metamorphic Age a b Figure 6.5. (a) Concordia diagram as it would have appeared at 3.0 Ga. Three zircons that ex-perience variable amounts of Pb loss move from the 4.0 Ga point on the concordia curve (their crystallization age) toward the origin. (b) The same three zircons as they would plot at pres??ent. The three define a cord between 3.0 Ga and 4.0 Ga. A possible interpretation of this result would be that 4.0 Ga is the crystallization age and 3.0 Ga is the metamorphic age. Figure 6.6. A concordia plot showing hypotheti-cal zircons that crystallized at 4.0 Ga and lost U during metamorphism at 3.0 Ga. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 63 February 4, 2009 these gneisses. Concordia diagrams for these gneisses are shown in Figure 6.7. Zircons having ages in the range of 4100-4260 Ma have been identified in quartzites at Mt. Narryer and the Jack Hills in western Australia (e.g., Compston and Pid-geon, 1986). The quartzites themselves are metamor-phosed sandstones that were probably deposited about 3100-3300 Ma. They contain zircons derived from a number of sources. A small fraction of these zircons has cores that are in the range of 4100-4200 Ma. The zircons were analyzed by a specially built high-resolution ion probe at the Australian National Uni-versity nicknamed ‘SHRIMP’ (subsequently, the instrument has been commercialized and similar instruments are available from competing compa-nies). The great advantage of these instrument over conventional analysis of zir-cons is not only that indi-vidual zircons can be ana-lyzed, but individual parts of the zircons can be ana-lyzed. As can be seen in Figure 6.2, zoning in zircon is not uncommon and this zoning often reflects multi-ple episodes of growth. The was the case with the Mt. Narryer zircons, which had complex histories suffering multiple metamorphic events between 4260 and 2600 Ma. The principal ef-fect was the growth of rims of new material on the older cores around 3500 Ma. Conventional analysis of these zircons would not have recognized the older ages. The cores of these zircons, however, proved to be nearly concordant at the older ages. These ages determined by ion probe were initially highly controversial. By and large, however, the community has come to accept them as reliable, when per-formed carefully. .
0.50 0.60 0.70 0.80 0.90 20.0 30.0 40.0 50.0 0.50 0.60 0.70 0.80 0.90 20.0 30.0 40.0 50.0 4000 3800 3600 3400 3200 4000 3800 3600 3400 3200 207Pb/235U 206Pb/238U 206Pb/238U Cores Altered Cores Altered Massive Massive Zoned Cores Clear Prismatic Equant Turbid Prismatic Banded Gneiss BGXM Porphyritic Gneiss SP-405 H H H H H H H H H Figure 6.7. Concordia diagrams showing ion probe Pb-U analyses of Acasta gneiss zircons. Size of point is proportional to 1 σ analytical uncertainty. Triangles are zircon analyses done by conventional mass spectrometry. From Bowring, et al, 1989. Geol. 655 Isotope Geochemistry Lecture 6 Spring 2009 64 February 4, 2009 Subsequently even older zircons (would be more correct to say parts of zircons) were discovered in the Jack Hills of Australia (Wilde, et al., 2001). An ion probe date on one part of one of these zircons (Figure 6.8) is shown 4.404 Ga ±8 Ma. Thus the oldest known terrestrial materials are approaching the oldest ages from other planetary bodies, including the Moon, Mars, and asteroids (as represented by meteorites). They remain, however, significantly younger than the age of the Solar System, which is 4.556 Ga. Nevertheless, these very old ages seem to demonstrate that it is zircons, not diamonds, that “are forever”. REFERENCES AND SUGGESTIONS FOR FURTHER READING Bowring, S. A., I. S. Williams, and W. Compston, 3.96 Ga gneisses from the Slave province, Northwest Territories, Canada, Geology, 17: 971-975, 1989. Compston, W. and R. T. Pidgeon, Jack Hills, evidence of more very old detrital zircons in Western Aus-tralia, Nature, 321:766-769, 1986. DePaolo, D. J. 1988. Neodymium Isotope Geochemistry, an Introduction, Berlin: Springer-Verlag. Dickin, A. 1995. Radiogenic Isotope Geochemistry. Cambridge: Cambridge University Press. Wilde, S. A., J. W. Valley, W. H. Peck and C. M. Graham, Evidence from detrital zircons for the exis-tence of continental crust and oceans on the Earth 4.4 Gyr ago, Nature, 409:175-178, 2001. Figure 6.8. Photomicrograph of the oldest zircon known. Light area are regions of ion probe analy-sis. Ages determined on these areas are shown. From Wilde et al. (2001). |
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Chromium(III) oxide
Chromium(III) oxide structure
CAS No.1308-38-9Chemical Name:Chromium(III) oxide Synonyms CHROMIUM OXIDE;CHROME OXIDE;CHROMIUM SESQUIOXIDE;ci77288;CHROMIA;CHROMIC OXIDE;Chromium(Ⅲ) oxide;CHROMIUM (III) OXIDE;okhp1;c-grun CBNumber:CB8394705 Molecular Formula:Cr2O3 Molecular Weight:151.99 MDL Number:MFCD00010949 MOL File:1308-38-9.molMSDS File:SDS
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Chromium(III) oxide Properties
| Melting point | 2435 °C |
| Boiling point | 4000 °C |
| Density | 5.21 |
| refractive index | 2.551 |
| Flash point | 3000°C |
| storage temp. | Room Temperature |
| solubility | Insoluble in all solvents |
| form | powder |
| color | Pale to dark green |
| Specific Gravity | 5.22 |
| Odor | at 100.00?%. odorless |
| Water Solubility | Insoluble |
| Crystal Structure | Trigonal |
| crystal system | Three sides |
| Merck | 14,2234 |
| Space group | R3c |
| Lattice constant | a/nm b/nm c/nm α/o β/o γ/o V/nm3 0.496 0.496 1.359 90 90 120 0.2895 |
| Exposure limits | NIOSH: IDLH 25 mg/m3; TWA 0.5 mg/m3 |
| Stability | Stable. |
| CAS DataBase Reference | 1308-38-9(CAS DataBase Reference) |
| FDA 21 CFR | 73.1327; 73.2327; 73.3111 |
| Indirect Additives used in Food Contact Substances | CHROMIUM OXIDE GREEN |
| EWG's Food Scores | 1 |
| FDA UNII | X5Z09SU859 |
| NIST Chemistry Reference | Chromium(iii) oxide(1308-38-9) |
| EPA Substance Registry System | Chromium(III) oxide (1308-38-9) |
| UNSPSC Code | 12352303 |
| NACRES | NA.55 |
SAFETY
Risk and Safety Statements
| Symbol(GHS) | GHS07 |
| Signal word | Danger |
| Hazard statements | H302-H317-H332 |
| Precautionary statements | P261-P280g-P301+P312a-P304+P340-P321-P501a |
| Safety Statements | 24/25 |
| WGK Germany
| RTECS | GB6475000 |
| TSCA | Yes |
| HS Code | 28199090 |
| Hazardous Substances Data | 1308-38-9(Hazardous Substances Data) |
| NFPA 704 | 0 1 1 |
Chromium(III) oxide price More Price(46)
| Manufacturer | Product number | Product description | CAS number | Packaging | Price | Updated | Buy |
--- --- --- --- |
| Sigma-Aldrich | 203068 | Chromium(III) oxide powder, 99.9% trace metals basis | 1308-38-9 | 5g | $69.92 | 2025-07-31 | Buy |
| Sigma-Aldrich | 203068 | Chromium(III) oxide powder, 99.9% trace metals basis | 1308-38-9 | 25g | $279 | 2025-07-31 | Buy |
| Sigma-Aldrich | 1.02483 | Chromium(III) oxide anhydrous, Technipur® | 1308-38-9 | 1 kg | $67.1 | 2025-07-31 | Buy |
| Sigma-Aldrich | 1.02483 | Chromium(III) oxide anhydrous, Technipur® | 1308-38-9 | 50 kg | $1890 | 2025-07-31 | Buy |
| TCI Chemical | C4007 | Chromium(III) Oxide | 1308-38-9 | 100G | $28 | 2025-07-31 | Buy |
| Product number | Packaging | Price | Buy |
--- --- |
| 203068 | 5g | $69.92 | Buy |
| 203068 | 25g | $279 | Buy |
| 1.02483 | 1 kg | $67.1 | Buy |
| 1.02483 | 50 kg | $1890 | Buy |
| C4007 | 100G | $28 | Buy |
Chromium(III) oxide Chemical Properties,Uses,Production
Description
Chromium(III) oxide is among the ten most abundant compounds in the Earth's crust. It is one of four oxides of chromium, chemical formula Cr2O3. It is commonly called "chrome green" when used as a pigment; however it was referred to as “viridian” when it was first discovered.
Chromium(III) oxide is a very refractory ceramic colorant (even a 50% mix with a high borax frit will not even begin to melt it in a crucible). Chrome oxide is the only stable oxide of the metal chromium. It is a bright to dark green crystalline powder insoluble in alkalis and acids. It is manufactured from the mineral Chromite mined in southern Africa, Asia, Turkey and Cuba. As with other powerful coloring agents, chrome must be milled fine enough to eliminate specking in glass or glaze.
Chromium is a "fast" colorant, meaning can produce strong green colors under all furnace conditions, slow or fast, reducing or oxidizing. It is also a flat colorant (due to its refractory nature), it usually produces an army helmet opaque green. It is powerful, typically only 2% will produce a dark color. It cannot be used to make a metallic glaze.
Chrome oxide is usually employed in raw glazes whereas potassium dichromate is used in fritted glazes.
Chemical Properties
Chromium oxide is a bright green, odorless powder. Chromium(III) oxide pigments are thermally stable and insoluble in water.
Chromium oxide pigments, also called chromium oxide green pigments, consist of chromium(III) oxide [1308-38-9], Cr2O3,Mr 151.99. Chromium oxide green is one of the few single-component pigments with green coloration. Chrome green is a blend of chrome yellow and iron blue pigments; phthalochrome green is a blend of chrome yellow and blue phthalocyanine pigments.
Alkali dichromates are used as starting materials for the production of chromium(III) oxide pigments. They are not classified as hazardous materials and are not subject to international transport regulations. As long as they are kept dry their utility as a pigment is practically unlimited.
Physical properties
Green hexagonal crystal system; corundum type structure; density 5.22 g/cm 3; melts at 2,330°C; vaporizes above 3,000°C; insoluble in water and alcohol.
Uses
Chromium oxide (Cr2O3) is a dull green synthetic inorganic pigment, which can be used in all types of paint systems where high chemical resistance and outstanding light-fastness are required.
Uses
In abrasives, refractory materials, electric semiconductors; as pigment, particularly in coloring glass; in alloys; printing fabrics and banknotes; as catalyst for organic and inorganic reactions.
Uses
Chromium(III) oxide is used as pigment for coloring green on glass and fabrics. Other important applications are in metallurgy; as a component of refractory bricks, abrasives and ceramics; and as a catalyst in hydrogenation, hydrogenolysis and many other organic conversion reactions. It also is used to prepare other chromium salts.
Preparation
Chromium(III) oxide can be obtained by thermal decomposition of ammonium dichromate. Above ca. 200 °C, a highly voluminous product is formed with elimination of nitrogen. The pigment is obtained after addition of alkali salts (e.g., sodium sulfate) and subsequent calcination.
In the industrial process, a mixture of ammonium sulfate or chloride and sodium dichromate is calcined:
Na2Cr2O7.2 H2O + (NH4)2SO4 →Cr2O3 + Na2SO4 + 6 H2O + N2
Definition
Chromium oxide (Cr2O3) also called green rouge is a dull yellowish-green pigment that may be prepared by blending an alkali dichromate with sulfur or with a carbonaceous material. Reduction to chrome (III) oxide is achieved in a kiln at 1000°C. And it is used chiefly for platinum and stainless steels.
Definition
chromium sesquioxide: A green crystallinewater-insoluble salt, Cr 2 O 3; r.d. 5.21;m.p. 2435°C; b.p. 4000°C. It is obtainedby heating chromium in astream of oxygen or by heating ammoniumdichromate. The industrialpreparation is by reduction ofsodium dichromate with carbon.Chromium(III) oxide is amphoteric,dissolving in acids to give chromium(III) ions and in concentratedsolutions of alkalis to give chromites.It is used as a green pigment in glass,porcelain, and oil paint.
Reactions
Chromium(III) oxide is amphoteric. Although insoluble in water, it dissolves in acid to produce hydrated chromium ion, [Cr(H2O)6]3+. It dissolves in concentrated alkali to yield chromite ion. When heated with finely divided aluminum or carbon it is reduced to chromium metal:
Cr2O3 + 3Al2Cr + Al2O3
Heating with chlorine and carbon yields chromium(III) chloride:
Cr2O3 + 3Cl2 + 3C2CrCl3 + 3CO
If chromium(III) oxide (also known as chrome green) is heated with potassium carbonate and potassium nitrate, the mixture slowly turns yellow. This colour change stems from the formation of potassium chromate, K2CrO4, in which chromium is found in oxidation state vi.
General Description
Chromium (III) oxide is a chromium complex in which the chromium ion is in +3 oxidation state. The synthesis of its sub-micron powder has been reported.? The IR and Raman spectra of chromium (III) oxide have been studied.? The impact of adding lysozyme on the stability of chromium (III) oxide (Cr 2 O 3) suspension has been evaluated.?
Hazard
Toxic by ingestion and inhalation.
Flammability and Explosibility
Not classified
Safety Profile
Confirmed carcinogen with experimental tumorigenic data. Mutation data reported. Probably a severe
Toxicology
Chromium(III) oxide, which forms the basis of chromium oxide pigments, crystallizes in a corundum lattice. Chromium oxide green pigments (Cr2O3, C.I. Pigment Green 17) contain only trivalent chrome. Acute toxicity: rat, oral, LD50>10 000 mg/kg.
Potential Exposure
Chromium(III) oxide is used as a paint pigment, a fixative for certain textile dyes; in the manufacture of chromium; and a catalyst.
First aid
If this chemical gets into the eyes, remove anycontact lenses at once and irrigate immediately for at least15 min, occasionally lifting upper and lower lids. Seek medical attention immediately. If this chemical contacts theskin, remove contaminated clothing and wash immediatelywith soap and water. Seek medical attention immediately.If this chemical has been inhaled, remove from exposure,begin rescue breathing (using universal precautions, including resuscitation mask) if breathing has stopped and CPRif heart action has stopped. Transfer promptly to a medicalfacility. When this chemical has been swallowed, getmedical attention. Give large quantities of water andinduce vomiting. Do not make an unconscious personvomit.
storage
(1) Color Code—Blue: Health Hazard/Poison:Store in a secure poison location. (2) Color Code—Yellow:Reactive Hazard; Store in a location separate from othermaterials, especially flammables and combustibles.Chromium(III) oxide must be stored to avoid contact withstrong oxidizers (such as chlorine, bromine, and fluorine),glycerol, and oxygen difluoride, since violent reactionsoccur. A regulated, marked area should be establishedwhere chromium(III) oxide is handled, used, or stored.Store in tightly closed containers in a cool, well-ventilatedarea.
Shipping
UN3086 Toxic solids, oxidizing, n.o.s., Hazard Class: 6.1; Labels: 6.1-Poisonous materials, 5.1-Oxidizer. Technical Name Required. Spill Handling: Evacuate persons not wearing protective equipment from the danger area of spill or leak until cleanup is complete. Remove all ignition sources. Collect powdered material in the most convenient and safe manner and deposit in sealed containers. Ventilate area after clean-up is complete. It may be necessary to contain and dispose of this chemical as a hazardous waste. If material or contaminated runoff enters waterways, notify downstream users of potentially contaminated waters. Contact your local or federal environmental protection agency for specific recommendations. If employees are required to clean-up spills, they must be properly trained and equipped. OSHA 1910.120(q) may be applicable.
Properties and Applications
TEST ITEMSSPECIFICATION
APPEARANCEGREEN POWDER
CONTENT OF Cr 2 O385% min
SHADEYELLOWISH
HYDROTROPE0.5% max
OIL ABSORPTION25% max
RESIDUE ON 45 MESH0.1% max
WATER SOLUBLE0.3% max
VOLATITE 105 °C0.05% max
TINTING STRENGTH95-100 %
pH VALUE OF AQUEOUS SUSPENSION5-8
Structure and conformation
Chromium oxide is a dense, crystalline material, density 5.2 g/cm3, of the corundum type. It is quite hard-abut 9 on the Moh scale-which makes it a good grinding material, but too abrasive for many pigment applications. The particle size depends on the manufacturing process, but it is distributed around mean values in the range 0.5-0.6 μm. The refractive index of chromium oxide is quite high, 2.5, which is almost as high as that of rutile TiO2, 2.7. Chromium oxide pigments are green with an olive green tint. Small particles are lighter green with a yellowish hue, whereas larger particles are darker, with a bluish tint. Chromium oxide pigments are very inert materials with outstanding lightfastness and excellent resistance to acids, alkalis, and high temperatures.
Incompatibilities
A strong oxidizer. Contact with reducing agents; organics, and combustibles may be violent
Toxics Screening Level
The initial threshold screening level (ITSL) for chromium Ill oxide is 0.5 μg/m3 based on a 24-hour averaging time.
Chromium(III) oxide Preparation Products And Raw materials
Raw materials
Chromic sulfateHydrogen SulfideHigh speed pulverizerPhthalocyanine blue pigmentfilter paper
ChroMiuM hydroxideSodium dichromate dihydratePrussian BlueSulfur dioxideChromium(VI) oxide
Sulfur trioxidePotassium dichromateMiddle chrome yellowLead chrome yellow pigmentCopper(II) phthalocyanine
Cadmium sulfideChromate, Ion chromatography standard solution, Specpure, CrO4ˉ2 1000μg/mlFilter pressSulphurBarium chloride
Potassium chloridePotassium sulfate
1 of 5
Preparation Products
Alkyd resin paintSodium dichromate dihydrateWater proof agent CRChromiumFurfuryl alcohol
Chromium(III) oxide Suppliers
Global( 550)Suppliers
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| Hebei Chuanghai Biotechnology Co., Ltd | 58 |
| Shaanxi Dideu Medichem Co. Ltd | 58 |
| Hebei Yanxi Chemical Co., Ltd. | 58 |
| Hebei Chuanghai Biotechnology Co,.LTD | 58 |
| Hebei Chuanghai Biotechnology Co., Ltd | 58 |
| Wuhan Han Sheng New Material Technology Co.，Ltd | 58 |
| Hebei Jingbo New Material Technology Co., Ltd | 58 |
| Henan Fengda Chemical Co., Ltd | 58 |
| Hebei Saisier Technology Co., LTD | 58 |
| Hebei Longbang Technology Co., LTD | 58 |
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View Lastest Price from Chromium(III) oxide manufacturers
| Image | Update time | Product | Price | Min. Order | Purity | Supply Ability | Manufacturer | |
--- --- --- ---
| | 2025-09-28 | Chromium oxide 1308-38-9 | | | 0.99 | | RongNa Biotechnology Co.,Ltd | |
| | 2025-09-26 | Chromium(III) oxide 1308-38-9 | US $11.00 / kg | 1kg | 99% | 300tons | Hebei Dangtong Import and export Co LTD | |
| | 2025-09-26 | Chromium(III) oxide 1308-38-9 | US $3.00 / kg | 1kg | 99% | 300tons | Hebei Dangtong Import and export Co LTD | |
Chromium oxide1308-38-9
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US $11.00 / kg
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Chromium(III) oxide Spectrum
Chromium(III) oxide(1308-38-9)IR1
1308-38-9(Chromium(III) oxide)Related Search:
ChroMiuM hydroxidePotassium chromateChromiumCarbon BlackCobalt oxide
Ammonium dichromateFerric oxideCOBALT(III) OXIDE BLACKLanthanum oxideMagnesium oxide
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Europium OxideBismuth trioxideMANGANESE (III) OXIDE
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4370 | https://www.gauthmath.com/solution/1801047806176261/following-by-rationalising-the-denominators-a-frac-12-square-root-of-24-square-r | Solved: following by rationalising the denominators. (a) 12/sqrt(24)-sqrt(6) [Math]
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following by rationalising the denominators. (a) 12/sqrt(24)-sqrt(6)
Gauth AI Solution
100%(6 rated)
Answer
2 6 2 \sqrt{6}2 6
Alternative forms:
≈4.898979\approx 4.898979≈4.898979
Explanation
12 24−6\dfrac{12}{\sqrt{24} - \sqrt{6}}24−612
Factor and rewrite the radicand in exponential form
12 2 2×6−6\dfrac{12}{\sqrt{2^{2} \times 6} - \sqrt{6}}2 2×6−612
Rewrite the expression using
a b n=a n⋅b n\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}n ab=n a⋅n b
12 2 2×6−6\dfrac{12}{\sqrt{2^{2}} \times \sqrt{6} - \sqrt{6}}2 2×6−612
Simplify the radical expression
12 2 6−6\dfrac{12}{2 \sqrt{6} - \sqrt{6}}2 6−612
Combine like terms
12 6\dfrac{12}{\sqrt{6}}612
Rationalize the denominator
12 6 6×6\dfrac{12 \sqrt{6}}{\sqrt{6} \times \sqrt{6}}6×612 6
Simplify the radical expression
12 6 6\dfrac{12 \sqrt{6}}{6}6 12 6
Cross out the common factor
2 6 2 \sqrt{6}2 6
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4371 | https://algony-tony.github.io/linear-programming/ | 线性规划及整数线性规划算法 – Algony Tony – etvdyn
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线性规划及整数线性规划算法
2022-12-04
类别:自然科学
标签:算法线性规划单纯形法运筹学
运筹学基础
线性规划
标准形式
转换到标准形式的步骤
基可行解
线性规划的解
单纯形法
大 M 法
两阶段法
退化情形与勃兰特准则
内点法
整数线性规划
分枝定界法
割平面法
隐枚举法
匈牙利法
蒙特卡洛法
Python 求解
线性规划问题求解
整数线性规划问题求解
0-1 规划问题求解
指派问题求解
参考链接
运筹学基础
运筹学中的一套完整的工作流程，可分为如下六个步骤：
问题定义：在实践中遇到的问题，最初可能是模糊的，不精确的，所以第一步就是要把问题明确下来。这一步需要确定研究目标，明确问题边界，列出问题约束和问题要素及其关系；
数据收集：数据收集有可能贯穿研究的全过程，根据研究目标明确需要收集的数据范围，数据既能加深对问题的理解，也是为下一步构建模型提供数据输入；
模型构建：模型是研究对象的抽象，好的模型是保真性、灵活性和成本之间的巧妙平衡，模型需要贴近现实世界的问题，当条件改变时模型也能适应变化，同时模型方便操作和维护；
模型求解：模型求解一般使用计算机软件来完成，需要熟悉成熟模型的算法才好灵活运用；
模型验证：一方面是检查模型是否得到了正确的解，另一方面验证模型的解是否正确反映了问题实际；
结论实施：一般要实施交互式决策服务的工具，同时帮助决策者能够使用和实施，包括提供操作手册和相关培训等，在使用过程中实施全周期管理，当发现原有假设严重偏离时，应重新检验模型并更新工具。
运筹学模型中的三要素：
决策变量（Decision Variables）：待优化的对象；
目标函数（Objective Function）：需要达成的目标的表达式，一般为寻求最大值或最小值，有时候可要求满意、可行、非劣等；
约束条件（Constraints）：表示进行优化时收到的限制，包括资源的限制，时间约束等；
还有模型中的常数项，称为模型的参数（Parameters），参数的取值一般需要收集数据后才能确定。对参数值的不确定性的研究称为灵敏度分析（Sensitivity Analysis）。
线性规划
线性规划用来解决这一类问题：如何使用有限的资源（人力、物力、财力或者时间等）达到一定的目标，很多时候还可能要求以最好的效果来达成目标。
线性指的是模型中所有表达式均为线性等式或不等式，线性规划问题的一般形式如下：
max(min)z=c 1 x 1+c 2 x 2+⋯+c n x n 或 无 约 束{a 11 x 1+a 12 x 2+⋯+a 1 n x n≤(=,≥)b 1 a 21 x 1+a 22 x 2+⋯+a 2 n x n≤(=,≥)b 2⋮a m 1 x 1+a m 2 x 2+⋯+a m n x n≤(=,≥)b m x 1,x 2,…,x n≤(=,≥)0,或无约束
标准形式
满足下面 3 个条件的形式为线性规划模型的标准形式，
目标函数最大化；
约束条件等式化；
决策变量非负化；
设线性规划模型中变量数为 n 个，非负约束条件共 m 个，标准形式可写为
max z=c 1 x 1+c 2 x 2+⋯+c n x n{a 11 x 1+a 12 x 2+⋯+a 1 n x n=b 1 a 21 x 1+a 22 x 2+⋯+a 2 n x n=b 2⋮a m 1 x 1+a m 2 x 2+⋯+a m n x n=b m x 1,x 2,…,x n≥0
改写成矩阵形式：
max z=C X{A X=b X≥0
其中
C=(c 1 c 2⋮c n)T,X=(x 1 x 2⋮x n),P j=(a 1 j a 2 j⋮a m j),b=(b 1 b 2⋮b m)A=A m×n=(a 11 a 12⋯a 1 n⋮⋮⋱⋮a m 1 a m 2⋯a m n)=(P 1,P 2,⋯,P n)
其中，A 称为 m×n系数矩阵，一般有 m<n，b 称为资源向量，C 称为价值向量， X 称为决策变量向量，P j 称为系数列向量。
转换到标准形式的步骤
所有一般形式的线性规划模型都可以转换为标准形式，步骤如下
决策变量非负化：若原变量不是非负化的约束，可以对原变量做加减一个常数值，或者乘以 -1 来转化为非负约束的形式；如果原来是无约束变量如 x k，可以转换成两个带有非负约束的变量 x k=x k′−x k″，其中 x k′,x k″≥0；
约束条件等式化：若约束方程为 ≤ 可在不等式的左端加入非负松弛变量将不等式转换为等式；若不等式为 ≥ 则在左端减去一个非负剩余变量，同样可以转换为等式；
目标函数最大化：若原目标函数是要求最小值，只需要取 z′=−z，就可以转换为求最大值的形式 max z′=(−C)X。
基可行解
区域 D={X∈R n|A X=b,X≥0} 称为可行域，若 X∈D，则称 X 为可行解（Feasible Solution），若对任意 X′∈D，都有 C X′≤C X，则称 X 为线性规划问题的最优解。
可以假设系数矩阵 A 是行满秩（如果不是说明约束条件是线性相关，可以通过变换消去部分约束条件），也即 A 的秩为 m，从而可以从 A 中找到 m 个线性无关的列向量 (P j 1,P j 2,⋯,P j m)，记为 B，称为线性规划问题的一个基（Basis），其中每个向量称为基向量，称对应的决策变量为基变量，其他的变量称为非基变量。
假设 B 是一个基，可以通过调整变量顺序使得基向量排在前面 B=(P 1,P 2,⋯,P m)，设 N=(P m+1,⋯,P n)，这样导出
(B,N)(X B X N)=b
因为 B 是可逆矩阵，可以得出
X B=B−1 b−B−1 N X N
在上式中令非基变量为 0，这样得到
X=(X B X N)=(B−1 b 0)
满足条件 A X=b，称为基解（basic solution），此时如果有 B−1 b≥0，则有 X≥0 满足条件，这样的 X 称为基可行解（Basic feasible solution，BFS），对应的基称为可行基（feasible basis）。他们的关系如下图。
线性规划问题的基最多有 (n m) 个（假定任意从 n 个列向量中抽取 m 个都线性无关，则基的数量取到最大值），基解对应于约束条件对应的多面体的交角（corner of the polyhedron），基可行解是位于可行域中的基解。在两个变量的二维平面中，基解就是任意两个约束条件的交点，如下面五个约束条件两两相交对应图中的 A,B,C,D,E,F,G,H 总共 8 个点（有两对直线平行（线性相关）无交点，所以 (5 2)−2=8 ），其中位于可行域中的 A,B,C,D,E 是基可行解，再结合目标函数至少可以找到一个最优解。
{x 1+2 x 2≤8 x 1≤4 x 2≤3 x 1≥0 x 2≥0
线性规划的解
关于线性规划问题有下面几个结论：
线性规划问题的可行域如果非空，则一定是凸集；
如果是有界凸集，由于目标函数的连续性，则一定可以在可行域的有界凸集上取得极值，也就是最优解；
在有界可行域的顶点处一定可以找到至少一个最优解；
线性规划问题的基可行解对应于可行域的顶点；（从代数角度定义的“基可行解”和几何角度定义的“顶点”两者本质上是一回事。）
线性规划的解有几种情况：
唯一的最优解；
无穷多最优解；
无界解；
无可行解；
当线性规划问题的可行域非空时，它是有界或无界凸多边形；若线性规划问题存在最优解，它一定在可行域的某个顶点得到；若在两个顶点同时得到最优解，则它们连线上的任意一点都是最优解，即有无穷最优解。
单纯形法
单纯形法（Simplex Method ）是建立在线性规划模型的“标准形式”之上的，总体来说可以分为 5 步：
寻找一个初始解；
构建单纯性表；
判断当前解是否最优；
如果是最优解就结束，不是最优解就换到一个新解上；
在单纯性表中计算相关值，返回第三步重复直到解出最优解或者其他终止条件；
如求解这个例子的单纯行表步骤如下：
max z=2 x 1+3 x 2{x 1+2 x 2≤8 x 1≤4 x 2≤3 x 1,x 2≥0
先转换成标准形式如下，以及容易找到一个初始解 (0,0,8,4,3)T
max z=2 x 1+3 x 2{x 1+2 x 2+x 3=8 x 1+x 4=4 x 2+x 5=3 x 1,x 2,x 3,x 4,x 5≥0
上面步骤中的检验数从下面可以推导出，它代表着目标函数会随着该变量的变化方向，以及变化的快慢，它是资源的影子价格，也被称为边际价格。
z=C X=C B X B+C N X N=C B(B−1 b−B−1 N X N)+C N X N=C B B−1 b+(C N−C B B−1 N)X N
从而对于非基变量 x j 的检验数为
∂z∂x j=c j−C B B−1 P j=σ j
大 M 法
单纯形法开始迭代时需要找到一个初始基可行解，但是上面把线性规划问题转换成标准形式的过程中不能自然得到一个单位矩阵， 大 M 法是在上面转换标准形式的过程中加入人工变量（artificial variables）和惩罚系数如下：
若约束条件为“≤”，在不等式左边加入非负松弛变量；
若约束条件为等式，在等式左边加入一个非负人工变量；
若约束条件为“≥”，在不等式左边减去非负剩余变量，再加入一个非负人工变量；
在目标函数中，加入的松弛变量和剩余变量的系数为 0，加入的人工变量系数为 −M，是一个任意大的正数，惩罚系数。
如下面的线性规划问题
min z=−3 x 1−2 x 2{x 1+4 x 2−x 3≤5−x 1+x 2−x 3≥−1−x 2+x 3=1 x 1,x 2,x 3≥0
用大 M 法转成标准形式：
max z=3 x 1+2 x 2−M x 6{2 x 1+4 x 2−x 3+x 4=5 x 1−x 2+x 3+x 5=1−x 2+x 3+x 6=1 x 1,x 2,x 3,x 4,x 5,x 6≥0
其中 x 6 是人工变量，这样可以得到 x 4,x 5,x 6 的系数列向量构成单位矩阵 B，初始基解 X=(0,0,0,5,1,1)T 为基可行解，从而开始单纯形法计算。
这里可以得到线性规划问题四种解的判别准则：
唯一的最优解：对于所有非基变量，检验数都有 σ j<0；
无穷多最优解：对于所有非基变量，检验数都有 σ j≤0，且存在某个非基变量的检验数等于 0；
无界解：存在一个非基变量的检验数 σ m+k>0，且该变量在系数矩阵中对应的列 P m+k≤0；
无可行解：满足上面的迭代终止条件，但是存在取非零值的人工变量，则此线性规划问题无可行解；
两阶段法
大 M 法在用计算机求解的过程中 M 值一般用很大的数来近似替代，这会造成计算上的累计误差，两阶段法就是用拆分成两个阶段的办法来解决这个问题，使得计算过程中不出现大 M。
第一阶段用来寻找一个不包含人工变量的初始基可行解。步骤是在原问题引入松弛变量和人工变量进行标准化，构造仅含人工变量的目标函数，将其价值系数设置为 -1，使用单纯形法求解，若得到了目标函数为 0，也即是得到了不包含人工变量的一个基可行解，从而转入第二阶段，否则原问题无可行解。
第二阶段，使用第一阶段得到的最终单纯形表，去掉其中的人工变量，并将目标函数还原成原问题的目标函数，继续使用单纯形法求解即可。
例子，线性规划问题如下
max z=−4 x 1−x 2+2 x 3{x 1−2 x 2+x 3≤11−4 x 1+x 2+2 x 3≥3−2 x 1+x 3=1 x 1,x 2,x 3≥0
第一阶段，转换成带人工变量的标准形式，用单纯形法第一次迭代 x 3 替换掉 x 7，第二次 x 2 替换掉 x 6，这样第一阶段就结束了。
max w=−x 6−x 7{x 1−2 x 2+x 3+x 4=11−4 x 1+x 2+2 x 3−x 5+x 6=3−2 x 1+x 3+x 7=1 x 1,x 2,x 3,x 4,x 5,x 6,x 7≥0
退化情形与勃兰特准则
在单纯形法的求解中可能出现 θ 为 0 的情形，即该行的资源系数为 0，称这种情况是退化的。出现退化的情形是因为约束条件中存在线性相关的条件。
出现退化的情况可能导致迭代出现局部循环，而最优解不在其中。解决的办法就是使用勃兰特准则（Bland’s rule）：
换入变量不用检验数最大的那个，而是用大于 0 的检验数中下标最小的决策变量；
换出变量还是使用 θ 最小的那个，如果有出现两个以上的相同最小值，使用下标最小的决策变量作为换出变量。
内点法
1984 年 AT&T Bell 实验室一位年轻的数学家 Narendra Karmarkar 发明了求解线性规划的问题的新算法，它在解特大型线性规划问题方面有巨大的潜力。 Karmakar 算法也是一种迭代算法，它从得出一个可行的实验解开始，在每次迭代中，它在可行域内从当前实验解开始移动到另一个更好的实验解，然后继续这个过程直到实验解（基本上）达到最优解为止。 对单纯形法所有的移动都是在可行域的边界上进行的，而 Karmarkar 算法实验解是内点（interior point），所以也被称为内点法（interior-point algorithm）。
因为内点法获得专利的早期版本被称为障碍点算法，故内点法也被称为障碍算法（barrier algorithm），“障碍”一词是因为从搜索的角度看所找的实验解都是内点，每一个约束边界都被作为障碍对待。
内点法比单纯形法复杂，它的每一次迭代都需要大量的迭代去寻找下一个实验解，因此内点法每次迭代的时间是单纯形法的好几倍。 对于非常小的问题，内点法需要的迭代次数和单纯形法差不多，这样对于小规模问题内点法速度上不如单纯形法。但是对于大规模的问题，比如 10000 个约束条件的问题可能只需要 100 次以下迭代，相反单纯形法可能需要 20000 次迭代，这样考虑上内点法每次迭代上多消耗的时间，总体时间内点法还可能更短一些。内点法和单纯形法也可以结合使用。
整数线性规划
在上述的线性规划问题的基础上，如果要求某些决策变量或者全部决策变量要求为整数，则称这样的问题为整数规划问题（Integer Programming，IP）， 除了整数约束条件外其他的部分称为相应的松弛问题（Slack Problem）。如果所有决策变量要求为整数称为纯整数规划（Pure Integer Programming）或全整数规划（All Integer Programming）。 如果只是部分决策变量要求为整数，则称为混合整数规划（Mixed Integer Programming，MIP）。 整数规划中如果所有决策变量要求取值只能是 0 或者 1，称为 0-1 型整数规划问题。有一类特殊的 0-1 整数规划问题称为指派问题（assignment problem），有 n 项任务要完成，有 n 项资源（可以理解为人、机器设备等）可以完成任务，并且每项任务交给一个对象完成，每个对象也只能完成一种任务。由于每个对象的特点与能力不同，故其完成各项任务的效率也不同。那么，如何分配资源，才能使完成各项任务的总效率最高（或总消耗最少）？ 指派问题的一般形式如下：
min z=∑i=1 n∑j=1 n c i j x i j 第 项 任 务 只 能 由 一 人 完 成 第 人 只 能 完 成 一 项 任 务{∑i=1 n x i j=1,j=1,2,⋯,n 第 j 项任务只能由一人完成∑j=1 n x i j=1,i=1,2,⋯,n 第 i 人只能完成一项任务 x i j∈{0,1}
其中的矩阵 C 称为效率矩阵或者系数矩阵。
下面是整数线性规划问题的几个主要求解算法：
分枝定界法：可求纯或混合整数线性规划。
割平面法：可求纯或混合整数线性规划。
隐枚举法：用于求解 0-1 整数规划。
匈牙利法：解决指派问题（0-1 规划特殊情形）。
蒙特卡罗法：求解各种类型规划。
分枝定界法
分枝定界法（Branch and Bound, B&B）在 20 世纪 60 年代初由 A.H.Land 和 A.G.Doig 两位学者提出，用于解纯整数或混合整数规划问题。
它的基本思路是考虑到整数规划可行解中任何相邻整数之间的区域均不含整数解，这样就可以将松弛问题的可行域分为多个分枝，同时将求解整数规划问题转化为求解多个线性规划的问题。
假设整数线性规划问题 A，去掉其中的整数约束得到的松弛问题 B，分枝定界法的基本步骤如下：
解出问题 B，可能出现下面解的情况
B 无可行解，则 A 无可行解，停止计算；
B 有最优解，并符合 A 的整数约束条件，则此最优解就是 A 的最优解，停止计算；
B 有最优解，但不符合 A 的整数约束条件，记此最优解的目标函数值为 z―；
分枝：选出 B 最优解中任一不符合整数约束的变量 x j，其值为 b j，构造两个约束条件 x j≤[b j] 和 x j≥[b j]+1 添加到 B 中，形成两个分枝问题 B1 和 B2；
定界：求解 B1 和 B2 问题，从最优目标函数值最大者作为新的上界 z―，从已符合整数约束的分枝中找出目标函数最大者作为新的下界 z―；
剪枝：各分枝的最优目标如果小于 z―，则剪掉这个分枝，若大于 z― 且不符合整数条件则重复第二步继续分枝；
| 问题 1 | 问题 2 | 说明 |
---
| 无可行解 | 无可行解 | 整数规划无可行解 |
| 无可行解 | 整数解 | 此整数解即最优解 |
| 无可行解 | 非整数最优解 | 对问题 2 继续分枝 |
| 整数解 | 整数解 | 目标值最大的为最优 |
| 整数解，目标值优于问题 2 | 非整数最优解 | 问题 1 整数解即为最优解 |
| 整数解，目标值不如问题 2 | 非整数最优解 | 问题 1 停止分枝，它的目标值作为下界，对问题 2 继续分枝 |
如用分枝定界法求解下面例子
max z=x 1+5 x 2 且 全 为 整 数{x 1−x 2≥−2 5 x 1+6 x 2≤30 x 1≤4 x 1,x 2≥0 且全为整数
第一步求解松弛问题，得到最优解： x 1=18 11,x 2=40 11,z=218 11
显然有 x 1=0,x 2=0 是满足规划问题的一个整数解，这样得到一个上界和下届，0≤z≤218 11。
对 x 1=18 11 进行分枝，构造两个约束条件 x 1≤1 和 x 1≥2，得到两个问题 LP1 和 LP2，求解得到两个最优解：
对 应 问 题 x 1=1,x 2=3,z=16 对应问题 LP1 对 应 问 题 x 1=2,x 2=10 3,z=56 3 对应问题 LP2
分枝 LP1 停止分枝，得到新的下界 16，分枝 LP2 给出新的上界 56/3，选择 x 2 对 LP2 继续分枝，如下：
割平面法
割平面法是 1958 年 R.E.Gomory 提出的，也称为 Gomory 割平面法，它的基本思想是不考虑整数要求，先求解松弛问题的解，如果没有得到满足整数要求的解，那么逐次增加一个新约束（即割平面）， 割掉原可行域的一部分（只含非整数解），使得切割后最终得到这样的可行域（不一定一次性得到），它的一个有整数坐标的顶点恰好是问题的最优解。切割方程由松散问题的最终单纯形表中含非整数解基变量的等式约束演变而来。
割平面法是在松弛问题的可行域的边缘不断地添加割平面切掉不含整数解的部分，从而不断逼近整数解，而分枝定界则是直接将可行域从整数处切成两个分枝部分再分别求解。
隐枚举法
隐枚举法（implict enumeration）：只检查变量取值组合的一部分，就能求得问题的最优解的方法。
例子如求解如下问题
max z=3 x 1−2 x 2+5 x 3{x 1+2 x 2−x 3≤2 x 1+4 x 2+x 3≤4 x 1+x 2≤3 4 x 1+x 3≤6 x 1,x 2,x 3∈{0,1}
先试探得到 (1,0,0) 是一个可行解，计算出目标函数值 z=3，结合目标函数可以增加约束条件 3 x 1−2 x 2+5 x 3≥3，这个条件称为过滤条件（filtering constraint），这样，原问题的线性约束条件就变成 5 个。用穷举法，3 个变量共有 8 个解。对每个解，依次代入 5 个约束条件左侧，先判断是否满足过滤条件，如果不满足就可以直接跳过，满足后再检查后面约束条件，如果都满足且得到了更好的目标值则更新过滤条件的值，这样继续如下表所示：
匈牙利法
匈牙利法是基于指派问题的标准型，标准型需要满足下面 3 个条件：
目标函数求最小 min；
效率矩阵为 n 阶方阵；
效率矩阵所有元素 c i j≥0，且为常数；
对于非标准形的指派问题，可以转换成标准形式：
蒙特卡洛法
蒙特卡罗法就是选择部分穷举法，随机取样来得到在有限取样下的一个最优解。
Python 求解
线性规划问题的求解主要依赖下面两个包：
SciPy：Python 的一个通用科学计算包；
PuLP：提供 Python 的定义线性规划问题的 API 接口，并调用外部的求解器求解；
线性规划问题求解
用 SciPy 来求解上面的线性规划问题：
```python
import scipy
scipy.version
'1.9.1'
import numpy as np
from scipy import optimize
定义参数，因为是算法是求最小值，需要把价值系数乘以 -1
c= np.array([-2, -3])
A_ub=np.array()
b_ub=np.array([8,4,3])
x0_bound=(0,None)
x1_bound=(0,None)
用单纯型法求解，得到最优解 [4,2] 和最优目标值 14
res = optimize.linprog(c, A_ub=A_ub, b_ub=b_ub,bounds=[x0_bound, x1_bound], method='simplex')
print(res)
con: array([], dtype=float64)
fun: -14.0
message: 'Optimization terminated successfully.'
nit: 3
slack: array([0., 0., 1.])
status: 0
success: True
x: array([4., 2.])
使用默认的 HiGHS 算法求解
optimize.linprog(c, A_ub=A_ub, b_ub=b_ub,bounds=[x0_bound, x1_bound])
con: array([], dtype=float64)
crossover_nit: 0
eqlin: marginals: array([], dtype=float64)
residual: array([], dtype=float64)
fun: -14.0
ineqlin: marginals: array([-1.5, -0.5, -0. ])
residual: array([0., 0., 1.])
lower: marginals: array([0., 0.])
residual: array([4., 2.])
message: 'Optimization terminated successfully. (HiGHS Status 7: Optimal)'
nit: 1
slack: array([0., 0., 1.])
status: 0
success: True
upper: marginals: array([0., 0.])
residual: array([inf, inf])
x: array([4., 2.])
```
现在 SciPy 的线性规划问题默认使用 HiGHS 求解，HIGHS 是用 C++ 写的一个求解软件，提供 C, C#, FORTRAN, Julia and Python 的接口。
HiGHS - high performance software for linear optimization
Open source serial and parallel solvers for large-scale sparse linear programming (LP), mixed-integer programming (MIP), and quadratic programming (QP) models
HiGHS 的 GitHub 地址，Wiki 地址，论文地址。它主要是两个并行对偶单纯形求解器（PAMI 和 SIP）的设计和实现。
整数线性规划问题求解
用 SciPy 来求解上面的整数线性规划问题，加上
plaintext
integrality
参数即可， 0 表示连续变量，无整型约束，1 表示整数约束，其他参考文档。
```python
import numpy as np
from scipy import optimize
定义参数，因为是算法是求最小值，需要把价值系数乘以 -1
ip_c= np.array([-1, -5])
ip_A_ub=np.array()
ip_b_ub=np.array([2,30,4])
ip_x0_bound=(0,None)
ip_x1_bound=(0,None)
ip_integrality=np.array([1,1])
加上整数约束求解，得到最优解 [2,3] 和最优目标值 17
optimize.linprog(ip_c, A_ub=ip_A_ub, b_ub=ip_b_ub,bounds=[ip_x0_bound, ip_x1_bound], integrality=ip_integrality)
con: array([], dtype=float64)
crossover_nit: -1
eqlin: marginals: array([], dtype=float64)
residual: array([], dtype=float64)
fun: -17.0
ineqlin: marginals: array([0., 0., 0.])
residual: array([1., 2., 2.])
lower: marginals: array([0., 0.])
residual: array([2., 3.])
message: 'Optimization terminated successfully. (HiGHS Status 7: Optimal)'
nit: -1
slack: array([1., 2., 2.])
status: 0
success: True
upper: marginals: array([0., 0.])
residual: array([inf, inf])
x: array([2., 3.])
```
0-1 规划问题求解
用 SciPy 来求解上面的 0-1 规划问题。
```python
import numpy as np
from scipy import optimize
定义参数，因为是算法是求最小值，需要把价值系数乘以 -1
zo_c= np.array([-3, 2,-5])
zo_A_ub=np.array()
zo_b_ub=np.array([2,4,3,6])
zo_x0_bound=(0,1)
zo_x1_bound=(0,1)
zo_x2_bound=(0,1)
zo_integrality=np.array([1,1])
加上整数约束求解，得到最优解 [1,0,1] 和最优目标值 8
optimize.linprog(zo_c, A_ub=zo_A_ub, b_ub=zo_b_ub,bounds=[zo_x0_bound, zo_x1_bound,zo_x2_bound], integrality=zo_integrality)
con: array([], dtype=float64)
crossover_nit: -1
eqlin: marginals: array([], dtype=float64)
residual: array([], dtype=float64)
fun: -8.0
ineqlin: marginals: array([0., 0., 0., 0.])
residual: array([2., 2., 2., 1.])
lower: marginals: array([0., 0., 0.])
residual: array([1., 0., 1.])
message: 'Optimization terminated successfully. (HiGHS Status 7: Optimal)'
nit: -1
slack: array([2., 2., 2., 1.])
status: 0
success: True
upper: marginals: array([0., 0., 0.])
residual: array([0., 1., 0.])
x: array([1., 0., 1.])
```
指派问题求解
用 SciPy 来求解上面的指派问题：
```python
from scipy.optimize import linear_sum_assignment
定义效率矩阵
cost = np.array()
指派问题求解
row_ind, col_ind = linear_sum_assignment(cost, maximize=False)
col_ind
array([1, 0, 2, 3], dtype=int64)
cost[row_ind, col_ind].sum()
34
```
参考链接
运筹学基础，李志猛著
Simplex method calculator
实用运筹学：案例、方法及应用，邢光军著
[学习笔记] 整数规划之割平面法 How and why?
运筹学-指派问题-匈牙利法
scipy.optimize.linear_sum_assignment
运筹学导论，（美）希利尔（Hillier,F.S.）/（美）利伯曼（Lieberman,G.J.）
运筹学基础
线性规划
标准形式
转换到标准形式的步骤
基可行解
线性规划的解
单纯形法
大 M 法
两阶段法
退化情形与勃兰特准则
内点法
整数线性规划
分枝定界法
割平面法
隐枚举法
匈牙利法
蒙特卡洛法
Python 求解
线性规划问题求解
整数线性规划问题求解
0-1 规划问题求解
指派问题求解
参考链接
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4372 | https://artofproblemsolving.com/community/c296035h1272533_invariants?srsltid=AfmBOoplSxm8HUIDaFOPNy6lUMNrJdeyxRvfrTwdMfo4kgKxD80YQm5E | m1234567's Mathematical blog : Invariants
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m1234567's Mathematical blog ============================
Invariants
by m1234567, Jul 15, 2016, 1:27 AM
So here is our first informative post, and today it will be on . Invariants is just a fancy word for 'not-variable' or 'not-changing'. In a mathematical sense, an invariant is something in a problem that remains the same. Here is my favorite example to demonstrate this:
Suppose the positive integer n is odd. First Al writes the numbers 1, 2, . . . , 2n on the blackboard. Then he picks any two numbers a, b, erases them, and writes, instead, |a − b|. Prove that an odd number will remain at the end.
Solution: (I think it's easier to learn from if answers come in solutions rather than rigorous proofs)
Let 1+...+2n = x
1) As we are on the topics of invariants, why don't we look for one (again, something that always stays the same).
Let's look at 1,2,...2n, and notice that 1+2+...2n will be odd. Now let's see if we can find our invariant. Let's pick our numbers, a and b, with a and b having same or different parities, to see if there is an invariant we can find. If a and b are both odd, then x - a - b + |a − b| is odd. Likewise, if a and b are even, the sum of x will still be odd. If a and b are different parities, we see that x is again odd.
2) Now that we see x is odd for all possibilities a and b, we see that at the end the sum will again be odd.
We see that this principle made this proof a lot easier than it could have been. The invariant showed that x will always be odd and let us make the easy connection to the conclusion of the proof
Now that we have a grasp let's try an Problem -- I won't give a solution, but I'll send it to anyone who Pms me.
Also here is a good handout on the subject:
In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room (Source IMO)
This post has been edited 2 times. Last edited by m1234567, Jul 15, 2016, 5:41 PM
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1100 views
by m1234567, Jun 20, 2017, 3:45 AM
Yay over 600 views!!!!!
by m1234567, Feb 12, 2017, 11:07 PM
I'm gonna post something tomorrow yay!!!!!!!!!!!!!!!
Lol school is long and hard but mostly long
And I have a practice Olympiad to do tomorrow -- wish me luck
by m1234567, Oct 11, 2016, 9:34 PM
Yes......
by m1234567, Aug 13, 2016, 2:29 PM
Wait are you doing WOOT this year?
by Designerd, Aug 13, 2016, 3:42 AM
Cool my parents let me join WOOT
by m1234567, Aug 12, 2016, 9:00 PM
Yay 108 views
by m1234567, Aug 11, 2016, 6:16 AM
I agree with you on the music thing, I do the same thing.
by Designerd, Aug 9, 2016, 1:45 AM
Thank you
by m1234567, Aug 8, 2016, 7:10 PM
Nice blog!
by champion999, Aug 8, 2016, 5:53 PM
So I jab a suggestion for learning to material and solving hard problems: listen to music. It relieves that pressure to do your problems as fast as you can and might make you enjoy your math more. Just don't do it when you are taking tests
by m1234567, Aug 5, 2016, 5:35 AM
Sorry if anyone is actively reading this but I had went on vacation and had to do all aops from an outside medium and didn't have time for blog
by m1234567, Jul 31, 2016, 9:49 PM
Most topics I'm going to post will to your point be more advanced but my goal here is to build reader's knowledge of subjects gradually and sometimes add in advanced but stand alone topics like I did today. Don't worry problems will be hard
by m1234567, Jul 15, 2016, 3:49 AM
You could try out stewart's thereom, or explain radians... it's all fine..
by JaLu, Jul 15, 2016, 3:37 AM
Thanks guys!
by m1234567, Jul 15, 2016, 2:38 AM
YAY. Good for you, m1234567.
by JaLu, Jul 15, 2016, 2:20 AM
second shout >
by budu, Jul 15, 2016, 1:25 AM
first shout >
by math101010, Jul 15, 2016, 12:21 AM
18 shouts
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4373 | https://www.reddit.com/r/askscience/comments/41x4pj/why_is_the_circumference_of_a_circle_defined_at/ | Why is the circumference of a circle defined at 2Πr instead of Π(diameter)? : r/askscience
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4374 | https://labex.io/tutorials/python-how-to-work-with-modular-arithmetic-functions-421961 | Learn
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How to work with modular arithmetic functions
PythonBeginner
How to work with modular arithmetic functions
Practice Now
Practice Now
Introduction
This comprehensive tutorial explores modular arithmetic functions in Python, providing developers with essential techniques to manipulate and calculate mathematical operations using modulo principles. By understanding these powerful computational methods, programmers can solve complex mathematical problems, implement cryptographic algorithms, and optimize numerical computations across various programming domains.
Modular Arithmetic Basics
Introduction to Modular Arithmetic
Modular arithmetic is a fundamental mathematical concept that deals with the remainders after division. It is widely used in various fields, including computer science, cryptography, and number theory. In Python, modular arithmetic provides powerful tools for solving complex computational problems.
Core Concepts
Modular arithmetic operates on the principle of finding the remainder when one number is divided by another. The basic operation is represented by the modulo operator %.
Key Properties
Modulo Operation: Returns the remainder after division
Cyclic Nature: Numbers wrap around after reaching the modulus
Congruence: Numbers are considered equivalent within a given modulus
Mathematical Representation
The modular arithmetic operation can be expressed mathematically as:
a ≡ b (mod n) a b RUN
This means a and b have the same remainder when divided by n.
Python Modulo Basics
Simple Modulo Examples
```
Basic modulo operations
print(10 % 3) ## Returns 1
print(15 % 4) ## Returns 3
print(20 % 5) ## Returns 0
```
Practical Modulo Scenarios
```
Checking even/odd numbers
def is_even(number):
return number % 2 == 0
Cyclic indexing
days = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri']
print(days[7 % 5]) ## Wraps around the list
```
Modular Arithmetic Visualization
Common Use Cases
| Scenario | Description | Example |
---
| Cryptography | Key generation | RSA algorithm |
| Clock Arithmetic | Time calculations | 24-hour clock |
| Hash Functions | Data distribution | Hash table indexing |
Performance Considerations
Modular arithmetic in Python is highly efficient and built into the language's core operations. LabEx recommends using native modulo operations for optimal performance.
Advanced Techniques
```
Modular exponentiation
def power_mod(base, exponent, modulus):
return pow(base, exponent, modulus)
Example
print(power_mod(2, 10, 100)) ## Efficient large number calculation
```
Conclusion
Understanding modular arithmetic provides developers with powerful computational techniques applicable across multiple domains in software development.
Python Modulo Operations
Basic Modulo Operator Usage
The modulo operator % in Python is a fundamental tool for performing remainder calculations. It works with various numeric types and provides essential functionality for many programming tasks.
Fundamental Operations
Integer Modulo
```
Basic integer modulo operations
print(10 % 3) ## Returns 1
print(15 % 4) ## Returns 3
print(20 % 5) ## Returns 0
```
Negative Number Handling
```
Modulo with negative numbers
print(-10 % 3) ## Returns 2
print(10 % -3) ## Returns -2
```
Modulo Operation Types
Floating-Point Modulo
```
Modulo with floating-point numbers
print(10.5 % 3) ## Returns 1.5
print(7.8 % 2.5) ## Returns 2.8
```
Advanced Modulo Techniques
Cyclic Indexing
```
List indexing with modulo
days = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri']
print(days[7 % 5]) ## Wraps around the list
```
Periodic Patterns
```
Creating periodic sequences
def generate_periodic_sequence(length, period):
return [i % period for i in range(length)]
print(generate_periodic_sequence(10, 3))
```
Modulo Operation Visualization
Performance Considerations
| Operation | Performance | Recommendation |
---
| Integer Modulo | Very Fast | Preferred method |
| Floating-Point Modulo | Slower | Use sparingly |
| Large Number Modulo | Efficient | Use built-in methods |
Practical Applications
Validation and Checking
```
Credit card validation
def is_valid_credit_card(number):
return number % 10 == 0
Even/odd detection
def is_even(number):
return number % 2 == 0
```
Advanced Modular Arithmetic
Modular Exponentiation
```
Efficient large number exponentiation
def power_mod(base, exponent, modulus):
return pow(base, exponent, modulus)
Example in cryptography
print(power_mod(2, 10, 100))
```
LabEx Recommended Practices
Always consider the range of your modulo operations
Use built-in Python methods for complex calculations
Be aware of performance implications with large numbers
Common Pitfalls
```
Potential division by zero
try:
print(10 % 0) ## Raises ZeroDivisionError
except ZeroDivisionError:
print("Cannot divide by zero")
```
Conclusion
Mastering Python's modulo operations provides powerful tools for various computational tasks, from simple remainder calculations to complex algorithmic implementations.
Practical Modular Programming
Real-World Modular Arithmetic Applications
Modular arithmetic extends far beyond simple mathematical calculations, finding critical applications in various domains of software development and computer science.
Cryptography and Security
RSA Encryption Simulation
```
def generate_keypair(p, q):
n = p q
phi = (p-1) (q-1)
def mod_inverse(a, m):
for x in range(1, m):
if (a x) % m == 1:
return x
return None
## Public key generation
e = 65537
d = mod_inverse(e, phi)
return ((e, n), (d, n))
Example key generation
public, private = generate_keypair(61, 53)
print("Public Key:", public)
print("Private Key:", private)
```
Data Validation Techniques
Credit Card Number Validation
```
def luhn_algorithm(card_number):
digits = [int(x) for x in str(card_number)]
checksum = 0
for i in range(len(digits)-2, -1, -1):
digit = digits[i] 2
checksum += digit if digit < 10 else digit - 9
return (checksum + digits[-1]) % 10 == 0
Validation examples
print(luhn_algorithm(4111111111111111)) ## Valid card
print(luhn_algorithm(4111111111111112)) ## Invalid card
```
Algorithmic Optimization
Hash Table Implementation
```
class ModularHashTable:
def init(self, size=100):
self.size = size
self.table = [[] for _ in range(size)]
def _hash_function(self, key):
return hash(key) % self.size
def insert(self, key, value):
index = self._hash_function(key)
self.table[index].append((key, value))
def get(self, key):
index = self._hash_function(key)
for stored_key, value in self.table[index]:
if stored_key == key:
return value
raise KeyError(key)
```
Modular Arithmetic Visualization
Performance Comparison
| Technique | Time Complexity | Space Complexity |
---
| Standard Lookup | O(n) | O(n) |
| Modular Hashing | O(1) | O(n) |
| Collision Resolution | O(k) | O(1) |
Practical Use Cases
Cyclic Buffer Implementation
```
class CircularBuffer:
def init(self, capacity):
self.buffer = [None] capacity
self.capacity = capacity
self.head = 0
self.tail = 0
self.size = 0
def enqueue(self, item):
if self.is_full():
self.head = (self.head + 1) % self.capacity
else:
self.size += 1
self.buffer[self.tail] = item
self.tail = (self.tail + 1) % self.capacity
def is_full(self):
return self.size == self.capacity
```
Advanced Techniques
Time-Based Operations
```
def periodic_task_scheduler(interval, total_time):
for current_time in range(total_time):
if current_time % interval == 0:
print(f"Executing task at time {current_time}")
Run tasks every 5 time units
periodic_task_scheduler(5, 30)
```
LabEx Recommended Practices
Use modular arithmetic for efficient data distribution
Implement hash functions with modulo operations
Consider performance implications in large-scale systems
Conclusion
Practical modular programming demonstrates the versatility of modular arithmetic in solving complex computational problems efficiently and elegantly.
Summary
Through this tutorial, Python developers have gained valuable insights into modular arithmetic functions, learning how to leverage modulo operations for solving mathematical challenges, implementing efficient algorithms, and expanding their computational problem-solving skills. The techniques covered demonstrate the versatility and practical applications of modular arithmetic in modern programming environments.
Other Python Tutorials you may like
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Gtk3 Spreadsheet Sgskip
Creating Annotated Heatmaps
Scaling Large Datasets
Monty Hall Problem Simulation Using Tkinter
How to create runtime type checkers
How to debug f-string parsing problems
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How to format multiline f-strings correctly
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4376 | https://www.biologydiscussion.com/genetics/population-genetics/calculating-gene-allele-frequencies-in-a-population-genetics/84576 | Published Time: 2017-11-09T18:48:02+00:00
Calculating Gene (Allele) Frequencies in a Population | Genetics
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Calculating Gene (Allele) Frequencies in a Population | Genetics
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Application of Hardy-Weinberg law in calculating Gene (Allele) frequencies in a population.
The gene frequencies for the autosomal and sex-chromosomal allele can be determined by the help of Hardy-Weinberg law by the following method:
A. Calculation of Gene Frequencies of Autosomal Genes:
An autosomal gene locus may have codominant alleles, dominant and recessive alleles or multiple alleles. If one desires to determine the gene frequencies for each of these kinds of autosomal alleles in a given population, he has to adopt the different methods.
(i) Calculation of Gene Frequencies for Codominant Alleles:
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When codominant alleles are present in a two-allele system, each genotype has a characteristic phenotype. The numbers of each allele in both homozygous and heterozygous conditions may be counted in a sample of individuals from the population and expressed as a percentage of total number of alleles in a sample.
If the sample is representative of the entire population (containing proportionately same numbers as found in the entire population) then we can obtain an estimate of the allelic frequencies in the gene pool. If in a given sample of N individuals of which D are homozygous for one allele (A 1 A 1), H are heterozygous (A 1 A 2), and R are homozygous for the allele (A 2 A 2), then N D + H + R.
Since each of the N individuals are diploid at this locus, there are 2N alleles represented in the sample. Each A 1 A 1 genotype has two A 1 alleles. Heterozygotes have only one A 1 allele. Letting p represents the frequency of the A 1 allele and q the frequency of the A 2 allele, we have-
Example:
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The M-N blood type furnishes a useful example of a series of phenotypes due to a pair of codominant alleles. None of three possible phenotypes, M, MN and N, appears to have any selection value. The frequencies of the two alleles (viz., L M and L N) for a sample from a group of white Americans living in New York City, Boston, and Columbus, Ohio, can be calculated by the following ways-
The sample of 6,129 Caucasian people includes the following three groups according to phenotypes and genotypes on M-N system:
To calculate frequencies of the two codominant alleles, L M and L N, it should be kept in mind that these 6,129 persons possess a total of 6,129 x 2 = 12,258 genes. The number of L M alleles, for example, is 1,787 + 1,787 + 3,039. Thus, calculation of the frequency of L M and L N alleles is worked out in this way.
Thus, the frequencies of the two codominant alleles in this sample are almost equal, and this is reflected in the close approximation to a 1: 2: 1 ratio, which is a simple monohybrid ratio for codominant alleles in Mendelian genetics.
Gene frequencies expressed as decimals may be used directly to state probabilities (a probability is a function that represents the likelihood of occurrence of any particular form of an event).
If we can assume this sample to be representative of the population, then there is a probability of 0.5395 that of the chromosomes bearing this pair of alleles, any one selected randomly will bear gene L M, and 0.4605 that it will bear L N.
Let p represents genotypic frequency of L M allele and q represents frequency of L N allele, then the frequencies of three genotypes to be expected in the population are as follows:
(ii) Calculation of Gene (Allele) Frequencies for Dominant and Recessive Autosomal Alleles:
Calculation of the gene frequencies for alleles which exhibit dominance and recessive relationships requires a different approach from that used with codominant alleles. A dominant phenotype may have either of two genotypes, AA or Aa, but we have no way (other than by laboriously test- crossing each dominant phenotype) of distinguishing how many are homozygous or heterozygous in our sample.
The only phenotype whose genotype is known for certain is the recessive (aa). If the population is in equilibrium then we can obtain an estimate of q (the frequency of the recessive allele) from q^ (the frequency of the recessive genotype or phenotype).
Example 1:
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If 75% of a population was of the dominant phenotype (A-), then 25% would have recessive phenotype (aa). If the population is in equilibrium with respect to this gene locus, we expect q 2 = frequency of aa.
Then q 2 = 0.25, q = 0.5, p= 1 – q = 0.5.
Example 2:
An interesting pair of contrasting traits, which has been detected in human populations and has no known selective value is the ability or inability to taste the chemical phenylthiocarbamide (“PTC”, C 7 H 3 N 2 S), also called phenylthiourea. This was reported by Fox in 1932, who found a similar situation for several other thiocarbamides.
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The test is a simple one which can easily be performed by any genetics class. The usual procedure is to impregnate filter paper with a dilute solution of PTC (about 0.5 to 1 gram per litre), allow it to dry, then place a bit of the treated paper on the tip of the tongue.
About 70 per cent of an American white population can taste the substance, generally as very bitter, rarely as sweetish. Although the physiological basis is unknown, tasting ability does depend on a completely dominant gene, which we will designate as T. Thus tasters are T-(TT or Tt) and nontasters are tt.
From a group of 146 genetics students who tested themselves for tasting ability, 105 were tasters and 41 were nontasters. From these results the frequencies of alleles T and t in the sample may be calculated readily.
The 41 (28 per cent of the sample) nontasters are persons of tt genotype, and in the Hardy-Weinberg theorem may be represented by q 2 Therefore:
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q 2 = 0.28 and q = VO.28 = 0.53 (frequency of t).
Since p + q = 1, p = 1 – q; p = 1 – 0.53 = 0.47 (frequency of T). The frequency of homozygous and heterozygous tasters may now be computed, using the expression p 2 + 2pq + q 2 = 1.
P 2 = TT = (0.47)2 = 0.2209
2pq = Tt = 2(0.47 x 0.53) = 0.4983
q 2 = tt = (0.53)2 = 0.2809/1.0000
By testing representative samples of different populations, the frequencies of T and t in those groups may similarly be calculated.
(iii) Calculation of Gene (Allele) Frequencies for Autosomal Multiple Alleles:
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The binomial (p + q)2 = 1 applies when only two autosomal alleles occur at a given locus. For cases of multiple alleles we simply add more terms to the expression.
Example:
The four human blood types—A, B, AB, and O are determined by a series of three multiple alleles, L A or I A, L B or I B, and L 0 or i, if we neglect the various subtypes.
Hence, in a gene frequency analysis, we can here let:
p = frequency of I A
q = frequency of I B
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r = frequency of i
and p + q + r= 1
Thus, genotypes in a population under random mating will be given by (p + q + r)2.
In a sample of 23,787 persons from Rechester, New York, following frequencies for four blood types were recorded:
The frequency of each allele may now be calculated from these data, remembering that we have let p, q and r represent the frequencies of genes I A, I B, i respectively.
The value of r, that is, the frequency of gene i, is immediately evident from the figure given:
r 2 = 0.444, hence
r = √0.444 = 0.6663 (= frequency of i)
The sum of A and O phenotypes is given by (p + r) 2 = 0.418 + 0.444 = 0.862; therefore,
p + r = √0.862 =0.9284
So p = (p + r)-r = 0.9284 – 0.6663 = 0.2621 (= frequency of I A).
Because p + q +r= 1, q = 1 – (p + r) = 1 -0.9284 = 0.0716 (= frequency of I B) we can now calculate genotypic frequencies as shown in the following table:
B. Calculation of Gene Frequencies for Sex-Linked Genes:
Alleles in the sex chromosomes may occur in a different frequency than those in autosomes because of the difficult arrangements of sex chromosomes in the two sexes. The same techniques with one small modification, may be used in treating sex-linked genes.
However, since human males or Drosophila males being heterogametic sexes contain only one X chromosome, they cannot reflect a binomial distribution for random combination of pairs of sex-linked genes as females. Equilibrium distribution of genotypes for a sex-linked trait, where p + q = I, is given by-
♂p + q
♀p 2 + 2pq + q 2
Example:
In human population red-green colour blindness is a trait due to a sex-linked recessive, which we may designate r. About 8 per cent of males are colour-blind. This shows at once that q, the frequency of gene r, is 0.08 and p, the frequency of its normal allele, R is 0.92. Thus, the frequency of colour blind females is expected to be q 2 = 0.0064.
This is about what is found. Sex-linked dominants may be handled in a similar fashion; in the case of normal colour vision, with the value of p – 0.92, the incidence of normal women is p 2 + 2pq = 0.9936.
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Table of Contents
A. Calculation of Gene Frequencies of Autosomal Genes:
(i) Calculation of Gene Frequencies for Codominant Alleles:
(ii) Calculation of Gene (Allele) Frequencies for Dominant and Recessive Autosomal Alleles:
(iii) Calculation of Gene (Allele) Frequencies for Autosomal Multiple Alleles:
B. Calculation of Gene Frequencies for Sex-Linked Genes:
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Civil 3D Intersections – Fixing an Alignment that Crosses Itself
Civil 3D Intersections – Fixing an Alignment that Crosses Itself
If you have an alignment that loops around and creates an intersection on itself, you have a situation to say the least. These are rare but they do exist in the real world and in real design. The AutoCAD Civil 3D Intersection command doesn’t know what to do with this intersecting alignment. It doesn’t give you a reply saying, “I can’t resolve a self-intersecting alignment.” It doesn’t even crash or give you an error. The command just does nothing, as if it is waiting for you to pick another location so it can go to work. So, there are a few work-arounds.
The DIY Intersection
One work-around would have you create the intersection manually. Although challenging, it isn’t a good use of time. It takes longer and everything that the intersection tool does automatically, has to be done manually:
• Creating offset and curb return alignments.
• Creating offset and curb return profiles.
• Defining quadrant baselines.
• Defining quadrant regions.
• Creating the primary through region.
• Configuration of targets.
• Altering the secondary alignment to reflect the geometry of the primary.
• Inserting manual sections in quadrants at the intersection point.
That’s a lot of room for manual error.
Spackle with Grading
Another work around is to grade in the intersection with feature lines. This is a manual effort also:
• Create feature lines at the locations of edges (pavement lines, curb lines, sidewalk lines, crowns of the secondary; and crown or edge of pavement of the primary depending on intersection type).
• While creating the feature lines, be careful not to OSNAP to pavement layers that are not part of the top layer of the corridor.
• Match end of feature line elevations to adjacent corridor regions.
Maybe something magical could be done involving the extraction of corridor feature lines to “influence” the elevation of the intersection feature lines. However, staying on top of those endpoint elevations can be like herding cats.
The Short Alignment
Another method is to duplicate the loop alignment that does not involve herding cats but still takes advantage of much of the Intersection automation. This is not a copy-paste thing. This is something done diligently, but resulting with a quicker and more intelligent outcome. Here are the steps.
To duplicate the alignment:
Now we are ready to run the Intersection command.
Here is the intersection:
Here is the short profile:
The next step will be to superimpose the loop alignment’s design profile into this profile view. The superimposed profile will be dynamic with the loop alignment’s design profile. This will help us to match up the final PVI of the short alignment with the loop alignment.
In the short alignment’s profile view, we can see the superimposed version of the loop alignment displayed below highlighted. This provides a visual reference of the loop profile so we can ensure that the loop and the short profile match. Displayed on the left, the superimposed profile is displaying two markers, the start and end. Since 10+05 for the loop profile exists outside the profile view, the program is forcing it to display at the end of the limits of the short profile view. This can be remedied by applying a style that does not display the end profile marker, as depicted in the before after images below.
Now can grip edit the end of the short profile to match up with end of the superimposed loop profile and rebuild the corridor.
In AutoCAD Civil 3D, looped alignments introduce challenges in our model-based design workflows. Using the Short Alignment method, allows you to keep most of the automation of the intersection and at the same time allows you to keep an eye on the vertical design in both the looped versions and the short versions of the design profiles. Being able to adapt with sound processes to keep the design efficient and buildable is key to establishing best practices in your design environment.
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Arithmetic sequence
In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence.
For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies.
More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if .
Contents
[hide]
1 Properties
2 Sum
3 Problems
3.1 Introductory problems
3.2 Intermediate problems
4 See Also
Properties
Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, .
A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to .
Proof: Let the sequence have first term and common difference . Then using the above result, as desired.
Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences.
Sum
An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value.
The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to .
Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required.
The second is that if an arithmetic series has first term , common difference , and terms, it has value .
Proof: The final term has value . Then by the above formula, the series has value This completes the proof.
Problems
Here are some problems with solutions that utilize arithmetic sequences and series.
Introductory problems
2005 AMC 10A Problem 17
2006 AMC 10A Problem 19
2012 AIME I Problems/Problem 2
2004 AMC 10B Problems/Problem 10
2006 AMC 10A, Problem 9
2006 AMC 12A, Problem 12
Intermediate problems
2003 AIME I, Problem 2
Find the roots of the polynomial , given that the roots form an arithmetic progression.
See Also
Geometric sequence
Harmonic sequence
Sequence
Series
Retrieved from "
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Algebra
Sequences and series
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What would be a good textbook for learning Fourier analysis if one is comfortable with Measure Theory, Hilbert spaces and Lp spaces in general? Other sources like video lectures would also be helpful if available. I am not familiar with Complex Analysis.
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Katznelson's "Introduction to Harmonic analysis" is a classic.
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One of the best introductory Fourier analysis textbook in my eyes is
Fourier analysis by J. Duoandikoetxea
The author explains things extremely well and chooses just the right level of detail. However, there are unfortunately no exercises.
Another fairly recent one is
Classical and multilinear harmonic analysis by W. Schlag and C. Muscalu
This work consists of two volumes, the first deals with introductory classical Fourier analysis material, while the second is more modern and touches on ongoing research. It has lots of exercises.
The book by Katznelson mentioned in the comment I can also recommend. It focuses on Fourier series and the exercises are nice.
Then of course the classical texts by E. M. Stein are a must-read:
Introduction to Fourier analysis on Euclidean spaces
Singular integrals and differentiability properties of functions
Harmonic analysis: real-variable methods, orthogonality, and oscillatory integrals
They should be read in that order. The second and third are already quite advanced and may be a bit hard to digest for a beginner since the presentation is quite dense and fairly general. The third book in particular is considered by many to be the standard reference regarding many of the core topics in harmonic analysis.
To complement the other answer, here is my opinion on the Grafakos texts: while the books are very well and thoroughly written, they are arguably not a good single source to learn the subject. The focus is on formalities, precision and long-winded detailed calculations instead of ideas and concepts. The books are a good reference because they contain extremely detailed proofs, cover a lot of material and contain quite general statements of important results. There are also lots of exercises. So maybe use these alongside a more conceptual textbook to look up details you don't understand in the other book or to get some practice from doing the exercises. It's always good to have multiple sources anyways, because different authors often explain the same thing in slightly different ways.
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I would recommend the book Classical Fourier Analysis by Loukas Grafakos published by Springer. It is a vast, extensive and highly formal introduction to the subject which suffices for most purposes. However, it is one of the hardest analysis books I know, but this can also originate from the subject itself. One big plus of this book is the large amount of exercises provided.
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4380 | https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOopKcZXnDbTUMKWXQkN9drSxx1ItHzSXRZ5dVS713Vah9r9xeMPi | Art of Problem Solving
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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Inequalities
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4381 | https://www.youtube.com/watch?v=HLNSouzygw0 | Difference of squares intro | Mathematics II | High School Math | Khan Academy
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When an expression can be viewed as the difference of two perfect squares, i.e. a²-b², then we can factor it as (a+b)(a-b). For example, x²-25 can be factored as (x+5)(x-5). This method is based on the pattern (a+b)(a-b)=a²-b², which can be verified by expanding the parentheses in (a+b)(a-b).
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we're now going to explore factoring a type of expression called a difference of squares and the reason why it's called a difference of squares is because it's expressions like x^2 - 9 this is a difference we're subtracting between two quantities that are each squares this is literally x squared let me do that a different color this is x squar minus 3 squared it's the difference between two quantities that have been squared and it turns out that this is pretty straightforward to factor and to see how it can be factored let me pause there for a second and get a little bit of a review of multiplying binomials so put this on the back burner a little bit before I give you the answer of how do you factor this let's do a little bit of an exercise let's multiply x + a x - A where a is some number and we can use that do that using either the foil method but I like just thinking of this as a distributive property twice we could take X plus a and distribute it onto the X and onto the a so when we multiply it by X we would get x X is x^2 a X is plus a x and then when we multiply it by thetive a well it'll become a x minus a^ 2 so these middle two terms cancel out and you are left with x^2 minus a 2 you're left with a difference of squares x^2 minus a^ 2 so we have an interesting result right over here that x^2 - A 2 is equal to is equal to x + a x + a x - A and so we can use and this is for NE a so we could use this pattern now to factor this here what is our a our a is 3 this is x^2 - 3^ 2 or we could say minus our a 2qu if we say 3 is a and so to factor it this is just going to be equal to x + r a which is 3 x - R A which is 3 so x + 3 X minus 3 now let's do some examples to really reinforce this idea of factoring differences of squares so let's say we want to factor let me say y^2 minus 25 it has to be a difference of squares can't doesn't work with a sum of squares well in this case this is going to be Y and you have to confirm okay yeah 25 is 5^ squar and y^2 is well y^ 2ar so it's going to be y + something Yus something and what is that something well this right here is 5 squar so it's y + 5 y - 5 and the variable doesn't have to come first we could write 121 minus I'll introduce a new variable minus b^ 2 well this is a difference of squares because 121 is 11 SAR so this is going to be 11 + something 11 minus something and in this case that something is going to be the thing that was squared so 11 + B 11 minus B so in general if you see a if you see a difference of squares one square being subtracted from another square and it could be a numeric perfect square or it could be a variable that has been squared that can be that you could take the square root of well then you could say all right well that's just going to be the first thing that's squared plus the second thing that has that has been squared times the first thing that was squared minus the second thing that was squared now some common mistakes that I've seen people do including my son when they first learn this is they say okay it's easy to recognize the difference of squares but then they say oh is this y^2 + 25 y^ 2 - 25 no the important thing to realize is is that what is getting squared over here Y is the thing getting squared and over here it is five that is getting squared those are the things that are getting squared in this difference of squares and so it's going to be y + 5 y - 5 I encourage you to just try this out we have a whole practice section on KH Academy where you can do many many more of these to become familiar |
4382 | https://bmcnephrol.biomedcentral.com/articles/10.1186/s12882-023-03083-8 | Advertisement
The expression of respiratory tract virus in pediatric glomerular disease: a retrospective study of 45 renal biopsy in China
BMC Nephrology
volume 24, Article number: 36 (2023)
Cite this article
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Abstract
Background
More attention has been put on the relationship between pediatric glomerular disease and respiratory tract virus infection. Children with glomerular illness, however, are uncommonly found to have biopsy-proven pathological evidence of viral infection. The purpose of this study is to determine whether and what kind of respiratory viruses are found in renal biopsy from glomerular disorders.
Methods
We used a multiplex PCR to identify a wide range of respiratory tract viruses in the renal biopsy samples (n = 45) from children with glomerular disorders and a specific PCR to verify their expression.
Results
These case series included 45 of 47 renal biopsy specimens, with 37.8% of male and 62.2% of female patients. Indications for a kidney biopsy were present in all of the individuals. In 80% of the samples, respiratory syncytial virus was discovered. Following that, the RSV subtypes in several pediatric renal disorders were found. There were 16 RSVA positives, 5 RSVB positives, and 15 RSVA/B positives, accounting for 44.4%, 13.9%, and 41.7%, respectively. Nephrotic syndrome samples made up 62.5% of RSVA positive specimens. The RSVA/B-positive was detected in all pathological histological types.
Conclusions
Patients with glomerular disease exhibit respiratory tract viral expression in the renal tissues, especially respiratory syncytial virus. This research offers new information on the detection of respiratory tract viruses in renal tissue, which may facilitate the identification and treatment of pediatric glomerular diseases.
Peer Review reports
Introduction
Glomerular disease is a group of illnesses with increased morbidity and death, overwhelming China's healthcare system . Due to the variety of clinical presentations and treatment responses, there was not many high-quality clinical research on glomerular disease . The most common manifestation of glomerular disease in children is nephrotic syndrome (NS), which is characterized by proteinuria, hypoalbuminemia, edema, and hyperlipidemia . The most typical reason for a kidney biopsy in children is thought to be minimal-change nephrotic syndrome (MCNS) .
There have been some studies associating enteroviruses, Epstein–Barr virus (EBV), and cytomegalovirus (CMV) to nephropathy [5,6,7]. Infection, especially respiratory tract infection, is believed to be one of the risk factors for the onset, relapse, exacerbations, and development of end-stage renal disease (ESRD) in both adults and children with NS [8,9,10,11]. Previous studies have demonstrated that the respiratory syncytial virus (RSV), the most common respiratory virus, and its antibody were found in the urine, serum, epithelial cells of the respiratory tract, and peripheral blood mononuclear cells (PBMC) of patients with steroid-responsive nephrotic syndrome [12, 13], and a small percentage of intranasal inoculation of RSV can cause nephropathy in rats . However, to our knowledge, no publication has connected respiratory virus infection with pediatric glomerular diseases through biopsy-proven pathological evidence.
The purpose of this research is to determine whether and which types of viral infections are found in renal biopsies from patients with glomerular disease. This study aims to offer visible proof of respiratory virus infection in children with glomerular diseases.
Methods
Ethics statement
The study was approved by the Institutional Review Board/Ethics Committee affiliated with West China Second University Hospital, Sichuan University (2,020,111), and followed the guidelines of national/international/institutional or Declaration of Helsinki. The informed consent was obtained from all participants’ guardians.
Patients and clinical specimens
From January 2010 to December 2017, we collected data and renal biopsy specimens from 47 patients with pediatric glomerular disorders at Sichuan University's West China Second University Hospital.
The main indication of kidney biopsy was: (1) Steroid-resistant nephrotic syndrome (SRNS). (2) Steroid-dependent nephrotic syndrome (SDNS)/ Frequent relapsing NS (FRNS). (3) Nephrotic syndrome with hematuria, renal impairment, or persistent hypertension. (4) NS under the age of one year or over the age of ten years. (5) Proteinuria and glomerular hematuria. (6) Systemic illness with an abnormal urinalysis. (7) Unknown aetiology of acute or chronic renal failure. Pathological renal biopsy specimens were collected and maintained at -80 °C for subsequent detection of respiratory viruses.
Electronic medical records were reviewed retrospectively for clinical data. One subject was ruled out due to incomplete clinical data and 1 Alport syndrome were not included. The remaining 45 patients were admitted to our study (Fig. 1).
Flowchart of this study. NS, nephrotic syndrome; SRNS, steroid-resistant nephrotic syndrome; SDNS, steroid-dependent nephrotic syndrome; FRNS, frequent relapsing NS
Isolation of viral genomic RNA and cDNA synthesis
Viral genomic RNA was extracted from the patients’ renal biopsy specimens by using the QIAamp RNA extraction kit (Qiagen GmbH, Hilden, Germany), according to the protocol suggested by the manufacturer. Briefly, clinical samples were homogenized by vortexing for the 30 s, and 140 μl was used to extract viral genomic RNA. The RNA was eluted from the columns with 50 μl of elution buffer. The RNA was immediately stored at -80 °C for later detection. cDNA was synthesized by using MultiScribe reverse transcriptase and random hexamers (both from PE Applied Biosystems). The cDNA was stored at -80 °C before further use.
Multiple PCR
Due to the diversity of the respiratory virus, we utilized multiplex nested PCR to test the respiratory tract virus in the meantime. A second (nested) amplification followed a one-tube RT-PCR. Seven sets of oligonucleotide primers were designed for mPCR and amplification of adenovirus, rhinovirus, influenza virus, parainfluenza virus, coronavirus, RSV, and GAPDH for internal reference, according to nucleotide sequences available from GenBank (shown in Table 1). Multiplex RT-PCR was performed by using a One-Step RT-PCR kit (Qiagen). Briefly, 5 μl of extracted RNA was added to a mixture composed of an enzyme mixture (CPG Inc., Lincoln Park, N.J.) and a mixture of the primers, each at a final concentration of 20 pmol. The optimized profile in the thermal cycler (PTC-100; MJ Research Watertown, Mass.) was 50 °C for 30 min and 95 °C for 15 min, followed by 35 amplification cycles (with each cycle consisting of denaturation at 94 °C for 45 s, annealing at 56 °C for 45 s, and synthesis at 72 °C for 1 min). Amplification was completed with a prolonged synthesis at 72 °C for 10 min.
Specific PCR with gel detection
Confirmatory tests were run to verify the initial results with specific PCR. Nested PCR was performed following multiplex PCR on specimens suspected of having RSV or another virus. For each nested PCR, a 0.1 μM concentration of each primer was used, 2 μM of first-run mPCR product was added, and an annealing temperature of 50℃ was used. Amplified products were electrophoretically separated on 2% agarose gels to differentiate virus-specific bands. The gels were then stained with ethidium bromide and visualized under UV light.
Statistical analysis
The data management and statistical analysis were performed with GraphPad Prism 8 software (GraphPad Software, La Jolla, Calif).
Results
Baseline characteristics
This study included a total of 45 of 47 children with glomerular disease, with 17 (37.8%) males and 28 (62.2%) girls. Twenty patients had nephrotic syndrome, eight had primary IgA nephropathy, three had glomerulonephritis, five had lupus nephritis, and nine had purpura nephritis. NS was classified into two categories based on the clinical classification of glomerular disorders: simple type (15 patients) and nephritic type (5 patients). NS was categorized as nephritic type NS if at least one of the following criteria was met: 1) glomerular hematuria, 2) renal impairment, 3) chronic hypertension, and 4) low serum complement. Based on hormone sensitivity among the 15 patients with simple type NS, three children had SRNS and twelve children had SDNS/FRNS. A percutaneous kidney biopsy was performed on all 45 individuals since it was indicated. The average age at the time of kidney biopsy was 9.9 years old, ranging from 1.2-year-old to 18.3-year-old. Their median age at the commencement of the disease was 8.7 (range 1 to 17.7 years) years old. Twenty individuals suffered from respiratory tract infection when they had a kidney biopsy. Table 2 provided a description of the patients' clinical characteristics.
Respiratory virus infection was detected in renal biopsies
We utilized specific primers to identify the expression of adenovirus, rhinovirus, influenza virus, parainfluenza virus, coronavirus, and RSV in the samples. The result was shown on electrophoresis image. The expression of respiratory syncytial virus was detected in 36 (80%) and influenza virus was detected in 2 (4.4%) of the 45 kidney samples with pediatric glomerular disease. No common respiratory virus expression had been found in the rest of the sample (Fig. 2).
Respiratory virus expression in renal biopsy tissues. a The presence of RSVA in renal biopsy tissues was identified. The image showed that RSVA expression was present in lanes 1–3 and lanes 5–6. b RSVB and INFA/B was detected in renal biopsy tissues. The RSVB was expressed in lanes 1–2 and 4, INFA in lanes 6, and INFB in lanes 7. Viruses that are uncommon are not included
Type of respiratory virus identified in pediatric glomerular disease renal biopsies
RSV was a single-stranded pneumovirus that belonged to the Pneumoviridae family . RSV contained a negative-sense RNA genome that was separated into two genetic subtypes, RSV subgroups A (RSVA) and B (RSVB), based on antigenic and genetic diversity . In 45 children's renal biopsies with glomerular disease, we found 36 RSV positive specimens, including 16 RSVA positive, 5 RSVB positive, and 15 RSVA/B positive specimens, accounting for 44.4%, 13.9%, and 41.7%, respectively (shown in Fig. 3a and Table 3). Nephrotic syndrome involved 62.5% of the RSVA positive samples, comprising four nephritic type NS and five SDNS/FRNS. Five RSVB specimens were identified in kidney biopsies, three of which were SDNS/FRNS (60%). In addition, the positive detection rate of RSVA/B in the Henoch-Schonlein purpura (HSP) nephritis, lupus nephritis, and IgAN specimens was 26.7% (n = 4), 20% (n = 3), and 26.7% (n = 4), respectively (Fig. 3b). Minimal change disease (MCD) was the most common cause of pediatric glomerular diseases as displayed in Table 3. A half of the RSVA-positive samples were histopathologically diagnosed with MCD. Only MCD, HSP nephritis and mesangial proliferative glomerylonephritis (MsPGN) were detected positive for the presence of RSVB, of which MCD accounted for 60%. The RSVA/B + was detected in all pathological histological forms.
Respiratory virus subtype in pediatric glomerular disease renal biopsies. a Graphical representation of the classification of respiratory virus subtypes. b Various glomerular diseases and respiratory virus subtype
The RSVA/B positive had nearly similar amounts of males and females. We did, however, find 12 females and 4 males with RSVA positivity. 4 out of 5 male kidney samples tested positive for RSVB.
In conclusion, renal tissues of children with glomerular disease were shown to include respiratory viruses, particularly RSV. Then, we identified RSV subtypes in various kidney biopsy specimens.
Discussion
We retrospectively analyze renal biopsies from 45 pediatric patients with glomerular disease, over the course of seven years, in a tertiary care pediatric hospital in southwestern China. Kidney biopsy is regarded as the "gold standard" in the diagnosis of glomerular disorders. Based on their clinical profiles and the effectiveness of their treatments, these people need and successfully undergo a renal biopsy. The range of biopsy-verified pediatric glomerular disorders in China were characterized by Nie and Zheng [17, 18]. The most frequent reason for our patient's kidney biopsy (44.4%, 20/45) is nephrotic syndrome, which is consistent with their findings. Nevertheless, no study has linked respiratory viruses to glomerular disease in kidney biopsies. The purpose of this study is to identify the presence and kinds of respiratory tract virus expression in renal tissues in order to give pathological evidence of respiratory tract virus infection in pediatric glomerular disease.
Heterogeneity is one of the factors that contributed to the lack of high-quality clinical trials in primary glomerular disease . In this case, our research may be of poor quality. To minimize the negative impact of homogeneous scarcity, we separate those individuals into groups depending on clinical symptoms and steroid hormone sensitivity. There are two categories for patients with glomerular disease: 1) primary disease group: nephrotic syndrome (SRNS, SDNS/FRNS and nephritic type NS), primary IgA nephropathy and glomerulonephritis; 2) secondary disease group: lupus nephritis and purpura nephritis.
Previous study has shown that bacteria, viruses, fungi, and mycobacteria all play a significant role in renal disease . Individuals with glomerular disease are more prone to infection with hepatitis B, herpesviruses, parvovirus B19, respiratory syncytial virus, influenza virus (Flu), parainfluenza, enterovirus, and COVID-19 [20, 21]. According to our knowledge, respiratory virus infections play a prominent part in the initiation, aggravation, and recurrence of childhood NS and primary IgA nephropathy [10, 22]. According to the findings, when higher levels of corticosteroids are provided at the onset of a viral upper respiratory infection, relapses of NS are reduced . We observe that respiratory virus is present in the majority of glomerular disease samples, implying that glomerular disease is related with viral infection of the respiratory tract. In previous research, respiratory tract viruses were detected in the airway epithelial cells, serum, urine, PBMC, and kidney tissue (only two specimens) of MCNS children in the active stage [12, 13, 24]. The virus in renal tissues, on the other hand, has attracted little study. In this study, multiplex PCR is utilized to directly detect the expression of possible respiratory viruses in pathologic renal biopsy specimens, resulting in visualized results indicating the presence of respiratory tract virus in pediatric glomerular disease. At the time of the biopsy, 45% of the subjects have an acute upper respiratory tract infection. We analyze that it has a moderate influence on the high percentage of respiratory tract virus positive (particularly RSV).
One of the study's advantages is the detection of RSV subgroups in numerous kidney biopsy specimens. RSVA is the most prevalent subtype in NS, while RSVA/B is the most common subtype in IgAN and secondary glomerular disease. Children with SDNS/FRNS are more prone to developing ESRD . RSV was found in 83.3 percent of SDNS/FRNS tissues, implying that RSV infection may be a risk factor for SDNS/FRNS pathogenesis. We have learned that the type of renal histopathology is various in glomerular disease. MCD is the leading cause of pediatric glomerular disease in our study. RSVA/B-positive is found in a broad range of renal pathology specimens. There is also a low level of influenza virus expression, according to our data. It is unclear whether it has an impact on the incidence of pediatric nephropathy due to the small sample size.
Previous research has suggested that defective cell-mediated immunity may play an important role in the etiology of renal disease [26, 27]. RSV infection is a primary cause of proteinuria and renal dysfunction in the rat model of RSV reinfection. After a respiratory tract infection, primary T lymphocyte cells could impair cellular immune function and produce abnormal cytokines. RSV-induced anti-virus responses, in particular, can lead to uncontrolled cytokine production [28, 29]. According to a recent study, children with NS suffer renal damage and an inflammatory response . We also found minor deterioration of renal function and low concentrations of immunoglobulin G (IgG) in glomerular disease patients, especially in NS, which is consistent with those results. However, we were unable to detect the expression of the apoptosis signal regulating kinase test, due to an insufficient testing facility in the early stages of the investigation.
It is important to note that this study has a number of limitations. First, even though this research is based on a retrospective case series, it is recommended to include some relevant mechanisms of RSV infection and pediatric glomerular disease. Furthermore, presenting the results of several control groups may help better reveal the association between viral infections and glomerular disease. Second, further investigation may focus on the connection between viral load and the severity of glomerular disease.
Conclusion
In conclusion, our results support that respiratory tract viruses, with RSV being the most prevalent virus, can be detected in the renal tissues of individuals with glomerular disease. Our study adds to the knowledge of how important a role RSV infection may play in the pediatric glomerular disease.
Availability of data and materials
All data generated or analysed during this study are included in this published article and its supplementary information files.
Abbreviations
Cytomegalovirus
Crescentic glomerulonephritis
Epstein–Barr virus
End-stage renal disease
Influenza virus
Frequent relapsing NS
Focal segmental glomerulosclerosis
Henoch-Schonlein purpura
Henoch-Schonlein purpura nephritis
IgA nephropathy
Immunoglobulin G
Lupus nephritis
Minimal change disease
Minimal-change nephrotic syndrome
Membrane nephrosis
Mesangial proliferative glomerylonephritis
Nephrotic syndrome
Peripheral blood mononuclear cells
Respiratory syncytial virus
RSV subgroup A
RSVA and B positive
RSV subgroup B
Steroid-dependent nephrotic syndrome
Steroid-resistant nephrotic syndrome
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Acknowledgements
We thank all of the participants.
Funding
This study was supported by the Fourth Baili Pediatric Scientific Research Fund of the Chinese Journal of Pediatrics (grant no. B2016-C-3) and the Sichuan Provincial Pediatric Clinical and Medical Research Center (Sichuan Scientific Research 2017–46-4).
Author information
Li Lin and Lu Li contributed equally to this work.
Authors and Affiliations
Department of Pediatrics, West China Second University Hospital, Sichuan University, No. 20, Section 3, South Renmin Road, Chengdu, 610041, Sichuan Province, China
Li Lin, Yao Cao, Xin Peng, Yi Wu & LiQun Dong
National Center for Birth Defect Monitoring, West China Second University Hospital, Sichuan University, No. 17, Section 3, South Renmin Road, Chengdu, 610041, Sichuan Province, China
Lu Li & Ping Yu
Key Laboratory of Birth Defects and Related Diseases of Women and Children (Sichuan University), Ministry of Education, No. 17, Section 3, South Renmin Road, Chengdu, 610041, Sichuan Province, China
Lu Li & Ping Yu
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Contributions
All authors contributed to the study's conception and design. Material preparation was performed by Yao Cao, Xin Peng, and Yi Wu. Li Lin and Lu Li performed data collection and analysis. The first draft of the manuscript was written by Li Lin and Lu Li. Ping Yu and LiQun Dong gave academic feedback and revised the manuscript. All authors commented on previous versions of the manuscript. All authors read and approved the final manuscript.
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Ping Yu or LiQun Dong.
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The study was approved by the Institutional Review Board/Ethics Committee affiliated with West China Second University Hospital, Sichuan University (2020111), and followed the guidelines of national/international/institutional or Declaration of Helsinki. The informed consent was obtained from all participants’ guardians.
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Lin, L., Li, L., Cao, Y. et al. The expression of respiratory tract virus in pediatric glomerular disease: a retrospective study of 45 renal biopsy in China.
BMC Nephrol 24, 36 (2023).
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4383 | https://www.wyzant.com/resources/answers/699549/maths-olympiad-minimum-possible-number-of-points | WYZANT TUTORING
Louis M.
Maths olympiad, minimum possible number of points
Six players compete in a tournament. Each player plays exactly two games against every other player. In each game, the winning player earns 2 points and the losing player earns 0 points. If the game results a draw (tie), each player earns 1 point. What is the minimum possible number of points that a player needs ti earn in order to guarantee that he/she will be champion (i.e he/she has more points than every other player)?
1 Expert Answer
Michael D.
answered • 07/17/19
Versatile STEM tutor eager to teach
In order to figure this out, we need to assume that 2 different players are basically dominating the other 4 and then play each other. Each player versus the other 5 players twice. Thus, each player plays 10 games. Max number of points is 20.
Let's say Player 1 plays Players 3-6 (4 different players) and wins all the games. Player 1 would then have 16 points (4 players (2 games / player) (2 points / game)
Let's say Player 2 does the same. Plays Players 3-6 (4 different players) and wins all the games. Player 2 would then have 16 points (4 players (2 games / player) (2 points / game)
This means, that going into the final 2 games (which are between Players 1 & 2), both players 1 & 2 have 16 points... The outcomes of their 2 games will crown the winner. Players 3-6 cannot compete any longer since they lost both matches to 2 players it is impossible for them to have more than either player 1 or 2.
Now between the last 2 games, there are only 4 points that can be given out. In order to ensure victory (and not end the contest in a tie of 18 to 18 points), either player 1 or 2 must get at least 3 points, resulting in a 19-17 or a 20-16 victory.
Thus, the minimum amount of points required to guarantee winning the tournament is 19 points.
You can also think about it like this... If you Win every match except for 1 match, which you tie, then noone can have as much or more than you since everyone person would have lost matches to you and not been able to beat you at all. Winning 9 games and tieing 1 gives 19 points.
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4384 | https://www.scribd.com/document/519293072/1-WHO-boys | WHO Growth Standards for Boys | PDF | Human Size | Human Weight
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4385 | https://math.stackexchange.com/questions/3483017/compute-telescopic-sum-of-binomial-coefficients | Stack Exchange Network
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Compute telescopic sum of binomial coefficients
Is there a nice or simple form for a sum of the following form?
$$ 1 + \sum_{i=1}^k \binom{n-1+i}{i} - \binom{n-1+i}{i-1}$$
Motivation: Due to a computation in the formalism of Schubert calculus the above sum with $k = \lceil n/2 \rceil -1$ is equal to the number of lines intersecting $2n-4$ general subspaces $H_j\subseteq \mathbb{P}^n$ of dimension $n-2$.
2 Answers 2
Hint
The following method does not use telescopic series but it makes use of elementary binomial theorem along with some geometric progression to arrive at the answer.
$$\begin{aligned}S&=\sum_{i=1}^{k}{n-1+i\choose n-1}-\sum_{i=1}^{k}{n-1+i\choose n}\&=\left(\text{coeff. of } x^{n-1} \text{ in } \sum_{i=1}^{k}(1+x)^{n-1+i}\right)-\left(\text{coeff. of } x^{n} \text{ in } \sum_{i=1}^{k}(1+x)^{n-1+i}\right)\end{aligned}$$
No need for a power tool when a manual tool does the trick.
Hockey-stick identity:
$$
\sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k}
$$
Applying to our expression:
$$
1 + \sum_{i=1}^k \binom{n-1+i}{i} - \binom{n-1+i}{i-1} \
1 + \sum_{i=1}^k \binom{n-1+i}{i} - \sum_{i=1}^k \binom{n-1+i}{i-1} \
1 - \binom{n-1}{0} + \sum_{i=0}^k \binom{n-1+i}{i} - \sum_{i=0}^k \binom{n+i}{i} \
\sum_{i=0}^k \binom{n-1+i}{i} - \sum_{i=0}^k \binom{n+i}{i} \
\binom{n+k}{k} - \binom{n+k+1}{k} \
\binom{n+k}{k} - \left(\binom{n+k}{k-1}+\binom{n+k}{k}\right) \
-\binom{n+k}{k-1}
$$
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4386 | https://www.youtube.com/watch?v=JgzhgoQyD4w | How to Evaluate Piecewise Functions with Given Values: f(-3), f(-2), f(-1), and f(0)
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For the following exercises, given each function f, evaluate f (−3), f (−2), f (−1), and f (0).
f(x) = { x + 1 if x is less than -2
f(x) = { -2x - 3 if x ≥ -2
f(x) = { 1 if x ≤ -3
f(x) = { 0 if x is greater than -3
f(x) = -2x2 + 3 if x ≤ -1
f(x) = 5x - 7 if x is greater than -1
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Transcript:
for the following exercises given the function f evaluate f of negative three f of negative two f of negative one and f of zero all right so just keep in mind the general idea down here that whenever they ask you to evaluate f of a number all right just like they're doing here they're asking us to evaluate f of negative 3 and negative 2 etc it basically just means to plug in the number for x into the equation that you're given right here are all the equations all right into the equation that's given and just simplify all right that's all it is so let's take a look at the first one so here is a piecewise function all that means is that there are two different equations that govern this overall function at different values of x okay so if i want to evaluate the um f of negative three here what that means is my x value is really negative three and i have to then consider which of these two equations should i plug in the value of negative three into for x you have to be aware of where negative 3 falls negative 3 is less than negative 2 right and therefore you have to use the first equation to find the function value of negative 3. okay because x lies on this continuum right where it's less than negative two so what i'm going to do is use the first equation okay so it says x plus one right but instead of writing x here i'm going to substitute that on out for negative three and then add one so what's the answer here so the answer is simply right negative two that is the value that is the f of negative three this is the answer all we have to do is after we plug in the x value just simplify right fairly straightforward now we can run through this okay we just got to be careful about which equation to plug into the next one said f of negative 2. so negative 2 if you notice here we would use now the second equation because this says you're going to this is telling me use this equation when x is either greater than or equal to negative two so we do have that scenario for the second part here so basically now take this equation of negative 2x minus 3 and plug in negative 2 for your x value and then all you have to do is just simplify this so this becomes positive four minus three is obviously a positive one that's over and then do the same thing for negative one notice negative one is greater than negative two so we're gonna use the second equation again so it's negative two times negative one minus three right this is now positive two minus three so this becomes a negative one and then last but not least here f of zero this is going to be same same thing we're using the last equation negative two times uh zero minus 3 and now is a negative 3. so those would be all of the values all right let's do the next one okay so the next one is just as easy except it's actually even easier but it might be a little confusing at the start so we're going to take our f of negative 3 value okay now consider which of these two equations you might be saying well this is an equation these are exact numbers yes that's true however just consider that these are the two functional values that we'll be talking about where does negative 3 lie which domain does negative 3 lie in well it's going to lie in this one right because this is telling us to you're going to plug in x for well there's no x here so that's the trick right but you're going to be using this value for the function when x is less than excuse me less than or equal to negative 3. so i would take my x value here and plug it in for whatever x value was here but there isn't any right so this is even easier than you think it's just one it's just whatever this number is there that's all there's nothing to do simple right so f now of negative two which which value will i be using this time well negative two is greater than negative three so i'm going to be using now the second value okay so that's just zero how about then f of negative one again same thing it's greater than negative 3 so that's just 0. and i think you can see the pattern here right this is just 0. great easy now let's do the last one okay same concept let's see if we can run through it so this is f of negative 3. consider where negative 3 lies which domain negative 3 is going to be less than or equal to right negative 1. it's going to be less than negative 1. so i'm going to be using this first function great so let's write it out so it's negative 2 times now negative 3 because that's the x value squared plus 3. all you have to do now is simplify so this becomes a positive 9 right positive 9 then times a negative 2 is going to be a negative 18. then when you add 3 to it we get a negative 15 as our final result easy so f of now negative 2 again it's still going to be less than negative 1 so we're still using the top function so it's negative 2 times negative 2 squared plus 3. do not forget your parentheses here all right so this becomes a positive 4 and a positive 4 multiplied by a negative 2 will be negative 8 and then we're going to add 3 to that so that's going to now be a negative five okay great how about now f of negative one so negative one is equal to negative one so we're still using the top function all right so this is negative two times negative one squared plus three negative one times negative one is positive one positive one times negative two is negative two and then negative two plus a three is going to be a positive one there we go and now f of zero we now change the function because zero is now greater than negative one so i'm going to use this function okay so this is five times zero plus excuse me minus seven and obviously that's just gonna be negative seven there you go ladies and gentlemen all right hope this helped all you got to do is basically just plug in the values and simplify so look forward to helping you know the next problems please remember to subscribe i'll see you then |
4387 | https://acsjournals.onlinelibrary.wiley.com/doi/pdf/10.1002/1097-0142(19840101)53:1%3C58::AID-CNCR2820530111%3E3.0.CO;2-Q | Gestational Trophoblastic Tumors Metastatic to the Lung Radiologic-Clinical Correlations
ALAN S. HENDIN, MD
The relationship between human chorionic gonadotropin (HCG) titers in urine and the number of discrete metastatic pulmonary nodules was studied in patients with persistent trophoblastic disease, after evacuation of hydatidiform mole, and before starting chemotherapy. A significant difference in HCG titers was found between patients with 0 to 2 nodules and patients with 5 or more nodules. A weak linear relationship was found. The distribution of 57 discrete pulmonary nodules in 13 patients was plotted by lung zones (upper, middle, and lower thirds). Twenty-eight percent of the nodules were in the upper third of the lungs. Two patients had solitary apical nodules. This differs from the characteristic predominantly basilar distribution of blood-borne metastases of other neoplasms. Pulmonary spread may have occurred during curettage of moles, when the patients were recumbent and pulmonary blood flow was redistributed to the upper portions of the lungs.
Cancer
5358-61, 1984. H E LUNG is the most common site of metastasis in
T patients with gestational trophoblastic disease (hy-datidiform mole, invasive mole, and chonocarcinoma). ’
The major forms of pulmonary involvement are discrete nodular metastases, military or “snowstorm” opacities, and pulmonary embolism by trophoblastic tissue emboli2 The most reliable indicator of the presence and extent of trophoblastic tissue in a patient is the level of human chononic gonadotropin (HCG) in blood or urine.3 The relationship between HCG titers and the form or number of pulmonary metastases has not previously been de-scribed. The author investigated: ( I ) the relationship be-tween HCG titers and the number of discrete metastatic pulmonary nodules in patients with persistent tropho-blastic disease, after curettage of hydatidiform mole, and before initiation of chemotherapy; and (2) the anatomic distribution of pulmonary metastatic nodules in these patients. This portion of the study was prompted by ob-servation of two patients with discrete nodular metastases limited to the lung apices, a distribution which appeared atypical for blood-borne metastases. From the Department of Radiology, University of California School of Medicine, San Francisco. California. Address for reprints: Alan S. Hendin. MD, 745 Escalona Drive. Santa Cruz. CA, 95060. The author thanks Carolina Braga. MD. for allowing him t o review the clinical records of the patients in this study, and for her helpful criticism of the manuscript. Accepted for publication October 20, 1982.
Methods
The records of all patients with gestational trophoblastic disease seen at the University of California-San Francisco Medical Center between 1968 and 1976 were reviewed. All patients with extrauterine trophoblastic metastases, and all with persistently elevated HCG titers in urine or blood, 1 month or more following evacuation of hyda-tidiform mole, were selected for inclusion in this study. Pulmonary nodules were considered to be metastases if they regressed or enlarged during therapy. Palpable vaginal or extrauterine pelvic masses, and focal abnormalities on brain or liver scintiphoto studies were recorded as me-tastases. Patients with hydatidiform moles, not yet evac-uated, or evacuated within the past month were not in-cluded, if they did not have palpable or radiographically demonstrable metastases. Pretreatment urinary HCG was measured in all patients by hemagglutination inhibition on a 24-hour urine col-lection. We recorded the number of pulmonary nodules on the pretreatment chest film of each patient. In all but five patients, the presence and number of nodules were confirmed by tomography. For each patient the number of pulmonary nodules was plotted against the urinary HCG titer. The significance of the relationship was evaluated by linear regression analysis by the method of least squares. The difference between HCG titers between groups of patients with ( I ) no visible nodules, (2) one or two nodules, and ( 3 ) five
5 8 No. 1
1 1 -
10 -
9 -
0 -7 -
6 - 05 -
4 -
3 -
2 - 0 0 0
1 - 0 0 00 0 0 0
I I I I l l I I I I I 1
< 950 1,000 2,000 4,000 10,000 20,000 40,000 100,000 600,000
TROPHOBLASTIC TUMOR METASTASES - Ifendin
0 000 0 0
24 HOUR URINARY H C G , INTERNATIONAL UNITS/LITER
FIG. 1. Relationship between urinary HCG and the number of metastatic pulmonary nodules.
or more nodules, was evaluated by unpaired t test. Sta-tistical calculations were performed on an H P 9100B
computer (Hewlett-Packard, Palo Alto, CA). The distribution of all nodules by lung thirds (upper, middle, or lower) was recorded for the pretreatment chest films. Each lung was divided into thirds using the distance between costophrenic angle and apex on the postero-anterior (PA) chest film or anteroposterior (AP) whole lung tomogram. Horizontal lines were drawn at the junc-tion of thirds. The position of the center of each nodule was used to determine location, since several large nodules overlapped two lung zones. Results Twenty-four patients were included, 13 with pulmonary metastases, and 1 I with persistent elevation of HCG in urine or blood without evident metastases. No patient had extrauterine pelvic metastasis determined by pal-pation. A weak direct relationship was found between the number of radiographically detected pulmonary nodules and the pretreatment titer of urinary HCG (linear cor-relation coefficient = 0.69) (Fig. 1). A significant difference in titers was found between patients with 0 to 2 nodules and patients with 5 or more nodules (t = 2.65, p = 0.02). No significant difference was found between other groups. The distribution of 57 discrete pulmonary nodules in 13 patients is shown in Figure 2. Sixteen nodules (28%) were in the upper third of the lungs, 22 (39%) were in the middle third, and 19 (33%) were in the lower third. 59 One patient had snowstorm or military opacities, and another developed a pulmonary embolus presumed to be due to trophoblastic tissue, both without nodules.
0..
o....
.o...
0
..... ..... o....
\ ....
FIG. 2. Distribution of 57 metastatic trophoblastic pulmonary nodules in 13 patients. Twenty-eight percent of the nodules were detected in the upper third of the lungs. 60 CANCER Junuurv 1 1984 Vol. 53
Discussion
Hydatidiform mole is an uncommon complication of pregnancy, occumng approximately once in 2000 preg-nancies in western countries. The placental villi become edematous and avascular, and there is abnormal tro-phoblastic proliferation, usually without an intact fetus. Invasive mole is invariably secondary to hydatidiform mole: gestational choriocarcinoma is usually but not al-ways secondary to mole.4 Trophoblastic cells typically spread to the lung and vaginal wall hematogenously; less often to the extrauterine soft tissues via lymphatics or blood vessels; and to the brain and liver via the blood stream.’ Following such spread, prior to the use of che-motherapy. the disease was fatal in 8 5 % to 95% of cases. The use of chemotherapy. pioneered by Li and associate^,^ has resulted in a dramatic reversal of the course of illness in most patients, so that approximately 90% of patients with metastatic trophoblastic disease are now cured.‘ Vi-able trophoblastic cells produce chorionic gonadotropins; HCG assays detect the presence of viable trophoblastic tissue in a patient, but do not indicate the location of the tissue. The plasma HCG level and the daily rate of urinary excretion correlate broadly with tumor size in patients with moles.’ Some degree of invasion occurred in approximately 70% of British women with primary hydatidiform moles in a follow-up study by Bagshawe and Walden.’ In most cases the invasive trophoblastic tissue regressed after cu-rettage: in 5% to 6% of patients, chemotherapy was re-quired for management of recurrence, invasion, or me-tastasis. Curry and associates,’ in a large study of American patients. found that 205% of patients with hydatidiform developed sequellae requiring chemotherapy. Metastases were detected in 4%, whereas 6% had rising HCG levels and 10% had prolonged elevation of HCG titers. Pul-monary metastases were found in 45% of 78 patients reviewed by Libshitz and coworkers.” Thirty-three pa-tients had well-defined pulmonary nodules, and 2 had “fluffy alveolar” opacities. Chest films and whole lung tomograms are helpful for defining the location and extent of metastases in patients whose HCG titers are persistently elevated or falling at a slower rate than expected after evacuation of hydati-diform mole. Prompt recognition of malignant sequellae and prompt institution of chemotherapy result in im-proved prognosi~.~,’ Necrotic tumor tissue and blood oc-cupy the central portions of pulmonary metastatic nod-ules.” The volume of a nodule may not be proportional to the number of viable trophoblastic cells in that nodule, because of variation in extent of necrosis and hemorrhage. The volumes of pulmonary nodules were not measured in this study, but it was noted that several patients with one or two large nodules had lower HCG titers than other patients with many smaller nodules. The number of nod-ules detected is possibly an underestimate of the number of actual metastases, since nodules smaller than 3 mm may not have been found on the tomograms. Despite this, it is plausible to expect a correlation between the number of radiographically detected pulmonary nodules with the HCG titer; the data are consistent with this re-lationship. The level of urinary HCG is not the most sensitive indicator of the presence of trophoblastic tissue but is reliable when elevated. One patient in this study had a normal 24-hour urinary HCG titer despite the presence of six pulmonary nodules. The Beta-subunit of HCG. determined in serum by radioimmunoassay, is more sen-sitive for detecting early metastasis. Pretreatment serum HCG titers were not available for many of the patients in this series. The presence of pulmonary nodules in a patient with previously treated metastatic trophoblastic disease does not necessarily indicate viable metastases, if the serum HCG titer is persistently normal.” Nodules may persist after eradication of viable trophoblastic tissue, due to fibrotic organization of residual hemorrhage and necrosis, and the nodules may calcify.’ I,’’ All pulmonary nodules regressed during chemotherapy in these patients, evidence that they were active metastases. The lung apices were involved almost as frequently
(28%) as the bases (33%) in these patients, differing from the typical lower zone predilection described for blood-borne metastases of other origins.13 The observation of a higher than expected frequency of apical metastases of gestational trophoblastic neoplasms has not previously been reported. Dissemination of trophoblastic cells may have occurred during curettage of some of the moles, when the patients were recumbent, and pulmonary blood flow was posturally redistributed to the upper portions of the lungs. Approximately uniform involvement of up-per, middle, and lower thirds of the lungs would be ex-pected if pulmonary spread occurred when patients were recumbent. Wagner has demonstrated that trophoblastic cells enter the maternal circulation during mechanical manipulation of the uterus (in both normal and abnormal pregnancies), and has shown trophoblastic spread to occur during curettage of a mole.14 REFERENCES I . Acosta-Sison H. The relative frequency of various anatomic sites as the point of first metastasis in 32 cases of chorionepithelioma. . A H I J Ohsrcr G j . n c ~ o / 1958: 75: I 149- I 152. 2. Bagshawe KD, Garnett BS. Radiological changes in the lungs of
patients with trophoblastic tumors. Br .l Radio/ 1963: 36:673-679.
No. I TROPHOBLASTIC TUMOR METASTASES - llcndin 61 3. Hertz R, Lewis J, Lipsett MB. Five years' experience with the chemotherapy of metastatic chonocarcinoma and related trophoblastic tumours in women. Am J Ohsic/ Gyni.col 196 1 ; 82:63 1-640. 4. Greenhill JP, Friedman EA. Biological Principles and Modern Practice of Obstetrics, chapter 53. Philadelphia: WB Saunders, 1974: 5. Li MC, Hertz R. Spencer DB. Effect of methotrexate therapy upon choriocarcinoma and chorioadenoma. Proc Soc Exp B i d M i d 1956; 6. Lewis JL. Treatment of metastatic gestational trophoblastic neo-plasms. Am J Ohsid Gyecol 1980; 136:163- 172. 7. Bagshawe KD, Walden PAM. Monitoring ofchonocarcinoma. Br
J Radio1 1976; 49:291-292. 8. Bagshawe KD, Wilson H, Dublon P, Smith A, Baldwin M, Kardana
A. Follow-up after hydatidiform mole: Studies using radioimmunoassay 556-566. 931361-366. for urinary human chorionic gonadotrophin (HCG). BrJ Oh.sic,i (;wwol.
Curry SL, Hammond CB, Tyrey L, Creasman WT, Parker RT. Hydatidiform mole: Diagnosis, management, and long-term follow-up of 347 patients. Ohsier Gyncwl 1975; 45: 1-8. 10. Libshitz HI, Baber CE, Hammond CB. The pulmonary metastases of choriocarcinoma. Ohslef Gynecol 1977; 49:4 12-4 16.
1
I . Swett HA, Westcott JL. Residual nonmalignant pulmonary nod-ules in choriocarcinoma. Che.s/ 1974; 65560-562. 12. Cockshott WP, Hendrickse J P de V. Pulmonary calcification at the site of trophoblastic metastases. Br J Radio1 1969: 42: 17-20,
Fraser RG, Pare JAP. Diagnosis of Disease of the Chest, vol. 4. Philadelphia: WB Saunders, 1979; 2198. 14. Wagner D. Trophoblastic cells in the blood stream in normal and abnormal pregnancy. Acia C j i i o l (Bu1iimorc.l 1968: 12: I 37- I 39. 1973; 80:461-468. |
4388 | https://www.physicsclassroom.com/class/waves/lesson-4/nodes-and-anti-nodes | Physics Tutorial: Nodes and Anti-nodes
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The Physics Classroom » Physics Tutorial » Vibrations and Waves » Nodes and Anti-nodes
Vibrations and Waves - Lesson 4 - Standing Waves
Nodes and Anti-nodes
Traveling Waves vs. Standing Waves
Formation of Standing Waves
Nodes and Anti-nodes
Harmonics and Patterns
Mathematics of Standing Waves
Getting your Trinity Audio player ready...
Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers.
As mentioned earlier in Lesson 4, a standing wave pattern is an interference phenomenon. It is formed as the result of the perfectly timed interference of two waves passing through the same medium. A standing wave pattern is not actually a wave; rather it is the pattern resulting from the presence of two waves of the same frequency with different directions of travel within the same medium.
What are Nodes and Antinodes?
One characteristic of every standing wave pattern is that there are points along the medium that appear to be standing still. These points, sometimes described as points of no displacement, are referred to as nodes. There are other points along the medium that undergo vibrations between a large positive and large negative displacement. These are the points that undergo the maximum displacement during each vibrational cycle of the standing wave. In a sense, these points are the opposite of nodes, and so they are called antinodes. A standing wave pattern always consists of an alternating pattern of nodes and antinodes. The animation shown below depicts a rope vibrating with a standing wave pattern. The nodes and antinodes are labeled on the diagram. When a standing wave pattern is established in a medium, the nodes and the antinodes are always located at the same position along the medium; they are standing still. It is this characteristic that has earned the pattern the name standing wav e.
Flickr Physics Photo
A standing wave is established upon a vibrating string using a harmonic oscillator and a frequency generator. A strobe is used to illuminate the string several times during each cycle. The finger is pointing at a nodal position.
Standing Wave Diagrams
The positioning of the nodes and antinodes in a standing wave pattern can be explained by focusing on the interference of the two waves. The nodes are produced at locations where destructive interference occurs. For instance, nodes form at locations where a crest of one wave meets a trough of a second wave; or a half-crest of one wave meets a half-trough of a second wave; or a quarter-crest of one wave meets a quarter-trough of a second wave; etc. Antinodes, on the other hand, are produced at locations where constructive interference occurs. For instance, if a crest of one wave meets a crest of a second wave, a point of large positive displacement results. Similarly, if a trough of one wave meets a trough of a second wave, a point of large negative displacement results. Antinodes are always vibrating back and forth between these points of large positive and large negative displacement; this is because during a complete cycle of vibration, a crest will meet a crest; and then one-half cycle later, a trough will meet a trough. Because antinodes are vibrating back and forth between a large positive and large negative displacement, a diagram of a standing wave is sometimes depicted by drawing the shape of the medium at an instant in time and at an instant one-half vibrational cycle later. This is done in the diagram below.
Nodes and antinodes should not be confused with crests and troughs. When the motion of a traveling wave is discussed, it is customary to refer to a point of large maximum displacement as a crest and a point of large negative displacement as a trough. These represent points of the disturbance that travel from one location to another through the medium. An antinode on the other hand is a point on the medium that is staying in the same location. Furthermore, an antinode vibrates back and forth between a large upward and a large downward displacement. And finally, nodes and antinodes are not actually part of a wave. Recall that a standing wave is not actually a wave but rather a pattern that results from the interference of two or more waves. Since a standing wave is not technically a wave, an antinode is not technically a point on a wave. The nodes and antinodes are merely unique points on the medium that make up the wave pattern.
Watch It!
A physics instructor demonstrates and explains the formation of a longitudinal standing wave in a spring.
We Would Like to Suggest ...
Why just read about it and when you could be interacting with it? Interact - that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of ourStanding Wave Patterns Interactive. You can find it in the Physics Interactives section of our website. TheStanding Wave Patterns Interactiveprovides the learner an environment for exploring the formation of standing waves, standing wave patterns, and mathematical relationships for standing wave patterns.
Visit: Standing Wave Patterns Interactive
Check Your Understanding
Suppose that there was a ride at an amusement park that was titled The Standing Wave. Which location - node or antinode - on the ride would give the greatest thrill?
See Answer
Answer: The antinode
The antinode is continually vibrating from a high to a low displacement - now that would be a ride.
A standing wave is formed when ____.
a. a wave refracts due to changes in the properties of the medium.
b. a wave reflects off a canyon wall and is heard shortly after it is formed.
c. red, orange, and yellow wavelengths bend around suspended atmospheric particles.
d. two identical waves moving different directions along the same medium interfere.
See Answer
Answer: D
If still uncertain, then review the previous page of Lesson 4.
The number of nodes in the standing wave shown in the diagram at the right is ____. > a. 6 b. 7 > c. 8 d. 14
See Answer
Answer: C (8 nodes)
There are eight positions along the medium which have no displacement. Be sure to avoid the common mistake of not counting the end positions.
The number of antinodes in the standing wave shown in the diagram above right is ____.
a. 6 b. 7 c. 8 d. 14
See Answer
Answer: B (7 antinodes)
There are seven positions along the medium which have vibrate between a large positive and a large negative displacement.
Be sure to avoid the common mistake of counting the antinodal positions twice. An antinode is simply a point along a medium which undergoes maximum displacement above and below the rest position. Do not count these positions twice.
Consider the standing wave pattern at the right in answering these next two questions.
The number of nodes in the entire pattern is ___. > a. 7 b. 8 > c. 9 d. 16 > > > See Answer > > Answer: C (9 nodes) > > > There are nine positions along the medium which have no displacement. (Be sure to avoid the common mistake of not counting the end positions.)
Of all the labeled points, destructive interference occurs at point(s) ____.
a. B, C, and D b. A, E, and F c. A only
d. C only e. all points
See Answer
Answer: A
Destructive interference has occurred at points B, C and D to produce the nodes which are seen at these points.
Next Section:
Harmonics and Patterns
Mathematics of Standing Waves
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4389 | https://books.google.com/books/about/Campbell_Walsh_Urology.html?id=fu3BBwAAQBAJ | Campbell-Walsh Urology: Expert Consult Premium Edition: Enhanced Online ... - Alan J. Wein, Louis R. Kavoussi, Andrew C. Novick, Alan W. Partin, Craig A. Peters - Google Books
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Campbell-Walsh Urology: Expert Consult Premium Edition: Enhanced Online Features and Print, 4-Volume Set ======================================================================================================== Alan J. Wein, Louis R. Kavoussi, Andrew C. Novick, Alan W. Partin, Craig A. Peters Elsevier Health Sciences, Aug 25, 2011 - Medical - 4320 pages Since 1954, Campbell-Walsh Urology has been internationally recognized as the pre-eminent text in its field. Edited by Alan J. Wein, MD, PhD(hon), Louis R. Kavoussi, MD, Alan W. Partin, MD, PhD, Craig A. Peters, MD, FACS, FAAP, and the late Andrew C. Novick, MD, it provides you with everything you need to know at every stage of your career, covering the entire breadth and depth of urology - from anatomy and physiology through the latest diagnostic approaches and medical and surgical treatments. Be certain with expert, dependable, accurate answers for every stage of your career from the most comprehensive, definitive text in the field! Required reading for all urology residents, Campbell-Walsh Urology is the predominant reference used by The American Board of Urology for its board examination questions. Visually grasp and better understand critical information with the aid of algorithms, photographs, radiographs, and line drawings to illustrate essential concepts, nuances of clinical presentation and technique, and decision making. Stay on the cutting edge with online updates. Get trusted perspectives and insights from hundreds of well-respected global contributors, all of whom are at the top and the cutting edge of their respective fields. Stay current with the latest knowledge and practices. Brand-new chapters and comprehensive updates throughout include new information on perioperative care in adults and children, premature ejaculation, retroperitoneal tumors, nocturia, and more! Meticulously revised chapters cover the most recent advancements in robotic and laparoscopic bladder surgery, open surgery of the kidney, management of metastic and invasive bladder cancer, and many other hot topics!Reference information quickly thanks to a new, streamlined print format and easily searchable online access to supplemental figures, tables, additional references, and expanded discussions as well as procedural videos and more at www.expertconsult.com. The new edition of Campbell-Walsh Urology is the must have reference for practitioners and residents! More » Preview this book »
Contents
Videos xxxiii
VIII Renal Physiology and Pathophysiology 1023
IX Upper Urinary Tract Obstruction and Trauma 1086
X Renal Failure and Transplantation 1192
XI Urinary Lithiasis and Endourology 1256
XII Neoplasms of the Upper Urinary Tract 1412
XIII The Adrenals 1684
Index 101
Copyright page iv
Dedication v
Contributors vii
Preface xxv
Table of Contentsxxvii
Videos xxxiii
XIV Urine Transport Storage and Emptying 1753
XV Benign and Malignant Bladder Disorders 2308
More
Online Index E-47
Endsheet2 E-143
Endsheet3 E-144
Front cover E-145
Half title page i
Editors ii
CAMPBELLWALSH Urology iii
XVI Prostate 2531
Index101
Online Index E-47
Endsheet2 E-143
Endsheet3 E-144
Copyright
Less
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Campbell-Walsh Urology
Alan J. Wein,Louis R. Kavoussi,Andrew C. Novick,Alan W. Partin,Craig A. Peters
Limited preview - 2011
Common terms and phrases
acidosisacuteaneurysmangiotensinassociatedbladderbloodcalcium oxalatecalycescathetercellschroniccitrateClinclinicalcolleaguescollecting systemcomplicationscreatininecystinedecreaseDepartment of UrologydiagnosisdialysisdilationdistaldiureticdonoreffectendopyelotomyEndourolevaluationexcretionfactorsglomerularholmiumhorseshoe kidneyhydronephrosishypercalciuriahypertensionimagingincreasedinfectioninhibitorsintravenousKidney Intlaparoscopiclaserlithotripsylower poleMedicalmetabolicNephrolnephrolithiasisnephropathynephrostomynephrostomy tubenormalpatientspelvicpercutaneouspercutaneous accesspercutaneous nephrolithotomyphosphatePhysiolpostoperativepotassiumpressureprocedureproteinproximalpyeloplastyreabsorptionreceptorrecurrentrenal artery stenosisrenal failurerenal functionrenal pelvisrenal transplantationrenal tubularreninrenovascularreportedretrograderetroperitoneal fibrosisriskserumshockwavesodiumstone diseasestone formationsurgerysurgicaltechniquetherapytiontraumatreatmenttubuleupper urinary tractureterureteral injuryureteral obstructionureteral stentureteral stricturesureteroscopicuric acidurineUrolUrologyvascular
About the author(2011)
Alan J. Wein, MDProfessor and Chair, Division of UrologyUniversity of Pennsylvania Health System Louis R. Kavoussi, MD, MBA, is Chairman of Urology for Northwell Health and Waldbaum-Gardner Distinguished Professor of Urology at the newly established Zucker School of Medicine. He heads the Arthur Smith Institute for Urology, which is dedicated to the treatment of urological disease through innovative surgical procedures, diagnostics and medical care Alan W. Partin, MD, PhDDavid Hall McConnell Professor and ChairUrologist -in-Chief, Department of Urology, OncologyThe Johns Hopkins Medical Institutions Craig A. Peters, MDChief, Division of Surgical Innovation, Technology and TranslationPrincipal Investigator, Sheikh Zayed Institute for Pediatric Surgical InnovationChildren's National Medical CenterProfessor of Urology and Pediatrics
Bibliographic information
Title Campbell-Walsh Urology: Expert Consult Premium Edition: Enhanced Online Features and Print, 4-Volume Set
Campbell's Urology
Saunders W.B
AuthorsAlan J. Wein, Louis R. Kavoussi, Andrew C. Novick, Alan W. Partin, Craig A. Peters
Edition illustrated, reprint
Publisher Elsevier Health Sciences, 2011
ISBN 1416069119, 9781416069119
Length 4320 pages
SubjectsMedical
›
Urology
Medical / Urology
Export CitationBiBTeXEndNoteRefMan
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4390 | https://www.analog.com/media/en/training-seminars/tutorials/mt-031.pdf | Rev.A, 10/08, WK Page 1 of 17 MT-031 TUTORIAL Grounding Data Converters and Solving the Mystery of "AGND" and "DGND" by Walt Kester, James Bryant, and Mike Byrne INTRODUCTION Today's signal processing systems generally require mixed-signal devices such as analog-to-digital converters (ADCs) and digital-to-analog converters (DACs) as well as fast digital signal processors (DSPs). Requirements for processing analog signals having wide dynamic ranges increases the importance of high performance ADCs and DACs. Maintaining wide dynamic range with low noise in hostile digital environments is dependent upon using good high-speed circuit design techniques including proper signal routing, decoupling, and grounding. In the past, "high precision, low-speed" circuits have generally been viewed differently than so-called "high-speed" circuits. With respect to ADCs and DACs, the sampling (or update) frequency has generally been used as the distinguishing speed criteria. However, the following two examples show that in practice, most of today's signal processing ICs are really "high-speed," and must therefore be treated as such in order to maintain high performance. This is certainly true of DSPs, and also true of ADCs and DACs. All sampling ADCs (ADCs with an internal sample-and-hold circuit) suitable for signal processing applications operate with relatively high speed clocks with fast rise and fall times (generally a few nanoseconds) and must be treated as high speed devices, even though throughput rates may appear low. For example, a medium-speed 12-bit successive approximation (SAR) ADC may operate on a 10-MHz internal clock, while the sampling rate is only 500 kSPS. Sigma-delta (Σ-Δ) ADCs also require high speed clocks because of their high oversampling ratios. Even high resolution, so-called "low frequency" Σ-Δ industrial measurement ADCs (having throughputs of 10 Hz to 7.5 kHz) operate on 5-MHz or higher clocks and offer resolution to 24-bits (for example, the Analog Devices AD77xx-series). To further complicate the issue, mixed-signal ICs have both analog and digital ports, and because of this, much confusion has resulted with respect to proper grounding techniques. In addition, some mixed-signal ICs have relatively low digital currents, while others have high digital currents. In many cases, these two types must be treated differently with respect to optimum grounding. Digital and analog design engineers tend to view mixed-signal devices from different perspectives, and the purpose of this tutorial is to develop a general grounding philosophy that will work for most mixed signal devices, without having to know the specific details of their internal circuits. MT-031 GROUND AND POWER PLANES The importance of maintaining a low impedance large area ground plane is critical to all analog and digital circuits today. The ground plane not only acts as a low impedance return path for decoupling high frequency currents (caused by fast digital logic) but also minimizes EMI/RFI emissions. Because of the shielding action of the ground plane, the circuit's susceptibility to external EMI/RFI is also reduced. Ground planes also allow the transmission of high speed digital or analog signals using transmission line techniques (microstrip or stripline) where controlled impedances are required. The use of "buss wire" is totally unacceptable as a "ground" because of its impedance at the equivalent frequency of most logic transitions. For instance, #22 gauge wire has about 20 nH/inch inductance. A transient current having a slew rate of 10 mA/ns created by a logic signal would develop an unwanted voltage drop of 200 mV at this frequency flowing through 1 inch of this wire: .
mV 200 ns mA 10 nH 20 t i L v = × = Δ Δ = Δ Eq. 1 For a signal having a 2-V peak-to-peak range, this translates into an error of about 10% (approximately 3.5-bit accuracy). Even in all-digital circuits, this error would result in considerable degradation of logic noise margins. Figure 1 shows the classic illustration of a situation where the digital return current modulates the analog return current (top figure). The ground return wire inductance and resistance is shared between the analog and digital circuits, and this is what causes the interaction and resulting error. A possible solution is to make the digital return current path flow directly to the GND REF as shown in the bottom figure. This is the fundamental concept of a "star," or single-point ground system. Implementing the true single-point ground in a system which contains multiple high frequency return paths is difficult because the physical length of the individual return current wires will introduce parasitic resistance and inductance which can make obtaining a low impedance high frequency ground difficult. In practice, the current returns must consist of large area ground planes for low impedance to high frequency currents. Without a low-impedance ground plane, it is therefore almost impossible to avoid these shared impedances, especially at high frequencies. Page 2 of 17 MT-031 ANALOG CIRCUITS DIGITAL CIRCUITS ANALOG CIRCUITS DIGITAL CIRCUITS VD VD VA VA + + + + ID IA ID IA + ID VIN VIN ID IA ID IA GND REF GND REF INCORRECT CORRECT Figure 1: Digital Currents Flowing in Analog Return Path Create Error Voltages All integrated circuit ground pins should be soldered directly to the low-impedance ground plane to minimize series inductance and resistance. The use of traditional IC sockets is not recommended with high-speed devices. The extra inductance and capacitance of even "low profile" sockets may corrupt the device performance by introducing unwanted shared paths. If sockets must be used with DIP packages, as in prototyping, individual "pin sockets" or "cage jacks" may be acceptable. Both capped and uncapped versions of these pin sockets are available (AMP part numbers 5-330808-3, and 5-330808-6). They have spring-loaded gold contacts which make good electrical and mechanical connection to the IC pins. Multiple insertions, however, may degrade their performance. LOW AND HIGH FREQUENCY DECOUPLING Each power supply should be decoupled to the low-impedance ground plane with a high quality electrolytic capacitor at the point it enters the PC board. This minimizes low frequency noise on the supply runs. At each individual analog stage, further local, high-frequency-only filtering is required at the individual IC package power pins. Figure 2 shows this technique, in both correct (left) as well as incorrect example implementations (right). In the left example, a typical 0.1-μF chip ceramic capacitor goes directly to the opposite PCB side ground plane, by virtue of the via, and on to the IC's GND pin by a second via. In contrast, the less desirable setup at the right adds additional PCB trace inductance in the ground path of the decoupling cap, reducing effectiveness. Page 3 of 17 MT-031 V+ GND VIAS TO GROUND PLANE DECOUPLING CAPACITOR V+ GND DECOUPLING CAPACITOR VIA TO GROUND PLANE PCB TRACE IC IC POWER SUPPLY TRACE POWER SUPPLY TRACE CORRECT INCORRECT OPTIONAL FERRITE BEADS Figure 2: Localized High Frequency Supply Filter(s) Provides Optimum Filtering and Decoupling Via Short Low-Inductance Path (Ground Plane) All high frequency (i.e., ≥10 MHz) ICs should use a bypassing scheme similar to Figure 2 for best performance. The ferrite beads aren't 100% necessary, but they will add extra high frequency noise isolation and decoupling, which is often desirable. Possible caveats here would be to verify that the beads never saturate, when the ICs are handling high currents. Note that with some ferrites, even before full saturation occurs, some beads can be non-linear, so if a power stage is required to operate with a low distortion output, this should also be checked. DOUBLE-SIDED VS. MULTILAYER PRINTED CIRCUIT BOARDS Each PCB in the system should have at least one complete layer dedicated to the ground plane. Ideally, a double-sided board should have one side completely dedicated to ground and the other side for interconnections. In practice, this is not possible, since some of the ground plane will certainly have to be removed to allow for signal and power crossovers, vias, and through-holes. Nevertheless, as much area as possible should be preserved, and at least 75% should remain. After completing an initial layout, the ground layer should be checked carefully to make sure there are no isolated ground "islands," because IC ground pins located in a ground "island" have no current return path to the ground plane. Also, the ground plane should be checked for "skinny" connections between adjacent large areas which may significantly reduce the effectiveness of the ground plane. Needless to say, auto-routing board layout techniques will generally lead to a layout disaster on a mixed-signal board, so manual intervention is highly recommended. Systems that are densely packed with surface mount ICs will have a large number of interconnections; therefore multilayer boards are mandatory. This allows at least one complete layer to be dedicated to ground. A simple 4-layer board would have internal ground and power plane layers with the outer two layers used for interconnections between the surface mount components. Placing the power and ground planes adjacent to each other provides additional Page 4 of 17 MT-031 inter-plane capacitance which helps high frequency decoupling of the power supply. In most systems, 4-layers are not enough, and additional layers are required for routing signals as well as power. MULTICARD MIXED-SIGNAL SYSTEMS The best way of minimizing ground impedance in a multicard system is to use a "motherboard" PCB as a backplane for interconnections between cards, thus providing a continuous ground plane to the backplane. The PCB connector should have at least 30-40% of its pins devoted to ground, and these pins should be connected to the ground plane on the backplane motherboard. To complete the overall system grounding scheme there are two possibilities: 1. The backplane ground plane can be connected to chassis ground at numerous points, thereby diffusing the various ground current return paths. This is commonly referred to as a "multipoint" grounding system and is shown in Figure 3. 2. The ground plane can be connected to a single system "star ground" point (generally at the power supply). The first approach is most often used in all-digital systems, but can be used in mixed-signal systems provided the ground currents due to digital circuits are sufficiently low and diffused over a large area. The low ground impedance is maintained all the way through the PC boards, the backplane, and ultimately the chassis. However, it is critical that good electrical contact be made where the grounds are connected to the sheet metal chassis. This requires self-tapping sheet metal screws or "biting" washers. Special care must be taken where anodized aluminum is used for the chassis material, since its surface acts as an insulator. POWER SUPPLIES GROUND PLANE VA VD VA VD GROUND PLANE BACKPLANE PCB GROUND PLANE VA VD PCB CHASSIS GROUND Figure 3: Multipoint Ground Concept Page 5 of 17 MT-031 The second approach ("star ground") is often used in high speed mixed-signal systems having separate analog and digital ground systems and warrants further discussion. SEPARATING ANALOG AND DIGITAL GROUND PLANES In mixed-signal systems with large amounts of digital circuitry, it is highly desirable to physically separate sensitive analog components from noisy digital components. It may also be beneficial to use separate ground planes for the analog and the digital circuitry. These planes should not overlap in order to minimize capacitive coupling between the two. The separate analog and digital ground planes are continued on the backplane using either motherboard ground planes or "ground screens" which are made up of a series of wired interconnections between the connector ground pins. The arrangement shown in Figure 4 illustrates that the two planes are kept separate all the way back to a common system "star" ground, generally located at the power supplies. The connections between the ground planes, the power supplies, and the "star" should be made up of multiple bus bars or wide copper braids for minimum resistance and inductance. The back-to-back Schottky diodes on each PCB are inserted to prevent accidental dc voltage from developing between the two ground systems when cards are plugged and unplugged. This voltage should be kept less than 300 mV to prevent damage to ICs which have connections to both the analog and digital ground planes. Schottky diodes are preferable because of their low capacitance and low forward voltage drop. The low capacitance prevents ac coupling between the analog and digital ground planes. Schottky diodes begin to conduct at about 300 mV, and several parallel diodes in parallel may be required if high currents are expected. In some cases, ferrite beads can be used instead of Schottky diodes, however they introduce dc ground loops which can be troublesome in precision systems. POWER SUPPLIES ANALOG GROUND PLANE DIGITAL GROUND PLANE A D ANALOG GROUND PLANE DIGITAL GROUND PLANE A D VA VD VA VA VD VD ANALOG GROUND PLANE DIGITAL GROUND PLANE BACKPLANE PCB PCB SYSTEM STAR GROUND Figure 4: Separating Analog and Digital Ground Planes Page 6 of 17 MT-031 It is mandatory that the impedance of the ground planes be kept as low as possible, all the way back to the system star ground. DC or ac voltages of more than 300 mV between the two ground planes can not only damage ICs but cause false triggering of logic gates and possible latchup. GROUNDING AND DECOUPLING MIXED-SIGNAL ICs WITH LOW DIGITALCURRENTS Sensitive analog components such as amplifiers and voltage references are always referenced and decoupled to the analog ground plane. The ADCs and DACs (and other mixed-signal ICs) with low digital currents should generally be treated as analog components and also grounded and decoupled to the analog ground plane. At first glance, this may seem somewhat contradictory, since a converter has an analog and digital interface and usually has pins designated as analog ground (AGND) and digital ground (DGND). The diagram shown in Figure 5 will help to explain this seeming dilemma. ANALOG CIRCUITS DIGITAL CIRCUITS BUFFER GATE OR REGISTER VA A B VD CSTRAY CSTRAY R A A A D D VNOISE VA AIN/ OUT AGND DGND DATA BUS FERRITE BEAD DATA VD A = ANALOG GROUND PLANE D = DIGITAL GROUND PLANE CIN ≈10pF LP LP LP LP RP RP RP RP SHORT CONNECTIONS IA ID SEE TEXT Figure 5: Proper Grounding of Mixed-signal ICs With Low Internal Digital Currents Inside an IC that has both analog and digital circuits, such as an ADC or a DAC, the grounds are usually kept separate to avoid coupling digital signals into the analog circuits. Figure 5 shows a simple model of a converter. There is nothing the IC designer can do about the wirebond inductance and resistance associated with connecting the bond pads on the chip to the package pins except to realize it's there. The rapidly changing digital currents produce a voltage at point B which will inevitably couple into point A of the analog circuits through the stray capacitance, CSTRAY. In addition, there is approximately 0.2-pF unavoidable stray capacitance between every pin of the IC package! It's the IC designer's job to make the chip work in spite of this. However, in order to prevent further coupling, the AGND and DGND pins should be joined Page 7 of 17 MT-031 together externally to the analog ground plane with minimum lead lengths. Any extra impedance in the DGND connection will cause more digital noise to be developed at point B; it will, in turn, couple more digital noise into the analog circuit through the stray capacitance. Note that connecting DGND to the digital ground plane applies VNOISE across the AGND and DGND pins and invites disaster! The name "DGND" on an IC tells us that this pin connects to the digital ground of the IC. This does not imply that this pin must be connected to the digital ground of the system. It is true that this arrangement may inject a small amount of digital noise onto the analog ground plane. These currents should be quite small, and can be minimized by ensuring that the converter output does not drive a large fanout (they normally can't, by design). Minimizing the fanout on the converter's digital port will also keep the converter logic transitions relatively free from ringing and minimize digital switching currents, and thereby reducing any potential coupling into the analog port of the converter. The logic supply pin (VD) can be further isolated from the analog supply by the insertion of a small lossy ferrite bead as shown in Figure 5. The internal transient digital currents of the converter will flow in the small loop from VD through the decoupling capacitor and to DGND (this path is shown with a heavy line on the diagram). The transient digital currents will therefore not appear on the external analog ground plane, but are confined to the loop. The VD pin decoupling capacitor should be mounted as close to the converter as possible to minimize parasitic inductance. These decoupling capacitors should be low inductance ceramic types, typically between 0.01 µF and 0.1 µF. TREAT THE ADC DIGITAL OUTPUTS WITH CARE It is always a good idea (as shown in Figure 5) to place a buffer register adjacent to the converter to isolate the converter's digital lines from noise on the data bus. The register also serves to minimize loading on the digital outputs of the converter and acts as a Faraday shield between the digital outputs and the data bus. Even though many converters have three-state outputs/inputs, this isolation register still represents good design practice. In some cases it may be desirable to add an additional buffer register on the analog ground plane next to the converter output to provide greater isolation. The series resistors (labeled "R" in Figure 5) between the ADC output and the buffer register input help to minimize the digital transient currents which may affect converter performance. The resistors isolate the digital output drivers from the capacitance of the buffer register inputs. In addition, the RC network formed by the series resistor and the buffer register input capacitance acts as a lowpass filter to slow down the fast edges. A typical CMOS gate combined with PCB trace and a through-hole will create a load of approximately 10 pF. A logic output slew rate of 1 V/ns will produce 10 mA of dynamic current if there is no isolation resistor: mA 10 ns V 1 pF 10 t v C I = × = Δ Δ = Δ . Eq. 2 Page 8 of 17 MT-031 A 500-Ω series resistors will minimize this output current and result in a rise and fall time of approximately 11 ns when driving the 10-pF input capacitance of the register: .
ns 11 pF 10 500 2 .
2 C R 2 .
2 2 .
2 tr = × Ω × = ⋅ × = τ × = Eq. 3 TTL registers should be avoided, since they can appreciably add to the dynamic switching currents because of their higher input capacitance. The buffer register and other digital circuits should be grounded and decoupled to the digital ground plane of the PC board. Notice that any noise between the analog and digital ground plane reduces the noise margin at the converter digital interface. Since digital noise immunity is of the orders of hundreds or thousands of millivolts, this is unlikely to matter. The analog ground plane will generally not be very noisy, but if the noise on the digital ground plane (relative to the analog ground plane) exceeds a few hundred millivolts, then steps should be taken to reduce the digital ground plane impedance, thereby maintaining the digital noise margins at an acceptable level. Under no circumstances should the voltage between the two ground planes exceed 300 mV, or the ICs may be damaged. Separate power supplies for analog and digital circuits are also highly desirable, even if the voltages are the same. The analog supply should be used to power the converter. If the converter has a pin designated as a digital supply pin (VD), it should either be powered from a separate analog supply, or filtered as shown in the diagram. All converter power pins should be decoupled to the analog ground plane, and all logic circuit power pins should be decoupled to the digital ground plane as shown in Figure 6. AMP VA VD VA A A AGND DGND ADC OR DAC VA A VOLTAGE REFERENCE VA A SAMPLING CLOCK GENERATOR A A VA A BUFFER GATE OR REGISTER VD D D A A R R A ANALOG GROUND PLANE D DIGITAL GROUND PLANE TO OTHER DIGITAL CIRCUITS FERRITE BEAD SEE TEXT Figure 6: Grounding and Decoupling Points Page 9 of 17 MT-031 In some cases it may not be possible to connect VD to the analog supply. Some of the newer, high speed ICs may have their analog circuits powered by +5 V, but the digital interface powered by +3 V to interface to 3 V logic. In this case, the +3 V pin of the IC should be decoupled directly to the analog ground plane. It is also advisable to connect a ferrite bead in series with the power trace that connects the pin to the +3 V digital logic supply. The sampling clock generation circuitry should be treated like analog circuitry and also be grounded and heavily-decoupled to the analog ground plane. Phase noise on the sampling clock produces degradation in system SNR as will be discussed shortly. SAMPLING CLOCK CONSIDERATIONS In a high performance sampled data system a low phase-noise oscillator should be used to generate the ADC (or DAC) sampling clock because sampling clock jitter modulates the analog input/output signal and raises the noise and distortion floor. The sampling clock generator should be isolated from noisy digital circuits and grounded and decoupled to the analog ground plane, as is true for the op amp and the ADC. The effect of sampling clock jitter on ADC signal-to-noise ratio (SNR) is given approximately by the equation: ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ π = j 10 ft 2 1 log 20 SNR , Eq. 4 where SNR is the SNR of a perfect ADC of infinite resolution where the only source of noise is that caused by the rms sampling clock jitter, tj. Note that f in the above equation is the analog input frequency. Just working through a simple example, if tj = 50 ps rms, f = 100 kHz, then SNR = 90 dB, equivalent to about 15-bit dynamic range. This effect of clock jitter on SNR is discussed in much more detail in Tutorial MT-007. It should be noted that tj in the above example is the root-sum-square (rss) value of the external clock jitter and the internal ADC clock jitter (called aperture jitter). However, in most high performance ADCs, the internal aperture jitter is negligible compared to the jitter on the sampling clock. Ideally, the sampling clock oscillator should be referenced to the analog ground plane in a split-ground system. However, this is not always possible because of system constraints. In many cases, the sampling clock must be derived from a higher frequency multi-purpose system clock which is generated on the digital ground plane. It must then pass from its origin on the digital ground plane to the ADC on the analog ground plane. Ground noise between the two planes adds directly to the clock signal and will produce excess jitter. The jitter can cause degradation in the signal-to-noise ratio and also produce unwanted harmonics. Page 10 of 17 MT-031 This can be remedied somewhat by transmitting the sampling clock signal as a differential signal using either a small RF transformer as shown in Figure 7 or a high speed differential driver and receiver IC. Many high-speed ADCs have differential sampling clock inputs to facilitate this approach. If an active differential driver and receiver are used, they should be ECL, low-level ECL, or LVDS to minimize phase jitter. In a single +5 V supply system, ECL logic can be connected between ground and +5 V (PECL), and the outputs ac coupled into the ADC sampling clock input. In either case, the original master system clock must be generated from a low phase noise oscillator, and not the clock output of a DSP, microprocessor, or microcontroller. In order to facilitate system clock management, a family clock generation and distribution products is available from Analog Devices as well as a complete selection of phase-locked loops (PLLs). LOW PHASE NOISE MASTER CLOCK SYSTEM CLOCK GENERATORS DSP OR MICROPROCESSOR D A D A VD VA VD VD VD D D D + _ SAMPLING CLOCK SAMPLING CLOCK METHOD 1 METHOD 2 DIGITAL GROUND PLANE ANALOG GROUND PLANE 1 2π f tj SNR = 20 log10 1 2π f tj SNR = 20 log10 tj = Sampling Clock Jitter f = Analog Input Frequency Figure 7: Sampling Clock Distribution From Digital to Analog Ground Planes THE ORIGINS OF THE CONFUSION ABOUT MIXED-SIGNAL GROUNDING: APPLYING SINGLE-CARD GROUNDING CONCEPTS TO MULTICARD SYSTEMS Most ADC, DAC, and other mixed-signal device data sheets discuss grounding relative to a single PCB, usually the manufacturer's own evaluation board. This has been a source of confusion when trying to apply these principles to multicard or multi-ADC/DAC systems. The recommendation is usually to split the PCB ground plane into an analog plane and a digital plane. It is then further recommended that the AGND and DGND pins of a converter be tied together and that the analog ground plane and digital ground planes be connected at that same point as shown in Figure 8. This essentially creates the system "star" ground at the mixed-signal device. Page 11 of 17 MT-031 All noisy digital currents flow through the digital power supply to the digital ground plane and back to the digital supply; they are isolated from the sensitive analog portion of the board. The system star ground occurs where the analog and digital ground planes are joined together at the mixed signal device. While this approach will generally work in a simple system with a single PCB and single ADC/DAC, it is not usually optimum for multicard mixed-signal systems. In systems having several ADCs or DACs on different PCBs (or on the same PCB, for that matter), the analog and digital ground planes become connected at several points, creating the possibility of ground loops and making a single-point "star" ground system impossible. For these reasons, this grounding approach is not recommended for multicard systems, and the approach previously discussed should be used for mixed signal ICs with low digital currents. ANALOG CIRCUITS DIGITAL CIRCUITS A A D D D VA VD ANALOG GROUND PLANE DIGITAL GROUND PLANE AGND DGND MIXED SIGNAL DEVICE A DIGITAL SUPPLY ANALOG SUPPLY SYSTEM STAR GROUND VA VD Figure 8: Grounding Mixed Signal ICs : Single PC Board (Typical Evaluation/Test Board) SUMMARY: GROUNDING MIXED SIGNAL DEVICES WITH LOW DIGITAL CURRENTS IN A MULTICARD SYSTEM Figure 9 summarizes the approach previously described for grounding a mixed signal device which has low digital currents. The analog ground plane is not corrupted because the small digital transient currents flow in the small loop between VD, the decoupling capacitor, and DGND (shown as a heavy line). The mixed signal device is for all intents and purposes treated as an analog component. The noise VN between the ground planes reduces the noise margin at the digital interface but is generally not harmful if kept less than 300 mV by using a low impedance digital ground plane all the way back to the system star ground. Page 12 of 17 MT-031 However, mixed signal devices such as sigma-delta ADCs, codecs, and DSPs with on-chip analog functions are becoming more and more digitally intensive. Along with the additional digital circuitry come larger digital currents and noise. For example, a sigma-delta ADC or DAC contains a complex digital filter which adds considerably to the digital current in the device. The method previously discussed depends on the decoupling capacitor between VD and DGND to keep the digital transient currents isolated in a small loop. However, if the digital currents are significant enough and have components at dc or low frequencies, the decoupling capacitor may have to be so large that it is impractical. Any digital current which flows outside the loop between VD and DGND must flow through the analog ground plane. This may degrade performance, especially in high resolution systems. ANALOG CIRCUITS DIGITAL CIRCUITS A A D D VA VD ANALOG GROUND PLANE DIGITAL GROUND PLANE MIXED SIGNAL DEVICE AGND DGND A A D D TO SYSTEM STAR GROUND TO SYSTEM DIGITAL SUPPLY TO SYSTEM ANALOG SUPPLY A VA VD BUFFER LATCH FILTER VN BUS R VN = NOISE BETWEEN GROUND PLANES Figure 9: Grounding Mixed Signal ICs with Low Internal Digital Currents: Multiple PC Boards It is difficult to predict what level of digital current flowing into the analog ground plane will become unacceptable in a system. All we can do at this point is to suggest an alternative grounding method which may yield better performance. SUMMARY: GROUNDING MIXED SIGNAL DEVICES WITH HIGH DIGITAL CURRENTS IN A MULTICARD SYSTEM (USE THIS METHOD WITH CAUTION!) An alternative grounding method for a mixed signal device with high levels of digital currents is shown in Figure 10. The AGND of the mixed signal device is connected to the analog ground plane, and the DGND of the device is connected to the digital ground plane. The digital currents are isolated from the analog ground plane, but the noise between the two ground planes is applied directly between the AGND and DGND pins of the device. For this method to be successful, the analog and digital circuits within the mixed signal device must be well isolated. Page 13 of 17 MT-031 The noise between AGND and DGND pins must not be large enough to reduce internal noise margins or cause corruption of the internal analog circuits. Figure 10 shows optional Schottky diodes (back-to-back) or a ferrite bead connecting the analog and digital ground planes. The Schottky diodes prevent large dc voltages or low frequency voltage spikes from developing across the two planes. These voltages can potentially damage the mixed signal IC if they exceed 300 mV because they appear directly between the AGND and DGND pins. As an alternative to the back-to-back Schottky diodes, a ferrite bead provides a dc connection between the two planes but isolates them at frequencies above a few MHz where the ferrite bead becomes resistive. This protects the IC from dc voltages between AGND and DGND, but the dc connection provided by the ferrite bead can introduce unwanted dc ground loops and may not be suitable for high resolution systems. ANALOG CIRCUITS DIGITAL CIRCUITS A A D D VA VD ANALOG GROUND PLANE DIGITAL GROUND PLANE MIXED SIGNAL DEVICE AGND DGND A A D D BACK-TO-BACK SCHOTTKY DIODES OR FERRITE BEAD (SEE TEXT) TO SYSTEM STAR GROUND TO SYSTEM DIGITAL SUPPLY TO SYSTEM ANALOG SUPPLY VA VD VN VN = NOISE BETWEEN GROUND PLANES Figure 10: Grounding Alternative for Mixed-Signal ICs with High Digital Currents: Multiple PC Boards Whenever AGND and DGND pins are separated in the special case of ICs with high digital currents, provisions should be made to connect them together if necessary. Jumpers and/or strap options allow both methods to be tried to verify which gives the best overall performance in the system. GROUNDING SUMMARY There is no single grounding method which will guarantee optimum performance 100% of the time! This section has presented a number of possible options depending upon the characteristics of the particular mixed signal devices in question. It is helpful, however, to provide for as many options as possible when laying out the initial PC board. Page 14 of 17 MT-031 It is mandatory that at least one layer of the PC board be dedicated to ground plane! The initial board layout should provide for non-overlapping analog and digital ground planes, but pads and vias should be provided at several locations for the installation of back-to-back Schottky diodes or ferrite beads, if required. It is also extremely important that pads and vias be provided so that the analog and digital ground planes can be connected together with jumpers if required. It is difficult to predict whether the "multi-point" (single ground plane) or the "star" ground (separate analog and digital ground planes) method will give best overall system performance; therefore, some experimentation with the final PC board using the jumpers may be required. When in doubt, it is always better to start out with a split analog and digital ground plane and later connect them with jumpers, rather than to start out with a single ground plane and try and later try and split it! SOME GENERAL PC BOARD LAYOUT GUIDELINES FOR MIXED-SIGNAL SYSTEMS It is evident that noise can be minimized by paying attention to the system layout and preventing different signals from interfering with each other. High level analog signals should be separated from low level analog signals, and both should be kept away from digital signals. We have seen elsewhere that in waveform sampling and reconstruction systems the sampling clock (which is a digital signal) is as vulnerable to noise as any analog signal, but is as liable to cause noise as any digital signal, and so must be kept isolated from both analog and digital systems. If clock driver packages are used in clock distribution, only one frequency clock should be passed through a single package. Sharing drivers between clocks of different frequencies in the same package will produce excess jitter and crosstalk and degrade performance. The ground plane can act as a shield where sensitive signals cross. Figure 11 shows a good layout for a data acquisition board where all sensitive areas are isolated from each other and signal paths are kept as short as possible. While real life is rarely as tidy as this, the principle remains a valid one. There are a number of important points to be considered when making signal and power connections. First of all a connector is one of the few places in the system where all signal conductors must run in parallel—it is therefore imperative to separate them with ground pins (creating a faraday shield) to reduce coupling between them. Multiple ground pins are important for another reason: they keep down the ground impedance at the junction between the board and the backplane. The contact resistance of a single pin of a PCB connector is quite low (of the order of 10 mΩ) when the board is new—as the board gets older the contact resistance is likely to rise, and the board's performance may be compromised. It is therefore well worthwhile to allocate extra PCB connector pins so that there are many ground connections (perhaps 30-40% of all the pins on the PCB connector should be ground pins). For similar reasons there should be several pins for each power connection, although there is no need to have as many as there are ground pins. Page 15 of 17 MT-031 REFERENCE ADC FILTER AMPLIFIER SAMPLING CLOCK GENERATOR TIMING CIRCUITS BUFFER REGISTER DSP OR µP CONTROL LOGIC DEMULTIPLEXER BUFFER MEMORY POWER ANALOG INPUT MULTIPLE GROUNDS DATA BUS ADDRESS BUS MULTIPLE GROUNDS ANALOG DIGITAL Figure 11: Analog and Digital Circuits Should be Partitioned on PCB Layout Analog Devices and other manufacturers of high performance mixed-signal ICs offer evaluation boards to assist customers in their initial evaluations and layout. ADC evaluation boards generally contain an on-board low-jitter sampling clock oscillator, output registers, and appropriate power and signal connectors. They also may have additional support circuitry such as the ADC input buffer amplifier and external reference. The layout of the evaluation board is optimized in terms of grounding, decoupling, and signal routing and can be used as a model when laying out the ADC PC board in the system. The actual evaluation board layout is usually available from the ADC manufacturer in the form of computer CAD files (Gerber files). In many cases, the layout of the various layers appears on the data sheet for the device. Page 16 of 17 Page 17 of 17 MT-031 REFERENCES 1. Ralph Morrison, Grounding and Shielding Techniques, 4th Edition, John Wiley, Inc., 1998, ISBN: 0471245186. 2. Henry W. Ott, Noise Reduction Techniques in Electronic Systems, 2nd Edition, John Wiley, Inc., 1988, ISBN: 0-471-85068-3. 3. Paul Brokaw, "An IC Amplifier User's Guide to Decoupling, Grounding and Making Things Go Right for a Change", Analog Devices Application Note AN-202. 4. Paul Brokaw and Jeff Barrow, "Grounding for Low- and High-Frequency Circuits," Analog Devices Application Note AN-345. 5. Howard W. Johnson and Martin Graham, High-Speed Digital Design, PTR Prentice Hall, 1993, ISBN: 0133957241. 6. Ralph Morrison, Solving Interference Problems in Electronics, John Wiley, 1995. 7. Crystal Oscillators: MF Electronics, 10 Commerce Drive, New Rochelle, NY, 10801, 914-576-6570. 8. Mark Montrose, EMC and the Printed Circuit Board, IEEE Press, 1999 (IEEE Order Number PC5756). 9. John Ardizzoni, "A Practical Guide to High-Speed Printed-Circuit-Board Layout," Analog Dialogue, Vol. 39, Sept. 2005. 10. Grant, Doug and Scott Wurcer, “Avoiding Passive-Component Pitfalls,” Analog Devices Application Note AN-348 11. Walt Kester, Analog-Digital Conversion, Analog Devices, 2004, ISBN 0-916550-27-3, Chapter 9. Also available as The Data Conversion Handbook, Elsevier/Newnes, 2005, ISBN 0-7506-7841-0, Chapter 9. Copyright 2009, Analog Devices, Inc. All rights reserved. Analog Devices assumes no responsibility for customer product design or the use or application of customers’ products or for any infringements of patents or rights of others which may result from Analog Devices assistance. All trademarks and logos are property of their respective holders. Information furnished by Analog Devices applications and development tools engineers is believed to be accurate and reliable, however no responsibility is assumed by Analog Devices regarding technical accuracy and topicality of the content provided in Analog Devices Tutorials. |
4391 | https://www.uptodate.com/contents/cervical-pregnancy-diagnosis-and-management | Cervical pregnancy: Diagnosis and management - UpToDate
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Outline
SUMMARY AND RECOMMENDATIONS
INTRODUCTION
CLINICAL FEATURES
EVALUATION
Physical examination
Transvaginal ultrasound
DIAGNOSIS
DIFFERENTIAL DIAGNOSIS
TREATMENT
Initial therapy
- Preferred: Methotrexate alone
- Alternate: Methotrexate plus balloon catheter
Persistent bleeding
- Endocervical curettage
Role of balloon catheter
- Uterine artery embolization
- Hysterectomy
PREGNANCY OUTCOME IN SUBSEQUENT GESTATIONS
SOCIETY GUIDELINE LINKS
SUMMARY AND RECOMMENDATIONS
REFERENCES
GRAPHICS
Pictures
- Transcervical balloon catheter
Diagnostic Images
- Cervical pregnancy: Transabdominal ultrasound
- Cervical pregnancy
- Cervical pregnancy with cardiac activity
- Cervical pregnancy: MR image
RELATED TOPICS
Abdominal pregnancy
Cesarean scar pregnancy
Ectopic pregnancy: Choosing a treatment
Ectopic pregnancy: Clinical manifestations and diagnosis
Ectopic pregnancy: Epidemiology, risk factors, and anatomic sites
Ectopic pregnancy: Methotrexate therapy
Gynecologic surgery: Overview of preoperative evaluation and preparation
Overview of postpartum hemorrhage
Postpartum hemorrhage: Medical and minimally invasive management
Pregnancy loss (miscarriage): Ultrasound diagnosis
Society guideline links: Ectopic pregnancy
Uterine fibroids (leiomyomas): Treatment with uterine artery embolization
Cervical pregnancy: Diagnosis and management
Author:Togas Tulandi, MD, MHCM, FRCSC, FACOG, FCAHSSection Editors:Courtney A Schreiber, MD, MPHLiina Poder, MDDeputy Editor:Alana Chakrabarti, MD, FACOG
Literature review current through:Aug 2025.
This topic last updated:Feb 04, 2025.
INTRODUCTION
Cervical pregnancy is a rare form of ectopic pregnancy in which the pregnancy implants in the lining of the endocervical canal. This entity accounts for less than 1 percent of ectopic pregnancies . The incidence is approximately 1 in 9000 pregnancies. In a 10-year, population-based study of 1800 cases, no cervical pregnancies were encountered . Cervical pregnancy may be more common in pregnancies achieved through assisted reproductive technologies; it occurs in 2 percent of in vitro fertilization ectopic gestations . Diagnosis and treatment early in pregnancy are important since there is a high risk of severe hemorrhage, and prevention of the need for hysterectomy is a high priority.
The cause of cervical pregnancy is unknown; local pathology related to previous cervical or uterine surgery may play a role given an apparent association with a prior history of curettage or cesarean birth . Another theory is rapid transport of the fertilized ovum into the endocervical canal before it is capable of nidation or because of an unreceptive endometrium.
The diagnosis and management of cervical pregnancy will be reviewed here. Related topics regarding ectopic pregnancy are discussed in detail separately, including:
●Epidemiology and risk factors (see "Ectopic pregnancy: Epidemiology, risk factors, and anatomic sites")
●Clinical manifestations and diagnosis (see "Ectopic pregnancy: Clinical manifestations and diagnosis")
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Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. It does NOT include all information about conditions, treatments, medications, side effects, or risks that may apply to a specific patient. It is not intended to be medical advice or a substitute for the medical advice, diagnosis, or treatment of a health care provider based on the health care provider's examination and assessment of a patient's specific and unique circumstances. Patients must speak with a health care provider for complete information about their health, medical questions, and treatment options, including any risks or benefits regarding use of medications. This information does not endorse any treatments or medications as safe, effective, or approved for treating a specific patient. UpToDate, Inc. and its affiliates disclaim any warranty or liability relating to this information or the use thereof. The use of this information is governed by the Terms of Use, available at 2025© UpToDate, Inc. and its affiliates and/or licensors. All rights reserved.
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Transcervical balloon catheter
Transcervical balloon catheter
Diagnostic Images
Cervical pregnancy: Transabdominal ultrasoundCervical pregnancyCervical pregnancy with cardiac activityCervical pregnancy: Magnetic resonance image
Cervical pregnancy: Transabdominal ultrasoundCervical pregnancyCervical pregnancy with cardiac activityCervical pregnancy: Magnetic resonance image
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4392 | https://artofproblemsolving.com/wiki/index.php/Constructive_counting?srsltid=AfmBOop6_SQMSE1F32JmNJl5IPV_ZGYY6q1OtyJpXRPGBFbNxObXZaa8 | Art of Problem Solving
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Constructive counting
In combinatorics, constructive counting is a counting technique that involves constructing an item belonging to a set. Along with the construction, one counts the total possibilities of each step and assembles these to enumerate the full set.
Along with casework and complementary counting, constructive counting is among the most fundamental techniques in counting. Familiarity with constructive counting is essential in combinatorics, especially in intermediate competitions.
Contents
[hide]
1 Introductory Examples
1.1 Example 1
1.2 Example 2
1.3 Example 3
1.4 Example 4
1.5 Example 5
2 Intermediate Examples
2.1 Example 1
3 More Problems
4 Resources
5 See also
Introductory Examples
Example 1
How many four-digit numbers are there?
Solution: We can construct a four-digit by picking the first digit, then the second, and so on until the fourth. The first digit can be any number from one to nine — zero excluded, or else it ceases to have four digits — so it has possible choices. The other three digits can be any number from zero to nine (total 10 digits), so they all have possibilities. We multiply the possibilities for each digit to arrive at our answer: .
Constructive counting is a simple concept, more so than its definition might lead one to think. All we did was think about constructing a four-digit number by choosing its digits, compute the possible numbers each digit can be, and multiply the resulting numbers.
This is a problem where constructive counting is not the simplest way to proceed. This next example is one where constructive counting is essential:
Example 2
How many lists of seven numbers are there such that each entry is between and inclusive and no two consecutive entries are equal?
Solution: We can model this situation as a row of 7 boxes, like this: which we must populate with numbers between and including and ; the key restriction here is that no two boxes right next to each other can have the same number.
The first digit can be any number from to , of which there are such options. The second can be any of to too, with the sole exception of the previous digit. Regardless of whatever the first digit is, we know that it removes one option, so there are options for the second digit.
The exact same logic applies to the third digit; it can be any of the digits except the one before it, so it has options. This continues until the seventh digit, which means that digits to all have options. Hence, our answer is , as desired.
These two examples use options specifically based on digits, but this isn't the entire picture of constructive counting. The following example uses constructive counting in a different context.
Example 3
How many permutations of are there?
Solution: We can model the question as a row of seven boxes, like this: which we have to populate with s and s. Using constructive counting is an idea, but there are multiple ways one might proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.
Instead, one might think to break it down by first placing the s separately, then placing the s. Starting with the s, we must choose the boxes of their placement; because all the s are indistinguishable, this is given by , where is a combination. One example among many placements of the s is For the s, their position is actually predetermined by choosing the s; the only place the three s can go is in the three empty boxes, so we don't have to account for them after choosing the s. Thus, there are different permutations of , as required. .
Like in this problem, there are sometimes multiple independent ways to construct a set. In others, however, an alternative method is not apparent, as with the next example:
Example 4
2001 AMC 12 Problem 16: A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?
Solution: Note that each leg has one designated shoe and a designated sock; each leg's shoe and sock belong to only that leg. The question is then only asking about the order in time in which it puts on all 16 socks and shoes. We can model the different orders as 16 boxes, where each box is populated with a certain sock or shoe, like this: Breaking up the problem by each box leads to a dead end. Instead, we can use a similar approach to example three to solve this problem — we can first choose two boxes in which each leg's sock-shoe pair is located, then we permute them. For the first leg, the location of its sock and shoe's two boxes are given by ; but on permuting them, we know that the sock appears first in the list, which implies that only one permutation exists.
So, there are just different places in which the first leg's sock and shoe can be located in the 16 boxes. By similar logic, the second leg has places in the boxes, the third has , and so on. The final answer is then the product of the leg's choices, which is Thus, our final answer is . .
This problem was much more challenging than the others mentioned thus far, but it's a lovely illustration of just how effective a cleverly chosen construction can be to counting problems.
Example 5
2004 AIME I Problem 6: An integer is called snakelike if its decimal representation satisfies if is odd and if is even. How many snakelike integers between 1000 and 9999 have four distinct digits?
Solution: We construct the set of snakelike integers. All the recursive requirement is saying is that the second digit must be greater than the first and third, and the fourth must be greater than the third. But before we start construction, we must divide our investigation into several cases, based on whether we allow zeroes
Case 1: Snakelike integers with no zero. First, we choose the four integers. They must be between and inclusive, and no digit can be repeated, so the combination that describes this is . Next, we find out how many permutations of these digits keep the number snakelike. For simplicity, consider the case of , , , and . The total possibilities are This applies to all snakelike integers with no zero, which means there arrangements that keep the number snakelike. Thus we have snakelike integers with four distinct digits and no zero.
Case 2: Snakelike integers with one zero. First, we choose the other three integers, which by similar logic above is . Next, we permute these digits. Without loss of generality, let our other three digits be , , and . The total possibilities are Hence, there are snakelike integers with four distinct digits and one zero.
It's easy to see that introducing a second zero would mandate them being the first and third digits. However, this breaks our requirement that our integers must be between and , so there are no four-digit snakelike integers with two or more zeroes. Thus, our total is snakelike integers with four distinct digits, as desired. .
It's worth noting that as problem-solving ability becomes more advanced, there are fewer problems that can be solved by constructive counting alone; the list of examples terminates at a late-introductory level consequentially. At the same time, there are many, many more problems at a higher level where constructive counting is a crucial intermediate step, combined with other counting strategies to reach an answer.
Intermediate Examples
These problems are more advanced than the introductory examples, requiring greater creativity to solve.
Example 1
Russia 1998: A 10-digit number is said to be interesting if its digits are all distinct and it is a multiple of 11111. How many interesting numbers are there?
Solution: There doesn't seem to be a useful pattern in multiples of 11111 And even if there was one, it's not clear how to mediate it with the distinct digits condition. Thus, we look for additional restrictions that will help us count.
Let be any interesting number. Because has ten distinct digits, its digits must be an ordered arrangement of to used exactly once. From here, note that the sum of its digits, , is divisible by . This implies through divisibility rules that is a multiple of ; therefore, is a multiple of .
An idea from here is to utilize the special properties of — namely, that . The compells us to express as for some five-digit numbers and ; doing so gives that Both sides of this equation must be divisible by , which implies that is divisible by . It's easy to verify that even without this condition, the maximum of is , which is less than . Therefore, must be equal to .
This also interacts with the distinct digits property; namely, that the corresponding digits of and add to . For example , where . Thus, corresponding digits must come in pairs — which now enables us to construct and . We first place these pairs in (which places it in ), of which they can be in any order. There are then options for the pairs' positions. The only thing left is which number in each pair goes to and which to ; there are ways we can divvy a pair up, which means there are total possibilities.
But wait! This construction counts numbers that start with zero — something that violates the 10-digit condition — as being interesting numbers. Note that in our count, each digit is equally likely to be first; thus, of our numbers start with , and only of our count keeps the number ten-digit. Putting this all together, there are interesting numbers.
More Problems
2005 AIME I Problems/Problem 5
Resources
AoPS Constructive Counting Part 1
AoPS Casework Counting Part 1
See also
Casework
Complementary counting
Overcounting
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4393 | http://web.ecs.baylor.edu/faculty/lee/ELC4340/Lecture%20note/Chapter8_GSO5.pdf | Generator stator showing completed windings for a 757-MVA, 3600-RPM, 60-Hz synchronous generator (Courtesy of General Electric) 8 SYMMETRICAL COMPONENTS The method of symmetrical components, first developed by C. L. Fortescue in 1918, is a powerful technique for analyzing unbalanced three-phase sys-tems. Fortescue defined a linear transformation from phase components to a new set of components called symmetrical components. The advantage of this transformation is that for balanced three-phase networks the equivalent cir-cuits obtained for the symmetrical components, called sequence networks, are separated into three uncoupled networks. Furthermore, for unbalanced three-phase systems, the three sequence networks are connected only at points of unbalance. As a result, sequence networks for many cases of unbalanced three-phase systems are relatively easy to analyze.
The symmetrical component method is basically a modeling technique that permits systematic analysis and design of three-phase systems. Decou-pling a detailed three-phase network into three simpler sequence networks reveals complicated phenomena in more simplistic terms. Sequence network 419 results can then be superposed to obtain three-phase network results. As an example, the application of symmetrical components to unsymmetrical short-circuit studies (see Chapter 9) is indispensable.
The objective of this chapter is to introduce the concept of symmetrical components in order to lay a foundation and provide a framework for later chapters covering both equipment models as well as power system analysis and design methods. In Section 8.1, we define symmetrical components. In Sections 8.2–8.7, we present sequence networks of loads, series impedances, transmission lines, rotating machines, and transformers. We discuss complex power in sequence networks in Section 8.8. Although Fortescue’s original work is valid for polyphase systems with n phases, we will consider only three-phase systems here.
C A S E S T U DY The following article provides an overview of circuit breakers with high voltage ratings at or above 72.5 kV . Circuit breakers are broadly classified by the medium used to extinguish the arc: bulk oil, minimum oil, air-blast, vacuum, and sulfur hexafluoride (SF6).
For high voltages, oil circuit breakers dominated in the early 1900s through the 1950s for applications up to 362 kV, with minimum oil circuit breakers developed up to 380 kV.
The development of air-blast circuit breakers started in Europe in the 1920s and became prevalent in the 1950s. Air-blast circuit breakers, which use air under high pressure that is blown between the circuit breaker contacts to extinguish the arc, have been used at voltages up to 800 kV and many are still in operation today. Air-blast circuit breakers were manufactured until the 1980s when they were supplanted by lower cost and simpler SF6 puffer-type circuit breakers. SF6 gas possesses exceptional arc-interrupting properties that have led to a worldwide change to SF6 high-voltage circuit breakers, which are more reliable, more efficient and more compact than other types of circuit breakers. Vacuum circuit breakers are commonly used at medium voltages between 1 and 72.5 kV.
Circuit Breakers Go High Voltage: The Low Operating Energy of SF6 Circuit Breakers Improves Reliability and Reduces Wear and Tear DENIS DUFOURNET The first sulfur hexafluoride (SF6) gas industrial de-velopments were in the medium voltage range. This equipment confirmed the advantages of a technique that uses SF6 at a low-pressure level concurrently with the auto-pneumatic blast system to interrupt the arc that was called later puffer.
High-voltage SF6 circuit breakers with self-blast interrupters have found worldwide acceptance because their high current interrupting capability is obtained with a low operating energy that can be provided by low-cost, spring-operated mechanisms.
The low-operating energy required reduces the stress and wear of the mechanical components and significantly improves the overall reliability of the circuit breaker. This switching principle was first introduced in the high-voltage area about 20 years 420 CHAPTER 8 SYMMETRICAL COMPONENTS (‘‘Circuit Breakers Go High Voltage’’ by Denis Dufournet.
2009 IEEE. Reprinted, with permission, from IEEE Power & Energy Magazine, January/February 2009) ago, starting with the voltage level of 72.5 kV. Today this technique is available up to 800 kV. Furthermore it is used for generator circuit breaker applications with short circuit currents of 63 kA and above.
Service experience shows that when the SF6 circuit breakers of the self-blast technology were first designed, the expectations of the designers had been fulfilled completely with respect to reliability and day-to-day operation.
A HISTORY OF CIRCUIT BREAKERS Bulk oil circuit breakers dominated in the early 1900s and remained in use throughout the 1950s, for appli-cations up to 362 kV for which they had eight breaks in series. They were replaced by minimum oil and air-blast circuit breakers for high-voltage applications.
Minimum oil circuit breakers, as shown in Figure 1, have arc control structures that improve the arc cooling process and significantly reduce the volume of oil. They were developed up to 380 kV, in particular for the first 380 kV network in the world (Harspra ˚nget–Halsberg line in Sweden in 1952).
There were tentative extensions to 765 kV, 50 kA, but minimum oil circuit breakers were supplanted in the EHV range by air-blast circuit breakers that were the first to be applied in 525, 735, and 765 kV networks, respectively in Russia (1960), Canada (1965), and the United States (1969).
Air-blast circuit breakers, as shown in Figure 2, use air under high pressure that is blown through the arc space between the opening contacts to ex-tinguish the arc. The development of air-blast circuit breakers started in Europe in the 1920s, with fur-ther development in 1930s and 1940s, and became prevalent in the 1950s.
Air-blast circuit breakers were very successful in North America and Europe. They had an inter-rupting capability of 63 kA, later increased to 90 kA in the 1970s. Many circuit breakers of this type are still in operation today, in particular in North America, at 550 and 800 kV.
Air-blast circuit breakers were manufactured until the 1980s when they were supplanted by the lower cost and less complex SF6 puffer-type circuit breakers.
Figure 1 Minimum oil circuit breaker 145 kV type orthojector (Courtesy of Alstom Grid) Figure 2 Air-blast circuit breaker type PK12 applied to 765 kV in North America (Courtesy of Alstom Grid) CASE STUDY 421 The first industrial application of SF6 dates from 1937 when it was used in the United States as an insulating medium for cables (patent by F.S. Cooper of General Electric). With the advent of the nuclear power industry in the 1950s, SF6 was produced in large quantities and its use extended to circuit breakers as a quenching medium.
The first application of SF6 for current interrup-tion was done in 1953 when 15–161 kV switches were developed by Westinghouse. The first high-voltage SF6 circuit breakers were built also by Westinghouse in 1956, the interrupting capability was then limited to 5 kA under 115 kV, with each pole having six interrupting units in series. In 1959, Westinghouse produced the first SF6 circuit break-ers with high current interrupting capabilities: 41.8 kA under 138 kV (10,000 MVA) and 37.8 kA under 230 kV (15,000 MVA). These circuit breakers were of the dual pressure type based on the axial blast principles used in air-blast circuit breakers.
They were supplanted by the SF6 puffer circuit breakers.
In 1967, the puffer-type technique was in-troduced for high-voltage circuit breakers where the relative movement of a piston and a cylinder linked to the moving contact produced the pressure build-up necessary to blast the arc. The puffer technique, shown in Figure 3, was applied in the first 245 kV metal-enclosed gas insulated circuit breaker installed in France in 1969.
The excellent properties of SF6 lead to the fast extension of this technique in the 1970s and to its use for the development of circuit breakers with high current interrupting capability, up to 800 kV.
The achievement, around 1983, of the first single-break 245 kV and the corresponding 420 kV, 550 kV, and 800 kV, with, respectively, two, three, and four chambers per pole, lead to the domi-nance of SF6 circuit breakers in the complete high-voltage range.
Several characteristics of SF6 puffer circuit breakers can explain their success: . simplicity of the interrupting chamber which does not need an auxiliary chamber for breaking . autonomy provided by the puffer technique . the possibility to obtain the highest perform-ances, up to 63 kA, with a reduced number of interrupting chambers (Figure 4) Figure 3 Puffer-type circuit breaker Figure 4 800 kV 50 kA circuit breaker type FX with closing resistors (Courtesy of Alstom Grid) 422 CHAPTER 8 SYMMETRICAL COMPONENTS . short interrupting time of 2-2. 5 cycles at 60 Hz . high electrical endurance, allowing at least 25 years of operation without reconditioning . possible compact solutions when used for gas-insulated switchgear (GIS) or hybrid switchgears . integrated closing resistors or synchronized operations to reduce switching over voltages . reliability and availability . low noise level . no compressor for SF6 gas.
The reduction in the number of interrupting chambers per pole has led to a considerable sim-plification of circuit breakers as the number of parts as well as the number of seals was decreased. As a direct consequence, the reliability of circuit breakers was improved, as verified later by CIGRE surveys.
SELF-BLAST TECHNOLOGY The last 20 years have seen the development of the self-blast technique for SF6 interrupting chambers. This technique has proven to be very efficient and has been widely applied for high-voltage circuit breakers up to 800 kV. It has al-lowed the development of new ranges of circuit breakers operated by low energy spring-operated mechanisms.
Another aim of this evolution was to further in-crease the reliability by reducing dynamic forces in the pole and its mechanism.
These developments have been facilitated by the progress made in digital simulations that were widely used to optimize the geometry of the inter-rupting chamber and the mechanics between the poles and the mechanism.
The reduction of operating energy was achieved by lowering energy used for gas compression and by making a larger use of arc energy to produce the pressure necessary to quench the arc and obtain current interruption.
Low-current interruption, up to about 30% of rated short-circuit current, is obtained by a puffer blast where the overpressure necessary to quench the arc is produced by gas compression in a volume limited by a fixed piston and a moving cylinder.
Figure 5 shows the self-blast interruption princi-ple where a valve (V) was introduced between the expansion and the compression volume.
When interrupting low currents, the valve (V) opens under the effect of the overpressure gen-erated in the compression volume. The interrup-tion of the arc is made as in a puffer circuit breaker thanks to the compression of the gas obtained by the piston action.
In the case of high-current interruption, the arc energy produces a high overpressure in the expan-sion volume, which leads to the closure of the valve (V) and thus isolating the expansion volume from the compression volume. The overpressure neces-sary for breaking is obtained by the optimal use of the thermal effect and of the nozzle clogging effect produced whenever the cross-section of the arc significantly reduces the exhaust of gas in the nozzle.
This technique, known as self-blast, has been used extensively for more than 15 years for the development of many types of interrupting cham-bers and circuit breakers (Figure 6).
The better knowledge of arc interruption ob-tained by digital simulations and validation of performances by interrupting tests has con-tributed to a higher reliability of these self-blast circuit breakers. In addition, the reduction in Figure 5 Self blast (or double volume) interrupting chamber CASE STUDY 423 operating energy, allowed by the self-blast tech-nique, leads to a higher mechanical endurance.
DOUBLE MOTION PRINCIPLE The self-blast technology was further optimized by using the double-motion principle. This leads to further reduction of the operating energy by re-ducing the kinetic energy consumed during opening.
The method consists of displacing the two arcing contacts in opposite directions. With such a sys-tem, it was possible to reduce the necessary open-ing energy for circuit breakers drastically.
Figure 7 shows the arcing chamber of a circuit breaker with the double motion principle. The pole columns are equipped with helical springs mounted in the crankcase.
These springs contain the necessary energy for an opening operation. The energy of the spring is trans-mitted to the arcing chamber via an insulating rod.
To interrupt an arc, the contact system must have sufficient velocity to avoid reignitions. Fur-thermore, a pressure rise must be generated to establish a gas flow in the chamber.
The movable upper contact system is connected to the nozzle of the arcing chamber via a linkage system. This allows the movement of both arcing contacts in opposite directions. Therefore the ve-locity of one contact can be reduced by 50% be-cause the relative velocity of both contacts is still 100%. The necessary kinetic energy scales with the square of the velocity, allowing—theoretically—an energy reduction in the opening spring by a factor of 4. In reality, this value can’t be achieved because the moving mass has to be increased. As in the self-blast technique described previously, the arc itself mostly generates the pressure rise.
Because the pressure generation depends on the level of the short-circuit current, an additional small piston is necessary to interrupt small cur-rents (i.e., less than 30% of the rated short-circuit current). Smaller pistons mean less operating energy.
The combination of both double motion of con-tacts and self-blast technique allows for the signifi-cant reduction of opening energy.
GENERATOR CIRCUIT BREAKERS Generator circuit breakers are connected between a generator and the step-up voltage transformer.
They are generally used at the outlet of high-power generators (100–1,800 MVA) to protect them in a Figure 7 Double motion interrupting chamber Figure 6 Dead tank circuit breaker 145 kV with spring-operating mechanism and double motion self blast interrupting chambers (Courtesy of Alstom Grid) 424 CHAPTER 8 SYMMETRICAL COMPONENTS sure, quick, and economical manner. Such circuit breakers must be able to allow the passage of high permanent currents under continuous service (6,300–40,000 A), and have a high breaking capacity (63–275 kA).
They belong to the medium voltage range, but the transient recovery voltage (TRV) withstand ca-pability is such that the interrupting principles de-veloped for the high-voltage range has been used.
Two particular embodiments of the thermal blast and self-blast techniques have been developed and applied to generator circuit breakers.
Thermal Blast Chamber with Arc-Assisted Opening In this interruption principle arc energy is used, on the one hand to generate the blast by thermal ex-pansion and, on the other hand, to accelerate the moving part of the circuit breaker when interrupt-ing high currents (Figure 8).
The overpressure produced by the arc energy downstream of the interruption zone is applied on an auxiliary piston linked with the moving part. The resulting force accelerates the moving part, thus increasing the energy available for tripping.
It is possible with this interrupting principle to increase the tripping energy delivered by the oper-ating mechanism by about 30% and to maintain the opening speed irrespective of the short circuit current.
It is obviously better suited to circuit breakers with high breaking currents such as generator cir-cuit breakers that are required to interrupt cur-rents as high as 120 kA or even 160 kA.
Self-Blast Chamber with Rear Exhaust This principle works as follows (Figure 9): In the first phase, the relative movement of the piston and the blast cylinder is used to compress the gas in the compression volume Vc. This overpressure opens the valve C and is then transmitted to ex-pansion volume Vt.
In the second phase, gas in volume Vc is ex-hausted to the rear through openings (O).
The gas compression is sufficient for the inter-ruption of low currents. During high short-circuit current interruption, volume Vt is pressurized by the thermal energy of the arc. This high pressure closes valve C. The pressure in volume Vc on the other hand is limited by an outflow of gas through the openings (O). The high overpressure generated in volume Vt produces the quenching blast necessary to extinguish the arc at current zero.
In this principle the energy that has to be deliv-ered by the operating mechanism is limited and low energy spring operated mechanism can be used.
Figure 10 shows a generator circuit breaker with such type of interrupting chamber.
Figure 8 Thermal blast chamber with arc-assisted opening Figure 9 Self-blast chamber with rear exhaust CASE STUDY 425 EVOLUTION OF TRIPPING ENERGY Figure 11 summarizes the evolution of tripping energy for 245 and 420 kV, from 1974 to 2003. It shows that the operating energy has been divided by a factor of five–seven during this period of nearly three decades. This illustrates the great progress that has been made in interrupting tech-niques for high-voltage circuit breakers during that period.
Figure 12 shows the continuous reduction of the necessary operating energy obtained through the technological progress.
OUTLOOK FOR THE FUTURE Several interrupting techniques have been pre-sented that all aim to reduce the operating energy of high-voltage circuit breakers. To date they have been widely applied, resulting in the lowering of drive energy, as shown in Figures 11 and 12.
Present interrupting technologies can be applied to circuit breakers with the higher rated interrupt-ing currents (63–80 kA) required in some networks with increasing power generation (Figure 13).
Progress can still be made by the further indus-trialization of all components and by introducing new drive technologies. Following the remarkable evolution in chamber technology, the operating mechanism represents a not negligible contribution to the moving mass of circuit breakers, especially in the extra high-voltage range ( 420 kV). Therefore progress in high-voltage circuit breakers can still be expected with the implementation of the same in-terrupting principles.
If one looks further in the future, other tech-nology developments could possibly lead to a Figure 11 Evolution of tripping energy since 1974 of 245 and 420 kV circuit breakers Figure 12 Operating energy as function of interrupting principle Figure 10 Generator circuit breaker SF6 17, 5 kV 63 kA 60 Hz 426 CHAPTER 8 SYMMETRICAL COMPONENTS further reduction in the SF6 content of circuit breakers.
CONCLUSIONS Over the last 50 years, high-voltage circuit breakers have become more reliable, more efficient, and more compact because the interrupting capability per break has been increased dramatically. These developments have not only produced major savings, but they have also had a massive impact on the layout of substations with respect to space requirements.
New types of SF6 interrupting chambers, which implement innovative interrupting principles, have been developed during the last three decades with the objective of reducing the operating energy of the circuit breaker. This has led to reduced stress and wear of the mechanical components and con-sequently to an increased reliability of circuit breakers.
Service experience shows that the expectations of the designers, with respect to reliability and day-to-day operation, have been fulfilled.
FOR FURTHER READING W.M. Leeds, R.E. Friedrich, C.L. Wagner, and T.E.
Browne Jr, ‘‘Application of switching surge, arc and gas flow studies to the design of SF6 breakers,’’ presented at CIGRE Session 1970, paper 13-11.
E. Thuries, ‘‘Development of air-blast circuit-breakers,’’ presented at CIGRE Session 1972, paper 13-09.
D. Dufournet and E. Thuries ‘‘Recent develop-ment of HV circuit-breakers,’’ presented at 11th CEPSI Conference, Kuala Lumpur, Malaisia, Oct.
1996.
D. Dufournet, F. Sciullo, J. Ozil, and A. Ludwig, ‘‘New interrupting and drive techniques to increase high-voltage circuit breakers performance and reli-ability,’’ presented at CIGRE session, 1998, paper 13-104.
A. Ludwig, D. Dufournet, and E. Mikes, ‘‘Improved performance and reliability of high-voltage circuit breakers with spring mechanisms through new breaking and operating elements,’’ presented at 12th CEPSI Conference, Pattaya, Thailand, 1998.
D. Dufournet, J.M. Willieme, and G.F. Mon-tillet, ‘‘Design and implementation of a SF6 inter-rupting chamber applied to low range generator circuit breakers suitable for interruption of current having a non-zero passage,’’ IEEE Trans.
Power Delivery, vol. 17, no. 4, pp. 963–967, Oct. 2002.
D. Dufournet, ‘‘Generator circuit breakers: SF6 Breaking chamber–interruption of current with non-zero passage. Influence of cable con-nection on TRV of system fed faults,’’ presented at CIGRE 2002, Paris, France, Aug. 2002, paper 13-101.
D. Dufournet, C. Lindner, D. Johnson, and D. Vondereck, ‘‘Technical trends in circuit breaker switching technologies,’’ presented at CIGRE SC A3 Colloquium, Sarajevo, 2003.
BIOGRAPHY Denis Dufournet is with AREVA T&D.
Figure 13 GIS circuit breaker 550 kV 63 kA 50/60 Hz CASE STUDY 427 8.1 DEFINITION OF SYMMETRICAL COMPONENTS Assume that a set of three-phase voltages designated V a, Vb, and V c is given.
In accordance with Fortescue, these phase voltages are resolved into the fol-lowing three sets of sequence components: 1. Zero-sequence components, consisting of three phasors with equal mag-nitudes and with zero phase displacement, as shown in Figure 8.1(a) 2. Positive-sequence components, consisting of three phasors with equal magnitudes, G 120 phase displacement, and positive sequence, as in Figure 8.1(b) 3. Negative-sequence components, consisting of three phasors with equal magnitudes, G120 phase displacement, and negative sequence, as in Figure 8.1(c) In this text we will work only with the zero-, positive-, and negative-sequence components of phase a, which are V a0, V a1, and V a2, respectively.
For simplicity, we drop the subscript a and denote these sequence compo-nents as V0, V1, and V2. They are defined by the following transformation: 2 6 4 V a Vb V c 3 7 5¼ 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 2 6 4 V0 V1 V2 3 7 5 ð8:1:1Þ FIGURE 8.1 Resolving phase voltages into three sets of sequence components 428 CHAPTER 8 SYMMETRICAL COMPONENTS where a ¼ 1 120 ¼ 1 2 þ j ffiffiffi 3 p 2 ð8:1:2Þ Writing (8.1.1) as three separate equations: V a ¼ V0 þ V1 þ V2 ð8:1:3Þ Vb ¼ V0 þ a2V1 þ aV2 ð8:1:4Þ V c ¼ V0 þ aV1 þ a2V2 ð8:1:5Þ In (8.1.2), a is a complex number with unit magnitude and a 120 phase angle. When any phasor is multiplied by a, that phasor rotates by 120 (counterclockwise). Similarly, when any phasor is multiplied by a2 ¼ ð1 120Þ ð1 120Þ ¼ 1 240, the phasor rotates by 240. Table 8.1 lists some common identities involving a.
The complex number a is similar to the well-known complex number j ¼ ffiffiffiffiffiffi ffi 1 p ¼ 1 90. Thus the only di¤erence between j and a is that the angle of j is 90, and that of a is 120.
Equation (8.1.1) can be rewritten more compactly using matrix nota-tion. We define the following vectors V p and V s, and matrix A: V p ¼ 2 6 4 V a Vb V c 3 7 5 ð8:1:6Þ V s ¼ 2 6 4 V0 V1 V2 3 7 5 ð8:1:7Þ A ¼ 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 ð8:1:8Þ V p is the column vector of phase voltages, V s is the column vector of sequence voltages, and A is a 3 3 transformation matrix. Using these definitions, (8.1.1) becomes V p ¼ AV s ð8:1:9Þ The inverse of the A matrix is A1 ¼ 1 3 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 ð8:1:10Þ TABLE 8.1 Common identities involving a ¼ 1 120 a4 ¼ a ¼ 1 120 a2 ¼ 1 240 a3 ¼ 1 0 1 þ a þ a2 ¼ 0 1 a ¼ ffiffiffi 3 p 30 1 a2 ¼ ffiffiffi 3 p þ30 a2 a ¼ ffiffiffi 3 p 270 ja ¼ 1 210 1 þ a ¼ a2 ¼ 1 60 1 þ a2 ¼ a ¼ 1 60 a þ a2 ¼ 1 ¼ 1 180 SECTION 8.1 DEFINITION OF SYMMETRICAL COMPONENTS 429 Equation (8.1.10) can be verified by showing that the product AA1 is the unit matrix. Also, premultiplying (8.1.9) by A1 gives V s ¼ A1V p ð8:1:11Þ Using (8.1.6), (8.1.7), and (8.1.10), then (8.1.11) becomes 2 6 4 V0 V1 V2 3 7 5¼ 1 3 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 2 6 4 V a Vb V c 3 7 5 ð8:1:12Þ Writing (8.1.12) as three separate equations, V0 ¼ 1 3ðV a þ Vb þ V cÞ ð8:1:13Þ V1 ¼ 1 3ðV a þ aVb þ a2V cÞ ð8:1:14Þ V2 ¼ 1 3ðV a þ a2Vb þ aV cÞ ð8:1:15Þ Equation (8.1.13) shows that there is no zero-sequence voltage in a balanced three-phase system because the sum of three balanced phasors is zero. In an un-balanced three-phase system, line-to-neutral voltages may have a zero-sequence component. However, line-to-line voltages never have a zero-sequence compo-nent, since by KVL their sum is always zero.
The symmetrical component transformation can also be applied to cur-rents, as follows. Let Ip ¼ AIs ð8:1:16Þ where Ip is a vector of phase currents, Ip ¼ 2 6 4 Ia Ib Ic 3 7 5 ð8:1:17Þ and Is is a vector of sequence currents, Is ¼ 2 6 4 I0 I1 I2 3 7 5 ð8:1:18Þ Also, Is ¼ A1Ip ð8:1:19Þ Equations (8.1.16) and (8.1.19) can be written as separate equations as fol-lows. The phase currents are Ia ¼ I0 þ I1 þ I2 ð8:1:20Þ Ib ¼ I0 þ a2I1 þ aI2 ð8:1:21Þ Ic ¼ I0 þ aI1 þ a2I2 ð8:1:22Þ 430 CHAPTER 8 SYMMETRICAL COMPONENTS and the sequence currents are I0 ¼ 1 3ðIa þ Ib þ IcÞ ð8:1:23Þ I1 ¼ 1 3ðIa þ aIb þ a2IcÞ ð8:1:24Þ I2 ¼ 1 3ðIa þ a2Ib þ aIcÞ ð8:1:25Þ In a three-phase Y-connected system, the neutral current In is the sum of the line currents: In ¼ Ia þ Ib þ Ic ð8:1:26Þ Comparing (8.1.26) and (8.1.23), In ¼ 3I0 ð8:1:27Þ The neutral current equals three times the zero-sequence current. In a bal-anced Y-connected system, line currents have no zero-sequence component, since the neutral current is zero. Also, in any three-phase system with no neutral path, such as a D-connected system or a three-wire Y-connected system with an ungrounded neutral, line currents have no zero-sequence component.
The following three examples further illustrate symmetrical compo-nents.
EXAMPLE 8.1 Sequence components: balanced line-to-neutral voltages Calculate the sequence components of the following balanced line-to-neutral voltages with abc sequence: V p ¼ 2 6 4 V an Vbn V cn 3 7 5¼ 2 6 4 277 0 277 120 277 þ120 3 7 5 volts SOLUTION Using (8.1.13)–(8.1.15): V0 ¼ 1 3½277 0 þ 277 120 þ 277 þ120 ¼ 0 V1 ¼ 1 3½277 0 þ 277 ð120 þ 120Þ þ 277 ð120 þ 240Þ ¼ 277 0 volts ¼ V an V2 ¼ 1 3½277 0 þ 277 ð120 þ 240Þ þ 277 ð120 þ 120Þ ¼ 1 3½277 0 þ 277 120 þ 277 240 ¼ 0 SECTION 8.1 DEFINITION OF SYMMETRICAL COMPONENTS 431 This example illustrates the fact that balanced three-phase systems with abc sequence (or positive sequence) have no zero-sequence or negative-sequence components. For this example, the positive-sequence voltage V1 equals V an, and the zero-sequence and negative-sequence voltages are both zero.
9 EXAMPLE 8.2 Sequence components: balanced acb currents A Y-connected load has balanced currents with acb sequence given by Ip ¼ 2 6 4 Ia Ib Ic 3 7 5¼ 2 6 4 10 0 10 þ120 10 120 3 7 5 A Calculate the sequence currents.
SOLUTION Using (8.1.23)–(8.1.25): I0 ¼ 1 3½10 0 þ 10 120 þ 10 120 ¼ 0 I1 ¼ 1 3½10 0 þ 10 ð120 þ 120Þ þ 10 ð120 þ 240Þ ¼ 1 3½10 0 þ 10 240 þ 10 120 ¼ 0 I2 ¼ 1 3½10 0 þ 10 ð120 þ 240Þ þ 10 ð120 þ 120Þ ¼ 10 0 A ¼ Ia This example illustrates the fact that balanced three-phase systems with acb sequence (or negative sequence) have no zero-sequence or positive-sequence components. For this example the negative-sequence current I2 equals Ia, and the zero-sequence and positive-sequence currents are both zero.
9 EXAMPLE 8.3 Sequence components: unbalanced currents A three-phase line feeding a balanced-Y load has one of its phases (phase b) open. The load neutral is grounded, and the unbalanced line currents are Ip ¼ 2 6 4 Ia Ib Ic 3 7 5¼ 2 6 4 10 0 0 10 120 3 7 5 A Calculate the sequence currents and the neutral current.
432 CHAPTER 8 SYMMETRICAL COMPONENTS SOLUTION The circuit is shown in Figure 8.2. Using (8.1.23)–(8.1.25): I0 ¼ 1 3½10 0 þ 0 þ 10 120 ¼ 3:333 60 A I1 ¼ 1 3½10 0 þ 0 þ 10 ð120 þ 240Þ ¼ 6:667 0 A I2 ¼ 1 3½10 0 þ 0 þ 10 ð120 þ 120Þ ¼ 3:333 60 A Using (8.1.26) the neutral current is In ¼ ð10 0 þ 0 þ 10 120Þ ¼ 10 60 A ¼ 3I0 This example illustrates the fact that unbalanced three-phase systems may have nonzero values for all sequence components. Also, the neutral current equals three times the zero-sequence current, as given by (8.1.27).
9 8.2 SEQUENCE NETWORKS OF IMPEDANCE LOADS Figure 8.3 shows a balanced-Y impedance load. The impedance of each phase is designated ZY, and a neutral impedance Zn is connected between the load neutral and ground. Note from Figure 8.3 that the line-to-ground volt-age V ag is V ag ¼ ZYIa þ ZnIn ¼ ZYIa þ ZnðIa þ Ib þ IcÞ ¼ ðZY þ ZnÞIa þ ZnIb þ ZnIc ð8:2:1Þ FIGURE 8.2 Circuit for Example 8.3 SECTION 8.2 SEQUENCE NETWORKS OF IMPEDANCE LOADS 433 Similar equations can be written for Vbg and V cg: Vbg ¼ ZnIa þ ðZY þ ZnÞIb þ ZnIc ð8:2:2Þ V cg ¼ ZnIa þ ZnIb þ ðZY þ ZnÞIc ð8:2:3Þ Equations (8.2.1)–(8.2.3) can be rewritten in matrix format: 2 6 4 V ag Vbg V cg 3 7 5¼ 2 6 4 ðZY þ ZnÞ Zn Zn Zn ðZY þ ZnÞ Zn Zn Zn ðZY þ ZnÞ 3 7 5 2 6 4 Ia Ib Ic 3 7 5 ð8:2:4Þ Equation (8.2.4) is written more compactly as V p ¼ ZpIp ð8:2:5Þ where V p is the vector of line-to-ground voltages (or phase voltages), Ip is the vector of line currents (or phase currents), and Zp is the 3 3 phase imped-ance matrix shown in (8.2.4). Equations (8.1.9) and (8.1.16) can now be used in (8.2.5) to determine the relationship between the sequence voltages and currents, as follows: AV s ¼ ZpAIs ð8:2:6Þ Premultiplying both sides of (8.2.6) of A1 gives V s ¼ ðA1ZpAÞIs ð8:2:7Þ or V s ¼ ZsIs ð8:2:8Þ where Zs ¼ A1ZpA ð8:2:9Þ The impedance matrix Zs defined by (8.2.9) is called the sequence impedance matrix. Using the definition of A, its inverse A1, and Zp given FIGURE 8.3 Balanced-Y impedance load 434 CHAPTER 8 SYMMETRICAL COMPONENTS by (8.1.8), (8.1.10), and (8.2.4), the sequence impedance matrix Zs for the balanced-Y load is Zs ¼ 1 3 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 2 6 4 ðZY þ ZnÞ Zn Zn Zn ðZY þ ZnÞ Zn Zn Zn ðZY þ ZnÞ 3 7 5 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 ð8:2:10Þ Performing the indicated matrix multiplications in (8.2.10), and using the identity ð1 þ a þ a2Þ ¼ 0, Zs ¼ 1 3 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 2 6 4 ðZY þ 3ZnÞ ZY ZY ðZY þ 3ZnÞ a2ZY aZY ðZY þ 3ZnÞ aZY a2ZY 3 7 5 ¼ 2 6 4 ðZY þ 3ZnÞ 0 0 0 ZY 0 0 0 ZY 3 7 5 ð8:2:11Þ As shown in (8.2.11), the sequence impedance matrix Zs for the balanced-Y load of Figure 8.3 is a diagonal matrix. Since Zs is diagonal, (8.2.8) can be written as three uncoupled equations. Using (8.1.7), (8.1.18), and (8.2.11) in (8.2.8), 2 6 4 V0 V1 V2 3 7 5¼ 2 6 4 ðZY þ 3ZnÞ 0 0 0 ZY 0 0 0 ZY 3 7 5 2 6 4 I0 I1 I2 3 7 5 ð8:2:12Þ Rewriting (8.2.12) as three separate equations, V0 ¼ ðZY þ 3ZnÞI0 ¼ Z0I0 ð8:2:13Þ V1 ¼ ZYI1 ¼ Z1I1 ð8:2:14Þ V2 ¼ ZYI2 ¼ Z2I2 ð8:2:15Þ As shown in (8.2.13), the zero-sequence voltage V0 depends only on the zero-sequence current I0 and the impedance ðZY þ 3ZnÞ. This impedance is called the zero-sequence impedance and is designated Z0. Also, the positive-sequence voltage V1 depends only on the positive-sequence current I1 and an impedance Z1 ¼ ZY called the positive-sequence impedance. Similarly, V2 de-pends only on I2 and the negative-sequence impedance Z2 ¼ ZY.
Equations (8.2.13)–(8.2.15) can be represented by the three networks shown in Figure 8.4. These networks are called the zero-sequence, positive-sequence, and negative-sequence networks. As shown, each sequence network SECTION 8.2 SEQUENCE NETWORKS OF IMPEDANCE LOADS 435 is separate, uncoupled from the other two. The separation of these sequence networks is a consequence of the fact that Zs is a diagonal matrix for a balanced-Y load. This separation underlies the advantage of symmetrical components.
Note that the neutral impedance does not appear in the positive- and negative-sequence networks of Figure 8.4. This illustrates the fact that positive- and negative-sequence currents do not flow in neutral impedances.
However, the neutral impedance is multiplied by 3 and placed in the zero-sequence network of the figure. The voltage I0ð3ZnÞ across the impedance 3Zn is the voltage drop ðInZnÞ across the neutral impedance Zn in Figure 8.3, since In ¼ 3I0.
When the neutral of the Y load in Figure 8.3 has no return path, then the neutral impedance Zn is infinite and the term 3Zn in the zero-sequence network of Figure 8.4 becomes an open circuit. Under this condition of an open neutral, no zero-sequence current exists. However, when the neutral of the Y load is solidly grounded with a zero-ohm conductor, then the neutral impedance is zero and the term 3Zn in the zero-sequence network becomes a short circuit. Under this condition of a solidly grounded neutral, zero-sequence current I0 can exist when there is a zero-sequence voltage caused by unbalanced voltages applied to the load.
Figure 2.15 shows a balanced-D load and its equivalent balanced-Y load. Since the D load has no neutral connection, the equivalent Y load in Figure 2.15 has an open neutral. The sequence networks of the equivalent Y load corresponding to a balanced-D load are shown in Figure 8.5. As shown, FIGURE 8.4 Sequence networks of a balanced-Y load 436 CHAPTER 8 SYMMETRICAL COMPONENTS the equivalent Y impedance ZY ¼ ZD=3 appears in each of the sequence net-works. Also, the zero-sequence network has an open circuit, since Zn ¼ y corresponds to an open neutral. No zero-sequence current occurs in the equiv-alent Y load.
The sequence networks of Figure 8.5 represent the balanced-D load as viewed from its terminals, but they do not represent the internal load charac-teristics. The currents I0, I1, and I2 in Figure 8.5 are the sequence compo-nents of the line currents feeding the D load, not the load currents within the D. The D load currents, which are related to the line currents by (2.5.14), are not shown in Figure 8.5.
EXAMPLE 8.4 Sequence networks: balanced-Y and balanced-D loads A balanced-Y load is in parallel with a balanced-D-connected capacitor bank.
The Y load has an impedance ZY ¼ ð3 þ j4Þ W per phase, and its neutral is grounded through an inductive reactance Xn ¼ 2 W. The capacitor bank has a reactance Xc ¼ 30 W per phase. Draw the sequence networks for this load and calculate the load-sequence impedances.
SOLUTION The sequence networks are shown in Figure 8.6. As shown, the Y-load impedance in the zero-sequence network is in series with three times the neutral impedance. Also, the D-load branch in the zero-sequence network is open, since no zero-sequence current flows into the D load. In the positive-and negative-sequence circuits, the D-load impedance is divided by 3 and placed in parallel with the Y-load impedance. The equivalent sequence im-pedances are FIGURE 8.5 Sequence networks for an equivalent Y representation of a balanced-D load SECTION 8.2 SEQUENCE NETWORKS OF IMPEDANCE LOADS 437 Z0 ¼ ZY þ 3Zn ¼ 3 þ j4 þ 3ð j2Þ ¼ 3 þ j10 W Z1 ¼ ZYEðZD=3Þ ¼ ð3 þ j4Þð j30=3Þ 3 þ j4 jð30=3Þ ¼ ð5 53:13 Þð10 90 Þ 6:708 63:43 ¼ 7:454 26:57 W Z2 ¼ Z1 ¼ 7:454 26:57 W 9 Figure 8.7 shows a general three-phase linear impedance load. The load could represent a balanced load such as the balanced-Y or balanced-D load, or an unbalanced impedance load. The general relationship between the line-to-ground voltages and line currents for this load can be written as 2 6 4 V ag Vbg V cg 3 7 5¼ 2 6 4 Zaa Zab Zac Zab Zbb Zbc Zac Zbc Zcc 3 7 5 2 6 4 Ia Ib Ic 3 7 5 ð8:2:16Þ or V p ¼ ZpIp ð8:2:17Þ where V p is the vector of line-to-neutral (or phase) voltages, Ip is the vector of line (or phase) currents, and Zp is a 3 3 phase impedance matrix. It is assumed here that the load is nonrotating, and that Zp is a symmetric matrix, which corresponds to a bilateral network.
FIGURE 8.6 Sequence networks for Example 8.4 438 CHAPTER 8 SYMMETRICAL COMPONENTS Since (8.2.17) has the same form as (8.2.5), the relationship between the sequence voltages and currents for the general three-phase load of Figure 8.6 is the same as that of (8.2.8) and (8.2.9), which are rewritten here: V s ¼ ZsIs ð8:2:18Þ Zs ¼ A1ZpA ð8:2:19Þ The sequence impedance matrix Zs given by (8.2.19) is a 3 3 matrix with nine sequence impedances, defined as follows: Zs ¼ 2 6 4 Z0 Z01 Z02 Z10 Z1 Z12 Z20 Z21 Z2 3 7 5 ð8:2:20Þ The diagonal impedances Z0, Z1, and Z2 in this matrix are the self-impedances of the zero-, positive-, and negative-sequence networks. The o¤-diagonal impedances are the mutual impedances between sequence networks.
Using the definitions of A; A1; Zp, and Zs, (8.2.19) is 2 6 4 Z0 Z01 Z02 Z10 Z1 Z12 Z20 Z21 Z2 3 7 5¼ 1 3 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 2 6 4 Zaa Zab Zac Zab Zbb Zbc Zac Zbc Zcc 3 7 5 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 ð8:2:21Þ Performing the indicated multiplications in (8.2.21), and using the identity ð1 þ a þ a2Þ ¼ 0, the following separate equations can be obtained (see Problem 8.18): Diagonal sequence impedances Z0 ¼ 1 3ðZaa þ Zbb þ Zcc þ 2Zab þ 2Zac þ 2ZbcÞ ð8:2:22Þ Z1 ¼ Z2 ¼ 1 3ðZaa þ Zbb þ Zcc Zab Zac ZbcÞ ð8:2:23Þ FIGURE 8.7 General three-phase impedance load (linear, bilateral network, nonrotating equipment) SECTION 8.2 SEQUENCE NETWORKS OF IMPEDANCE LOADS 439 Off-diagonal sequence impedances Z01 ¼ Z20 ¼ 1 3ðZaa þ a2Zbb þ aZcc aZab a2Zac ZbcÞ ð8:2:24Þ Z02 ¼ Z10 ¼ 1 3ðZaa þ aZbb þ a2Zcc a2Zab aZac ZbcÞ ð8:2:25Þ Z12 ¼ 1 3ðZaa þ a2Zbb þ aZcc þ 2aZab þ 2a2Zac þ 2ZbcÞ ð8:2:26Þ Z21 ¼ 1 3ðZaa þ aZbb þ a2Zcc þ 2a2Zab þ 2aZac þ 2ZbcÞ ð8:2:27Þ A symmetrical load is defined as a load whose sequence impedance matrix is diagonal; that is, all the mutual impedances in (8.2.24)–(8.2.27) are zero. Equating these mutual impedances to zero and solving, the following conditions for a symmetrical load are determined. When both Zaa ¼ Zbb ¼ Zcc Zab ¼ Zac ¼ Zbc 9 > > = > > ; conditions for a symmetrical load ð8:2:28Þ and (8.2.29) then Z01 ¼ Z10 ¼ Z02 ¼ Z20 ¼ Z12 ¼ Z21 ¼ 0 ð8:2:30Þ Z0 ¼ Zaa þ 2Zab ð8:2:31Þ Z1 ¼ Z2 ¼ Zaa Zab ð8:2:32Þ The conditions for a symmetrical load are that the diagonal phase impedances be equal and that the o¤-diagonal phase impedances be equal.
These conditions can be verified by using (8.2.28) and (8.2.29) with the FIGURE 8.8 Sequence networks of a three-phase symmetrical impedance load (linear, bilateral network, nonrotating equipment) 440 CHAPTER 8 SYMMETRICAL COMPONENTS identity ð1 þ a þ a2Þ ¼ 0 in (8.2.24)–(8.2.27) to show that all the mutual se-quence impedances are zero. Note that the positive- and negative-sequence impedances are equal for a symmetrical load, as shown by (8.2.32), and for a nonsymmetrical load, as shown by (8.2.23). This is always true for linear, symmetric impedances that represent nonrotating equipment such as trans-formers and transmission lines. However, the positive- and negative-sequence impedances of rotating equipment such as generators and motors are gener-ally not equal. Note also that the zero-sequence impedance Z0 is not equal to the positive- and negative-sequence impedances of a symmetrical load unless the mutual phase impedances Zab ¼ Zac ¼ Zbc are zero.
The sequence networks of a symmetrical impedance load are shown in Figure 8.8. Since the sequence impedance matrix Zs is diagonal for a sym-metrical load, the sequence networks are separate or uncoupled.
8.3 SEQUENCE NETWORKS OF SERIES IMPEDANCES Figure 8.9 shows series impedances connected between two three-phase buses denoted abc and a0b0c0. Self-impedances of each phase are denoted Zaa, Zbb, and Zcc. In general, the series network may also have mutual impedances between phases. The voltage drops across the series-phase impedances are given by 2 6 4 V an V a 0n Vbn Vb0n V cn V c 0n 3 7 5¼ 2 6 4 V aa0 Vbb0 V cc 0 3 7 5¼ 2 6 4 Zaa Zab Zac Zab Zbb Zbc Zac Zcb Zcc 3 7 5 2 6 4 Ia Ib Ic 3 7 5 ð8:3:1Þ Both self-impedances and mutual impedances are included in (8.3.1). It is assumed that the impedance matrix is symmetric, which corresponds to a bilateral network. It is also assumed that these impedances represent FIGURE 8.9 Three-phase series impedances (linear, bilateral network, nonrotating equipment) SECTION 8.3 SEQUENCE NETWORKS OF SERIES IMPEDANCES 441 nonrotating equipment. Typical examples are series impedances of transmis-sion lines and of transformers. Equation (8.3.1) has the following form: V p V p 0 ¼ ZpIp ð8:3:2Þ where V p is the vector of line-to-neutral voltages at bus abc, V p 0 is the vector of line-to-neutral voltages at bus a0b0c0, Ip is the vector of line currents, and Zp is the 3 3 phase impedance matrix for the series network. Equation (8.3.2) is now transformed to the sequence domain in the same manner that the load-phase impedances were transformed in Section 8.2. Thus, V s V s0 ¼ ZsIs ð8:3:3Þ where Zs ¼ A1ZpA ð8:3:4Þ From the results of Section 8.2, this sequence impedance Zs matrix is diago-nal under the following conditions: Zaa ¼ Zbb ¼ Zcc Zab ¼ Zac ¼ Zbc 9 > > = > > ; conditions for symmetrical series impedances ð8:3:5Þ and When the phase impedance matrix Zp of (8.3.1) has both equal self-impedances and equal mutual impedances, then (8.3.4) becomes Zs ¼ 2 6 4 Z0 0 0 0 Z1 0 0 0 Z2 3 7 5 ð8:3:6Þ where Z0 ¼ Zaa þ 2Zab ð8:3:7Þ and Z1 ¼ Z2 ¼ Zaa Zab ð8:3:8Þ and (8.3.3) becomes three uncoupled equations, written as follows: V0 V0 0 ¼ Z0I0 ð8:3:9Þ V1 V1 0 ¼ Z1I1 ð8:3:10Þ V2 V2 0 ¼ Z2I2 ð8:3:11Þ Equations (8.3.9)–(8.3.11) are represented by the three uncoupled sequence networks shown in Figure 8.10. From the figure it is apparent that for symmetrical series impedances, positive-sequence currents produce only positive-sequence voltage drops. Similarly, negative-sequence currents produce only negative-sequence voltage drops, and zero-sequence currents produce only zero-sequence voltage drops. However, if the series impedances 442 CHAPTER 8 SYMMETRICAL COMPONENTS are not symmetrical, then Zs is not diagonal, the sequence networks are coupled, and the voltage drop across any one sequence network depends on all three sequence currents.
8.4 SEQUENCE NETWORKS OF THREE-PHASE LINES Section 4.7 develops equations suitable for computer calculation of the series phase impedances, including resistances and inductive reactances, of three-phase overhead transmission lines. The series phase impedance matrix ZP for an untransposed line is given by Equation (4.7.19), and ^ ZP for a completely transposed line is given by (4.7.21)–(4.7.23). Equation (4.7.19) can be trans-formed to the sequence domain to obtain ZS ¼ A1ZPA ð8:4:1Þ ZS is the 3 3 series sequence impedance matrix whose elements are ZS ¼ 2 6 4 Z0 Z01 Z02 Z10 Z1 Z12 Z20 Z21 Z2 3 7 5 W=m ð8:4:2Þ In general ZS is not diagonal. However, if the line is completely trans-posed, ^ ZS ¼ A1 ^ ZPA ¼ 2 6 4 ^ Z0 0 0 0 ^ Z1 0 0 0 ^ Z2 3 7 5 ð8:4:3Þ where, from (8.3.7) and (8.3.8), FIGURE 8.10 Sequence networks of three-phase symmetrical series impedances (linear, bilateral network, nonrotating equipment) SECTION 8.4 SEQUENCE NETWORKS OF THREE-PHASE LINES 443 ^ Z0 ¼ ^ Zaaeq þ 2 ^ Zabeq ð8:4:4Þ ^ Z1 ¼ ^ Z2 ¼ ^ Zaaeq ^ Zabeq ð8:4:5Þ A circuit representation of the series sequence impedances of a completely transposed three-phase line is shown in Figure 8.11.
Section 4.11 develops equations suitable for computer calculation of the shunt phase admittances of three-phase overhead transmission lines. The shunt admittance matrix YP for an untransposed line is given by Equation (4.11.16), and ^ YP for a completely transposed three-phase line is given by (4.11.17).
Equation (4.11.16) can be transformed to the sequence domain to obtain YS ¼ A1YPA ð8:4:6Þ where YS ¼ GS þ jð2pf ÞCS ð8:4:7Þ CS ¼ 2 6 4 C0 C01 C02 C10 C1 C12 C20 C21 C2 3 7 5 F=m ð8:4:8Þ In general, CS is not diagonal. However, for the completely transposed line, ^ YS ¼ A1 ^ YPA ¼ 2 6 4 ^ y0 0 0 0 ^ y1 0 0 0 ^ y2 3 7 5 ¼ jð2pf Þ 2 6 4 ^ C0 0 0 0 ^ C1 0 0 0 ^ C2 3 7 5 ð8:4:9Þ FIGURE 8.11 Circuit representation of the series sequence impedances of a completely transposed three-phase line 444 CHAPTER 8 SYMMETRICAL COMPONENTS where ^ C0 ¼ ^ Caa þ 2 ^ Cab F=m ð8:4:10Þ ^ C1 ¼ ^ C2 ¼ ^ Caa ^ Cab F=m ð8:4:11Þ Since ^ Cab is negative, the zero-sequence capacitance ^ C0 is usually much less than the positive- or negative-sequence capacitance.
Circuit representations of the phase and sequence capacitances of a completely transposed three-phase line are shown in Figure 8.12.
8.5 SEQUENCE NETWORKS OF ROTATING MACHINES A Y-connected synchronous generator grounded through a neutral impedance Zn is shown in Figure 8.13. The internal generator voltages are designated Ea, Eb, and Ec, and the generator line currents are designated Ia, Ib, and Ic.
FIGURE 8.12 Circuit representations of the capacitances of a completely transposed three-phase line FIGURE 8.13 Y-connected synchronous generator SECTION 8.5 SEQUENCE NETWORKS OF ROTATING MACHINES 445 The sequence networks of the generator are shown in Figure 8.14. Since a three-phase synchronous generator is designed to produce balanced internal phase voltages Ea, Eb, Ec with only a positive-sequence component, a source voltage Eg1 is included only in the positive-sequence network. The sequence components of the line-to-ground voltages at the generator terminals are de-noted V0, V1, and V2 in Figure 8.14.
The voltage drop in the generator neutral impedance is ZnIn, which can be written as ð3ZnÞI0, since, from (8.1.27), the neutral current is three times the zero-sequence current. Since this voltage drop is due only to zero-sequence current, an impedance ð3ZnÞ is placed in the zero-sequence network of Figure 8.14 in series with the generator zero-sequence impedance Zg0.
The sequence impedances of rotating machines are generally not equal.
A detailed analysis of machine-sequence impedances is given in machine theory texts. We give only a brief explanation here.
When a synchronous generator stator has balanced three-phase positive-sequence currents under steady-state conditions, the net mmf produced by these positive-sequence currents rotates at the synchronous rotor speed in the same direction as that of the rotor. Under this condition, a high value of mag-netic flux penetrates the rotor, and the positive-sequence impedance Zg1 has a high value. Under steady-state conditions, the positive-sequence generator impedance is called the synchronous impedance.
When a synchronous generator stator has balanced three-phase negative-sequence currents, the net mmf produced by these currents rotates at syn-chronous speed in the direction opposite to that of the rotor. With respect to the rotor, the net mmf is not stationary but rotates at twice synchronous speed. Under this condition, currents are induced in the rotor windings that prevent the magnetic flux from penetrating the rotor. As such, the negative-sequence impedance Zg2 is less than the positive-sequence synchronous im-pedance.
When a synchronous generator has only zero-sequence currents, which are line (or phase) currents with equal magnitude and phase, then the net mmf produced by these currents is theoretically zero. The generator zero-sequence impedance Zg0 is the smallest sequence impedance and is due to leakage flux, end turns, and harmonic flux from windings that do not pro-duce a perfectly sinusoidal mmf.
FIGURE 8.14 Sequence networks of a Y-connected synchronous generator 446 CHAPTER 8 SYMMETRICAL COMPONENTS Typical values of machine-sequence impedances are listed in Table A.1 in the Appendix. The positive-sequence machine impedance is synchronous, transient, or subtransient. Synchronous impedances are used for steady-state conditions, such as in power-flow studies, which are described in Chapter 6.
Transient impedances are used for stability studies, which are described in Chapter 13, and subtransient impedances are used for short-circuit studies, which are described in Chapters 7 and 9. Unlike the positive-sequence im-pedances, a machine has only one negative-sequence impedance and only one zero-sequence impedance.
The sequence networks for three-phase synchronous motors and for three-phase induction motors are shown in Figure 8.15. Synchronous motors have the same sequence networks as synchronous generators, except that the sequence currents for synchronous motors are referenced into rather than out of the sequence networks. Also, induction motors have the same sequence networks as synchronous motors, except that the positive-sequence voltage FIGURE 8.15 Sequence networks of three-phase motors SECTION 8.5 SEQUENCE NETWORKS OF ROTATING MACHINES 447 source Em1 is removed. Induction motors do not have a dc source of mag-netic flux in their rotor circuits, and therefore Em1 is zero (or a short circuit).
The sequence networks shown in Figures 8.14 and 8.15 are simplified networks for rotating machines. The networks do not take into account such phenomena as machine saliency, saturation e¤ects, and more complicated transient e¤ects. These simplified networks, however, are in many cases accu-rate enough for power system studies.
EXAMPLE 8.5 Currents in sequence networks Draw the sequence networks for the circuit of Example 2.5 and calculate the sequence components of the line current. Assume that the generator neutral is grounded through an impedance Zn ¼ j10 W, and that the generator se-quence impedances are Zg0 ¼ j1 W, Zg1 ¼ j15 W, and Zg2 ¼ j3 W.
SOLUTION The sequence networks are shown in Figure 8.16. They are ob-tained by interconnecting the sequence networks for a balanced-D load, for FIGURE 8.16 Sequence networks for Example 8.5 448 CHAPTER 8 SYMMETRICAL COMPONENTS series-line impedances, and for a synchronous generator, which are given in Figures 8.5, 8.10, and 8.14.
It is clear from Figure 8.16 that I0 ¼ I2 ¼ 0 since there are no sources in the zero- and negative-sequence networks. Also, the positive-sequence gen-erator terminal voltage V1 equals the generator line-to-neutral terminal volt-age. Therefore, from the positive-sequence network shown in the figure and from the results of Example 2.5, I1 ¼ V1 ZL1 þ 1 3 ZD ¼ 25:83 73:78 A ¼ Ia Note that from (8.1.20), I1 equals the line current Ia, since I0 ¼ I2 ¼ 0.
9 The following example illustrates the superiority of using symmetrical com-ponents for analyzing unbalanced systems.
EXAMPLE 8.6 Solving unbalanced three-phase networks using sequence components A Y-connected voltage source with the following unbalanced voltage is ap-plied to the balanced line and load of Example 2.5.
2 6 4 V ag Vbg V cg 3 7 5¼ 2 6 4 277 0 260 120 295 þ115 3 7 5 volts The source neutral is solidly grounded. Using the method of symmet-rical components, calculate the source currents Ia, Ib, and Ic.
SOLUTION Using (8.1.13)–(8.1.15), the sequence components of the source voltages are: V0 ¼ 1 3ð277 0 þ 260 120 þ 295 115Þ ¼ 7:4425 þ j14:065 ¼ 15:912 62:11 volts V1 ¼ 1 3ð227 0 þ 260 120 þ 120 þ 295 115 þ 240Þ ¼ 1 3ð277 0 þ 260 0 þ 295 5Þ ¼ 276:96 j8:5703 ¼ 277:1 1:772 volts V2 ¼ 1 3ð277 0 þ 260 120 þ 240 þ 295 115 þ 120Þ ¼ 1 3ð277 0 þ 260 120 þ 295 235Þ ¼ 7:4017 j5:4944 ¼ 9:218 216:59 volts These sequence voltages are applied to the sequence networks of the line and load, as shown in Figure 8.17. The sequence networks of this figure SECTION 8.5 SEQUENCE NETWORKS OF ROTATING MACHINES 449 are uncoupled, and the sequence components of the source currents are easily calculated as follows: I0 ¼ 0 I1 ¼ V1 ZL1 þ ZD 3 ¼ 277:1 1:772 10:73 43:78 ¼ 25:82 45:55 A I2 ¼ V2 ZL2 þ ZD 3 ¼ 9:218 216:59 10:73 43:78 ¼ 0:8591 172:81 A Using (8.1.20)–(8.1.22), the source currents are: Ia ¼ ð0 þ 25:82 45:55 þ 0:8591 172:81Þ ¼ 17:23 j18:32 ¼ 25:15 46:76 A Ib ¼ ð0 þ 25:82 45:55 þ 240 þ 0:8591 172:81 þ 120Þ ¼ ð25:82 194:45 þ 0:8591 292:81Þ ¼ 24:67 j7:235 ¼ 25:71 196:34 A FIGURE 8.17 Sequence networks for Example 8.6 450 CHAPTER 8 SYMMETRICAL COMPONENTS Ic ¼ ð0 þ 25:82 45:55 þ 120 þ 0:8591 172:81 þ 240Þ ¼ ð25:82 74:45 þ 0:8591 52:81Þ ¼ 7:441 þ j25:56 ¼ 26:62 73:77 A You should calculate the line currents for this example without using symmetrical components, in order to verify this result and to compare the two solution methods (see Problem 8.33). Without symmetrical components, coupled KVL equations must be solved. With symmetrical components, the conversion from phase to sequence components decouples the networks as well as the resulting KVL equations, as shown above.
9 8.6 PER-UNIT SEQUENCE MODELS OF THREE-PHASE TWO-WINDING TRANSFORMERS Figure 8.18(a) is a schematic representation of an ideal Y–Y transformer grounded through neutral impedances ZN and Zn. Figures 8.18(b–d) show the per-unit sequence networks of this ideal transformer.
When balanced positive-sequence currents or balanced negative-sequence currents are applied to the transformer, the neutral currents are zero and there are no voltage drops across the neutral impedances. Therefore, the per-unit positive- and negative-sequence networks of the ideal Y–Y trans-former, Figures 8.18(b) and (c), are the same as the per-unit single-phase ideal transformer, Figure 3.9(a).
Zero-sequence currents have equal magnitudes and equal phase angles.
When per-unit sequence currents IA0 ¼ IB0 ¼ IC0 ¼ I0 are applied to the high-voltage windings of an ideal Y–Y transformer, the neutral current IN ¼ 3I0 flows through the neutral impedance ZN, with a voltage drop ð3ZNÞI0. Also, per-unit zero-sequence current I0 flows in each low-voltage winding [from (3.3.9)], and therefore 3I0 flows through neutral impedance Zn, with a voltage drop ð3I0ÞZn. The per-unit zero-sequence network, which includes the im-pedances ð3ZNÞ and ð3ZnÞ, is shown in Figure 8.18(b).
Note that if either one of the neutrals of an ideal transformer is un-grounded, then no zero sequence can flow in either the high- or low-voltage windings. For example, if the high-voltage winding has an open neutral, then IN ¼ 3I0 ¼ 0, which in turn forces I0 ¼ 0 on the low-voltage side. This can be shown in the zero-sequence network of Figure 8.18(b) by making ZN ¼ y, which corresponds to an open circuit.
The per-unit sequence networks of a practical Y–Y transformer are shown in Figure 8.19(a). These networks are obtained by adding external im-pedances to the sequence networks of the ideal transformer, as follows. The leakage impedances of the high-voltage windings are series impedances like the series impedances shown in Figure 8.9, with no coupling between phases SECTION 8.6 THREE-PHASE TWO-WINDING TRANSFORMERS 451 ðZab ¼ 0Þ. If the phase a, b, and c windings have equal leakage impedances ZH ¼ RH þ jXH, then the series impedances are symmetrical with sequence networks, as shown in Figure 8.10, where ZH0 ¼ ZH1 ¼ ZH2 ¼ ZH. Similarly, the leakage impedances of the low-voltage windings are symmetrical series impedances with ZX0 ¼ ZX1 ¼ ZX2 ¼ ZX. These series leakage impedances are shown in per-unit in the sequence networks of Figure 8.19(a).
The shunt branches of the practical Y–Y transformer, which represent exciting current, are equivalent to the Y load of Figure 8.3. Each phase in Figure 8.3 represents a core loss resistor in parallel with a magnetizing induc-tance. Assuming these are the same for each phase, then the Y load is sym-metrical, and the sequence networks are shown in Figure 8.4. These shunt FIGURE 8.18 Ideal Y–Y transformer 452 CHAPTER 8 SYMMETRICAL COMPONENTS branches are also shown in Figure 8.19(a). Note that ð3ZNÞ and ð3ZnÞ have already been included in the zero-sequence network.
The per-unit positive- and negative-sequence transformer impedances of the practical Y–Y transformer in Figure 8.19(a) are identical, which is always true for nonrotating equipment. The per-unit zero-sequence network, how-ever, depends on the neutral impedances ZN and Zn.
FIGURE 8.19 Per-unit sequence networks of practical Y–Y, Y–D, and D–D transformers SECTION 8.6 THREE-PHASE TWO-WINDING TRANSFORMERS 453 The per-unit sequence networks of the Y–D transformer, shown in Figure 8.19(b), have the following features: 1. The per-unit impedances do not depend on the winding connections.
That is, the per-unit impedances of a transformer that is connected Y–Y, Y–D, D–Y, or D–D are the same. However, the base voltages do depend on the winding connections.
2. A phase shift is included in the per-unit positive- and negative-sequence networks. For the American standard, the positive-sequence voltages and currents on the high-voltage side of the Y–D trans-former lead the corresponding quantities on the low-voltage side by 30. For negative sequence, the high-voltage quantities lag by 30.
3. Zero-sequence currents can flow in the Y winding if there is a neutral connection, and corresponding zero-sequence currents flow within the D winding. However, no zero-sequence current enters or leaves the D winding.
The phase shifts in the positive- and negative-sequence networks of Figure 8.19(b) are represented by the phase-shifting transformer of Figure 3.4.
Also, the zero-sequence network of Figure 8.19(b) provides a path on the Y side for zero-sequence current to flow, but no zero-sequence current can enter or leave the D side.
The per-unit sequence networks of the D–D transformer, shown in Figure 8.19(c), have the following features: 1. The positive- and negative-sequence networks, which are identical, are the same as those for the Y–Y transformer. It is assumed that the wind-ings are labeled so there is no phase shift. Also, the per-unit impedances do not depend on the winding connections, but the base voltages do.
2. Zero-sequence currents cannot enter or leave either D winding, al-though they can circulate within the D windings.
EXAMPLE 8.7 Solving unbalanced three-phase networks with transformers using per-unit sequence components A 75-kVA, 480-volt D/208-volt Y transformer with a solidly grounded neutral is connected between the source and line of Example 8.6. The transformer leakage reactance is Xeq ¼ 0:10 per unit; winding resistances and exciting cur-rent are neglected. Using the transformer ratings as base quantities, draw the per-unit sequence networks and calculate the phase a source current Ia.
SOLUTION The base quantities are Sbase1f ¼ 75=3 ¼ 25 kVA, VbaseHLN ¼ 480= ffiffiffi 3 p ¼ 277:1 volts, VbaseXLN ¼ 208= ffiffiffi 3 p ¼ 120:1 volts, and ZbaseX ¼ ð120:1Þ2=25;000 ¼ 0:5770 W. The sequence components of the actual source voltages are given in Figure 8.17. In per-unit, these voltages are 454 CHAPTER 8 SYMMETRICAL COMPONENTS V0 ¼ 15:91 62:11 277:1 ¼ 0:05742 62:11 per unit V1 ¼ 277:1 1:772 277:1 ¼ 1:0 1:772 per unit V2 ¼ 9:218 216:59 277:1 ¼ 0:03327 216:59 per unit The per-unit line and load impedances, which are located on the low-voltage side of the transformer, are ZL0 ¼ ZL1 ¼ ZL2 ¼ 1 85 0:577 ¼ 1:733 85 per unit Zload1 ¼ Zload2 ¼ ZD 3ð0:577Þ ¼ 10 40 0:577 ¼ 17:33 40 per unit FIGURE 8.20 Per-unit sequence networks for Example 8.7 SECTION 8.6 THREE-PHASE THREE-WINDING TRANSFORMERS 455 The per-unit sequence networks are shown in Figure 8.20. Note that the per-unit line and load impedances, when referred to the high-voltage side of the phase-shifting transformer, do not change [(see (3.1.26)]. Therefore, from Figure 8.20, the sequence components of the source currents are I0 ¼ 0 I1 ¼ V1 jXeq þ ZL1 þ Zload1 ¼ 1:0 1:772 j0:10 þ 1:733 85 þ 17:33 40 ¼ 1:0 1:772 13:43 þ j12:97 ¼ 1:0 1:772 18:67 44:0 ¼ 0:05356 45:77 per unit I2 ¼ V2 jXeq þ ZL2 þ Zload2 ¼ 0:03327 216:59 18:67 44:0 ¼ 0:001782 172:59 per unit The phase a source current is then, using (8.1.20), Ia ¼ I0 þ I1 þ I2 ¼ 0 þ 0:05356 45:77 þ 0:001782 172:59 ¼ 0:03511 j0:03764 ¼ 0:05216 46:19 per unit Using IbaseH ¼ 75;000 480 ffiffiffi 3 p ¼ 90:21 A, Ia ¼ ð0:05216Þð90:21Þ 46:19 ¼ 4:705 46:19 A 9 8.7 PER-UNIT SEQUENCE MODELS OF THREE-PHASE THREE-WINDING TRANSFORMERS Three identical single-phase three-winding transformers can be connected to form a three-phase bank. Figure 8.21 shows the general per-unit sequence networks of a three-phase three-winding transformer. Instead of labeling the windings 1, 2, and 3, as was done for the single-phase transformer, the letters H, M, and X are used to denote the high-, medium-, and low-voltage wind-ings, respectively. By convention, a common Sbase is selected for the H, M, and X terminals, and voltage bases VbaseH, VbaseM, and VbaseX are selected in proportion to the rated line-to-line voltages of the transformer.
For the general zero-sequence network, Figure 8.21(a), the connection between terminals H and H0 depends on how the high-voltage windings are connected, as follows: 1. Solidly grounded Y—Short H to H0.
2. Grounded Y through ZN—Connect ð3ZNÞ from H to H0.
456 CHAPTER 8 SYMMETRICAL COMPONENTS 3. Ungrounded Y—Leave H–H0 open as shown.
4. D—Short H0 to the reference bus.
Terminals X–X0 and M–M0 are connected in a similar manner.
The impedances of the per-unit negative-sequence network are the same as those of the per-unit positive-sequence network, which is always true for non-rotating equipment. Phase-shifting transformers, not shown in Figure 8.21(b), can be included to model phase shift between D and Y windings.
EXAMPLE 8.8 Three-winding three-phase transformer: per-unit sequence networks Three transformers, each identical to that described in Example 3.9, are connected as a three-phase bank in order to feed power from a 900-MVA, 13.8-kV generator to a 345-kV transmission line and to a 34.5-kV distribu-tion line. The transformer windings are connected as follows: 13:8-kV windings ðXÞ: D; to generator 199:2-kV windings ðHÞ: solidly grounded Y; to 345-kV line 19:92-kV windings ðMÞ: grounded Y through Zn ¼ j0:10 W; to 34:5-kV line FIGURE 8.21 Per-unit sequence networks of a three-phase three-winding transformer SECTION 8.7 THREE-PHASE THREE-WINDING TRANSFORMERS 457 The positive-sequence voltages and currents of the high- and medium-voltage Y windings lead the corresponding quantities of the low-voltage D winding by 30. Draw the per-unit sequence networks, using a three-phase base of 900 MVA and 13.8 kV for terminal X.
SOLUTION The per-unit sequence networks are shown in Figure 8.22. Since VbaseX ¼ 13:8 kV is the rated line-to-line voltage of terminal X, VbaseM ¼ ffiffiffi 3 p ð19:92Þ ¼ 34:5 kV, which is the rated line-to-line voltage of terminal M.
The base impedance of the medium-voltage terminal is then ZbaseM ¼ ð34:5Þ2 900 ¼ 1:3225 W Therefore, the per-unit neutral impedance is Zn ¼ j0:10 1:3225 ¼ j0:07561 per unit FIGURE 8.22 Per-unit sequence networks for Example 8.8 458 CHAPTER 8 SYMMETRICAL COMPONENTS and ð3ZnÞ ¼ j0:2268 is connected from terminal M to M0 in the per-unit zero-sequence network. Since the high-voltage windings have a solidly grounded neutral, H to H0 is shorted in the zero-sequence network. Also, phase-shifting transformers are included in the positive- and negative-sequence networks. 9 8.8 POWER IN SEQUENCE NETWORKS The power delivered to a three-phase network can be determined from the power delivered to the sequence networks. Let Sp denote the total com-plex power delivered to the three-phase load of Figure 8.7, which can be calculated from Sp ¼ V agI a þ VbgI b þ V cgI c ð8:8:1Þ Equation (8.8.1) is also valid for the total complex power delivered by the three-phase generator of Figure 8.13, or for the complex power delivered to any three-phase bus. Rewriting (8.8.1) in matrix format, Sp ¼ ½V agVbgV cg 2 6 4 I a I b I c 3 7 5 ¼ V T p I p ð8:8:2Þ where T denotes transpose and denotes complex conjugate. Now, using (8.1.9) and (8.1.16), Sp ¼ ðAV sÞTðAIsÞ ¼ V T s ½ATAI s ð8:8:3Þ Using the definition of A, which is (8.1.8), to calculate the term within the brackets of (8.8.3), and noting that a and a2 are conjugates, ATA ¼ 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 T2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 ¼ 2 6 4 1 1 1 1 a2 a 1 a a2 3 7 5 2 6 4 1 1 1 1 a a2 1 a2 a 3 7 5 ¼ 2 6 4 3 0 0 0 3 0 0 0 3 3 7 5 ¼ 3U ð8:8:4Þ SECTION 8.8 POWER IN SEQUENCE NETWORKS 459 Equation (8.8.4) can now be used in (8.8.3) to obtain Sp ¼ 3V T s I s ¼ 3½V0 þ V1 þ V2 2 6 4 I 0 I 1 I 2 3 7 5 ð8:8:5Þ Sp ¼ 3ðV0I 0 þ V1I 1 þ V2I 2 Þ ¼ 3Ss ð8:8:6Þ Thus, the total complex power Sp delivered to a three-phase network equals three times the total complex power Ss delivered to the sequence networks.
The factor of 3 occurs in (8.8.6) because ATA ¼ 3U, as shown by (8.8.4). It is possible to eliminate this factor of 3 by defining a new transfor-mation matrix A1 ¼ ð1= ffiffiffi 3 p ÞA such that AT 1 A 1 ¼ U, which means that A1 is a unitary matrix. Using A1 instead of A, the total complex power delivered to three-phase networks would equal the total complex power delivered to the sequence networks. However, standard industry practice for symmetrical components is to use A, defined by (8.1.8).
EXAMPLE 8.9 Power in sequence networks Calculate Sp and Ss delivered by the three-phase source in Example 8.6.
Verify that Sp ¼ 3Ss.
SOLUTION Using (8.5.1), Sp ¼ ð277 0Þð25:15 þ46:76Þ þ ð260 120Þð25:71 196:34Þ þ ð295 115Þð26:62 73:77Þ ¼ 6967 46:76 þ 6685 43:66 þ 7853 41:23 ¼ 15;520 þ j14;870 ¼ 21;490 43:78 VA In the sequence domain, Ss ¼ V0I 0 þ V1I 1 þ V2I 2 ¼ 0 þ ð277:1 1:77Þð25:82 45:55Þ þ ð9:218 216:59Þð0:8591 172:81Þ ¼ 7155 43:78 þ 7:919 43:78 ¼ 5172 þ j4958 ¼ 7163 43:78 VA Also, 3Ss ¼ 3ð7163 43:78Þ ¼ 21;490 43:78 ¼ Sp 9 460 CHAPTER 8 SYMMETRICAL COMPONENTS M U L T I P L E C H O I C E Q U E S T I O N S SECTION 8.1 8.1 Positive-sequence components consist of three phasors with _ magnitudes, and _ phase displacement in positive sequence; negative-sequence compo-nents consist of three phasors with _ magnitudes, and _ phase dis-placement in negative sequence; and zero-sequence components consist of three phasors with _ magnitudes, and _ phase displacement. Fill in the Blanks.
8.2 In symmetrical-component theory, express the complex-number operator a ¼ 1 120 in exponential and rectangular forms.
8.3 In terms of sequence components of phase a given by Va0 ¼ V0; Va1 ¼ V1 and Va2 ¼ V2, give expressions for the phase voltages Va, Vb, and Vc.
Va ¼ ___; Vb ¼ __; Vc ¼ ___ 8.4 The sequence components V0, V1, and V2 can be expressed in terms of phase compo-nents Va, Vb, and Vc.
V0 ¼ __; V1 ¼ ___; V2 ¼ __ 8.5 In a balanced three-phase system, what is the zero-sequence voltage?
V0 ¼ ___ 8.6 In an unblanced three-phase system, line-to-neutral voltage _ have a zero-sequence component, whereas line-to-line voltages __ have a zero-sequence component. Fill in the Blanks.
8.7 Can the symmetrical component transformation be applied to currents, just as applied to voltages?
(a) Yes (b) No 8.8 In a three-phase Wye-connected system with a neutral, express the neutral current in terms of phase currents and sequence-component terms.
In ¼ __ ¼ ___ 8.9 In a balanced Wye-connected system, what is the zero-sequence component of the line currents?
8.10 In a delta-connected three-phase system, line currents have no zero-sequence component.
(a) True (b) False 8.11 Balanced three-phase systems with positive sequence do not have zero-sequence and negative-sequence components.
(a) True (b) False 8.12 Unbalanced three-phase systems may have nonzero values for all sequence components.
(a) True (b) False SECTION 8.2 8.13 For a balanced-Y impedance load with per-phase impedance of ZY and A neutral impedance Zn connected between the load neutral and the ground, the 3 3 phase-impedance matrix will consist of equal diagonal elements given by _, and equal nondiagonal elements given by _. Fill in the Blanks.
MULTIPLE CHOICE QUESTIONS 461 8.14 Express the sequence impedance matrix Zs in terms of the phase-impedance matrix Zp, and the transformation matrix A which relates Vp ¼ AVs and I p ¼ AIs.
Zs ¼ _. Fill in the Blank.
8.15 The sequence impedance matrix Zs for a balanced-Y load is a diagonal matrix and the sequence networks are uncoupled.
(a) True (b) False 8.16 For a balanced-Y impedance load with per-phase impedance of ZY and a neutral im-pedance Zn, the zero-sequence voltage V0 ¼ Z0 I0, where Z0 ¼ _. Fill in the Blank.
8.17 For a balanced- load with per-phase impedance of Z the equivalent Y-load will have an open neutral; for the corresponding uncoupled sequence networks, Z0 ¼ _, Z1 ¼ _, and Z2 ¼ _. Fill in the Blanks.
8.18 For a three-phase symmetrical impedance load, the sequence impedance matrix is _ and hence the sequence networks are coupled/uncoupled.
SECTION 8.3 8.19 Sequence networks for three-phase symmetrical series impedances are coupled/ uncoupled; positive-sequence currents produce only _ voltage drops.
SECTION 8.4 8.20 The series sequence impedance matrix of a completely transposed three-phase line is _, with its nondiagonal elements equal to _. Fill in the Blanks.
SECTION 8.5 8.21 A Y-connected synchronous generator grounded through a neutral impedance Zn, with a zero-sequence impedance Zg0, will have zero-sequence impedance Z0 ¼ _ in its zero-sequence network. Fill in the Blank.
8.22 In sequence networks, a Y-connected synchronous generator is represented by its source per-unit voltage only in _ network, while synchronous/transient/sub-transient impedance is used in positive-sequence network for short-circuit studies.
8.23 In the positive-sequence network of a synchronous motor, a source voltage is repre-sented, whereas in that of an induction motor, the source voltage does/does not come into picture.
8.24 With symmetrical components, the conversion from phase to sequence components decouples the networks and the resulting kVL equations.
(a) True (b) False SECTION 8.6 8.25 Consider the per-unit sequence networks of Y-Y, Y-, and transformers, with neutral impedances of ZN on the high-voltage Y-side, and Zn on the low-voltage Y-side. Answer the following: (i) Zero-sequence currents can/cannot flow in the Y winding with a neutral connection; corresponding zero-sequence currents do/do not flow within the delta winding; 462 CHAPTER 8 SYMMETRICAL COMPONENTS however zero-sequence current does/does not enter or leave the winding. In zero-sequence network, 1/2/3 times the neutral impedance comes into play in series.
(ii) In Y(HV)- (LV) transformers, if a phase shift is included as per the American-standard notation, the ratio _ is used in positive-sequence network, and the ratio _ is used in the negative-sequence network.
(iii) The base voltages depend on the winding connections; the per-unit impedances do/do not depend on the winding connections.
SECTION 8.7 8.26 In per-unit sequence models of three-phase three-winding transformers, for the general zero-sequence network, the connection between terminals H and H0 depends on how the high-voltage windings are connected: (i) For solidly grounded Y, _ H to H0: (ii) For grounded Y through Zn, connect _ from H to H0.
(iii) For ungrounded Y, leave HH0 _.
(iv) For , _ H0 to the reference bus.
SECTION 8.8 8.27 The total complex power delivered to a three-phase network equals 1/2/3 times the total complex power delivered to the sequence networks.
8.28 Express the complex power Ss Delivered to the sequence networks in terms of se-quence voltages and sequence currents.
Ss ¼ _ P R O B L E M S SECTION 8.1 8.1 Using the operator a ¼ 1 120, evaluate the following in polar form: (a) ða 1Þ= ð1 þ a a2Þ, (b) ða2 þ a þ jÞ=ð ja þ a2Þ, (c) ð1 þ aÞð1 þ a2Þ, (d) ða a2Þða2 1Þ.
8.2 Using a ¼ 1 120, evaluate the following in rectangular form: a. a10 b. ð jaÞ10 c. ð1 aÞ3 d. ea Hint for (d): eðxþ jyÞ ¼ exe jy ¼ ex y, where y is in radians.
8.3 Determine the symmetrical components of the following line currents: (a) Ia ¼ 5 90, Ib ¼ 5 320, Ic ¼ 5 220 A; (b) Ia ¼ j50, Ib ¼ 50, Ic ¼ 0 A.
8.4 Find the phase voltages V an, Vbn, and V cn whose sequence components are: V0 ¼ 50 80, V1 ¼ 100 0, V2 ¼ 50 90 V.
PROBLEMS 463 8.5 For the unbalanced three-phase system described by Ia ¼ 12 0A; Ib ¼ 6 90A; IC ¼ 8 150A compute the symmetrical components I0, I1, I2.
8.6 (a) Given the symmetrical components to be V0 ¼ 10 0V; V1 ¼ 80 30V; V2 ¼ 40 30V determine the unbalanced phase voltages Va, Vb, and Vc.
(b) Using the results of part (a), calculate the line-to-line voltages Vab, Vbc, and Vca.
Then determine the symmetrical components of these ling-to-line voltages, the sym-metrical components of the corresponding phase voltages, and the phase voltages.
Compare them with the result of part (a). Comment on why they are di¤erent, even though either set will result in the same line-to-line voltages.
8.7 One line of a three-phase generator is open circuited, while the other two are short-circuited to ground. The line currents are Ia ¼ 0, Ib ¼ 1000 150, and Ic ¼ 1000 þ30 A. Find the symmetrical components of these currents. Also find the current into the ground.
8.8 Let an unbalanced, three-phase, Wye-connected load (with phase impedances of Za, Zb, and Zc) be connected to a balanced three-phase supply, resulting in phase voltages of Va, Vb, and Vc across the corresponding phase impedances.
Choosing Vab as the reference, show that Vab; 0 ¼ 0; Vab; 1 ¼ ffiffiffi 3 p Va; 1e j30; Vab; 2 ¼ ffiffiffi 3 p Va; 2ej30: 8.9 Reconsider Problem 8.8 and choosing Vbc as the reference, show that Vbc; 0 ¼ 0; Vbc; 1 ¼ j ffiffiffi 3 p Va; 1; Vbc; 2 ¼ j ffiffiffi 3 p Va; 2: 8.10 Given the line-to-ground voltages V ag ¼ 280 0, Vbg ¼ 250 110, and V cg ¼ 290 130 volts, calculate (a) the sequence components of the line-to-ground voltages, denoted VLg0, VLg1, and VLg2; (b) line-to-line voltages V ab, Vbc, and V ca; and (c) sequence com-ponents of the line-to-line voltages VLL0, VLL1, and VLL2. Also, verify the following general relation: VLL0 ¼ 0, VLL1 ¼ ffiffiffi 3 p VLg1 þ30, and VLL2 ¼ ffiffiffi 3 p VLg2 30 volts.
8.11 A balanced D-connected load is fed by a three-phase supply for which phase C is open and phase A is carrying a current of 10 0 A. Find the symmetrical components of the line currents. (Note that zero-sequence currents are not present for any three-wire system.) 8.12 A Y-connected load bank with a three-phase rating of 500 kVA and 2300 V consists of three identical resistors of 10.58 W. The load bank has the following applied voltages: Vab ¼ 1840 82:8, Vbc ¼ 2760 41:4, and Vca ¼ 2300 180 V. Determine the symmetrical components of (a) the line-to-line voltages Vab0, Vab1, and Vab2; (b) the line-to-neutral voltages Van0, Van1, and Van2; (c) and the line currents Ia0, Ia1, and Ia2. (Note that the absence of a neutral connection means that zero-sequence cur-rents are not present.) SECTION 8.2 8.13 The currents in a D load are Iab ¼ 10 0, Ibc ¼ 15 90, and Ica ¼ 20 90 A. Calcu-late (a) the sequence components of the D-load currents, denoted ID0, ID1, ID2; (b) the line currents Ia, Ib, and Ic, which feed the D load; and (c) sequence components of the line currents IL0, IL1, and IL2. Also, verify the following general relation: IL0 ¼ 0, IL1 ¼ ffiffiffi 3 p ID1 30, and IL2 ¼ ffiffiffi 3 p ID2 þ30 A.
464 CHAPTER 8 SYMMETRICAL COMPONENTS 8.14 The voltages given in Problem 8.10 are applied to a balanced-Y load consisting of ð12 þ j16Þ ohms per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate I0, I1, and I2, the sequence components of the line currents.
Then calculate the line currents Ia, Ib, and Ic.
8.15 Repeat Problem 8.14 with the load neutral open.
8.16 Repeat Problem 8.14 for a balanced-D load consisting of ð12 þ j16Þ ohms per phase.
8.17 Repeat Problem 8.14 for the load shown in Example 8.4 (Figure 8.6).
8.18 Perform the indicated matrix multiplications in (8.2.21) and verify the sequence im-pedances given by (8.2.22)–(8.2.27).
8.19 The following unbalanced line-to-ground voltages are applied to the balanced-Y load shown in Figure 3.3: V ag ¼ 100 0, Vbg ¼ 75 180, and V cg ¼ 50 90 volts. The Y load has ZY ¼ 3 þ j4 W per phase with neutral impedance Zn ¼ j1 W. (a) Calculate the line currents Ia, Ib, and Ic without using symmetrical components. (b) Calculate the line currents Ia, Ib, and Ic using symmetrical components. Which method is easier?
8.20 (a) Consider three equal impedances of (j27) W connected in D. Obtain the sequence networks.
(b) Now, with a mutual impedance of (j6) W between each pair of adjacent branches in the D-connected load of part (a), how would the sequence networks change?
8.21 The three-phase impedance load shown in Figure 8.7 has the following phase imped-ance matrix: Zp ¼ 2 6 4 ð6 þ j10Þ 0 0 0 ð6 þ j10Þ 0 0 0 ð6 þ j10Þ 3 7 5 W Determine the sequence impedance matrix Zs for this load. Is the load symmetrical?
8.22 The three-phase impedance load shown in Figure 8.7 has the following sequence im-pedance matrix: ZS ¼ 2 6 4 ð8 þ j12Þ 0 0 0 4 0 0 0 4 3 7 5 W Determine the phase impedance matrix Zp for this load. Is the load symmetrical?
8.23 Consider a three-phase balanced Y-connected load with self and mutual impedances as shown in Figure 8.23. Let the load neutral be grounded through an impedance Zn.
Using Kirchho¤’s laws, develop the equations for line-to-neutral voltages, and then determine the elements of the phase impedance matrix. Also find the elements of the corresponding sequence impedance matrix.
8.24 A three-phase balanced voltage source is applied to a balanced Y-connected load with ungrounded neutral. The Y-connected load consists of three mutually coupled re-actances, where the reactance of each phase is j12 W and the mutual coupling between any two phases is j4 W. The line-to-line source voltage is 100 ffiffiffi 3 p V. Determine the line currents (a) by mesh analysis without using symmetrical components, and (b) using symmetrical components.
PROBLEMS 465 8.25 A three-phase balanced Y-connected load with series impedances of ð8 þ j24Þ W per phase and mutual impedance between any two phases of j4 W is supplied by a three-phase unbalanced source with line-to-neutral voltages of Van ¼ 200 25, Vbn ¼ 100 155, Vcn ¼ 80 100 V. The load and source neutrals are both solidly grounded. Determine: (a) the load sequence impedance matrix, (b) the symmetrical components of the line-to-neutral voltages, (c) the symmetrical components of the load currents, and (d) the load currents.
SECTION 8.3 8.26 Repeat Problem 8.14 but include balanced three-phase line impedances of ð3 þ j4Þ ohms per phase between the source and load.
8.27 Consider the flow of unbalanced currents in the symmetrical three-phase line sec-tion with neutral conductor as shown in Figure 8.24. (a) Express the voltage drops across the line conductors given by Vaa 0, Vbb 0, and Vcc 0 in terms of line currents, self-impedances defined by Zs ¼ Zaa þ Znn 2Zan, and mutual impedances defined by Zm ¼ Zab þ Znn 2Zan. (b) Show that the sequence components of the voltage drops between the ends of the line section can be written as Vaa 00 ¼ Z0Ia0, Vaa 01 ¼ Z1Ia1, and Vaa 02 ¼ Z2Ia2, where Z0 ¼ Zs þ 2Zm ¼ Zaa þ 2Zab þ 3Znn 6Zan and Z1 ¼ Z2 ¼ Zs Zm ¼ Zaa Zab.
FIGURE 8.23 Problem 8.23 FIGURE 8.24 Problem 8.27 466 CHAPTER 8 SYMMETRICAL COMPONENTS 8.28 Let the terminal voltages at the two ends of the line section shown in Figure 8.24 be given by: Van ¼ ð182 þ j70Þ kV Van 0 ¼ ð154 þ j28Þ kV Vbn ¼ ð72:24 j32:62Þ kV Vbn 0 ¼ ð44:24 þ j74:62Þ kV Vcn ¼ ð170:24 þ j88:62Þ kV Vcn 0 ¼ ð198:24 þ j46:62Þ kV The line impedances are given by: Zaa ¼ j60 W Zab ¼ j20 W Znn ¼ j80 W Zan ¼ 0 (a) Compute the line currents using symmetrical components. (Hint: See Problem 8.27.) (b) Compute the line currents without using symmetrical components.
8.29 A completely transposed three-phase transmission line of 200 km in length has the following symmetrical sequence impedances and sequence admittances: Z1 ¼ Z2 ¼ j0:5 W=km; Z0 ¼ j2 W=km Y1 ¼ Y2 ¼ j3 109 s=m; Y0 ¼ j1 109 s=m Set up the nominal P sequence circuits of this medium-length line.
SECTION 8.5 8.30 As shown in Figure 8.25, a balanced three-phase, positive-sequence source with VAB ¼ 480 0 volts is applied to an unbalanced D load. Note that one leg of the D is open. Determine: (a) the load currents IAB and IBC; (b) the line currents IA, IB, and IC, which feed the D load; and (c) the zero-, positive-, and negative-sequence components of the line currents.
8.31 A balanced Y-connected generator with terminal voltage Vbc ¼ 200 0 volts is con-nected to a balanced-D load whose impedance is 10 40 ohms per phase. The line im-pedance between the source and load is 0:5 80 ohm for each phase. The generator neutral is grounded through an impedance of j5 ohms. The generator sequence im-pedances are given by Zg0 ¼ j7, Zg1 ¼ j15, and Zg2 ¼ j10 ohms. Draw the sequence networks for this system and determine the sequence components of the line currents.
FIGURE 8.25 Problem 8.30 PROBLEMS 467 8.32 In a three-phase system, a synchronous generator supplies power to a 200-volt syn-chronous motor through a line having an impedance of 0:5 80 ohm per phase. The motor draws 5 kW at 0.8 p.f. leading and at rated voltage. The neutrals of both the generator and motor are grounded through impedances of j5 ohms. The sequence impedances of both machines are Z0 ¼ j5, Z1 ¼ j15, and Z2 ¼ j10 ohms. Draw the sequence networks for this system and find the line-to-line voltage at the generator terminals. Assume balanced three-phase operation.
8.33 Calculate the source currents in Example 8.6 without using symmetrical components.
Compare your solution method with that of Example 8.6. Which method is easier?
8.34 A Y-connected synchronous generator rated 20 MVA at 13.8 kV has a positive-sequence reactance of j2.38 W, negative-sequence reactance of j3.33 W, and zero-sequence reactance of j0.95 W. The generator neutral is solidly grounded. With the generator operating unloaded at rated voltage, a so-called single line-to-ground fault occurs at the machine terminals. During this fault, the line-to-ground voltages at the generator terminals are Vag ¼ 0, Vbg ¼ 8:071 102:25, and Vcg ¼ 8:071 102:25 kV.
Determine the sequence components of the generator fault currents and the generator fault currents. Draw a phasor diagram of the pre-fault and post-fault generator terminal voltages. (Note: For this fault, the sequence components of the generator fault currents are all equal to each other.) 8.35 Figure 8.26 shows a single-line diagram of a three-phase, interconnected generator-reactor system, in which the given per-unit reactances are based on the ratings of the individual pieces of equipment. If a three-phase short-circuit occurs at fault point F, obtain the fault MVA and fault current in kA, if the pre-fault busbar line-to-line volt-age is 13.2 kV. Choose 100 MVA as the base MVA for the system.
8.36 Consider Figures 8.13 and 8.14 of the text with reference to a Y-connected synchro-nous generator (grounded through a neutral impedance Zn) operating at no load. For a line-to-ground fault occurring on phase a of the generator, list the constraints on the currents and voltages in the phase domain, transform those into the sequence domain, and then obtain a sequence-network representation. Also, find the expression for the fault current in phase a.
FIGURE 8.26 One-line diagram for Problem 8.35 468 CHAPTER 8 SYMMETRICAL COMPONENTS 8.37 Reconsider the synchronous generator of Problem 8.36. Obtain sequence-network representations for the following fault conditions.
(a) A short-circuit between phases b and c.
(b) A double line-to-ground fault with phases b and c grounded.
SECTION 8.6 8.38 Three single-phase, two-winding transformers, each rated 450 MVA, 20 kV/288.7 kV, with leakage reactance Xeq ¼ 0:12 per unit, are connected to form a three-phase bank.
The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit zero-, positive-, and negative-sequence networks if the low-voltage wind-ings are connected: (a) in D with American standard phase shift, (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and excit-ing current are neglected.
8.39 The leakage reactance of a three-phase, 500-MVA, 345 Y/23 D-kV transformer is 0.09 per unit based on its own ratings. The Y winding has a solidly grounded neutral.
Draw the sequence networks. Neglect the exciting admittance and assume American standard phase shift.
8.40 Choosing system bases to be 360/24 kV and 100 MVA, redraw the sequence networks for Problem 8.39.
8.41 Draw the zero-sequence reactance diagram for the power system shown in Figure 3.33.
The zero-sequence reactance of each generator and of the synchronous motor is 0.05 per unit based on equipment ratings. Generator 2 is grounded through a neutral reactor of 0.06 per unit on a 100-MVA, 18-kV base. The zero-sequence reactance of each transmission line is assumed to be three times its positive-sequence reactance. Use the same base as in Problem 3.29.
8.42 Three identical Y-connected resistors of 1:0 0 per unit form a load bank, which is supplied from the low-voltage Y-side of a Y D transformer. The neutral of the load is not connected to the neutral of the system. The positive- and negative-sequence currents flowing toward the resistive load are given by Ia; 1 ¼ 1 4:5 per unit; Ia; 2 ¼ 0:25 250 per unit and the corresponding voltages on the low-voltage Y-side of the transformer are Van; 1 ¼ 1 45 per unit (Line-to-neutral voltage base) Van; 2 ¼ 0:25 250 per unit (Line-to-neutral voltage base) Determine the line-to-line voltages and the line currents in per unit on the high-voltage side of the transformer. Account for the phase shift.
SECTION 8.7 8.43 Draw the positive-, negative-, and zero-sequence circuits for the transformers shown in Figure 3.34. Include ideal phase-shifting transformers showing phase shifts deter-mined in Problem 3.32. Assume that all windings have the same kVA rating and that the equivalent leakage reactance of any two windings with the third winding open is 0.10 per unit. Neglect the exciting admittance.
8.44 A single-phase three-winding transformer has the following parameters: Z1 ¼ Z2 ¼ Z3 ¼ 0 þ j0:05, Gc ¼ 0, and Bm ¼ 0:2 per unit. Three identical transformers, as PROBLEMS 469 described, are connected with their primaries in Y (solidly grounded neutral) and with their secondaries and tertiaries in D. Draw the per-unit sequence networks of this transformer bank.
SECTION 8.8 8.45 For Problem 8.14, calculate the real and reactive power delivered to the three-phase load.
8.46 A three-phase impedance load consists of a balanced-D load in parallel with a balanced-Y load. The impedance of each leg of the D load is ZD ¼ 6 þ j6 W, and the impedance of each leg of the Y load is ZY ¼ 2 þ j2 W. The Y load is grounded through a neutral impedance Zn ¼ j1 W. Unbalanced line-to-ground source volt-ages Vag, Vbg, and Vcg with sequence components V0 ¼ 10 60, V1 ¼ 100 0, and V2 ¼ 15 200 volts are applied to the load. (a) Draw the zero-, positive-, and negative-sequence networks. (b) Determine the complex power delivered to each sequence network. (c) Determine the total complex power delivered to the three-phase load.
8.47 For Problem 8.12, compute the power absorbed by the load using symmetrical com-ponents. Then verify the answer by computing directly without using symmetrical components.
8.48 For Problem 8.25, determine the complex power delivered to the load in terms of symmetrical components. Verify the answer by adding up the complex power of each of the three phases.
8.49 Using the voltages of Problem 8.6(a) and the currents of Problem 8.5, compute the complex power dissipated based on (a) phase components, and (b) symmetrical com-ponents.
C AS E S T U DY QU E S T I ON S A.
What are the advantages of SF6 circuit breakers for applications at or above 72.5 kV?
B.
What are the properties of SF6 that make it make it advantageous as a medium for interrupting an electric arc?
R E F E R E N C E S 1.
Westinghouse Electric Corporation, Applied Protective Relaying (Newark, NJ: West-inghouse, 1976).
2.
P. M. Anderson, Analysis of Faulted Power Systems (Ames, IA: Iowa State University Press, 1973).
3.
W. D. Stevenson, Jr., Elements of Power System Analysis, 4th ed. (New York: McGraw-Hill, 1982).
4.
D. Dufournet, ‘‘Circuit Breakers Go High Voltage,’’ IEEE Power & Energy Maga-zine, 7, 1(January/February 2009), pp. 34–40.
470 CHAPTER 8 SYMMETRICAL COMPONENTS |
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Chicken McNugget Theorem
The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively primepositive integers, the greatest integer that cannot be written in the form for nonnegative integers is .
A consequence of the theorem is that there are exactly positive integers which cannot be expressed in the form . The proof is based on the fact that in each pair of the form , exactly one element is expressible.
Contents
1 Origins
2 Proof Without Words
3 Proof 1
4 Proof 2
5 Corollary
6 Generalization
7 Problems
7.1 Introductory
7.2 Intermediate
7.3 Olympiad
7.4 See Also
Origins
There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.
Proof Without Words
Example using m= and n=
Proof 1
Definition. An integer will be called purchasable if there exist nonnegative integers such that .
We would like to prove that is the largest non-purchasable integer. We are required to show that:
(1) is non-purchasable
(2) Every is purchasable
Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.
Lemma. Let be the set of solutions to . Then for any .
Proof: By Bezout's Lemma, there exist integers such that . Then . Hence is nonempty. It is easy to check that for all . We now prove that there are no others. Suppose and are solutions to . Then implies . Since and are coprime and divides , divides and . Similarly . Let be integers such that and . Then implies We have the desired result.
Lemma. For any integer , there exists unique such that .
Proof: By the division algorithm, there exists one and only one such that .
Lemma. is purchasable if and only if .
Proof: If , then we may simply pick so is purchasable. If , then if and if , hence at least one coordinate of is negative for all . Thus is not purchasable.
Thus the set of non-purchasable integers is . We would like to find the maximum of this set. Since both are positive, the maximum is achieved when and so that .
Proof 2
We start with this statement taken from Proof 2 of Fermat's Little Theorem:
"Let . Then, we claim that the set , consisting of the product of the elements of with , taken modulo , is simply a permutation of . In other words,
Clearly none of the for are divisible by , so it suffices to show that all of the elements in are distinct. Suppose that for . Since , by the cancellation rule, that reduces to , which is a contradiction."
Because and are coprime, we know that multiplying the residues of by simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form , is and is the original residue. We now prove the following lemma.
Lemma: For any nonnegative integer , is the least purchasable number .
Proof: Any number that is less than and congruent to it can be represented in the form , where is a positive integer. If this is purchasable, we can say for some nonnegative integers . This can be rearranged into , which implies that is a multiple of (since ). We can say that for some positive integer , and substitute to get . Because , , and . We divide by to get . However, we defined to be a positive integer, and all positive integers are greater than or equal to . Therefore, we have a contradiction, and is the least purchasable number congruent to .
This means that because is purchasable, every number that is greater than and congruent to it is also purchasable (because these numbers are in the form where ). Another result of this Lemma is that is the greatest number that is not purchasable. , so , which shows that is the greatest number in the form . Any number greater than this and congruent to some is purchasable, because that number is greater than . All numbers are congruent to some , and thus all numbers greater than are purchasable.
Putting it all together, we can say that for any coprime and , is the greatest number not representable in the form for nonnegative integers .
Corollary
This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.
Lemma: For any integer , exactly one of the integers , is not purchasable.
Proof: Because every number is congruent to some residue of permuted by , we can set for some . We can break this into two cases.
Case 1: . This implies that is not purchasable, and that . is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself that is purchasable. Therefore, and , so is purchasable.
Case 2: . This implies that is purchasable, and that . Again, because is the least number congruent to itself that is purchasable, and because and , is not purchasable.
We now limit the values of to all integers , which limits the values of to . Because and are coprime, only one of them can be a multiple of . Therefore, , showing that is not an integer and that and are integers. We can now set limits that are equivalent to the previous on the values of and so that they cover all integers form to without overlap: and . There are values of , and each is paired with a value of , so we can make different ordered pairs of the form . The coordinates of these ordered pairs cover all integers from to inclusive, and each contains exactly one not-purchasable integer, so that means that there are different not-purchasable integers from to . All integers greater than are purchasable, so that means there are a total of integers that are not purchasable.
In other words, for every pair of coprime integers , there are exactly nonnegative integers that cannot be represented in the form for nonnegative integers .
Generalization
If and are not relatively prime, then we can simply rearrange into the form and are relatively prime, so we apply Chicken McNugget to find a bound We can simply multiply back into the bound to get Therefore, all multiples of greater than are representable in the form for some non-negative integers .
Problems
Introductory
Marcy buys paint jars in containers of and . What's the largest number of paint jars that Marcy can't obtain?
Answer: containers
Bay Area Rapid food sells chicken nuggets. You can buy packages of or . What is the largest integer such that there is no way to buy exactly nuggets? Can you Generalize? (Source: The Art and Craft of Problem Solving)
Answer:
If a game of American Football has only scores of field goals ( points) and touchdowns with the extra point ( points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?
Answer: points
The town of Hamlet has people for each horse, sheep for each cow, and ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
(Source: AMC 10B 2015 Problem 15)
Answer:
In the state of Coinland, coins have values and cents. Suppose is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of (Source: 2023 AMC 12B Problems/Problem 16)
Answer:
Intermediate
Ninety-four bricks, each measuring are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes or or to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? (Source: AIME)
Find the sum of all positive integers such that, given an unlimited supply of stamps of denominations and cents, cents is the greatest postage that cannot be formed. (Source: AIME II 2019 Problem 14)
Olympiad
On the real number line, paint red all points that correspond to integers of the form , where and are positive integers. Paint the remaining integer points blue. Find a point on the line such that, for every integer point , the reflection of with respect to is an integer point of a different color than . (Source: India TST)
See Also
Theorem
Prime
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4395 | https://brainly.com/question/32919979 | [FREE] The heat of hydrogenation of an unsaturated compound can be used to estimate its stability. The heat of - brainly.com
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The heat of hydrogenation of an unsaturated compound can be used to estimate its stability. The heat of hydrogenation of benzene is _ than expected by comparison with cyclohexene and 1,3-cyclohexadiene. This difference indicates that benzene is much ___ stable than a system containing three isolated double bonds.
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Answer: The heat of hydrogenation of benzene is lower
Explanation: less, lower (since benzene is more stable than expected, it is already at a lower energy than an isolated triene. Less energy will therefore be released during hydrogenation).
Answer: This means that real benzene is about 150 kJ mol -1 more stable than the Kekulé structure gives it credit for. This increase in stability of benzene is known as the delocalization energy or resonance energy of benzene.
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The heat of hydrogenation of benzene is lower than expected, indicating that benzene is more stable than a system with three isolated double bonds due to resonance. This stability stems from the delocalization of electrons in the benzene ring. The substantial difference in heat of hydrogenation demonstrates the unique stability of aromatic compounds.
Explanation
The heat of hydrogenation of an unsaturated compound like benzene is lower than what would be expected when compared to cyclohexene and 1,3-cyclohexadiene. This difference indicates that benzene is much more stable than a system containing three isolated double bonds.
To understand this, we first need to define what heat of hydrogenation is. It refers to the amount of heat released when an unsaturated compound (like a compound with double or triple bonds) undergoes hydrogenation, which is the addition of hydrogen to convert it to a saturated compound (typically an alkane).
In the case of benzene, we observe that it has a heat of hydrogenation of approximately -208 kJ/mol. In contrast, if benzene were similar to a molecule with three isolated double bonds, its heat of hydrogenation would be expected to be much higher. Specifically, we can estimate that for 1,3,5-cyclohexatriene, where we would expect it to release roughly three times the heat of hydrogenation from cyclohexene (approximately 3 × 130 kJ/mol = 390 kJ/mol). However, the actual heat of hydrogenation observed for benzene is significantly lower, indicating that it doesn't release as much energy upon hydrogenation as predicted.
This remarkable stability of benzene is attributed to resonance and its aromatic character. Benzene’s structure allows for electron delocalization, where electrons are shared among all six carbon atoms, rather than being confined to specific bonds. This stabilization from resonance gives benzene about 150 kJ/mol more stability than predicted by its structure alone.
Hence, the comparison of the heat of hydrogenation reveals that benzene is much more stable than what one would initially expect based on the presence of three isolated double bonds. This enhanced stability is a characteristic trait of aromatic compounds and is referred to as resonance energy or delocalization energy.
Examples & Evidence
An example of the difference in stability can be seen when comparing benzene to 1,3-cyclohexadiene, where the latter is slightly more stable than expected due to conjugation, but benzene surpasses it significantly in stability due to resonance.
The stability of benzene has been confirmed through energy calculations, showing that the actual heat of hydrogenation is much lower than that predicted for a system of isolated double bonds, illustrating the concept of resonance energy.
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Community Answer 13. Hydrogenation of 1,3-cyclohexadiene and benzene are both exothermic, but benzene requires much more forcing conditions than 1,3-cyclohexadiene to get hydrogenated. when 1,3-cyclohexadiene is hydrogenated, more heat is released than with benzene, even though benzene adds three H2 molecules and 1,3-cyclohexadiene only ads two per molecule. Explain the differences in reactivity.
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4396 | https://brainly.com/question/25564500 | [FREE] Carefully evaluate f(1) . - brainly.com
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Carefully evaluate f(1).
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To evaluate f(1) for a function, substitute 1 for x in the function expression and simplify. Using the example, f(1)=4 for f(x)=2 x 2−3 x+5 .
To evaluate f(1)for a given function f(x) , substitute 1 for x in the function expression. Let's consider an example: f(x)=2 x 2−3 x+5 .
Substitute x=1 : Plug in 1 for x in the function.
f(1)=2(1)2−3(1)+5=2−3+5=4
Simplify the Expression: Perform the calculations to obtain the result.
f(1)=4
So, for the given function f(x)=2 x 2−3 x+5, f(1)equals 4.
The step-by-step process involves substitution and simplification, where the value of x is replaced with 1 in the function expression. This process allows us to find the specific function value at the given point.
In summary, to evaluate f(1)for a function, substitute 1 for x in the function expression and simplify. Using the example, f(1)=4 for f(x)=2 x 2−3 x+5.
The correct question maybe:-
How can the value f(1)be carefully evaluated for a given function f(x) ? Provide an example, and explain the step-by-step process involved in determining f(1) .
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To evaluate f(1) for a function, substitute 1 for x in the function expression and simplify. For example, if f(x)=x 2+3 x−4, then f(1)=0. Following this method allows you to determine the value of the function at any given point.
Explanation
To evaluate f(1) for a function, first, we need the expression of the function f(x). Once we have that, we can substitute the value x=1 into the function expression and simplify it to find the output.
Here are the steps:
Identify the Function: Assume we have a function defined as f(x)=x 2+3 x−4.
Substitute: Replace x with 1 in the function:
f(1)=(1)2+3(1)−4
Simplify: Calculate the expression:
f(1)=1+3−4=0
So, for the function f(x)=x 2+3 x−4, the value of f(1) is 0.
By following these steps of substitution and simplification, you can evaluate any function at a specific point. If you are provided with a different function, just apply the same method: substitute the desired value for x and simplify to get your answer.
Examples & Evidence
An example of evaluating a function is using f(x)=2 x+1 to find f(1). Substituting yields f(1)=2(1)+1=3, showing step-by-step evaluation for different functions.
The process of substitution in evaluating functions is foundational in mathematics, as outlined in many algebra textbooks, which teach this method as a standard approach to function evaluation.
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Factor. w 2+12 w+36
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m−9=3 2 m−12
Divide. 5×1 0−1 2×1 0 1
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4397 | https://pmc.ncbi.nlm.nih.gov/articles/PMC344520/ | Resolution of Discrepant Results for Candida Species Identification by Using DNA Probes - PMC
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. 2004 Feb;42(2):858–861. doi: 10.1128/JCM.42.2.858-861.2004
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Resolution of Discrepant Results for Candida Species Identification by Using DNA Probes
Catherine Coignard
Catherine Coignard
1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
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1 Mycotic Diseases Branch, Division of Bacterial and Mycotic Diseases, National Center for Infectious Diseases, Centers for Disease Control and Prevention, Atlanta, Georgia
Corresponding author. Mailing address: Mycotic Diseases Branch, Centers for Disease Control and Prevention, 1600 Clifton Rd., NE, Mailstop G-11, Atlanta, GA 30333. Phone: (404) 639-3098. Fax: (404) 639-3546. E-mail: cjm3@cdc.gov.
Received 2003 Sep 12; Revised 2003 Oct 14; Accepted 2003 Nov 3.
Copyright © 2004, American Society for Microbiology
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PMCID: PMC344520 PMID: 14766873
Abstract
Candida species bloodstream isolates were collected from institutions participating in an active, population-based surveillance for candidemia. Species identifications were performed locally and then confirmed at the Centers for Disease Control and Prevention (CDC) by phenotype-based methods. Discrepancies in species identification between the referring institution and the CDC were noted for 43 of 935 isolates (4.6%). A DNA probe-based species identification system (PCR-enzyme immunoassay [EIA]) was then used to resolve these discrepancies. The PCR-EIA result was identical to the CDC phenotypic identification method for 98% of the isolates tested. The most frequently misidentified species was Candida glabrata (37% of all discrepant identifications). Such misidentifications could lead to the administration of inappropriate therapy given the propensity of C. glabrata to develop resistance to azole antifungal drugs.
Candida species are the fourth most common cause of health care-associated bloodstream infections and are increasingly important causes of morbidity in hospitalized patients (5, 10). The emergence of non-Candida albicans species, including those innately or adaptively less susceptible to azole antifungals (14, 15), makes the identification of bloodstream isolates to the species level important for the implementation of appropriate antifungal therapy. Species identification is also important for an understanding of the epidemiology of candidemia, including trends in species distribution and antifungal drug susceptibility patterns.
Candida species have traditionally been identified by a combination of phenotypic tests that assess morphological characteristics and carbohydrate assimilation and fermentation patterns (9, 11). Whereas presumptive identification of C. albicans may be obtained in a few hours, identification of non-C. albicans species may require up to 72 h (8, 9, 17, 18). More recently, fungus-specific PCR primers and Candida species-specific DNA probes, directed to the internal transcribed spacer 2 (ITS2) region of ribosomal DNA (rDNA), have been used to detect PCR amplicons in a colorimetric enzyme immunoassay format (PCR-EIA) (6). This test has been shown to be highly specific, rapid, and easy to perform. Therefore, we used the PCR-EIA to resolve discrepancies in Candida species identifications between referring institutions and the Centers for Disease Control and Prevention (CDC) laboratory as part of an active, population-based surveillance for candidemia.
Collection and identification of bloodstream isolates.
Bloodstream isolates were obtained from institutions in the state of Connecticut and the city and county of Baltimore, Md., from October 1998 to September 2000 (10a). All blood cultures positive for Candida species were identified at the referring institution according to their standard methods. In describing their blood culture identification practices, 23 of 51 responding laboratories (45%) used the germ tube formation test to identify C. albicans. Forty-one respondents (80%) used some type of carbon assimilation/biochemical panel to identify non-C. albicans species, with 20 (39%) using the API 20C system (bioMerieux Vitek, Inc., Hazelwood, Mo.) and 13 (26%) using the Vitek yeast biochemical card (bioMerieux Vitek).
A total of 935 isolates were sent to the Mycotic Diseases Branch, CDC, for confirmation of species identification. At the CDC, the isolates were first subcultured onto Sabouraud dextrose agar (BBL Difco Laboratories, Detroit, Mich.) as well as onto CHROMagar Candida medium (DRG International, Mountainside, N.J.). Isolates were then identified to the species level with the API 20C AUX (bioMerieux Vitek) or RapID Yeast Plus system (Innovative Diagnostics, Norcross, Ga.) and by microscopic morphology on cornmeal-Tween 80 (Dalmau) plates.
Discrepancies in the phenotypic identification between the CDC and the referring institution were resolved by using species-specific DNA probes in an EIA detection format (PCR-EIA) (6). Candida species isolates were grown for 18 h at 35°C in 50-ml Erlenmeyer flasks containing 10 ml of YPD broth (1% yeast extract, 2% Bacto Peptone, 2% dextrose; BBL Difco Laboratories). DNA was isolated from these cultures with the PUREGENE DNA Purification kit for yeast and gram-positive bacteria (Gentra Systems, Inc., Minneapolis, Minn.) according to the manufacturer's directions. Universal fungus-specific primers ITS3 (5′ GCA TCG ATG AAG AAC GCA GC 3′) and ITS4 (5′ TCC TCC GCT TAT TGA TAT GC 3′) (19) were then used to amplify by PCR a portion of the 5.8S rDNA region, the entire ITS2 rDNA region, and a portion of the 28S rDNA region using a Perkin-Elmer (Emeryville, Calif.) model 9700 thermal cycler and Taq DNA polymerase (Roche Molecular Biochemicals, Inc., Indianapolis, Ind.). All other PCR reagents and thermal cycling conditions used were as previously described (6). Amplicons were captured onto a streptavidin-coated microtitration plate (Roche) with a biotinylated, all-Candida species DNA probe (5′ CAT GCC TGT TTG AGC GTC [GA]TT 3′) and were detected with digoxigenin-labeled species-specific DNA probes (C. albicans: 5′ AT TGC TTG CGG CGG TAA CGT CC 3′; C. glabrata: 5′ TT TAC CAA CTC GGT GTT GAT CT 3′; C. krusei: 5′ GG CCC GAG CGA ACT AGA CTT TT 3′; C. lusitaniae: 5′ CT CCG AAA TAT CAA CCG CGC TG 3′; C. parapsilosis: 5′ AC AAA CTC CAA AAC TTC TTC CA 3′; C. tropicalis: 5′ AA CGC TTA TTT TGC TAG TGG CC 3′) and horseradish peroxidase-labeled antidigoxigenin antibodies in a colorimetric EIA format (6). Oligonucleotide primers and probes were synthesized and labeled as described previously (12).
Discrepancies in species identification between the CDC's phenotypic methods and the PCR-EIA were resolved by rDNA sequencing on a Perkin-Elmer ABI Prism 310 automated capillary DNA sequencer as previously described (7). Briefly, universal fungus-specific primers ITS1 (5′ TCC GTA GGT GAA CCT GCG G 3′) (19) and ITS4 were used to amplify by PCR a portion of the 18S rDNA region; the entire ITS1, 5.8S, and ITS2 rDNA regions; and a portion of the 28S rDNA region. Amplicons were purified with the QIAquick PCR purification kit (Qiagen, Valencia, Calif.) and sequenced on both strands with primers ITS1 or ITS4 and the Big Dye Terminator Cycle Sequencing Ready Reaction kit (Applied Biosystems, Foster City, Calif.) (7). GenBank searches and comparative sequence analyses were assisted by using BLAST search tools (1) and GeneTool, version 1.0, software (BioTools, Inc., Edmonton, Alberta, Canada), respectively.
Distribution and resolution of discrepancies in Candida species identification.
Of the 935 isolates received at the CDC, there were discrepancies in species identification between the referring institution and the CDC laboratory for 43 (4.6%). In all but one case, the CDC identification based on biochemical and morphological criteria was validated by the PCR-EIA (Table 1). In the remaining case, an isolate reported as C. albicans by the referring institution and as C. parapsilosis or C. lusitaniae by the CDC was ultimately identified as C. lusitaniae by probe testing and DNA sequence analysis (GenBank accession number AY383555). All isolates identified at the CDC as C. albicans by phenotypic methods were subsequently screened by molecular identification methods (2, 10a ) to differentiate isolates of C. dubliniensis from those of C. albicans. Nine cases of C. dubliniensis candidemia were identified. All C. dubliniensis isolates were reported to be C. albicans by the referring institutions, but these identifications were not considered to be discrepant as C. dubliniensis is not routinely differentiated from C. albicans by most clinical laboratories.
TABLE 1.
Phenotypic identifications of Candida species by the referring hospital and by CDC compared to those by the PCR-EIA
| Hospital IDb | CDC ID | PCR-EIA ID | CHROMagar result | API 20C or RapID profile resulta (% IDc) | Morphologyd |
:---: :---: :---: |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. albicans | C. glabrata | C. glabrata | Pink | Implicit CG (99.0) | BSP w/o PSH |
| C. krusei | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. krusei | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. krusei | C. glabrata | C. glabrata | Purple | Very good CG (99.4) | BSP w/o PSH |
| C. parapsilosis | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. tropicalis | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. tropicalis | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. tropicalis | C. glabrata | C. glabrata | Pink | Very good CG (99.4) | BSP w/o PSH |
| C. tropicalis | C. glabrata | C. glabrata | Pink | Satisfactory CG (99.2) | BSP w/o PSH |
| C. tropicalis | C. glabrata | C. glabrata | Pink | Low discrim. PW (70.7), CG (29.3) | BSP w/o PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Excellent CP (99.9) | BSP and PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Good CP (99.1) | BSP and PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Acceptable CP (97.4) | BSP and PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Satisfactory CP (95.7) | BSP and PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Low discrim. CP (83.3), CN (13.3) | BSP and PSH |
| C. albicans | C. parapsilosis | C. parapsilosis | Pale pink | Acceptable genus CT (57.1), CA (27.6), CP (12.2) | BSP and PSH |
| C. glabrata | C. parapsilosis | C. parapsilosis | Pale pink | Good CP (97.8) | BSP and PSH |
| C. glabrata | C. parapsilosis | C. parapsilosis | Pale pink | Acceptable CP (90.5) | BSP and PSH |
| C. glabrata | C. parapsilosis | C. parapsilosis | Pale pink | Low discrim. CN (50.2), CP (47.5) | BSP and PSH |
| C. guilliermondii | C. parapsilosis | C. parapsilosis | Pale pink | Good CP (98.0) | BSP and PSH |
| C. lusitaniae | C. parapsilosis | C. parapsilosis | Pale pink | Very good CP (99.9) | BSP and PSH |
| C. lusitaniae | C. parapsilosis | C. parapsilosis | Pale pink | Low discrim. CP (71.9), CN (18.9), CT (4.9) | BSP and PSH |
| C. tropicalis | C. parapsilosis | C. parapsilosis | Pale pink | Very good CP (99.0) | BSP and PSH |
| C. tropicalis | C. parapsilosis | C. parapsilosis | Pale pink | Good CP (95.9) | BSP and PSH |
| C. tropicalis | C. parapsilosis | C. parapsilosis | Pale pink | Low discrim. CN (50.2), CP (47.5) | BSP and PSH |
| C. albicans | C. tropicalis | C. tropicalis | Blue | Low discrim. CT (71.1), CN (18.5), CL (7.6) | BSP and PSH |
| C. albicans | C. tropicalis | C. tropicalis | Blue | Low discrim. CT (71.1), CN (18.5), CL (7.6) | BSP and PSH |
| C. albicans | C. tropicalis | C. tropicalis | Blue | Low discrim. CT (71.1), CN (18.5), CL (7.6) | BSP and PSH |
| C. albicans | C. tropicalis | C. tropicalis | Blue | Low discrim. CT (71.1), CN (18.5), CL (7.6) | BSP and PSH |
| C. glabrata | C. tropicalis | C. tropicalis | Blue | Adequate CT (97.0) | BSP and PSH |
| C. glabrata | C. tropicalis | C. tropicalis | Blue | Good CT (95.9) | BSP and PSH |
| C. glabrata | C. tropicalis | C. tropicalis | Blue | Low discrim. CT (71.1), CN (18.5), CL (7.6) | BSP and PSH |
| C. glabrata | C. albicans | C. albicans | Green | Low discrim. CA (58.7), CT (26.0), CN (14.4) | CHL and PSH |
| C. tropicalis | C. albicans | C. albicans | Green | Good genus CT (48.4), CA (44.4) | CHL and PSH |
| C. tropicalis | C. albicans | C. albicans | Green | Unacceptable profile | CHL and PSH |
| C. albicans | C. lusitaniae | C. lusitaniae | Tan | Good genus CT (60.7), CL (29.6) | BSP and PSH |
| C. albicans | C. parapsilosis or C. lusitaniae | C. lusitaniae | Pale pink | Very good genus CP (53.5), CL (38.8), CT (7.1) | BSP and PSH |
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a
CA, C. albicans; CG, C. glabrata; CL, C. lusitaniae; CP, C. parapsilosis; CT, C. tropicalis; CN, Cryptococcus neoformans; PW, Prototheca wickerhamii. “Implicit CG” and “satisfactory CG” are RapID Yeast Plus results; all other results are from the API 20C AUX system. Low discrim., presumptive identification.
b
ID, identification.
c
%ID, percent identification likelihood.
d
BSP, blastospores; PSH, pseudohyphae; CHL, chlamydospores; w/o, without.
The most frequent misidentifications by the referring institutions were of C. glabrata (16 of 43 isolates or 37% of all discrepant identifications), followed by C. parapsilosis (15 of 43, 35%), C. tropicalis (7 of 43, 16%), C. albicans (3 of 43, 7%), and C. lusitaniae (2 of 43, 5%) (Table 1). These misidentifications represent 12 (15 of 123), 7 (16 of 226), 6 (7 of 118), and 0.7% (3 of 423) of all C. parapsilosis, C. glabrata, C. tropicalis, and C. albicans isolates, respectively. The most common misidentifications by the referring institutions were C. albicans for C. glabrata (7 of 43 isolates or 16% of all misidentifications), C. albicans for C. parapsilosis (6 of 43, 14%), C. tropicalis for C. glabrata (5 of 43, 12%), and C. albicans for C. tropicalis (4 of 43, 9%).
Of all isolates reidentified at the CDC that could be associated with a given type of institution, 618 (66%) were from nonacademic institutions and 316 (34%) were from academic institutions. Comparison of the misidentification rate between academic (university- or medical school-associated) and nonacademic institutions showed that, for the 41 of 43 isolates that could be associated with a particular category of institution, 34 (83%) of the misidentifications were from nonacademic institutions whereas 7 (17%) were from academic institutions. As a percentage of the total number of isolates received from academic versus nonacademic institutions, the overall misidentification rate for academic institutions was roughly one-half that for the nonacademic institutions (i.e., 2.2 versus 5.5%, respectively). Misidentifications were not associated with any one particular institution.
The vast majority (15 of 16, 94%) of misidentified isolates that were reidentified as C. glabrata in the CDC laboratory had typical biochemical profiles and morphologies on Dalmau plates and gave the expected colony color and appearance on CHROMagar Candida medium (Table 1). In contrast, 53% of C. parapsilosis isolates and 71% of C. tropicalis isolates showed profiles interpreted as “acceptable” to “low discrimination” by the API 20C AUX system; these isolates were differentiated from the alternative species choices listed in the API 20C AUX profile index by microscopic morphology and colony color on CHROMagar Candida medium (Table 1). C. albicans isolates could also be identified by their distinctive colony color on CHROMagar Candida medium and by their capacity to form chlamydospores (Table 1).
Specificity of DNA probes to identify Candida species.
The PCR-EIA generated results that were highly specific, the DNA did not cross-react with DNA from other Candida species tested, and the results were easy to interpret (Table 2). Mean positive EIA values ± standard errors (SE) ranged from 0.95 ± 0.10 for DNA from C. tropicalis to 0.38 ± 0.02 for DNA from C. lusitaniae. Inherent differences in absolute EIA values obtained for each of the probes may reflect differences in their G+C compositions and, as a result, their rate of denaturation and annealing during thermal cycling or their rate of hybridization during probe attachment to the PCR product. Nonetheless, the EIA values reported here were very similar to those reported previously for the identification of these same species (6) and were approximately 200 times above background values after subtraction of the water blank (Table 2). Testing of heterologous-species DNA gave no significant background reactivity (mean A 650 ± SE = 0.002 ± 0.0001), making discrimination of a positive from a negative result unequivocal (Table 2). All Candida species DNAs were also tested with a probe specific for C. krusei DNA, and no reactivity with heterologous DNA was observed (mean A 650 ± SE for the C. krusei-specific probe versus C. krusei DNA and versus all other Candida species DNAs, 0.49 ± 0.05 and 0.0014 ± 0.0004, respectively; n = 72).
TABLE 2.
Specificity of DNA probes for species identification by PCR-EIA
| DNA target species (no. of isolates tested) | Mean A 650 ± SEa after reaction with probe for: |
:---: |
| C. glabrata | C. parapsilosis | C. tropicalis | C. albicans | C. lusitaniae |
| C. glabrata (16) | 0.91 ± 0.07 | 0 | 0 | 0 | 0 |
| C. parapsilosis (15) | 0 | 0.79 ± 0.05 | 0 | 0 | 0 |
| C. tropicalis (7) | 0 | 0 | 0.95 ± 0.10 | 0 | 0 |
| C. albicans (3) | 0 | 0 | 0 | 0.40 ± 0.05 | 0 |
| C. lusitaniae (2) | 0 | 0 | 0 | 0 | 0.38 ± 0.02b |
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a
Mean A 650 ± SE was calculated from spectrophotometric readings after target DNA was reacted with the DNA probes listed above. All samples were run in duplicate, and reagent blanks were run on each plate for each probe. Reagent blank values have been subtracted from test sample values above (mean reagent blank A 650 = 0.038 ± 0.001; n = 72). Mean A 650 ± SE for all control samples after subtraction of the reagent blanks for all probes was 0.002 ± 0.0001 (n = 226) and is represented in this table as 0 for ease of presentation.
b
Includes one C. lusitaniae isolate identified as C. albicans by the referring institution and as C. parapsilosis or C. lusitaniae by CDC phenotypic methods.
Conclusions.
Several population-based and sentinel surveillance studies have noted an increase in the proportion of Candida bloodstream infections caused by species other than C. albicans, and, in particular, an increase in the frequency of candidemia due to C. glabrata (reviewed in reference 15). Given the known propensity of C. glabrata to develop resistance to azole antifungals, the fact that this species was most frequently misidentified in this study is disturbing. Because each referring institution used its own method(s) for species identification, the reasons for the high rate of C. glabrata misidentification are not clear. However, for the 28% of non-C. glabrata isolates that were misidentified as C. albicans by the referring institution, two factors may provide some insight. First, in many referring institutions (45% of those surveyed), the germ tube test was performed as a primary screen for the identification of C. albicans. Second, a disproportionately greater number of misidentifications were received from nonuniversity laboratories than from university or medical school laboratories. These data suggest that the nonuniversity institutions in our study may have employed fewer specialists in mycology, who in turn had less experience in interpreting the germ tube test. This hypothesis is supported by the work of others (4) who found that, when the germ tube test was performed on a series of isolates tested in a blinded fashion by technicians who were not specialists in mycology, germ tube test specificity declined. Misinterpretation of results, particularly of pseudohyphal production, accounted for this drop in specificity (4). This might account for those isolates originally identified by the referring institution as C. albicans but ultimately identified as C. lusitaniae, C. parapsilosis, or C. tropicalis by the CDC. In addition, the CDC laboratory routinely employs CHROMagar Candida medium (16) as an adjunct to biochemical and morphological tests whereas none of the institutions polled in the surveillance area reported the use of this medium. The distinctive color of each species on this medium may be helpful in cases where the biochemical results are equivocal and the expertise for distinguishing various species based on morphology on cornmeal-Tween agar is lacking.
Most clinical laboratories do not differentiate isolates of C. albicans from those of C. dubliniensis. Therefore, the nine cases of C. dubliniensis candidemia identified at the CDC by molecular identification methods and reported as C. albicans by the referring institutions were not considered to be discrepant identifications. Nonetheless, it has been demonstrated previously that the PCR-EIA can unequivocally differentiate isolates of C. dubliniensis from those of C. albicans (6, 7).
Unlike current phenotypic identification methods, which may require a series of tests to confirm the identity of a given Candida species (8, 9, 17), the PCR-EIA is a single test that can be used to identify the majority of medically important Candida species (6). In contrast to phenotype-based identification methods, the PCR-EIA can be performed in a single day and the results are very easy to interpret. Use of a commercial kit for the isolation of Candida species DNA (6) makes this test a fast and reliable method for Candida species identification. Although other PCR-based methods for the identification of Candida species have been described (reviewed in references 3 and 13), these methods did not use a combination of (i) an EIA detection format; (ii) universal fungal, multicopy rRNA gene targets to increase test sensitivity; (iii) a commercial kit for rapid and simple sample preparation; and (iv) probes that could detect more than one or a few Candida species. This study demonstrates the usefulness of the PCR-EIA for the resolution of discrepant phenotype-based species identifications. Conversion of the DNA probes described in this study into either a real-time PCR format or an automated microarray format would reduce postamplification manipulation steps and further reduce the time required for accurate species identification.
Acknowledgments
We thank Hans Peter Hinrikson for assistance with the C. lusitaniae DNA sequence analysis.
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This book may be purchased from the publisher at eureka-math.org 10 9 8 7 6 5 4 3 2 1 Eureka Math™ Grade 6, Module 4 Teacher Edition GRADE 6 • MODULE 4 Module 4: Expressions and Equations 6 GRADE Mathematics Curriculum Table of Contents1 Expressions and Equations Module Overview .................................................................................................................................................. 3 Topic A: Relationships of the Operations (6.EE.A.3) ........................................................................................... 13 Lesson 1: The Relationship of Addition and Subtraction ........................................................................ 15 Lesson 2: The Relationship of Multiplication and Division ..................................................................... 25 Lesson 3: The Relationship of Multiplication and Addition .................................................................... 34 Lesson 4: The Relationship of Division and Subtraction ......................................................................... 41 Topic B: Special Notations of Operations (6.EE.A.1, 6.EE.A.2c) .......................................................................... 50 Lesson 5: Exponents ............................................................................................................................... 52 Lesson 6: The Order of Operations ......................................................................................................... 62 Topic C: Replacing Letters and Numbers (6.EE.A.2c, 6.EE.A.4) ........................................................................... 73 Lesson 7: Replacing Letters with Numbers ............................................................................................. 75 Lesson 8: Replacing Numbers with Letters ............................................................................................. 85 Topic D: Expanding, Factoring, and Distributing Expressions (6.EE.A.2a, 6.EE.A.2b, 6.EE.A.3, 6.EE.A.4) .......... 97 Lesson 9: Writing Addition and Subtraction Expressions ....................................................................... 99 Lesson 10: Writing and Expanding Multiplication Expressions ............................................................ 106 Lesson 11: Factoring Expressions ......................................................................................................... 119 Lesson 12: Distributing Expressions ..................................................................................................... 132 Lessons 13–14: Writing Division Expressions ....................................................................................... 140 Topic E: Expressing Operations in Algebraic Form (6.EE.A.2a, 6.EE.A.2b) ........................................................ 154 Lesson 15: Read Expressions in Which Letters Stand for Numbers ..................................................... 155 Lessons 16–17: Write Expressions in Which Letters Stand for Numbers ............................................. 160 Mid-Module Assessment and Rubric ................................................................................................................ 180 Topics A through E (assessment 1 day, return 1 day, remediation or further applications 3 days) 1Each lesson is ONE day, and ONE day is considered a 45-minute period. A STORY OF RATIOS 1 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Topic F: Writing and Evaluating Expressions and Formulas (6.EE.A.2a, 6.EE.A.2c, 6.EE.B.6) ........................... 192 Lesson 18: Writing and Evaluating Expressions—Addition and Subtraction........................................ 194 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions ........................................ 200 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division ..................................... 213 Lesson 21: Writing and Evaluating Expressions—Multiplication and Addition .................................... 222 Lesson 22: Writing and Evaluating Expressions—Exponents ............................................................... 230 Topic G: Solving Equations (6.EE.B.5, 6.EE.B.6, 6.EE.B.7) ................................................................................. 239 Lessons 23–24: True and False Number Sentences ............................................................................. 241 Lesson 25: Finding Solutions to Make Equations True ........................................................................ 261 Lesson 26: One-Step Equations—Addition and Subtraction ................................................................ 275 Lesson 27: One-Step Equations—Multiplication and Division ............................................................. 288 Lesson 28: Two-Step Problems—All Operations .................................................................................. 304 Lesson 29: Multi-Step Problems—All Operations ................................................................................ 320 Topic H: Applications of Equations (6.EE.B.5, 6.EE.B.6, 6.EE.B.7, 6.EE.B.8, 6.EE.C.9) ...................................... 330 Lesson 30: One-Step Problems in the Real World ................................................................................ 332 Lesson 31: Problems in Mathematical Terms ....................................................................................... 346 Lesson 32: Multi-Step Problems in the Real World .............................................................................. 354 Lesson 33: From Equations to Inequalities ........................................................................................... 362 Lesson 34: Writing and Graphing Inequalities in Real-World Problems .............................................. 369 End-of-Module Assessment and Rubric ............................................................................................................ 377 Topics A through H (assessment 1 day, return 1 day, remediation or further applications 4 day) A STORY OF RATIOS 2 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Grade 6 • Module 4 Expressions and Equations OVERVIEW In Module 4, students extend their arithmetic work to include using letters to represent numbers. Students understand that letters are simply “stand-ins” for numbers and that arithmetic is carried out exactly as it is with numbers. Students explore operations in terms of verbal expressions and determine that arithmetic properties hold true with expressions because nothing has changed—they are still doing arithmetic with numbers. Students determine that letters are used to represent specific but unknown numbers and are used to make statements or identities that are true for all numbers or a range of numbers. Students understand the importance of specifying units when defining letters. Students say “Let 𝐾𝐾represent Karolyn’s weight in pounds” instead of “Let 𝐾𝐾represent Karolyn’s weight” because weight cannot be a specific number until it is associated with a unit, such as pounds, ounces, or grams. They also determine that it is inaccurate to define 𝐾𝐾 as Karolyn because Karolyn is not a number. Students conclude that in word problems, each letter (or variable) represents a number, and its meaning is clearly stated. To begin this module, students are introduced to important identities that are useful in solving equations and developing proficiency with solving problems algebraically. In Topic A, students understand the relationships of operations and use them to generate equivalent expressions (6.EE.A.3). By this time, students have had ample experience with the four operations since they have worked with them from kindergarten through Grade 5 (1.OA.B.3, 3.OA.B.5). The topic opens with the opportunity to clarify those relationships, providing students with the knowledge to build and evaluate identities that are important for solving equations. In this topic, students discover and work with the following identities: 𝑤𝑤−𝑥𝑥+ 𝑥𝑥= 𝑤𝑤, 𝑤𝑤+ 𝑥𝑥−𝑥𝑥= 𝑤𝑤, 𝑎𝑎÷ 𝑏𝑏∙𝑏𝑏= 𝑎𝑎, 𝑎𝑎∙𝑏𝑏÷ 𝑏𝑏= 𝑎𝑎 (when 𝑏𝑏≠0), and 3𝑥𝑥= 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥. Students also discover that if 12 ÷ 𝑥𝑥= 4, then 12 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0. In Topic B, students experience special notations of operations. They determine that 3𝑥𝑥= 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥 is not the same as 𝑥𝑥3, which is 𝑥𝑥∙𝑥𝑥∙𝑥𝑥. Applying their prior knowledge from Grade 5, where whole number exponents were used to express powers of ten (5.NBT.A.2), students examine exponents and carry out the order of operations, including exponents. Students demonstrate the meaning of exponents to write and evaluate numerical expressions with whole number exponents (6.EE.A.1). Students represent letters with numbers and numbers with letters in Topic C. In past grades, students discovered properties of operations through example (1.OA.B.3, 3.OA.B.5). Now, they use letters to represent numbers in order to write the properties precisely. Students realize that nothing has changed because the properties still remain statements about numbers. They are not properties of letters; nor are they new rules introduced for the first time. Now, students can extend arithmetic properties from manipulating numbers to manipulating expressions. In particular, they develop the following identities: 𝑎𝑎∙ 𝑏𝑏= 𝑏𝑏∙𝑎𝑎, 𝑎𝑎+ 𝑏𝑏= 𝑏𝑏+ 𝑎𝑎, 𝑔𝑔∙1 = 𝑔𝑔, 𝑔𝑔+ 0 = 𝑔𝑔, 𝑔𝑔÷ 1 = 𝑔𝑔, 𝑔𝑔÷ 𝑔𝑔= 1, and 1 ÷ 𝑔𝑔= 1 𝑔𝑔. Students understand that a letter in an expression represents a number. When that number replaces that letter, the expression A STORY OF RATIOS 3 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations can be evaluated to one number. Similarly, they understand that a letter in an expression can represent a number. When that number is replaced by a letter, an expression is stated (6.EE.A.2). In Topic D, students become comfortable with new notations of multiplication and division and recognize their equivalence to the familiar notations of the prior grades. The expression 2 × 𝑏𝑏 is exactly the same as 2 ∙𝑏𝑏, and both are exactly the same as 2𝑏𝑏. Similarly, 6 ÷ 2 is exactly the same as 6 2. These new conventions are practiced to automaticity, both with and without variables. Students extend their knowledge of GCF and the distributive property from Module 2 to expand, factor, and distribute expressions using new notation (6.NS.B.4). In particular, students are introduced to factoring and distributing as algebraic identities. These include 𝑎𝑎+ 𝑎𝑎= 2 ∙𝑎𝑎= 2𝑎𝑎, (𝑎𝑎+ 𝑏𝑏) + (𝑎𝑎+ 𝑏𝑏) = 2 ∙(𝑎𝑎+ 𝑏𝑏) = 2(𝑎𝑎+ 𝑏𝑏) = 2𝑎𝑎+ 2𝑏𝑏, and 𝑎𝑎÷ 𝑏𝑏= 𝑎𝑎 𝑏𝑏. In Topic E, students express operations in algebraic form. They read and write expressions in which letters stand for and represent numbers (6.EE.A.2). They also learn to use the correct terminology for operation symbols when reading expressions. For example, the expression 3 2𝑥𝑥−4 is read as “the quotient of three and the difference of twice a number and four.” Similarly, students write algebraic expressions that record operations with numbers and letters that stand for numbers. Students determine that 3𝑎𝑎+ 𝑏𝑏 can represent the story: “Martina tripled her money and added it to her sister’s money” (6.EE.A.2b). Students write and evaluate expressions and formulas in Topic F. They use variables to write expressions and evaluate those expressions when given the value of the variable (6.EE.A.2). From there, students create formulas by setting expressions equal to another variable. For example, if there are 4 bags containing 𝑐𝑐 colored cubes in each bag with 3 additional cubes, students use this information to express the total number of cubes as 4𝑐𝑐+ 3. From this expression, students develop the formula 𝑡𝑡= 4𝑐𝑐+ 3, where 𝑡𝑡 is the total number of cubes. Once provided with a value for the amount of cubes in each bag (𝑐𝑐= 12 cubes), students can evaluate the formula for 𝑡𝑡: 𝑡𝑡= 4(12) + 3, 𝑡𝑡= 48 + 3, 𝑡𝑡= 51. Students continue to evaluate given formulas such as the volume of a cube, 𝑉𝑉= 𝑠𝑠3, given the side length, or the volume of a rectangular prism, 𝑉𝑉= 𝑙𝑙⋅𝑤𝑤⋅ℎ, given those dimensions (6.EE.A.2c). In Topic G, students are introduced to the fact that equations have a structure similar to some grammatical sentences. Some sentences are true: “George Washington was the first president of the United States,” or “2 + 3 = 5.” Some are clearly false: “Benjamin Franklin was a president of the United States,” or “7 + 3 = 5.” Sentences that are always true or always false are called closed sentences. Some sentences need additional information to determine whether they are true or false. The sentence “She is 42 years old” can be true or false, depending on who “she” is. Similarly, the sentence “𝑥𝑥+ 3 = 5” can be true or false, depending on the value of 𝑥𝑥. Such sentences are called open sentences. An equation with one or more variables is an open sentence. The beauty of an open sentence with one variable is that if the variable is replaced with a number, then the new sentence is no longer open: It is either clearly true or clearly false. For example, for the open sentence 𝑥𝑥+ 3 = 5: If 𝑥𝑥 is replaced by 7, the new closed sentence, 7 + 3 = 5, is false because 10 ≠5. If 𝑥𝑥 is replaced by 2, the new closed sentence, 2 + 3 = 5, is true because 5 = 5. From here, students conclude that solving an equation is the process of determining the number or numbers that, when substituted for the variable, result in a true sentence (6.EE.B.5). In the previous example, the solution for 𝑥𝑥+ 3 = 5 is obviously 2. The extensive use of bar diagrams in Grades K–5 makes solving equations in Topic G a fun and exciting adventure for students. Students solve many equations twice, once A STORY OF RATIOS 4 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations with a bar diagram and once using algebra. They use identities and properties of equality that were introduced earlier in the module to solve one-step, two-step, and multi-step equations. Students solve problems finding the measurements of missing angles represented by letters (4.MD.C.7) using what they learned in Grade 4 about the four operations and what they now know about equations. In Topic H, students use their prior knowledge from Module 1 to construct tables of independent and dependent values in order to analyze equations with two variables from real-life contexts. They represent equations by plotting the values from the table on a coordinate grid (5.G.A.1, 5.G.A.2, 6.RP.A.3a, 6.RP.A.3b, 6.EE.C.9). The module concludes with students referring to true and false number sentences in order to move from solving equations to writing inequalities that represent a constraint or condition in real-life or mathematical problems (6.EE.B.5, 6.EE.B.8). Students understand that inequalities have infinitely many solutions and represent those solutions on number line diagrams. The 45-day module consists of 34 lessons; 11 days are reserved for administering the Mid- and End-of-Module Assessments, returning assessments, and remediating or providing further applications of the concepts. The Mid-Module Assessment follows Topic E, and the End-of-Module Assessment follows Topic H. Focus Standards Apply and extend previous understandings of arithmetic to algebraic expressions.2 6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. a. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract 𝑦𝑦 from 5” as 5 −𝑦𝑦. b. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2(8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. c. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas 𝑉𝑉= 𝑠𝑠3 and 𝐴𝐴= 6𝑠𝑠2 to find the volume and surface area of a cube with sides of length 𝑠𝑠= 1/2. 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + 𝑥𝑥) to produce the equivalent expression 6 + 3𝑥𝑥; apply the distributive property to the expression 24𝑥𝑥+ 18𝑦𝑦 to produce the equivalent expression 6(4𝑥𝑥+ 3𝑦𝑦); apply properties of operations to 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 to produce the equivalent expression 3𝑦𝑦. 26.EE.A.2c is also taught in Module 4 in the context of geometry. A STORY OF RATIOS 5 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations 6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 and 3𝑦𝑦 are equivalent because they name the same number regardless of which number 𝑦𝑦 stands for. Reason about and solve one-variable equations and inequalities.3 6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form 𝑥𝑥+ 𝑝𝑝= 𝑞𝑞 and 𝑝𝑝𝑝𝑝= 𝑞𝑞 for cases in which 𝑝𝑝, 𝑞𝑞, and 𝑥𝑥 are all nonnegative rational numbers. 6.EE.B.8 Write an inequality of the form 𝑥𝑥> 𝑐𝑐 or 𝑥𝑥< 𝑐𝑐 to represent a constraint or condition in a real-world mathematical problem. Recognize that inequalities of the form 𝑥𝑥> 𝑐𝑐 or 𝑥𝑥< 𝑐𝑐 have infinitely many solutions; represent solutions of such inequalities on number line diagrams. Represent and analyze quantitative relationships between dependent and independent variables. 6.EE.C.9 Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation 𝑑𝑑= 65𝑡𝑡 to represent the relationship between distance and time. Foundational Standards Understand and apply properties of operations and the relationship between addition and subtraction. 1.OA.B.3 Apply properties of operations as strategies to add and subtract.4 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.) 3Except for 6.EE.B.8, this cluster is also taught in Module 4 in the context of geometry. 4Students need not use formal terms for these properties. A STORY OF RATIOS 6 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Understand properties of multiplication and the relationship between multiplication and division. 3.OA.B.5 Apply properties of operations as strategies to multiply and divide.5 Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.) Gain familiarity with factors and multiples. 4.OA.B.4 Find all factor pairs for a whole number in the range 1–100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1– 100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1–100 is prime or composite. Geometric measurement: understand concepts of angle and measure angles. 4.MD.C.5 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement: a. An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles. b. An angle that turns through 𝑛𝑛 one-degree angles is said to have an angle measure of 𝑛𝑛 degrees. 4.MD.C.6 Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. 4.MD.C.7 Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. Write and interpret numerical expressions. 5.OA.A.2 Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them. For example, express the calculation “add 8 and 7, then multiply by 2” as 2 × (8 + 7). Recognize that 3 × (18932 + 921) is three times as large as 18932 + 921, without having to calculate the indicated sum or product. 5Students need not use formal terms for these properties. A STORY OF RATIOS 7 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Analyze patterns and relationships. 5.OA.B.3 Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule “Add 3” and the starting number 0, and given the rule “Add 6” and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so. Understand the place value system. 5.NBT.A.2 Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. Graph points on the coordinate plane to solve real-world and mathematical problems. 5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., 𝑥𝑥-axis and 𝑥𝑥-coordinate, 𝑦𝑦-axis and 𝑦𝑦-coordinate). 5.G.A.2 Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation. Understand ratio concepts and use ratio reasoning to solve problems. 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. a. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. b. Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed? Compute fluently with multi-digit numbers and find common factors and multiples. 6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4(9 + 2). A STORY OF RATIOS 8 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Focus Standards for Mathematical Practice MP.2 Reason abstractly and quantitatively. Students connect symbols to their numerical referents. They understand exponential notation as repeated multiplication of the base number. Students realize that 32 is represented as 3 × 3, with a product of 9, and explain how 32 differs from 3 × 2, where the product is 6. Students determine the meaning of a variable within a real-life context. They write equations and inequalities to represent mathematical situations. Students manipulate equations using the properties so that the meaning of the symbols and variables can be more closely related to the real-world context. For example, given the expression 12𝑥𝑥 represents how many beads are available to make necklaces, students rewrite 12𝑥𝑥 as 4𝑥𝑥+ 4𝑥𝑥+ 4𝑥𝑥 when trying to show the portion each person gets if there are three people or rewrite 12𝑥𝑥 as 6𝑥𝑥+ 6𝑥𝑥 if there are two people sharing. Students recognize that these expressions are equivalent. Students can also use equivalent expressions to express the area of rectangles and to calculate the dimensions of a rectangle when the area is given. Also, students make connections between a table of ordered pairs of numbers and the graph of those data. MP.6 Attend to precision. Students are precise in defining variables. They understand that a variable represents one number. For example, students understand that in the equation 𝑎𝑎+ 4 = 12, the variable 𝑎𝑎 can only represent one number to make the equation true. That number is 8, so 𝑎𝑎= 8. When variables are represented in a real-world problem, students precisely define the variables. In the equation 2𝑤𝑤= 18, students define the variable as weight in pounds (or some other unit) rather than just weight. Students are precise in using operation symbols and can connect between previously learned symbols and new symbols (3 × 2 can be represented with parentheses, 3(2), or with the multiplication dot 3 ∙2; similarly, 3 ÷ 2 is also represented with the fraction bar 3 2). In addition, students use appropriate vocabulary and terminology when communicating about expressions, equations, and inequalities. For example, students write expressions, equations, and inequalities from verbal or written descriptions. “A number increased by 7 is equal to 11” can be written as 𝑥𝑥+ 7 = 11. Students refer to 7𝑦𝑦 as a term or an expression, whereas 7𝑦𝑦= 56 is referred to as an equation. MP.7 Look for and make use of structure. Students look for structure in expressions by deconstructing them into a sequence of operations. They make use of structure to interpret an expression’s meaning in terms of the quantities represented by the variables. In addition, students make use of structure by creating equivalent expressions using properties. For example, students write 6𝑥𝑥 as 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥, 4𝑥𝑥+ 2𝑥𝑥, 3(2𝑥𝑥), or other equivalent expressions. Students also make sense of algebraic solutions when solving an equation for the value of the variable through connections to bar diagrams and properties. For example, when there are two copies of 𝑎𝑎+ 𝑏𝑏, this can be expressed as either (𝑎𝑎+ 𝑏𝑏) + (𝑎𝑎+ 𝑏𝑏), 2𝑎𝑎+ 2𝑏𝑏, or 2(𝑎𝑎+ 𝑏𝑏). Students use tables and graphs to compare different expressions or equations to make decisions in real-world scenarios. These models also create structure as students gain knowledge in writing expressions and equations. A STORY OF RATIOS 9 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations MP.8 Look for and express regularity in repeated reasoning. Students look for regularity in a repeated calculation and express it with a general formula. Students work with variable expressions while focusing more on the patterns that develop than the actual numbers that the variable represents. For example, students move from an expression such as 3 + 3 + 3 + 3 = 4 ∙3 to the general form 𝑚𝑚+ 𝑚𝑚+ 𝑚𝑚+ 𝑚𝑚= 4 ∙𝑚𝑚, or 4𝑚𝑚. Similarly, students move from expressions such as 5 ∙5 ∙5 ∙5 = 54 to the general form 𝑚𝑚∙𝑚𝑚∙𝑚𝑚∙𝑚𝑚= 𝑚𝑚4. These are especially important when moving from the general form back to a specific value for the variable. Terminology New or Recently Introduced Terms Equation (An equation is a statement of equality between two expressions.) Equivalent Expressions (Two expressions are equivalent if both expressions evaluate to the same number for every substitution of numbers into all the variables in both expressions.) Exponential Notation for Whole Number Exponents (Let 𝑚𝑚 be a nonzero whole number. For any number 𝑎𝑎, the expression 𝑎𝑎𝑚𝑚 is the product of 𝑚𝑚 factors of 𝑎𝑎 (i.e., 𝑎𝑎𝑚𝑚= 𝑎𝑎∙𝑎𝑎∙ ⋅⋅⋅ ∙𝑎𝑎 ᇣᇧ ᇧᇤᇧ ᇧᇥ 𝑚𝑚 times ). The number 𝑎𝑎 is called the base, and 𝑚𝑚 is called the exponent or power of 𝑎𝑎.) Expression (An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables.) Linear Expression (A linear expression is an expression that is equivalent to the sum/difference of one or more expressions where each expression is either a number, a variable, or a product of a number and a variable.) Number Sentence (A number sentence is a statement of equality between two numerical expressions.) Numerical Expression (A numerical expression is a number, or it is any combination of sums, differences, products, or divisions of numbers that evaluates to a number.) Solution of an Equation (A solution to an equation with one variable is a number such that the number sentence resulting from substituting the number for all instances of the variable in both expressions is a true number sentence. If an equation has more than one variable, then a solution is an ordered tuple of numbers such that the number sentence resulting from substituting each number from the tuple into all instances of its corresponding variable is a true number sentence.) Truth Values of a Number Sentence (A number sentence is said to be true if both numerical expressions evaluate to the same number; it is said to be false otherwise. True and false are called truth values.) Value of a Numerical Expression (The value of a numerical expression is the number found by evaluating the expression.) Variable (A variable is a symbol (such as a letter) that is a placeholder for a number.) A STORY OF RATIOS 10 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations Familiar Terms and Symbols6 Distribute Expand Factor Number Sentence Product Properties of Operations (distributive, commutative, associative) Quotient Sum Term True or False Number Sentence Variable or Unknown Number Suggested Tools and Representations Bar model Geometric figures Protractors Rapid White Board Exchanges Implementing an RWBE requires that each student be provided with a personal white board, a white board marker, and a means of erasing his or her work. An economic choice for these materials is to place sheets of card stock inside sheet protectors to use as the personal white boards and to cut sheets of felt into small squares to use as erasers. An RWBE consists of a sequence of 10 to 20 problems on a specific topic or skill that starts out with a relatively simple problem and progressively gets more difficult. The teacher should prepare the problems in a way that allows him or her to reveal them to the class one at a time. A flip chart or PowerPoint presentation can be used, or the teacher can write the problems on the board and either cover some with paper or simply write only one problem on the board at a time. The teacher reveals, and possibly reads aloud, the first problem in the list and announces, “Go.” Students work the problem on their personal white boards as quickly as possible and hold their work up for their teacher to see their answers as soon as they have the answer ready. The teacher gives immediate feedback to each student, pointing and/or making eye contact with the student and responding with an affirmation for correct work, such as “Good job!”, “Yes!”, or “Correct!”, or responding with guidance for incorrect work such as “Look again,” “Try again,” “Check your work,” and so on. In the case of the RWBE, it is not recommended that the feedback include the name of the student receiving the feedback. 6These are terms and symbols students have seen previously. A STORY OF RATIOS 11 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Module Overview Module 4: Expressions and Equations If many students have struggled to get the answer correct, go through the solution of that problem as a class before moving on to the next problem in the sequence. Fluency in the skill has been established when the class is able to go through each problem in quick succession without pausing to go through the solution of each problem individually. If only one or two students have not been able to successfully complete a problem, it is appropriate to move the class forward to the next problem without further delay; in this case, find a time to provide remediation to those students before the next fluency exercise on this skill is given. Assessment Summary Assessment Type Administered Format Standards Addressed Mid-Module Assessment Task After Topic E Constructed response with rubric 6.EE.A.1, 6.EE.A.2, 6.EE.A.3, 6.EE.A.4 End-of-Module Assessment Task After Topic H Constructed response with rubric 6.EE.A.2, 6.EE.B.5, 6.EE.B.6, 6.EE.B.7, 6.EE.B.8, 6.EE.C.9 A STORY OF RATIOS 12 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 GRADE 6 • MODULE 4 Topic A: Relationships of the Operations 6 GRADE Mathematics Curriculum Topic A Relationships of the Operations 6.EE.A.3 Focus Standard: 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + 𝑥𝑥) to produce the equivalent expression 6 + 3𝑥𝑥; apply the distributive property to the expression 24𝑥𝑥+ 18𝑦𝑦 to produce the equivalent expression 6(4𝑥𝑥+ 3𝑦𝑦); apply properties of operations to 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 to produce the equivalent expression 3𝑦𝑦. Instructional Days: 4 Lesson 1: The Relationship of Addition and Subtraction (S)1 Lesson 2: The Relationship of Multiplication and Division (E) Lesson 3: The Relationship of Multiplication and Addition (S) Lesson 4: The Relationship of Division and Subtraction (S) Prior to this module, students have worked with numbers and operations from Kindergarten through Grade 5. In Topic A, students further discover and clarify the relationships of the operations using models. From these models, students build and evaluate identities that are useful in solving equations and developing proficiency with solving problems algebraically. To begin, students use models to discover the relationship between addition and subtraction. In Lesson 1, for example, a model could represent the number three. Students notice that if two are taken away, there is a remainder of one. However, when students replace the two units, they notice the answer is back to the original three. Hence, students first discover the identity 𝑤𝑤−𝑥𝑥+ 𝑥𝑥= 𝑤𝑤 and later discover that 𝑤𝑤+ 𝑥𝑥−𝑥𝑥= 𝑤𝑤. In Lesson 2, students model the relationship between multiplication and division. They note that when they divide eight units into two equal groups, they find a quotient of four. They discover that if they multiply that quotient by the number of groups, then they return to their original number, eight, and ultimately build the identities 𝑎𝑎÷ 𝑏𝑏∙𝑏𝑏= 𝑎𝑎 and 𝑎𝑎∙𝑏𝑏÷ 𝑏𝑏= 𝑎𝑎, when 𝑏𝑏≠𝑎𝑎. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 13 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic A Topic A: Relationships of the Operations Students continue to discover identities in Lesson 3, where they determine the relationship between multiplication and addition. Using tape diagrams from previous modules, students are assigned a diagram with three equal parts, where one part is assigned a value of four. They note that since there are three equal parts, they can add four three times to determine the total amount. They relate to multiplication and note that three groups with four items in each group produces a product of twelve, determining that 3 ∙𝑔𝑔= 𝑔𝑔+ 𝑔𝑔+ 𝑔𝑔. Finally, in Lesson 4, students relate division to subtraction. They notice that dividing eight by two produces a quotient of four. They experiment and find that if they subtract the divisor from the dividend four times (the quotient), they find a remainder of zero. They continue to investigate with other examples and prove that if they continually subtract the divisor from the dividend, they determine a difference, or remainder, of zero. Hence, 12 ÷ 𝑥𝑥= 4 means 12 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0. A STORY OF RATIOS 14 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Lesson 1: The Relationship of Addition and Subtraction Student Outcomes Students build and clarify the relationship of addition and subtraction by evaluating identities such as 𝑤𝑤−𝑥𝑥+ 𝑥𝑥= 𝑤𝑤 and 𝑤𝑤+ 𝑥𝑥−𝑥𝑥= 𝑤𝑤. Lesson Notes Teachers need to create square pieces of paper in order for students to build tape diagrams. Each pair of students needs 10 squares to complete the activities. If the teacher has square tiles, these can be used in place of paper squares. The template for the squares and other shapes used in the lesson are provided at the end of the lesson. Teachers need to cut out the shapes. Classwork Fluency Exercise (5 minutes): Multiplication of Decimals RWBE: Refer to the Rapid White Board Exchanges sections in the Module Overview for directions to administer an RWBE. Opening Exercise (5 minutes) Opening Exercise a. Draw a tape diagram to represent the following expression: 𝟓𝟓+ 𝟒𝟒. b. Write an expression for each tape diagram. i. ii. Discuss the answers with the class. If students struggled with the Opening Exercise, provide more examples before moving into the Discussion. 𝟓𝟓 𝟒𝟒 + 𝟐𝟐 𝟐𝟐 + 𝟑𝟑 𝟒𝟒 + A STORY OF RATIOS 15 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Discussion (15 minutes) Provide each pair of students with a collection of 10 squares, so they can use these squares to create tape diagrams throughout the lesson. If each of the squares represents 1 unit, represent the number 3 using the squares provided. Add two more squares to your tape diagram. Write an expression to represent how we created a tape diagram with five squares. Remove two squares from the tape diagram. Alter our original expression 3 + 2 to create an expression that represents what we did with the tape diagram. 3 + 2 −2 Evaluate the expression. 3 Let’s start a new diagram. This time, create a tape diagram with six squares. Use your squares to demonstrate the expression 6 + 4. Remove four squares from the tape diagram. Alter the expression 6 + 4 to create an expression to represent the tape diagram. 6 + 4 −4 How many squares are left on your desk? 6 Evaluate the expression. 6 How many squares did we start with? 6 What effect did adding four squares and then subtracting the four squares have on the number of squares? Adding and then subtracting the same number of squares resulted in the same number that we started with. What if I asked you to add 215 squares to the six squares we started with and then subtract 215 squares? Do you need to actually add and remove these squares to know what the result will be? Why is that? We do not actually need to do the addition and subtraction because we now know that it will result in the same amount of squares that we started with; when you add and then subtract the same number, the results will be the original number. 3 2 + A STORY OF RATIOS 16 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction What do you notice about the expressions we created with the tape diagrams? Possible answer: When we add one number and then subtract the same number, we get our original number. Write an equation, using variables, to represent what we just demonstrated with tape diagrams. Remember that a variable is a letter that represents a number. Use the shapes provided to create tape diagrams to demonstrate this equation. Provide students time to work with their partners to write an equation. Possible answer: 𝑤𝑤+ 𝑥𝑥−𝑥𝑥= 𝑤𝑤. Emphasize that both 𝑤𝑤’s represent the same number, and the same rule applies to the 𝑥𝑥’s. Why is the equation 𝑤𝑤+ 𝑥𝑥−𝑥𝑥= 𝑤𝑤 called an identity? Possible answer: The equation is called an identity because the variables can be replaced with any numbers, and after completing the operations, I returned to the original value. Exercises (12 minutes) Students use their knowledge gained in the Discussion to create another equation using identities. Allow students to continue to work with their partners and 10 squares. Exercises 1. Predict what will happen when a tape diagram has a large number of squares, some squares are removed, and then the same amount of squares are added back on. Possible answer: When some squares are removed from a tape diagram, and then the same amount of squares are added back on, the tape diagram will end up with the same amount of squares that it started with. 2. Build a tape diagram with 𝟏𝟏𝟏𝟏 squares. a. Remove six squares. Write an expression to represent the tape diagram. 𝟏𝟏𝟏𝟏−𝟔𝟔 b. Add six squares onto the tape diagram. Alter the original expression to represent the current tape diagram. 𝟏𝟏𝟏𝟏−𝟔𝟔+ 𝟔𝟔 Scaffolding: The exercise could be completed as a class if students are struggling with the concept. 𝑤𝑤 𝑤𝑤+ 𝑥𝑥−𝑥𝑥 𝑤𝑤+ 𝑥𝑥−𝑥𝑥 = 𝑤𝑤 𝑤𝑤+ 𝑥𝑥 MP.7 MP.2 A STORY OF RATIOS 17 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction c. Evaluate the expression. 𝟏𝟏𝟏𝟏 3. Write an equation, using variables, to represent the identities we demonstrated with tape diagrams. Possible answer: 𝒘𝒘−𝒙𝒙+ 𝒙𝒙= 𝒘𝒘 4. Using your knowledge of identities, fill in each of the blanks. a. 𝟒𝟒+ 𝟓𝟓−_ = 𝟒𝟒 𝟓𝟓 b. 𝟐𝟐𝟐𝟐− + 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 c. _ +𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 d. 𝟓𝟓𝟓𝟓−𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓 5. Using your knowledge of identities, fill in each of the blanks. a. 𝒂𝒂+ 𝒃𝒃− _ = 𝒂𝒂 𝒃𝒃 b. 𝒄𝒄−𝒅𝒅+ 𝒅𝒅= 𝒄𝒄 c. 𝒆𝒆+ −𝒇𝒇= 𝒆𝒆 𝒇𝒇 d. _ − 𝒉𝒉+ 𝒉𝒉= 𝒈𝒈 𝒈𝒈 A STORY OF RATIOS 18 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Closing (3 minutes) In every problem we did today, why did the final value of the expression equal the initial expression? The overall change to the expression was 0. Initially, we added an amount and then subtracted the same amount. Later in the lesson, we subtracted an amount and then added the same amount. Did this alter the outcome? This did not alter the outcome; in both cases, we still ended with our initial value. Why were we able to evaluate the final expression even when we did not know the amount we were adding and subtracting? If we add and subtract the same value, it is similar to adding 0 to an expression because the two numbers are opposites, which have a sum of 0. Exit Ticket (5 minutes) A STORY OF RATIOS 19 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Name Date Lesson 1: The Relationship of Addition and Subtraction Exit Ticket 1. Draw tape diagrams to represent each of the following number sentences. a. 3 + 5 −5 = 3 b. 8 −2 + 2 = 8 2. Fill in each blank. a. 65 + −15 = 65 b. + 𝑔𝑔−𝑔𝑔= 𝑘𝑘 c. 𝑎𝑎+ 𝑏𝑏− _ = 𝑎𝑎 d. 367 −93 + 93 = A STORY OF RATIOS 20 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Exit Ticket Sample Solutions 1. Draw a series of tape diagrams to represent the following number sentences. a. 𝟑𝟑+ 𝟓𝟓−𝟓𝟓= 𝟑𝟑 b. 𝟖𝟖−𝟐𝟐+ 𝟐𝟐= 𝟖𝟖 2. Fill in each blank. a. 𝟔𝟔𝟔𝟔+ − 𝟏𝟏𝟏𝟏= 𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏 b. _ + 𝒈𝒈−𝒈𝒈= 𝒌𝒌 𝒌𝒌 c. 𝒂𝒂+ 𝒃𝒃− = 𝒂𝒂 𝒃𝒃 d. 𝟑𝟑𝟑𝟑𝟑𝟑−𝟗𝟗𝟗𝟗+ 𝟗𝟗𝟗𝟗= _ 𝟑𝟑𝟑𝟑𝟑𝟑 Problem Set Sample Solutions 1. Fill in each blank. a. + 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 b. 𝟒𝟒𝟒𝟒𝟒𝟒−𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒𝟒𝟒 c. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏− _ + 𝟖𝟖𝟖𝟖𝟖𝟖= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖𝟖𝟖 2. Why are the equations 𝒘𝒘−𝒙𝒙+ 𝒙𝒙= 𝒘𝒘 and 𝒘𝒘+ 𝒙𝒙−𝒙𝒙= 𝒘𝒘 called identities? Possible answer: These equations are called identities because the variables can be replaced with any numbers, and after completing the operations, I returned to the original value. A STORY OF RATIOS 21 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction Multiplication of Decimals Progression of Exercises 1. 0.5 × 0.5 = 𝟎𝟎. 𝟐𝟐𝟐𝟐 2. 0.6 × 0.6 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 3. 0.7 × 0.7 = 𝟎𝟎. 𝟒𝟒𝟒𝟒 4. 0.5 × 0.6 = 𝟎𝟎. 𝟑𝟑 5. 1.5 × 1.5 = 𝟐𝟐. 𝟐𝟐𝟐𝟐 6. 2.5 × 2.5 = 𝟔𝟔. 𝟐𝟐𝟐𝟐 7. 0.25 × 0.25 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 8. 0.1 × 0.1 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 9. 0.1 × 123.4 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 10. 0.01 × 123.4 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 A STORY OF RATIOS 22 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction 𝑤𝑤+ 𝑥𝑥 𝑤𝑤 𝑥𝑥 𝑤𝑤 𝑥𝑥 𝑤𝑤+ 𝑥𝑥 𝑤𝑤 𝑥𝑥 𝑤𝑤 𝑥𝑥 A STORY OF RATIOS 23 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 1 Lesson 1: The Relationship of Addition and Subtraction A STORY OF RATIOS 24 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Lesson 2: The Relationship of Multiplication and Division Student Outcomes Students build and clarify the relationship of multiplication and division by evaluating identities such as 𝑎𝑎÷ 𝑏𝑏∙𝑏𝑏= 𝑎𝑎 and 𝑎𝑎∙𝑏𝑏÷ 𝑏𝑏= 𝑎𝑎. Lesson Notes Students use the squares that were used in Lesson 1; however, each pair of students should receive 20 squares for this lesson. Also, students need large paper to complete the Exploratory Challenge. Classwork Fluency Exercise (5 minutes): Division of Fractions I Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions to administer a Sprint. Opening (2 minutes) Remind students of the identities they learned the previous day. Discuss the relationship between addition and subtraction. Inform students that the relationship between multiplication and division is discussed today. Have students make predictions about this relationship using their knowledge gained in the previous lesson. Opening Exercise (5 minutes) Opening Exercise Draw a pictorial representation of the division and multiplication problems using a tape diagram. a. 𝟖𝟖÷ 𝟐𝟐 b. 𝟑𝟑× 𝟐𝟐 𝟑𝟑 𝟑𝟑 𝟔𝟔 MP.2 𝟒𝟒 𝟒𝟒 𝟖𝟖 𝟐𝟐 𝟖𝟖 𝟐𝟐 𝟐𝟐 𝟐𝟐 A STORY OF RATIOS 25 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Discussion (optional—see Scaffolding notes) Provide each pair of students with a collection of 20 squares, which they use to create tape diagrams throughout the lesson. Build a tape diagram to represent 9 units. Divide the 9 units into three equal groups. Write an expression to represent the process you modeled with the tape diagram. 9 ÷ 3 Evaluate the expression. 3 Use your squares to demonstrate what it would look like to multiply 3 by 3. Alter our original expression, 9 ÷ 3, to create an expression that represents what we did with the tape diagram. 9 ÷ 3 × 3 Evaluate the expression. 9 What do you notice about the expression of the tape diagram? Possible answer: When we divide by one number and then multiply by the same number, we end up with our original number. Write an equation, using variables, to represent the identities we demonstrated with tape diagrams. Draw a series of tape diagrams to demonstrate this equation. Provide students time to work in pairs. Possible answer: 𝑎𝑎÷ 𝑏𝑏× 𝑏𝑏= 𝑎𝑎. Emphasize that both 𝑎𝑎’s represent the same number, and the same rule applies to the 𝑏𝑏’s. Scaffolding: The Discussion is provided if students struggled during Lesson 1. If the Discussion is included in the lesson, the Exploratory Challenge is shortened because students only develop one number sentence. MP.7 𝑎𝑎 𝑏𝑏 𝑎𝑎÷ 𝑏𝑏 𝑎𝑎÷ 𝑏𝑏× 𝑏𝑏 𝑎𝑎÷ 𝑏𝑏× 𝑏𝑏 A STORY OF RATIOS 26 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Exploratory Challenge (23 minutes) Students work in pairs or small groups to determine equations to show the relationship between multiplication and division. They use tape diagrams to provide support for their findings. Exploratory Challenge Work in pairs or small groups to determine equations to show the relationship between multiplication and division. Use tape diagrams to provide support for your findings. 1. Create two equations to show the relationship between multiplication and division. These equations should be identities and include variables. Use the squares to develop these equations. 2. Write your equations on large paper. Show a series of tape diagrams to defend each of your equations. Only one number sentence is shown there; the second number sentence and series of tape diagrams are included in the optional Discussion. Possible answer: 𝒂𝒂× 𝒃𝒃÷ 𝒃𝒃= 𝒂𝒂 Possible answer: 𝒂𝒂÷ 𝒃𝒃× 𝒃𝒃= 𝒂𝒂 When students complete their work on the large paper, hang the papers around the room. Provide time for students to walk around and critique their peers’ work. While examining the other posters, students should be comparing the equations and tape diagrams to their own. Use the following rubric to critique other posters. 1. Name of the group you are critiquing 2. Equation you are critiquing 3. Whether or not you believe the equations are true and reasons why Closing (5 minutes) What did you determine about the relationship of multiplication and division? When a number is multiplied and divided by the same number, the result is the original number. What equations can be used to show the relationship of multiplication and division? 𝑎𝑎÷ 𝑏𝑏∙𝑏𝑏= 𝑎𝑎 and 𝑎𝑎∙𝑏𝑏÷ 𝑏𝑏= 𝑎𝑎 Exit Ticket (5 minutes) 𝒂𝒂 𝒂𝒂× 𝒃𝒃 𝒂𝒂 𝒂𝒂 𝒂𝒂 𝒂𝒂× 𝒃𝒃÷ 𝒃𝒃 Scaffolding: If students struggle with getting started, show them the identity equations for addition and subtraction learned in Lesson 1. MP.7 MP.3 A STORY OF RATIOS 27 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Name Date Lesson 2: The Relationship of Multiplication and Division Exit Ticket 1. Fill in the blanks to make each equation true. a. 12 ÷ 3 × _ = 12 b. 𝑓𝑓× ℎ÷ ℎ= c. 45 × ÷ 15 = 45 d. ÷ 𝑟𝑟× 𝑟𝑟= 𝑝𝑝 2. Draw a series of tape diagrams to represent the following number sentences. a. 12 ÷ 3 × 3 = 12 b. 4 × 5 ÷ 5 = 4 A STORY OF RATIOS 28 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Exit Ticket Sample Solutions 1. Fill in the blanks to make each equation true. a. 𝟏𝟏𝟏𝟏÷ 𝟑𝟑× = 𝟏𝟏𝟏𝟏 𝟑𝟑 b. 𝒇𝒇× 𝒉𝒉÷ 𝒉𝒉= _ 𝒇𝒇 c. 𝟒𝟒𝟒𝟒× ÷ 𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏 d. ÷ 𝒓𝒓× 𝒓𝒓= 𝒑𝒑 𝒑𝒑 2. Draw a series of tape diagrams to represent the following number sentences. a. 𝟏𝟏𝟏𝟏÷ 𝟑𝟑× 𝟑𝟑= 𝟏𝟏𝟏𝟏 b. 𝟒𝟒× 𝟓𝟓÷ 𝟓𝟓= 𝟒𝟒 Problem Set Sample Solutions 1. Fill in each blank to make each equation true. a. 𝟏𝟏𝟏𝟏𝟏𝟏÷ 𝟑𝟑× 𝟑𝟑= _ 𝟏𝟏𝟏𝟏𝟏𝟏 b. ÷ 𝟐𝟐𝟐𝟐× 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 c. 𝟓𝟓𝟓𝟓× ÷ 𝟖𝟖= 𝟓𝟓𝟓𝟓 𝟖𝟖 d. 𝟒𝟒𝟒𝟒𝟒𝟒× 𝟏𝟏𝟏𝟏÷ __ = 𝟒𝟒𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏 2. How is the relationship of addition and subtraction similar to the relationship of multiplication and division? Possible answer: Both relationships create identities. A STORY OF RATIOS 29 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Number Correct: _ Division of Fractions—Round 1 Directions: Evaluate each expression and simplify. Use blank spaces to create like units, where applicable. 1. 9 ones ÷ 3 ones 23. 6 10 ÷ 4 10 2. 9 ÷ 3 24. 6 10 ÷ 2 5 = 6 10 ÷ 10 3. 9 tens ÷ 3 tens 25. 10 12 ÷ 5 12 4. 90 ÷ 30 26. 5 6 ÷ 5 12 = 12 ÷ 5 12 5. 9 hundreds ÷ 3 hundreds 27. 10 12 ÷ 3 12 6. 900 ÷ 300 28. 10 12 ÷ 1 4 = 10 12 ÷ 12 7. 9 halves ÷ 3 halves 29. 5 6 ÷ 3 12 = 12 ÷ 3 12 8. 9 2 ÷ 3 2 30. 5 10 ÷ 2 10 9. 9 fourths ÷ 3 fourths 31. 5 10 ÷ 1 5 = 5 10 ÷ 10 10. 9 4 ÷ 3 4 32. 1 2 ÷ 2 10 = 10 ÷ 2 10 11. 9 8 ÷ 3 8 33. 1 2 ÷ 2 4 12. 2 3 ÷ 1 3 34. 3 4 ÷ 2 8 13. 1 3 ÷ 2 3 35. 1 2 ÷ 3 8 14. 6 7 ÷ 2 7 36. 1 2 ÷ 1 5 = 10 ÷ 10 15. 5 7 ÷ 2 7 37. 2 4 ÷ 1 3 16. 3 7 ÷ 4 7 38. 1 4 ÷ 4 6 17. 6 10 ÷ 2 10 39. 3 4 ÷ 2 6 18. 6 10 ÷ 4 10 40. 5 6 ÷ 1 4 19. 6 10 ÷ 8 10 41. 2 9 ÷ 5 6 20. 7 12 ÷ 2 12 42. 5 9 ÷ 1 6 21. 6 12 ÷ 9 12 43. 1 2 ÷ 1 7 22. 4 12 ÷ 11 12 44. 5 7 ÷ 1 2 A STORY OF RATIOS 30 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Division of Fractions—Round 1 [KEY] Directions: Evaluate each expression and simplify. Use blank spaces to create like units, where applicable. 1. 9 ones ÷ 3 ones 𝟗𝟗 𝟑𝟑= 𝟑𝟑 23. 6 10 ÷ 4 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 2. 9 ÷ 3 𝟗𝟗 𝟑𝟑= 𝟑𝟑 24. 6 10 ÷ 2 5 = 6 10 ÷ 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 3. 9 tens ÷ 3 tens 𝟗𝟗 𝟑𝟑= 𝟑𝟑 25. 10 12 ÷ 5 12 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 4. 90 ÷ 30 𝟗𝟗 𝟑𝟑= 𝟑𝟑 26. 5 6 ÷ 5 12 = 12 ÷ 5 12 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 5. 9 hundreds ÷ 3 hundreds 𝟗𝟗 𝟑𝟑= 𝟑𝟑 27. 10 12 ÷ 3 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 6. 900 ÷ 300 𝟗𝟗 𝟑𝟑= 𝟑𝟑 28. 10 12 ÷ 1 4 = 10 12 ÷ 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 7. 9 halves ÷ 3 halves 𝟗𝟗 𝟑𝟑= 𝟑𝟑 29. 5 6 ÷ 3 12 = 12 ÷ 3 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 8. 9 2 ÷ 3 2 𝟗𝟗 𝟑𝟑= 𝟑𝟑 30. 5 10 ÷ 2 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 9. 9 fourths ÷ 3 fourths 𝟗𝟗 𝟑𝟑= 𝟑𝟑 31. 5 10 ÷ 1 5 = 5 10 ÷ 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 10. 9 4 ÷ 3 4 𝟗𝟗 𝟑𝟑= 𝟑𝟑 32. 1 2 ÷ 2 10 = 10 ÷ 2 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 11. 9 8 ÷ 3 8 𝟗𝟗 𝟑𝟑= 𝟑𝟑 33. 1 2 ÷ 2 4 𝟐𝟐 𝟐𝟐= 𝟏𝟏 12. 2 3 ÷ 1 3 𝟐𝟐 𝟏𝟏= 𝟐𝟐 34. 3 4 ÷ 2 8 𝟑𝟑 13. 1 3 ÷ 2 3 𝟏𝟏 𝟐𝟐 35. 1 2 ÷ 3 8 𝟒𝟒 𝟑𝟑= 𝟏𝟏𝟏𝟏 𝟑𝟑 14. 6 7 ÷ 2 7 𝟔𝟔 𝟐𝟐= 𝟑𝟑 36. 1 2 ÷ 1 5 = 10 ÷ 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 15. 5 7 ÷ 2 7 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 37. 2 4 ÷ 1 3 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 16. 3 7 ÷ 4 7 𝟑𝟑 𝟒𝟒 38. 1 4 ÷ 4 6 𝟑𝟑 𝟖𝟖 17. 6 10 ÷ 2 10 𝟔𝟔 𝟐𝟐= 𝟑𝟑 39. 3 4 ÷ 2 6 𝟗𝟗 𝟒𝟒= 𝟐𝟐𝟏𝟏 𝟒𝟒 18. 6 10 ÷ 4 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 40. 5 6 ÷ 1 4 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 19. 6 10 ÷ 8 10 𝟔𝟔 𝟖𝟖= 𝟑𝟑 𝟒𝟒 41. 2 9 ÷ 5 6 𝟒𝟒 𝟏𝟏𝟏𝟏 20. 7 12 ÷ 2 12 𝟕𝟕 𝟐𝟐= 𝟑𝟑𝟏𝟏 𝟐𝟐 42. 5 9 ÷ 1 6 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟓𝟓 21. 6 12 ÷ 9 12 𝟔𝟔 𝟗𝟗= 𝟐𝟐 𝟑𝟑 43. 1 2 ÷ 1 7 𝟕𝟕 𝟐𝟐= 𝟑𝟑𝟏𝟏 𝟐𝟐 22. 4 12 ÷ 11 12 𝟒𝟒 𝟏𝟏𝟏𝟏 44. 5 7 ÷ 1 2 𝟏𝟏𝟏𝟏 𝟕𝟕= 𝟏𝟏𝟑𝟑 𝟕𝟕 A STORY OF RATIOS 31 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Number Correct: Improvement: Division of Fractions—Round 2 Directions: Evaluate each expression and simplify. Use blank spaces to create like units, where applicable. 1. 12 ones ÷ 2 ones 23. 6 12 ÷ 4 12 2. 12 ÷ 2 24. 6 12 ÷ 2 6 = 6 12 ÷ 12 3. 12 tens ÷ 2 tens 25. 8 14 ÷ 7 14 4. 120 ÷ 20 26. 8 14 ÷ 1 2 = 8 14 ÷ 14 5. 12 hundreds ÷ 2 hundreds 27. 11 14 ÷ 2 14 6. 1,200 ÷ 200 28. 11 14 ÷ 1 7 = 11 14 ÷ 14 7. 12 halves ÷ 2 halves 29. 1 7 ÷ 6 14 = 14 ÷ 6 14 8. 12 2 ÷ 2 2 30. 7 18 ÷ 3 18 9. 12 fourths ÷ 3 fourths 31. 7 18 ÷ 1 6 = 7 18 ÷ 18 10. 12 4 ÷ 3 4 32. 1 3 ÷ 12 18 = 18 ÷ 12 18 11. 12 8 ÷ 3 8 33. 1 6 ÷ 4 18 12. 2 4 ÷ 1 4 34. 4 12 ÷ 8 6 13. 1 4 ÷ 2 4 35. 1 3 ÷ 3 15 14. 4 5 ÷ 2 5 36. 2 6 ÷ 1 9 = 18 ÷ 18 15. 2 5 ÷ 4 5 37. 1 6 ÷ 4 9 16. 3 5 ÷ 4 5 38. 2 3 ÷ 3 4 17. 6 8 ÷ 2 8 39. 1 3 ÷ 3 5 18. 6 8 ÷ 4 8 40. 1 7 ÷ 1 2 19. 6 8 ÷ 5 8 41. 5 6 ÷ 2 9 20. 6 10 ÷ 2 10 42. 5 9 ÷ 2 6 21. 7 10 ÷ 8 10 43. 5 6 ÷ 4 9 22. 4 10 ÷ 7 10 44. 1 2 ÷ 4 5 A STORY OF RATIOS 32 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 2 Lesson 2: The Relationship of Multiplication and Division Division of Fractions—Round 2 [KEY] Directions: Evaluate each expression and simplify. Use blank spaces to create like units, where applicable. 1. 12 ones ÷ 2 ones 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 23. 6 12 ÷ 4 12 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 2. 12 ÷ 2 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 24. 6 12 ÷ 2 6 = 6 12 ÷ 12 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 3. 12 tens ÷ 2 tens 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 25. 8 14 ÷ 7 14 𝟖𝟖 𝟕𝟕= 𝟏𝟏𝟏𝟏 𝟕𝟕 4. 120 ÷ 20 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 26. 8 14 ÷ 1 2 = 8 14 ÷ 14 𝟖𝟖 𝟕𝟕= 𝟏𝟏𝟏𝟏 𝟕𝟕 5. 12 hundreds ÷ 2 hundreds 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 27. 11 14 ÷ 2 14 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟓𝟓𝟏𝟏 𝟐𝟐 6. 1,200 ÷ 200 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 28. 11 14 ÷ 1 7 = 11 14 ÷ 14 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟓𝟓𝟏𝟏 𝟐𝟐 7. 12 halves ÷ 2 halves 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 29. 1 7 ÷ 6 14 = 14 ÷ 6 14 𝟐𝟐 𝟔𝟔= 𝟏𝟏 𝟑𝟑 8. 12 2 ÷ 2 2 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 30. 7 18 ÷ 3 18 𝟕𝟕 𝟑𝟑= 𝟐𝟐𝟏𝟏 𝟑𝟑 9. 12 fourths ÷ 3 fourths 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 31. 7 18 ÷ 1 6 = 7 18 ÷ 18 𝟕𝟕 𝟑𝟑= 𝟐𝟐𝟏𝟏 𝟑𝟑 10. 12 4 ÷ 3 4 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 32. 1 3 ÷ 12 18 = 18 ÷ 12 18 𝟔𝟔 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟐𝟐 11. 12 8 ÷ 3 8 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 33. 1 6 ÷ 4 18 𝟑𝟑 𝟒𝟒 12. 2 4 ÷ 1 4 𝟐𝟐 𝟏𝟏= 𝟐𝟐 34. 4 12 ÷ 8 6 𝟒𝟒 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟒𝟒 13. 1 4 ÷ 2 4 𝟏𝟏 𝟐𝟐 35. 1 3 ÷ 3 15 𝟓𝟓 𝟑𝟑= 𝟏𝟏𝟐𝟐 𝟑𝟑 14. 4 5 ÷ 2 5 𝟒𝟒 𝟐𝟐= 𝟐𝟐 36. 2 6 ÷ 1 9 = 18 ÷ 18 𝟔𝟔 𝟐𝟐= 𝟑𝟑 15. 2 5 ÷ 4 5 𝟐𝟐 𝟒𝟒= 𝟏𝟏 𝟐𝟐 37. 1 6 ÷ 4 9 𝟑𝟑 𝟖𝟖 16. 3 5 ÷ 4 5 𝟑𝟑 𝟒𝟒 38. 2 3 ÷ 3 4 𝟖𝟖 𝟗𝟗 17. 6 8 ÷ 2 8 𝟔𝟔 𝟐𝟐= 𝟑𝟑 39. 1 3 ÷ 3 5 𝟓𝟓 𝟗𝟗 18. 6 8 ÷ 4 8 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 40. 1 7 ÷ 1 2 𝟐𝟐 𝟕𝟕 19. 6 8 ÷ 5 8 𝟔𝟔 𝟓𝟓= 𝟏𝟏𝟏𝟏 𝟓𝟓 41. 5 6 ÷ 2 9 𝟏𝟏𝟏𝟏 𝟒𝟒= 𝟑𝟑𝟑𝟑 𝟒𝟒 20. 6 10 ÷ 2 10 𝟔𝟔 𝟐𝟐= 𝟑𝟑 42. 5 9 ÷ 2 6 𝟏𝟏𝟏𝟏 𝟔𝟔= 𝟏𝟏𝟐𝟐 𝟑𝟑 21. 7 10 ÷ 8 10 𝟕𝟕 𝟖𝟖 43. 5 6 ÷ 4 9 𝟏𝟏𝟏𝟏 𝟖𝟖= 𝟏𝟏𝟕𝟕 𝟖𝟖 22. 4 10 ÷ 7 10 𝟒𝟒 𝟕𝟕 44. 1 2 ÷ 4 5 𝟓𝟓 𝟖𝟖 A STORY OF RATIOS 33 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition Lesson 3: The Relationship of Multiplication and Addition Student Outcomes Students build and clarify the relationship of multiplication and addition by evaluating identities such as 3 ∙𝑔𝑔= 𝑔𝑔+ 𝑔𝑔+ 𝑔𝑔. Lesson Notes Students continue to use the squares from Lessons 1 and 2 to create tape diagrams. Each pair of students needs 30 squares to complete the activities. Classwork Opening Exercise (5 minutes) Opening Exercise Write two different expressions that can be depicted by the tape diagram shown. One expression should include addition, while the other should include multiplication. a. Possible answers: 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑 or 𝟑𝟑× 𝟑𝟑 b. Possible answers: 𝟖𝟖+ 𝟖𝟖 or 𝟐𝟐× 𝟖𝟖 c. Possible answers: 𝟓𝟓+ 𝟓𝟓+ 𝟓𝟓 or 𝟑𝟑× 𝟓𝟓 Discussion (17 minutes) Provide each pair of students with a collection of 30 squares, which they use to create tape diagrams throughout the lesson. One partner builds a tape diagram to represent the expression 2 + 2 + 2 + 2, while the other partner builds a tape diagram to represent 4 × 2. 2 2 2 2 + + + 4 4 A STORY OF RATIOS 34 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition What do you notice about the two tape diagrams you created? Possible answer: Although the tape diagrams represent two different expressions, they each have the same number of squares. Why are the two tape diagrams the same? What does it say about the value of the expressions? The two tape diagrams are the same because the values of the expressions are equivalent. If both expressions yield the same value, is there an advantage to using one over the other? Answers will vary. Since each tape diagram has the same number of squares, can we say the two expressions are equivalent? Why or why not? Possible answer: The two expressions are equivalent because they represent the same value. When evaluated, both expressions will equal 8. Therefore, 2 + 2 + 2 + 2 = 4 × 2. Let’s build a new set of tape diagrams. One partner builds a tape diagram to represent the expression 3 × 4, while the other partner builds a tape diagram to represent the expression 4 + 4 + 4. Is 3 × 4 equivalent to 4 + 4 + 4? Why or why not? Possible answer: The two expressions are equivalent because when each of them is evaluated, they equal 12, as we can see with our tape diagrams. Using variables, write an equation to show the relationship of multiplication and addition. Provide students with time to create an equation. Possible answer: 3𝑔𝑔= 𝑔𝑔+ 𝑔𝑔+ 𝑔𝑔. Emphasize that each 𝑔𝑔 represents the same number. 3𝑔𝑔 is the same as writing 3 × 𝑔𝑔, but we no longer use the × for multiplication because it looks like a variable and can become confusing. When a number is next to a variable with no sign, multiplication is implied. In the two previous lessons, we talked about identities. Is the equation 3𝑔𝑔= 𝑔𝑔+ 𝑔𝑔+ 𝑔𝑔 also an identity? Why or why not? Possible answer: The equation 3𝑔𝑔= 𝑔𝑔+ 𝑔𝑔+ 𝑔𝑔 is an identity because we can replace 𝑔𝑔 with any number, and the equation will always be true. 𝑔𝑔 + 𝑔𝑔 + 𝑔𝑔 𝑔𝑔 4 1st group 2nd group 3rd group MP.2 A STORY OF RATIOS 35 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition Exercises (15 minutes) Students can continue to work with the given squares and with their partners to answer the following questions. Exercises 1. Write the addition sentence that describes the model and the multiplication sentence that describes the model. 𝟓𝟓+ 𝟓𝟓+ 𝟓𝟓 and 𝟑𝟑× 𝟓𝟓 2. Write an equivalent expression to demonstrate the relationship of multiplication and addition. a. 𝟔𝟔+ 𝟔𝟔 𝟐𝟐× 𝟔𝟔 b. 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑 𝟔𝟔× 𝟑𝟑 c. 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒 𝟓𝟓× 𝟒𝟒 d. 𝟔𝟔× 𝟐𝟐 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐 e. 𝟒𝟒× 𝟔𝟔 𝟔𝟔+ 𝟔𝟔+ 𝟔𝟔+ 𝟔𝟔 f. 𝟑𝟑× 𝟗𝟗 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗 g. 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉 𝟓𝟓𝟓𝟓 h. 𝟔𝟔𝟔𝟔 𝒚𝒚+ 𝒚𝒚+ 𝒚𝒚+ 𝒚𝒚+ 𝒚𝒚+ 𝒚𝒚 A STORY OF RATIOS 36 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition 3. Roberto is not familiar with tape diagrams and believes that he can show the relationship of multiplication and addition on a number line. Help Roberto demonstrate that the expression 𝟑𝟑× 𝟐𝟐 is equivalent to 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐 on a number line. Possible answer: The first number line shows that there are 𝟑𝟑 groups of 𝟐𝟐, resulting in 𝟔𝟔. The second number line shows the sum of 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐, resulting in 𝟔𝟔. Since both number lines start at 𝟎𝟎 and end at 𝟔𝟔, the expressions are equivalent. 4. Tell whether the following equations are true or false. Then, explain your reasoning. a. 𝒙𝒙+ 𝟔𝟔𝟔𝟔−𝟔𝟔𝟔𝟔= 𝒙𝒙 The equation is true because it demonstrates the addition identity. b. 𝟐𝟐𝟐𝟐−𝟒𝟒𝟒𝟒+ 𝟒𝟒𝟒𝟒= 𝟐𝟐𝟐𝟐 The equation is true because it demonstrates the subtraction identity. 5. Write an equivalent expression to demonstrate the relationship between addition and multiplication. a. 𝟔𝟔+ 𝟔𝟔+ 𝟔𝟔+ 𝟔𝟔+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒 𝟒𝟒× 𝟔𝟔+ 𝟑𝟑× 𝟒𝟒 b. 𝒅𝒅+ 𝒅𝒅+ 𝒅𝒅+ 𝒘𝒘+ 𝒘𝒘+ 𝒘𝒘+ 𝒘𝒘+ 𝒘𝒘 𝟑𝟑𝟑𝟑+ 𝟓𝟓𝟓𝟓 c. 𝒂𝒂+ 𝒂𝒂+ 𝒃𝒃+ 𝒃𝒃+ 𝒃𝒃+ 𝒄𝒄+ 𝒄𝒄+ 𝒄𝒄+ 𝒄𝒄 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 A STORY OF RATIOS 37 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition Closing (4 minutes) Create a diagram that models 3 groups of size 𝑏𝑏. Write two equivalent expressions that represent this model. Possible answers: 3𝑏𝑏, 𝑏𝑏+ 𝑏𝑏+ 𝑏𝑏 Peter says that since the addition expression yields the same value as the multiplication expression, he will always choose to use the addition expression when solving these types of problems. Convince Peter that he may want to reconsider his position. Answers will vary but should include the idea that when the group size is large, it is more advantageous to multiply instead of add. Exit Ticket (4 minutes) 𝑏𝑏 𝑏𝑏 𝑏𝑏 A STORY OF RATIOS 38 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition Name Date Lesson 3: The Relationship of Multiplication and Addition Exit Ticket Write an equivalent expression to show the relationship of multiplication and addition. 1. 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 2. 4 × 9 3. 6 + 6 + 6 4. 7ℎ 5. 𝑗𝑗+ 𝑗𝑗+ 𝑗𝑗+ 𝑗𝑗+ 𝑗𝑗 6. 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢+ 𝑢𝑢 A STORY OF RATIOS 39 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 3 Lesson 3: The Relationship of Multiplication and Addition Exit Ticket Sample Solutions Write an equivalent expression to show the relationship of multiplication and addition. 1. 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖+ 𝟖𝟖 𝟗𝟗× 𝟖𝟖 2. 𝟒𝟒× 𝟗𝟗 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗 3. 𝟔𝟔+ 𝟔𝟔+ 𝟔𝟔 𝟑𝟑× 𝟔𝟔 4. 𝟕𝟕𝟕𝟕 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉+ 𝒉𝒉 5. 𝒋𝒋+ 𝒋𝒋+ 𝒋𝒋+ 𝒋𝒋+ 𝒋𝒋 𝟓𝟓𝟓𝟓 6. 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖+ 𝒖𝒖 𝟏𝟏𝟏𝟏𝟏𝟏 Problem Set Sample Solutions Write an equivalent expression to show the relationship of multiplication and addition. 1. 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 𝟑𝟑× 𝟏𝟏𝟏𝟏 2. 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒 𝟕𝟕× 𝟒𝟒 3. 𝟖𝟖× 𝟐𝟐 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐+ 𝟐𝟐 4. 𝟑𝟑× 𝟗𝟗 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗 5. 𝟔𝟔𝟔𝟔 𝒎𝒎+ 𝒎𝒎+ 𝒎𝒎+ 𝒎𝒎+ 𝒎𝒎+ 𝒎𝒎 6. 𝒅𝒅+ 𝒅𝒅+ 𝒅𝒅+ 𝒅𝒅+ 𝒅𝒅 𝟓𝟓𝟓𝟓 A STORY OF RATIOS 40 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Lesson 4: The Relationship of Division and Subtraction Student Outcomes Students build and clarify the relationship of division and subtraction by determining that 12 ÷ 𝑥𝑥= 4 means 12 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0. Lesson Notes Students continue to use the squares from Lessons 1–3 to create tape diagrams. Each pair of students needs 30 squares to complete the activities. Classwork Discussion (20 minutes) Provide each pair of students with a collection of 30 squares so they can use these squares to create tape diagrams throughout the lesson. Build a tape diagram that has 20 squares. Divide the tape diagram into 4 equal sections. How many squares are in each of the 4 sections? 5 Write a number sentence to demonstrate what happened. 20 ÷ 4 = 5 Combine your squares again to have a tape diagram with 20 squares. Now, subtract 4 squares from your tape diagram. Write an expression to demonstrate what happened. 20 −4 Subtract 4 more squares, and alter your expression to represent the new tape diagram. 20 −4 −4 Subtract 4 more squares, and alter your expression to represent the new tape diagram. 20 −4 −4 −4 A STORY OF RATIOS 41 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 20 ÷ 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 20 −𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 20 −𝑥𝑥−𝑥𝑥 𝑥𝑥 𝑥𝑥 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥 𝑥𝑥 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥 Subtract 4 more squares, and alter your expression to represent the new tape diagram. 20 −4 −4 −4 −4 Last time. Subtract 4 more squares, and alter your expression to an equation in order to represent a number sentence showing the complete transformation of the tape diagram. No squares should remain. 20 −4 −4 −4 −4 −4 = 0 Let’s take a look at the process we took to determine the difference to be zero. Discuss the process step-by-step to determine that the number of times the divisor was subtracted from the dividend is the same number as the quotient. Do you recognize a relationship between 20 ÷ 4 = 5 and 20 −4 −4 −4 −4 −4 = 0? If so, what is it? Possible answer: If you subtract the divisor from the dividend 5 times (the quotient), there will be no remaining squares. Let’s take a look at a similar problem, 20 ÷ 5 = 4, to see if the quotient is the number of times the divisor is subtracted from the dividend. Let’s create a number sentence when we subtract the divisor. 20 −5 −5 −5 −5 = 0 Discuss the process to determine that the number of times the divisor is subtracted from the dividend is the same number as the quotient. Determine the relationship between 20 ÷ 5 = 4 and 20 −5 −5 −5 −5 = 0. 20 ÷ 5 = 4 can be interpreted as subtracting 4 fives from 20 is 0, or 20 −5 −5 −5 −5 = 0. Is this relationship always true? Let’s try to prove that it is. Model the following set of tape diagrams with leading questions for discussion. 𝑥𝑥 is a number. What does 20 ÷ 𝑥𝑥= 5 mean? Exactly five 𝑥𝑥’s can be subtracted from twenty. What must 𝑥𝑥 be in this division sentence? 4 MP.8 MP.8 A STORY OF RATIOS 42 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 20 ÷ 𝑥𝑥= 5 𝑥𝑥 𝑥𝑥 𝑥𝑥 20 −𝑥𝑥 or 20 −5 𝑥𝑥 𝑥𝑥 20 −𝑥𝑥−𝑥𝑥 or 20 −5 −5 𝑥𝑥 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥 or 20 −5 −5 −5 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0 or 20 −5 −5 −5 −5 = 0 Let’s keep taking 𝑥𝑥 away until we reach zero. Model taking each 𝑥𝑥 away and creating subtraction expressions to record. Build a subtraction expression. 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0 Is 20 −4 −4 −4 −4 −4 = 0? Yes Develop two equations using numbers and letters to show the relationship of division and subtraction. Possible answers: 20 ÷ 𝑥𝑥= 5 and 20 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0 Or 20 ÷ 𝑥𝑥= 5 means that 5 can be subtracted exactly 𝑥𝑥 number of times from 20. Is it true when 𝑥𝑥= 4? To determine if 𝑥𝑥= 4, let’s keep taking 𝑥𝑥 away until we reach zero. Model taking each 𝑥𝑥 away and creating subtraction expressions to record by following the diagram to the right. Build a subtraction equation. 20 −5 −5 −5 −5 = 0 What two operations are we relating in the problems we completed? Division and subtraction Exercise 1 (10 minutes) Students work in pairs to answer the following questions. Exercise 1 Build subtraction equations using the indicated equations. The first example has been completed for you. Division Equation Divisor Indicates the Size of the Unit Tape Diagram What is 𝒙𝒙, 𝒚𝒚, 𝒛𝒛? 𝟏𝟏𝟏𝟏÷ 𝒙𝒙= 𝟒𝟒 𝟏𝟏𝟏𝟏−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙= 𝟎𝟎 𝒙𝒙= 𝟑𝟑 𝟏𝟏𝟏𝟏÷ 𝒙𝒙= 𝟑𝟑 𝟏𝟏𝟏𝟏−𝒙𝒙−𝒙𝒙−𝒙𝒙= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 𝟑𝟑𝟑𝟑÷ 𝒚𝒚= 𝟓𝟓 𝟑𝟑𝟑𝟑−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚= 𝟎𝟎 𝒚𝒚= 𝟕𝟕 𝟒𝟒𝟒𝟒÷ 𝒛𝒛= 𝟔𝟔 𝟒𝟒𝟒𝟒−𝒛𝒛−𝒛𝒛−𝒛𝒛−𝒛𝒛−𝒛𝒛−𝒛𝒛= 𝟎𝟎 𝒛𝒛= 𝟕𝟕 MP.2 A STORY OF RATIOS 43 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Division Equation Divisor Indicates the Number of Units Tape Diagram What is 𝒙𝒙, 𝒚𝒚, 𝒛𝒛? 𝟏𝟏𝟏𝟏÷ 𝒙𝒙= 𝟒𝟒 𝟏𝟏𝟏𝟏−𝟒𝟒−𝟒𝟒−𝟒𝟒= 𝟎𝟎 𝒙𝒙= 𝟑𝟑 𝟏𝟏𝟏𝟏÷ 𝒙𝒙= 𝟑𝟑 𝟏𝟏𝟏𝟏−𝟑𝟑−𝟑𝟑−𝟑𝟑−𝟑𝟑−𝟑𝟑−𝟑𝟑= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 𝟑𝟑𝟑𝟑÷ 𝒚𝒚= 𝟓𝟓 𝟑𝟑𝟑𝟑−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓= 𝟎𝟎 𝒚𝒚= 𝟕𝟕 𝟒𝟒𝟒𝟒÷ 𝒛𝒛= 𝟔𝟔 𝟒𝟒𝟒𝟒−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔= 𝟎𝟎 𝒛𝒛= 𝟕𝟕 Exercise 2 (5 minutes) Exercise 2 Answer each question using what you have learned about the relationship of division and subtraction. a. If 𝟏𝟏𝟏𝟏÷ 𝒙𝒙= 𝟑𝟑, how many times would 𝒙𝒙 have to be subtracted from 𝟏𝟏𝟏𝟏 in order for the answer to be zero? What is the value of 𝒙𝒙? 𝟑𝟑; 𝒙𝒙= 𝟒𝟒 b. 𝟑𝟑𝟑𝟑−𝒇𝒇−𝒇𝒇−𝒇𝒇−𝒇𝒇= 𝟎𝟎. Write a division sentence for this repeated subtraction sentence. What is the value of 𝒇𝒇? 𝟑𝟑𝟑𝟑÷ 𝟒𝟒= 𝒇𝒇 or 𝟑𝟑𝟑𝟑÷ 𝒇𝒇= 𝟒𝟒; 𝒇𝒇= 𝟗𝟗 c. If 𝟐𝟐𝟐𝟐÷ 𝒃𝒃= 𝟏𝟏𝟏𝟏, which number is being subtracted 𝟏𝟏𝟏𝟏 times in order for the answer to be zero? Two A STORY OF RATIOS 44 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Closing (5 minutes) Display the graphic organizer provided at the end of the lesson. Copies of the organizer can be made for students to follow along and record. In each of the circles, we can place an operation to satisfy the organizer. In the last four lessons, we have discovered that each operation has a relationship with other operations, whether they are inverse operations or they are repeats of another. Place the addition symbol in the upper left-hand circle. Let’s start with addition. What is the inverse operation of addition? Subtraction Place the subtraction symbol in the upper right-hand circle. After our discussion today, repeated subtraction can be represented by which operation? Division Place the division symbol in the lower right-hand circle. Which operation is the inverse of division? Multiplication Place the multiplication symbol in the lower left-hand circle. Let’s see if this is correct. Is multiplication the repeat operation of addition? Yes Understanding the relationships of operations is going to be instrumental when solving equations later in this unit. Exit Ticket (5 minutes) A STORY OF RATIOS 45 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Name Date Lesson 4: The Relationship of Division and Subtraction Exit Ticket 1. Represent 56 ÷ 8 = 7 using subtraction. Explain your reasoning. 2. Explain why 30 ÷ 𝑥𝑥= 6 is the same as 30 −𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥−𝑥𝑥= 0. What is the value of 𝑥𝑥 in this example? A STORY OF RATIOS 46 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Exit Ticket Sample Solutions 1. Represent 𝟓𝟓𝟓𝟓÷ 𝟖𝟖= 𝟕𝟕 using subtraction. Explain your reasoning. 𝟓𝟓𝟓𝟓−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕= 𝟎𝟎 because 𝟓𝟓𝟓𝟓−𝟕𝟕= 𝟒𝟒𝟒𝟒; 𝟒𝟒𝟒𝟒−𝟕𝟕= 𝟒𝟒𝟒𝟒; 𝟒𝟒𝟒𝟒−𝟕𝟕= 𝟑𝟑𝟑𝟑; 𝟑𝟑𝟑𝟑−𝟕𝟕= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐−𝟕𝟕= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐−𝟕𝟕= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏−𝟕𝟕= 𝟕𝟕; 𝟕𝟕−𝟕𝟕= 𝟎𝟎. OR 𝟓𝟓𝟓𝟓−𝟖𝟖−𝟖𝟖−𝟖𝟖−𝟖𝟖−𝟖𝟖−𝟖𝟖−𝟖𝟖= 𝟎𝟎 because 𝟓𝟓𝟓𝟓−𝟖𝟖= 𝟒𝟒𝟒𝟒; 𝟒𝟒𝟒𝟒−𝟖𝟖= 𝟒𝟒𝟒𝟒; 𝟒𝟒𝟒𝟒−𝟖𝟖= 𝟑𝟑𝟑𝟑; 𝟑𝟑𝟑𝟑−𝟖𝟖= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐−𝟖𝟖= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏−𝟖𝟖= 𝟖𝟖; 𝟖𝟖−𝟖𝟖= 𝟎𝟎. 2. Explain why 𝟑𝟑𝟑𝟑÷ 𝒙𝒙= 𝟔𝟔 is the same as 𝟑𝟑𝟑𝟑−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙= 𝟎𝟎. What is the value of 𝒙𝒙 in this example? 𝟑𝟑𝟑𝟑÷ 𝟓𝟓= 𝟔𝟔, so 𝒙𝒙= 𝟓𝟓. When I subtract 𝟓𝟓 from 𝟑𝟑𝟑𝟑 six times, the result is zero. Division is a repeat operation of subtraction. Problem Set Sample Solutions Build subtraction equations using the indicated equations. Division Equation Divisor Indicates the Size of the Unit Tape Diagram What is 𝒙𝒙, 𝒚𝒚, 𝒛𝒛? 1. 𝟐𝟐𝟐𝟐÷ 𝒙𝒙= 𝟒𝟒 𝟐𝟐𝟐𝟐−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 2. 𝟑𝟑𝟑𝟑÷ 𝒙𝒙= 𝟔𝟔 𝟑𝟑𝟑𝟑−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙−𝒙𝒙= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 3. 𝟐𝟐𝟐𝟐÷ 𝒚𝒚= 𝟕𝟕 𝟐𝟐𝟐𝟐−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚= 𝟎𝟎 𝒚𝒚= 𝟒𝟒 4. 𝟑𝟑𝟑𝟑÷ 𝒚𝒚= 𝟓𝟓 𝟑𝟑𝟑𝟑−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚−𝒚𝒚= 𝟎𝟎 𝒚𝒚= 𝟔𝟔 5. 𝟏𝟏𝟏𝟏÷ 𝒛𝒛= 𝟒𝟒 𝟏𝟏𝟏𝟏−𝒛𝒛−𝒛𝒛−𝒛𝒛−𝒛𝒛= 𝟎𝟎 𝒛𝒛= 𝟒𝟒 A STORY OF RATIOS 47 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Division Equation Divisor Indicates the Number of Units Tape Diagram What is 𝒙𝒙, 𝒚𝒚, 𝒛𝒛? 1. 𝟐𝟐𝟐𝟐÷ 𝒙𝒙= 𝟒𝟒 𝟐𝟐𝟐𝟐−𝟒𝟒−𝟒𝟒−𝟒𝟒−𝟒𝟒−𝟒𝟒−𝟒𝟒= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 2. 𝟑𝟑𝟑𝟑÷ 𝒙𝒙= 𝟔𝟔 𝟑𝟑𝟑𝟑−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔−𝟔𝟔= 𝟎𝟎 𝒙𝒙= 𝟔𝟔 3. 𝟐𝟐𝟐𝟐÷ 𝒚𝒚= 𝟕𝟕 𝟐𝟐𝟐𝟐−𝟕𝟕−𝟕𝟕−𝟕𝟕−𝟕𝟕= 𝟎𝟎 𝒚𝒚= 𝟒𝟒 4. 𝟑𝟑𝟑𝟑÷ 𝒚𝒚= 𝟓𝟓 𝟑𝟑𝟑𝟑−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓−𝟓𝟓= 𝟎𝟎 𝒚𝒚= 𝟔𝟔 5. 𝟏𝟏𝟏𝟏÷ 𝒛𝒛= 𝟒𝟒 𝟏𝟏𝟏𝟏−𝟒𝟒−𝟒𝟒−𝟒𝟒−𝟒𝟒= 𝟎𝟎 𝒛𝒛= 𝟒𝟒 A STORY OF RATIOS 48 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 4 Lesson 4: The Relationship of Division and Subtraction Graphic Organizer Reproducible A STORY OF RATIOS 49 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 GRADE 6 • MODULE 4 Topic B: Special Notations of Operations 6 GRADE Mathematics Curriculum Topic B Special Notations of Operations 6.EE.A.1, 6.EE.A.2c Focus Standards: 6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. 6.EE.A.2c Write, read, and evaluate expressions in which letters stand for numbers. c. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas 𝑉𝑉= 𝑠𝑠3 and 𝐴𝐴= 6𝑠𝑠2 to find the volume and surface area of a cube with sides of length 𝑠𝑠= 1/2. Instructional Days: 2 Lesson 5: Exponents (S)1 Lesson 6: The Order of Operations (P) In Topic B, students differentiate between the product of two numbers and whole numbers with exponents. They differentiate between the two through exploration of patterns, specifically noting how squares grow from a 1 × 1 measure. They determine that a square with a length and width of three units in measure is constructed with nine square units. This expression is represented as 32 and is evaluated as the product of 3 × 3 = 9, not the product of the base and exponent, 6. They further differentiate between the two by comparing the areas of two models with similar measures, as shown below. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 50 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic B Topic B: Special Notations of Operations Once students understand that the base is multiplied by itself the number of times as stated by the exponent, they make a smooth transition into bases that are represented with positive fractions and decimals. They know that for any number 𝑎𝑎, 𝑎𝑎𝑚𝑚 is defined as the product of 𝑚𝑚 factors of 𝑎𝑎. The number 𝑎𝑎 is the base, and 𝑚𝑚 is called the exponent (or the power) of 𝑎𝑎. In Lesson 6, students build on their previous understanding of the order of operations by including work with exponents. They follow the order of operations to evaluate numerical expressions. They recognize that, in the absence of parentheses, exponents are evaluated first. Students identify when the order of operations is incorrectly applied and determine the applicable course to correctly evaluate expressions. They understand that the placement of parentheses can alter the final solution when evaluating expressions, as in the following example: 24 ∙(2 + 8) −16 24 ∙10 −16 16 ∙10 −16 160 −16 144 24 ∙2 + 8 −16 16 ∙2 + 8 −16 32 + 8 −16 40 −16 24 Students continue to apply the order of operations throughout the module as they evaluate numerical and algebraic expressions. A STORY OF RATIOS 51 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Lesson 5: Exponents Student Outcomes Students discover that 3𝑥𝑥= 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥 is not the same thing as 𝑥𝑥3, which is 𝑥𝑥· 𝑥𝑥· 𝑥𝑥. Students understand that a base number can be represented with a positive whole number, positive fraction, or positive decimal and that for any number 𝑎𝑎, 𝑎𝑎𝑚𝑚 is defined as the product of 𝑚𝑚 factors of 𝑎𝑎. The number 𝑎𝑎 is the base, and 𝑚𝑚 is called the exponent or power of 𝑎𝑎. Lesson Notes In Grade 5, students are introduced to exponents. Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole number exponents to denote powers of 10 (5.NBT.A.2). In this lesson, students use new terminology (base, squared, and cubed) and practice moving between exponential notation, expanded notation, and standard notation. The following terms should be displayed, defined, and emphasized throughout Lesson 5: base, exponent, power, squared, and cubed. Classwork Fluency Exercise (5 minutes): Multiplication of Decimals RWBE: Refer to the Rapid White Board Exchanges section in the Module Overview for directions on how to administer an RWBE. Opening Exercise (2 minutes) Opening Exercise As you evaluate these expressions, pay attention to how you arrive at your answers. 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒+ 𝟒𝟒 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗+ 𝟗𝟗 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 Discussion (15 minutes) How many of you solved the problems by counting on? That is, starting with 4, you counted on 4 more each time (5, 6, 7, 𝟖𝟖, 9, 10, 11, 𝟏𝟏𝟏𝟏, 13, 14, 15, 𝟏𝟏𝟏𝟏, … ). If you did not find the answer that way, could you have done so? Yes, but it is time consuming and cumbersome. Addition is a faster way of counting on. A STORY OF RATIOS 52 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Scaffolding: When teaching students how to write an exponent as a superscript, compare and contrast the notation with how to write a subscript, as in the molecular formula for water, H2O, or carbon dioxide, CO2. Here the number is again half as tall as the capital letters, and the top of the 2 is halfway down it. The bottom of the subscript can extend a little lower than the bottom of the letter. Ignore the meaning of a chemical subscript. How else could you find the sums using addition? Count by 4, 9, or 10. How else could you solve the problems? Multiply 4 times 10; multiply 9 times 5; or multiply 10 times 5. Multiplication is a faster way to add numbers when the addends are the same. When we add five groups of 10, we use an abbreviation and a different notation, called multiplication. 10 + 10 + 10 + 10 + 10 = 5 × 10 If multiplication is a more efficient way to represent addition problems involving the repeated addition of the same addend, do you think there might be a more efficient way to represent the repeated multiplication of the same factor, as in 10 × 10 × 10 × 10 × 10 = ? Allow students to make suggestions; some may recall this from previous lessons. 10 × 10 × 10 × 10 × 10 = 105 We see that when we add five groups of 10, we write 5 × 10, but when we multiply five copies of 10, we write 105. So, multiplication by 5 in the context of addition corresponds exactly to the exponent 5 in the context of multiplication. Make students aware of the correspondence between addition and multiplication because what they know about repeated addition helps them learn exponents as repeated multiplication going forward. The little 5 we write is called an exponent and is written as a superscript. The numeral 5 is written only half as tall and half as wide as the 10, and the bottom of the 5 should be halfway up the number 10. The top of the 5 can extend a little higher than the top of the zero in 10. Why do you think we write exponents so carefully? It reduces the chance that a reader will confuse 105 with 105. Examples 1–5 (5 minutes) Work through Examples 1–5 as a group; supplement with additional examples if needed. Examples 1–10 Write each expression in exponential form. 1. 𝟓𝟓× 𝟓𝟓× 𝟓𝟓× 𝟓𝟓× 𝟓𝟓= 𝟓𝟓𝟓𝟓 2. 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐= 𝟐𝟐𝟒𝟒 Write each expression in expanded form. 3. 𝟖𝟖𝟑𝟑= 𝟖𝟖× 𝟖𝟖× 𝟖𝟖 4. 𝟏𝟏𝟏𝟏𝟔𝟔= 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏 5. 𝒈𝒈𝟑𝟑= 𝒈𝒈× 𝒈𝒈× 𝒈𝒈 The repeated factor is called the base, and the exponent is also called the power. Say the numbers in Examples 1–5 to a partner. MP.2 & MP.7 A STORY OF RATIOS 53 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Check to make sure students read the examples correctly: Five to the fifth power, two to the fourth power, eight to the third power, ten to the sixth power, and 𝑔𝑔 to the third power. Go back to Examples 1–4, and use a calculator to evaluate the expressions. 1. 𝟓𝟓× 𝟓𝟓× 𝟓𝟓× 𝟓𝟓× 𝟓𝟓= 𝟓𝟓𝟓𝟓= 𝟑𝟑, 𝟏𝟏𝟏𝟏𝟏𝟏 2. 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐= 𝟐𝟐𝟒𝟒= 𝟏𝟏𝟏𝟏 3. 𝟖𝟖𝟑𝟑= 𝟖𝟖× 𝟖𝟖× 𝟖𝟖= 𝟓𝟓𝟓𝟓𝟓𝟓 4. 𝟏𝟏𝟏𝟏𝟔𝟔= 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏= 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟎𝟎𝟎𝟎𝟎𝟎 What is the difference between 𝟑𝟑𝟑𝟑 and 𝒈𝒈𝟑𝟑? 𝟑𝟑𝟑𝟑= 𝒈𝒈+ 𝒈𝒈+ 𝒈𝒈 or 𝟑𝟑 times 𝒈𝒈; 𝒈𝒈𝟑𝟑= 𝒈𝒈× 𝒈𝒈× 𝒈𝒈 Take time to clarify this important distinction. The base number can also be written in decimal or fraction form. Try Examples 6, 7, and 8. Use a calculator to evaluate the expressions. Examples 6–8 (4 minutes) 6. Write the expression in expanded form, and then evaluate. (𝟑𝟑. 𝟖𝟖)𝟒𝟒= 𝟑𝟑. 𝟖𝟖× 𝟑𝟑. 𝟖𝟖× 𝟑𝟑. 𝟖𝟖× 𝟑𝟑. 𝟖𝟖= 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 7. Write the expression in exponential form, and then evaluate. 𝟐𝟐. 𝟏𝟏× 𝟐𝟐. 𝟏𝟏= (𝟐𝟐. 𝟏𝟏)𝟐𝟐= 𝟒𝟒. 𝟒𝟒𝟒𝟒 8. Write the expression in exponential form, and then evaluate. 𝟎𝟎. 𝟕𝟕𝟕𝟕× 𝟎𝟎. 𝟕𝟕𝟕𝟕× 𝟎𝟎. 𝟕𝟕𝟕𝟕= (𝟎𝟎. 𝟕𝟕𝟕𝟕)𝟑𝟑= 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 𝟖𝟖𝟖𝟖𝟖𝟖 The base number can also be a fraction. Convert the decimals to fractions in Examples 7 and 8 and evaluate. Leave your answer as a fraction. Remember how to multiply fractions! Example 7: 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏× 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏= ൬𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏൰ 𝟐𝟐 = 𝟒𝟒𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 Example 8: 𝟑𝟑 𝟒𝟒× 𝟑𝟑 𝟒𝟒× 𝟑𝟑 𝟒𝟒= ൬𝟑𝟑 𝟒𝟒൰ 𝟑𝟑 = 𝟐𝟐𝟐𝟐 𝟔𝟔𝟔𝟔 Note to teacher: If students need additional help multiplying fractions, refer to the first four modules of Grade 5. A STORY OF RATIOS 54 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Examples 9–10 (1 minute) 9. Write the expression in exponential form, and then evaluate. 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐= ൬𝟏𝟏 𝟐𝟐൰ 𝟑𝟑 = 𝟏𝟏 𝟖𝟖 10. Write the expression in expanded form, and then evaluate. ൬𝟐𝟐 𝟑𝟑൰ 𝟐𝟐 = 𝟐𝟐 𝟑𝟑× 𝟐𝟐 𝟑𝟑= 𝟒𝟒 𝟗𝟗 There is a special name for numbers raised to the second power. When a number is raised to the second power, it is called squared. Remember that in geometry, squares have the same two dimensions: length and width. For 𝑏𝑏> 0, 𝑏𝑏2 is the area of a square with side length 𝑏𝑏. What is the value of 5 squared? 25 What is the value of 7 squared? 49 What is the value of 8 squared? 64 What is the value of 1 squared? 1 A multiplication chart is included at the end of this lesson. Post or project it as needed. Where are square numbers found on the multiplication table? On the diagonal There is also a special name for numbers raised to the third power. When a number is raised to the third power, it is called cubed. Remember that in geometry, cubes have the same three dimensions: length, width, and height. For 𝑏𝑏> 0, 𝑏𝑏3 is the volume of a cube with edge length 𝑏𝑏. What is the value of 1 cubed? 1 × 1 × 1 = 1 What is the value of 2 cubed? 2 × 2 × 2 = 8 What is the value of 3 cubed? 3 × 3 × 3 = 27 In general, for any number 𝑥𝑥,𝑥𝑥1 = 𝑥𝑥, and for any positive integer 𝑛𝑛> 1, 𝑥𝑥𝑛𝑛 is, by definition, 𝑥𝑥𝑛𝑛= (𝑥𝑥∙𝑥𝑥⋯𝑥𝑥) ᇣᇧ ᇧᇤᇧ ᇧᇥ 𝑛𝑛 times . What does the 𝑥𝑥 represent in this equation? The 𝑥𝑥 represents the factor that will be repeatedly multiplied by itself. What does the 𝑛𝑛 represent in this expression? 𝑛𝑛 represents the number of times 𝑥𝑥 will be multiplied. MP.6 A STORY OF RATIOS 55 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Let’s look at this with some numbers. How would we represent 4𝑛𝑛? 4𝑛𝑛= (4 ∙4 ⋯4) ᇣᇧ ᇧᇤᇧ ᇧᇥ 𝑛𝑛 times What does the 4 represent in this expression? The 4 represents the factor that will be repeatedly multiplied by itself. What does the 𝑛𝑛 represent in this expression? 𝑛𝑛 represents the number of times 4 will be multiplied. What if we were simply multiplying? How would we represent 4𝑛𝑛? Because multiplication is repeated addition, 4𝑛𝑛= (4 + 4 ⋯4) ᇣᇧᇧᇤᇧᇧᇥ 𝑛𝑛 times . What does the 4 represent in this expression? The 4 represents the addend that will be repeatedly added to itself. What does the 𝑛𝑛 represent in this expression? 𝑛𝑛 represents the number of times 4 will be added. Exercises (8 minutes) Ask students to fill in the chart, supplying the missing expressions. Exercises 1. Fill in the missing expressions for each row. For whole number and decimal bases, use a calculator to find the standard form of the number. For fraction bases, leave your answer as a fraction. Exponential Form Expanded Form Standard Form 𝟑𝟑𝟐𝟐 𝟑𝟑× 𝟑𝟑 𝟗𝟗 𝟐𝟐𝟔𝟔 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐 𝟔𝟔𝟔𝟔 𝟒𝟒𝟓𝟓 𝟒𝟒× 𝟒𝟒× 𝟒𝟒× 𝟒𝟒× 𝟒𝟒 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 ൬𝟑𝟑 𝟒𝟒൰ 𝟐𝟐 𝟑𝟑 𝟒𝟒× 𝟑𝟑 𝟒𝟒 𝟗𝟗 𝟏𝟏𝟏𝟏 (𝟏𝟏. 𝟓𝟓)𝟐𝟐 𝟏𝟏. 𝟓𝟓× 𝟏𝟏. 𝟓𝟓 𝟐𝟐. 𝟐𝟐𝟐𝟐 2. Write five cubed in all three forms: exponential form, expanded form, and standard form. 𝟓𝟓𝟑𝟑; 𝟓𝟓× 𝟓𝟓× 𝟓𝟓; 𝟏𝟏𝟏𝟏𝟏𝟏 3. Write fourteen and seven-tenths squared in all three forms. (𝟏𝟏𝟏𝟏. 𝟕𝟕)𝟐𝟐; 𝟏𝟏𝟏𝟏. 𝟕𝟕× 𝟏𝟏𝟏𝟏. 𝟕𝟕; 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 4. One student thought two to the third power was equal to six. What mistake do you think he made, and how would you help him fix his mistake? The student multiplied the base, 𝟐𝟐, by the exponent, 𝟑𝟑. This is wrong because the exponent never multiplies the base; the exponent tells how many copies of the base are to be used as factors. MP.6 A STORY OF RATIOS 56 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Closing (2 minutes) We use multiplication as a quicker way to do repeated addition if the addends are the same. We use exponents as a quicker way to multiply if the factors are the same. Carefully write exponents as superscripts to avoid confusion. Exit Ticket (3 minutes) Lesson Summary EXPONENTIAL NOTATION FOR WHOLE NUMBER EXPONENTS: Let 𝒎𝒎 be a nonzero whole number. For any number 𝒂𝒂, the expression 𝒂𝒂𝒎𝒎 is the product of 𝒎𝒎 factors of 𝒂𝒂, i.e., 𝒂𝒂𝒎𝒎= 𝒂𝒂∙𝒂𝒂∙ ⋅⋅⋅ ∙𝒂𝒂 ᇣᇧ ᇧᇤᇧ ᇧᇥ 𝒎𝒎 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 . The number 𝒂𝒂 is called the base, and 𝒎𝒎 is called the exponent or power of 𝒂𝒂. When 𝒎𝒎 is 𝟏𝟏, “the product of one factor of 𝒂𝒂” just means 𝒂𝒂 (i.e., 𝒂𝒂𝟏𝟏= 𝒂𝒂). Raising any nonzero number 𝒂𝒂 to the power of 𝟎𝟎 is defined to be 𝟏𝟏 (i.e., 𝒂𝒂𝟎𝟎= 𝟏𝟏 for all 𝒂𝒂≠𝟎𝟎). A STORY OF RATIOS 57 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Name Date Lesson 5: Exponents Exit Ticket 1. What is the difference between 6𝑧𝑧 and 𝑧𝑧6? 2. Write 103 as a multiplication expression having repeated factors. 3. Write 8 × 8 × 8 × 8 using an exponent. A STORY OF RATIOS 58 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Exit Ticket Sample Solutions 1. What is the difference between 𝟔𝟔𝒛𝒛 and 𝒛𝒛𝟔𝟔? 𝟔𝟔𝒛𝒛= 𝒛𝒛+ 𝒛𝒛+ 𝒛𝒛+ 𝒛𝒛+ 𝒛𝒛+ 𝒛𝒛 or 𝟔𝟔 times 𝒛𝒛; 𝒛𝒛𝟔𝟔= 𝒛𝒛× 𝒛𝒛× 𝒛𝒛× 𝒛𝒛× 𝒛𝒛× 𝒛𝒛 2. Write 𝟏𝟏𝟏𝟏𝟑𝟑 as a multiplication expression having repeated factors. 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏 3. Write 𝟖𝟖× 𝟖𝟖× 𝟖𝟖× 𝟖𝟖 using an exponent. 𝟖𝟖𝟒𝟒 Problem Set Sample Solutions 1. Complete the table by filling in the blank cells. Use a calculator when needed. Exponential Form Expanded Form Standard Form 𝟑𝟑𝟓𝟓 𝟑𝟑× 𝟑𝟑× 𝟑𝟑× 𝟑𝟑× 𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟑𝟑 𝟒𝟒× 𝟒𝟒× 𝟒𝟒 𝟔𝟔𝟔𝟔 (𝟏𝟏. 𝟗𝟗)𝟐𝟐 𝟏𝟏. 𝟗𝟗× 𝟏𝟏. 𝟗𝟗 𝟑𝟑. 𝟔𝟔𝟔𝟔 ൬𝟏𝟏 𝟐𝟐൰ 𝟓𝟓 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐× 𝟏𝟏 𝟐𝟐 𝟏𝟏 𝟑𝟑𝟑𝟑 2. Why do whole numbers raised to an exponent get greater, while fractions raised to an exponent get smaller? As whole numbers are multiplied by themselves, products are larger because there are more groups. As fractions of fractions are taken, the product is smaller. A part of a part is less than how much we started with. 3. The powers of 𝟐𝟐 that are in the range 𝟐𝟐 through 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 are 𝟐𝟐, 𝟒𝟒, 𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑, 𝟔𝟔𝟔𝟔, 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐, and 𝟓𝟓𝟓𝟓𝟓𝟓. Find all the powers of 𝟑𝟑 that are in the range 𝟑𝟑 through 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎. 𝟑𝟑, 𝟗𝟗, 𝟐𝟐𝟐𝟐, 𝟖𝟖𝟖𝟖, 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟕𝟕𝟕𝟕𝟕𝟕 4. Find all the powers of 𝟒𝟒 in the range 𝟒𝟒 through 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎. 𝟒𝟒, 𝟏𝟏𝟏𝟏, 𝟔𝟔𝟔𝟔, 𝟐𝟐𝟐𝟐𝟐𝟐 5. Write an equivalent expression for 𝒏𝒏× 𝒂𝒂 using only addition. (𝒂𝒂+ 𝒂𝒂+ ⋯𝒂𝒂) ᇣᇧᇧᇧᇤᇧᇧᇧᇥ 𝒏𝒏 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 A STORY OF RATIOS 59 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents 6. Write an equivalent expression for 𝒘𝒘𝒃𝒃 using only multiplication. 𝒘𝒘𝒃𝒃= (𝒘𝒘∙𝒘𝒘∙⋯∙𝒘𝒘) ᇣᇧᇧᇧᇤᇧᇧᇧᇥ 𝒃𝒃 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 a. Explain what 𝒘𝒘 is in this new expression. 𝒘𝒘 is the factor that will be repeatedly multiplied by itself. b. Explain what 𝒃𝒃 is in this new expression. 𝒃𝒃 is the number of times 𝒘𝒘 will be multiplied. 7. What is the advantage of using exponential notation? It is a shorthand way of writing a multiplication expression if the factors are all the same. 8. What is the difference between 𝟒𝟒𝒙𝒙 and 𝒙𝒙𝟒𝟒? Evaluate both of these expressions when 𝒙𝒙= 𝟐𝟐. 𝟒𝟒𝟒𝟒 means four times 𝒙𝒙; this is the same as 𝒙𝒙+ 𝒙𝒙+ 𝒙𝒙+ 𝒙𝒙. On the other hand, 𝒙𝒙𝟒𝟒means 𝒙𝒙 to the fourth power, or 𝒙𝒙× 𝒙𝒙× 𝒙𝒙× 𝒙𝒙. When 𝒙𝒙= 𝟐𝟐, 𝟒𝟒𝟒𝟒= 𝟒𝟒× 𝟐𝟐= 𝟖𝟖. When 𝒙𝒙= 𝟐𝟐, 𝒙𝒙𝟒𝟒= 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝟐𝟐= 𝟏𝟏𝟏𝟏. A STORY OF RATIOS 60 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 5 Lesson 5: Exponents Multiplication of Decimals Progression of Exercises 1. 0.5 × 0.5 = 𝟎𝟎. 𝟐𝟐𝟐𝟐 2. 0.6 × 0.6 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 3. 0.7 × 0.7 = 𝟎𝟎. 𝟒𝟒𝟒𝟒 4. 0.5 × 0.6 = 𝟎𝟎. 𝟑𝟑 5. 1.5 × 1.5 = 𝟐𝟐. 𝟐𝟐𝟐𝟐 6. 2.5 × 2.5 = 𝟔𝟔. 𝟐𝟐𝟐𝟐 7. 0.25 × 0.25 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 8. 0.1 × 0.1 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 9. 0.1 × 123.4 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 10. 0.01 × 123.4 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 A STORY OF RATIOS 61 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Lesson 6: The Order of Operations Student Outcomes Students evaluate numerical expressions. They recognize that in the absence of parentheses, exponents are evaluated first.
Classwork Opening (5 minutes) Take a few minutes to review the Problem Set from the previous lesson. Clarify any misconceptions about the use and evaluation of exponents. Opening Exercise (5 minutes) Post the following expression on the board, and ask students to evaluate it. 3 + 4 × 2 Ask students to record their answers and report them using personal white boards, cards, or electronic vote devices. Students who arrive at an answer other than 11 or 14 should recheck their work. Discussion (5 minutes) How did you evaluate the expression 3 + 4 × 2?
I added 3 + 4 first for a sum of 7; then, I multiplied 7 × 2 for a product of 14.
I multiplied 4 × 2 first for a product of 8; then, I added 8 + 3 for a sum of 11.
Only one of these answers can be correct. When we evaluate expressions, we must agree to use one set of rules so that everyone arrives at the same correct answer.
During the last lesson, we said that addition was a shortcut to counting on. How could you think about subtraction?
Subtraction is a shortcut to “counting back.” These were the first operations that you learned because they are the least complicated. Next, you learned about multiplication and division.
Multiplication can be thought of as repeated addition. Thinking back on Lesson 4, how could you think about division?
Division is repeated subtraction.
Multiplication and division are more powerful than addition and subtraction, which led mathematicians to develop the order of operations in this way. When we evaluate expressions that have any of these four operations, we always calculate multiplication and division before doing any addition or subtraction. Since multiplication and division are equally powerful, we simply evaluate these two operations as they are written in the expression, from left to right. A STORY OF RATIOS 62 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Addition and subtraction are at the same level in the order of operations and are evaluated from left to right in an expression. Now that these rules of order of operations are clear, can you go back and evaluate the expression 3 + 4 × 2 as 11? The diagram correctly models the expression 3 + 4 × 2. With addition, we are finding the sum of two addends. In this example, the first addend is the number 3. The second addend happens to be the number that is the value of the expression 4 × 2; so, before we can add, we must determine the value of the second addend. Example 1 (5 minutes): Expressions with Only Addition, Subtraction, Multiplication, and Division Example 1: Expressions with Only Addition, Subtraction, Multiplication, and Division What operations are evaluated first? Multiplication and division are evaluated first, from left to right. What operations are always evaluated last? Addition and subtraction are always evaluated last, from left to right. Ask students to evaluate the expressions. Exercises 1–3 1. 𝟒𝟒+ 𝟐𝟐× 𝟕𝟕 𝟒𝟒+ 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 2. 𝟑𝟑𝟑𝟑÷ 𝟑𝟑× 𝟒𝟒 𝟏𝟏𝟏𝟏× 𝟒𝟒 𝟒𝟒𝟒𝟒 3. 𝟐𝟐𝟐𝟐−𝟓𝟓× 𝟐𝟐 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 63 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations In the last lesson, you learned about exponents, which are a way of writing repeated multiplication. So, exponents are more powerful than multiplication or division. If exponents are present in an expression, they are evaluated before any multiplication or division. We now know that when we evaluate expressions, we must agree to use one set of rules so that everyone arrives at the same correct answer. These rules are based on doing the most powerful operations first (exponents), then the less powerful ones (multiplication and division, going from left to right), and finally, the least powerful ones last (addition and subtraction, going from left to right). Evaluate the expression 4 + 6 × 6 ÷ 8. 4 + (6 × 6) ÷ 8 4 + (36 ÷ 8) 4 + 4.5 8.5 Now, evaluate the expression 4 + 62 ÷ 8. 4 + (62) ÷ 8 4 + (36 ÷ 8) 4 + 4.5 8.5 Why was your first step to find the value of 62? Because exponents are evaluated first. Example 2 (5 minutes): Expressions with Four Operations and Exponents Display the following expression. Example 2: Expressions with Four Operations and Exponents 𝟒𝟒+ 𝟗𝟗𝟐𝟐÷ 𝟑𝟑× 𝟐𝟐−𝟐𝟐 What operation is evaluated first? Exponents (𝟗𝟗𝟐𝟐= 𝟗𝟗× 𝟗𝟗= 𝟖𝟖𝟖𝟖) What operations are evaluated next? Multiplication and division, from left to right (𝟖𝟖𝟖𝟖÷ 𝟑𝟑= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐× 𝟐𝟐= 𝟓𝟓𝟓𝟓) What operations are always evaluated last? Addition and subtraction, from left to right (𝟒𝟒+ 𝟓𝟓𝟓𝟓= 𝟓𝟓𝟓𝟓; 𝟓𝟓𝟓𝟓−𝟐𝟐= 𝟓𝟓𝟓𝟓) What is the final answer? 𝟓𝟓𝟓𝟓 Scaffolding: Some students may benefit from rewriting the expression on successive lines, evaluating only one or two operations on each line. A STORY OF RATIOS 64 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Evaluate the next two exercises. While the answers are provided, it is extremely important to circulate to ensure that students are using the correct order of operations to achieve the answer. For example, in Exercise 5, they should show 43 first, followed by 2 × 8. Exercises 4–5 4. 𝟗𝟗𝟗𝟗−𝟓𝟓𝟐𝟐× 𝟑𝟑 𝟗𝟗𝟗𝟗−𝟐𝟐𝟐𝟐× 𝟑𝟑 𝟗𝟗𝟗𝟗−𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏 5. 𝟒𝟒𝟑𝟑+ 𝟐𝟐× 𝟖𝟖 𝟔𝟔𝟔𝟔+ 𝟐𝟐× 𝟖𝟖 𝟔𝟔𝟔𝟔+ 𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖 Example 3 (5 minutes): Expressions with Parentheses The last important rule in the order of operations involves grouping symbols, usually parentheses. These tell us that in certain circumstances or scenarios, we need to do things out of the usual order. Operations inside grouping symbols are always evaluated first, before exponents and any operations. Example 3: Expressions with Parentheses Consider a family of 𝟒𝟒 that goes to a soccer game. Tickets are $𝟓𝟓. 𝟎𝟎𝟎𝟎 each. The mom also buys a soft drink for $𝟐𝟐. 𝟎𝟎𝟎𝟎. How would you write this expression? 𝟒𝟒× 𝟓𝟓+ 𝟐𝟐 How much will this outing cost? $𝟐𝟐𝟐𝟐 Here is a model of the scenario: Consider a different scenario: The same family goes to the game as before, but each of the family members wants a drink. How would you write this expression? 𝟒𝟒× (𝟓𝟓+ 𝟐𝟐) A STORY OF RATIOS 65 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Why would you add the 𝟓𝟓 and 𝟐𝟐 first? We need to determine how much each person spends. Each person spends $𝟕𝟕; then, we multiply by 𝟒𝟒 people to figure out the total cost. How much will this outing cost? $𝟐𝟐𝟐𝟐 Here is a model of the second scenario: How many groups are there? 𝟒𝟒 What does each group comprise? $𝟓𝟓+ $𝟐𝟐, or $𝟕𝟕 The last complication that can arise is if two or more sets of parentheses are ever needed; evaluate the innermost parentheses first, and then work outward. Try Exercises 6 and 7. Exercises 6–7 6. 𝟐𝟐+ (𝟗𝟗𝟐𝟐−𝟒𝟒) 𝟐𝟐+ (𝟖𝟖𝟖𝟖−𝟒𝟒) 𝟐𝟐+ 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕 7. 𝟐𝟐· ൫𝟏𝟏𝟏𝟏+ 𝟓𝟓−𝟏𝟏𝟏𝟏÷ (𝟑𝟑+ 𝟒𝟒)൯ 𝟐𝟐∙(𝟏𝟏𝟏𝟏+ 𝟓𝟓−𝟏𝟏𝟏𝟏÷ 𝟕𝟕) 𝟐𝟐∙(𝟏𝟏𝟏𝟏+ 𝟓𝟓−𝟐𝟐) 𝟐𝟐∙𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 If students are confused trying to divide 14 by 3, reiterate the rule about nested parentheses. A STORY OF RATIOS 66 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Example 4 (5 minutes): Expressions with Parentheses and Exponents Let’s take a look at how parentheses and exponents work together. Sometimes a problem will have parentheses, and the values inside the parentheses have an exponent. Let’s evaluate the following expression. Place the expression on the board. We will evaluate the parentheses first. Example 4: Expressions with Parentheses and Exponents 𝟐𝟐× (𝟑𝟑+ 𝟒𝟒𝟐𝟐) Which value will we evaluate first within the parentheses? Evaluate. First, evaluate 𝟒𝟒𝟐𝟐, which is 𝟏𝟏𝟏𝟏; then, add 𝟑𝟑. The value of the parentheses is 𝟏𝟏𝟏𝟏. 𝟐𝟐× (𝟑𝟑+ 𝟒𝟒𝟐𝟐) 𝟐𝟐× (𝟑𝟑+ 𝟏𝟏𝟏𝟏) 𝟐𝟐× 𝟏𝟏𝟏𝟏 Evaluate the rest of the expression. 𝟐𝟐× 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 Place the expression on the board: What do you think will happen when the exponent in this expression is outside of the parentheses? 𝟐𝟐× (𝟑𝟑+ 𝟒𝟒)𝟐𝟐 Will the answer be the same? Answers will vary. Which should we evaluate first? Evaluate. Parentheses 𝟐𝟐× (𝟑𝟑+ 𝟒𝟒)𝟐𝟐 𝟐𝟐× (𝟕𝟕)𝟐𝟐 What happened differently here than in our last example? The 𝟒𝟒 was not raised to the second power because it did not have an exponent. We simply added the values inside the parentheses. A STORY OF RATIOS 67 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations What should our next step be? We need to evaluate the exponent next. 𝟕𝟕𝟐𝟐= 𝟕𝟕× 𝟕𝟕= 𝟒𝟒𝟒𝟒 Evaluate to find the final answer. 𝟐𝟐× 𝟒𝟒𝟒𝟒 𝟗𝟗𝟗𝟗 What do you notice about the two answers? The final answers were not the same. What was different between the two expressions? Answers may vary. In the first problem, a value inside the parentheses had an exponent, and that value was evaluated first because it was inside of the parentheses. In the second problem, the exponent was outside of the parentheses, which made us evaluate what was in the parentheses first; then, we raised that value to the power of the exponent. What conclusions can you draw about evaluating expressions with parentheses and exponents? Answers may vary. Regardless of the location of the exponent in the expression, evaluate the parentheses first. Sometimes there will be values with exponents inside the parentheses. If the exponent is outside the parentheses, evaluate the parentheses first, and then evaluate to the power of the exponent. Try Exercises 8 and 9. Exercises 8–9 8. 𝟕𝟕+ (𝟏𝟏𝟏𝟏−𝟑𝟑𝟐𝟐) 𝟕𝟕+ (𝟏𝟏𝟏𝟏−𝟗𝟗) 𝟕𝟕+ 𝟑𝟑 𝟏𝟏𝟏𝟏 9. 𝟕𝟕+ (𝟏𝟏𝟏𝟏−𝟑𝟑)𝟐𝟐 𝟕𝟕+ 𝟗𝟗𝟐𝟐 𝟕𝟕+ 𝟖𝟖𝟖𝟖 𝟖𝟖𝟖𝟖 A STORY OF RATIOS 68 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Closing (5 minutes) When we evaluate expressions, we use one set of rules so that everyone arrives at the same correct answer. Grouping symbols, like parentheses, tell us to evaluate whatever is inside them before moving on. These rules are based on doing the most powerful operations first (exponents), then the less powerful ones (multiplication and division, going from left to right), and finally, the least powerful ones last (addition and subtraction, going from left to right). Note: Please do not stress words over meaning here. It is okay to talk about the number computed, computation, calculation, and so on to refer to the value as well. Exit Ticket (5 minutes) Lesson Summary NUMERICAL EXPRESSION: A numerical expression is a number, or it is any combination of sums, differences, products, or divisions of numbers that evaluates to a number. Statements like “𝟑𝟑+” or “𝟑𝟑÷ 𝟎𝟎” are not numerical expressions because neither represents a point on the number line. Note: Raising numbers to whole number powers are considered numerical expressions as well since the operation is just an abbreviated form of multiplication, e.g., 𝟐𝟐𝟑𝟑= 𝟐𝟐∙𝟐𝟐∙𝟐𝟐. VALUE OF A NUMERICAL EXPRESSION: The value of a numerical expression is the number found by evaluating the expression. For example: 𝟏𝟏 𝟑𝟑∙(𝟐𝟐+ 𝟒𝟒) + 𝟕𝟕 is a numerical expression, and its value is 𝟗𝟗. A STORY OF RATIOS 69 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Name Date Lesson 6: The Order of Operations Exit Ticket 1. Evaluate this expression: 39 ÷ (2 + 1) −2 × (4 + 1). 2. Evaluate this expression: 12 × (3 + 22) ÷ 2 −10. 3. Evaluate this expression: 12 × (3 + 2)2 ÷ 2 −10. A STORY OF RATIOS 70 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations Exit Ticket Sample Solutions 1. Evaluate this expression: 𝟑𝟑𝟑𝟑÷ (𝟐𝟐+ 𝟏𝟏) −𝟐𝟐× (𝟒𝟒+ 𝟏𝟏). 𝟑𝟑𝟑𝟑÷ 𝟑𝟑−𝟐𝟐× 𝟓𝟓 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 𝟑𝟑 2. Evaluate this expression: 𝟏𝟏𝟏𝟏× (𝟑𝟑+ 𝟐𝟐𝟐𝟐) ÷ 𝟐𝟐−𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏× (𝟑𝟑+ 𝟒𝟒) ÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏× 𝟕𝟕÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒−𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 3. Evaluate this expression: 𝟏𝟏𝟏𝟏× (𝟑𝟑+ 𝟐𝟐)𝟐𝟐÷ 𝟐𝟐−𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏× 𝟓𝟓𝟐𝟐÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏× 𝟐𝟐𝟐𝟐÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑𝟑𝟑÷ 𝟐𝟐−𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 Problem Set Sample Solutions Evaluate each expression. 1. 𝟑𝟑× 𝟓𝟓+ 𝟐𝟐× 𝟖𝟖+ 𝟐𝟐 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏+ 𝟐𝟐 𝟑𝟑𝟑𝟑 2. ($𝟏𝟏. 𝟕𝟕𝟕𝟕+ 𝟐𝟐× $𝟎𝟎. 𝟐𝟐𝟐𝟐+ 𝟓𝟓× $𝟎𝟎. 𝟎𝟎𝟎𝟎) × 𝟐𝟐𝟐𝟐 ($𝟏𝟏. 𝟕𝟕𝟕𝟕+ $𝟎𝟎. 𝟓𝟓𝟓𝟓+ $𝟎𝟎. 𝟐𝟐𝟐𝟐) × 𝟐𝟐𝟐𝟐 $𝟐𝟐. 𝟓𝟓𝟓𝟓× 𝟐𝟐𝟐𝟐 $𝟔𝟔𝟔𝟔. 𝟎𝟎𝟎𝟎 3. (𝟐𝟐× 𝟔𝟔) + (𝟖𝟖× 𝟒𝟒) + 𝟏𝟏 𝟏𝟏𝟏𝟏+ 𝟑𝟑𝟑𝟑+ 𝟏𝟏 𝟒𝟒𝟒𝟒 A STORY OF RATIOS 71 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 6 Lesson 6: The Order of Operations 4. ൫(𝟖𝟖× 𝟏𝟏. 𝟗𝟗𝟗𝟗) + (𝟑𝟑× 𝟐𝟐. 𝟗𝟗𝟗𝟗) + 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗൯× 𝟏𝟏. 𝟎𝟎𝟎𝟎 (𝟏𝟏𝟏𝟏. 𝟔𝟔+ 𝟖𝟖. 𝟖𝟖𝟖𝟖+ 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗) × 𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟑𝟑𝟑𝟑. 𝟒𝟒× 𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓𝟓𝟓 5. ൫(𝟏𝟏𝟏𝟏÷ 𝟑𝟑)𝟐𝟐−(𝟏𝟏𝟏𝟏÷ 𝟑𝟑𝟐𝟐)൯× (𝟒𝟒÷ 𝟐𝟐) (𝟒𝟒𝟐𝟐−(𝟏𝟏𝟏𝟏÷ 𝟗𝟗)) × (𝟒𝟒÷ 𝟐𝟐) (𝟏𝟏𝟏𝟏−𝟐𝟐) × 𝟐𝟐 𝟏𝟏𝟏𝟏× 𝟐𝟐 𝟐𝟐𝟐𝟐 A STORY OF RATIOS 72 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 GRADE 6 • MODULE 4 Topic C: Replacing Letters and Numbers 6 GRADE Mathematics Curriculum Topic C Replacing Letters and Numbers 6.EE.A.2c, 6.EE.A.4 Focus Standards: 6.EE.A.2c Write, read, and evaluate expressions in which letters stand for numbers. c. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas 𝑉𝑉= 𝑠𝑠3 and 𝐴𝐴= 6𝑠𝑠2 to find the volume and surface area of a cube with sides of length 𝑠𝑠= 1/2. 6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 and 3𝑦𝑦 are equivalent because they name the same number regardless of which number 𝑦𝑦 stands for. Instructional Days: 2 Lesson 7: Replacing Letters with Numbers (P)1 Lesson 8: Replacing Numbers with Letters (S) Students begin substituting, or replacing, letters with numbers and numbers with letters in Topic C in order to evaluate expressions with a given number and to determine expressions to create identities. In Lesson 7, students replace letters with a given number in order to evaluate the expression to one number. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 73 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic C Topic C: Replacing Letters and Numbers They continue to practice with exponents in this lesson in order to determine the area of squares and rectangles as shown below. In Lesson 8, students understand that a number in an expression can be replaced with a letter to determine identities. Through replacement of numbers, students discover and build identities such as 𝑎𝑎+ 𝑏𝑏= 𝑏𝑏+ 𝑎𝑎, 𝑎𝑎× 𝑏𝑏= 𝑏𝑏× 𝑎𝑎, 𝑔𝑔× 1 = 𝑔𝑔, 𝑔𝑔+ 0 = 𝑔𝑔, 𝑔𝑔÷ 1 = 𝑔𝑔, 𝑔𝑔÷ 𝑔𝑔= 1, 1 ÷ 𝑔𝑔= 1 𝑔𝑔. These identities aid in solving equations with variables, as well as problem solving with equations. 4 × 1 = 4 4 ÷ 1 = 4 4 × 0 = 0 1 ÷ 4 = 1 4 𝑔𝑔× 1 = 𝑔𝑔 𝑔𝑔÷ 1 = 𝑔𝑔 𝑔𝑔× 0 = 0 1 ÷ 𝑔𝑔= 1 𝑔𝑔 3 + 4 = 4 + 3 3 × 4 = 4 × 3 3 + 3 + 3 + 3 = 4 × 3 3 ÷ 4 = 3 4 𝑎𝑎+ 4 = 4 + 𝑎𝑎 𝑎𝑎× 4 = 4 × 𝑎𝑎 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎= 4 × 𝑎𝑎 𝑎𝑎÷ 4 = 𝑎𝑎 4 A STORY OF RATIOS 74 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Lesson 7: Replacing Letters with Numbers Student Outcomes Students understand that a letter represents one number in an expression. When that number replaces the letter, the expression can be evaluated to one number. Lesson Notes Before this lesson, make it clear to students that, just like 3 × 3 is 32 or three squared, units × units is units2 or units squared (also called square units). It may be helpful to cut and paste some of the figures from this lesson onto either paper or an interactive white board application. Each of the basic figures is depicted two ways: One has side lengths that can be counted, and the other is a similar figure without grid lines. Also, ahead of time, draw a 23 cm square on a chalkboard, a white board, or an interactive board. There is a square in the student materials that is approximately 23 mm square, or 529 mm2. Classwork Example 1 (10 minutes) Draw or project the square shown. Example 1 What is the length of one side of this square? 𝟑𝟑 units What is the formula for the area of a square? 𝑨𝑨= 𝒔𝒔𝟐𝟐 What is the square’s area as a multiplication expression? 𝟑𝟑 units × 𝟑𝟑 units A STORY OF RATIOS 75 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers What is the square’s area? 𝟗𝟗 square units We can count the units. However, look at this other square. Its side length is 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜. That is just too many tiny units to draw. What expression can we build to find this square’s area? 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 What is the area of the square? Use a calculator if you need to. 𝟓𝟓𝟓𝟓𝟓𝟓 𝐜𝐜𝐦𝐦𝟐𝟐 A letter represents one number in an expression. That number was 3 in our first square and 23 in our second square. When that number replaces the letter, the expression can be evaluated to one number. In our first example, the expression was evaluated to be 9, and in the second example, the expression was evaluated to be 529. Make sure students understand that 9 is one number, but 529 is also one number. (It happens to have 3 digits, but it is still one number.) Exercise 1 (5 minutes) Ask students to work both problems from Exercise 1 in their student materials. Make clear to students that these drawings are not to scale. Exercise 1 Complete the table below for both squares. Note: These drawings are not to scale. 𝒔𝒔= 𝟒𝟒 𝒔𝒔= 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 A STORY OF RATIOS 76 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Length of One Side of the Square Square’s Area Written as an Expression Square’s Area Written as a Number 𝟒𝟒 units 𝟒𝟒 units × 𝟒𝟒 units 𝟏𝟏𝟏𝟏 square units 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢.× 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. 𝟔𝟔𝟔𝟔𝟔𝟔 𝐢𝐢𝐧𝐧𝟐𝟐 Make sure students have the units correctly recorded in each of the cells of the table. When units are not specified, keep the label unit or square unit. Example 2 (10 minutes) Example 2 The formula 𝐴𝐴= 𝑙𝑙× 𝑤𝑤 is an efficient way to find the area of a rectangle without being required to count the area units in a rectangle. What does the letter 𝒃𝒃 represent in this blue rectangle? 𝒃𝒃= 𝟖𝟖 Give students a short time for discussion of the next question among partners, and then ask for an answer and an explanation. With a partner, answer the following question: Given that the second rectangle is divided into four equal parts, what number does the 𝒙𝒙 represent? 𝒙𝒙= 𝟖𝟖 How did you arrive at this answer? We reasoned that each width of the 𝟒𝟒 congruent rectangles must be the same. Two 𝟒𝟒 𝐜𝐜𝐜𝐜 lengths equals 𝟖𝟖 𝐜𝐜𝐜𝐜. What is the total length of the second rectangle? Tell a partner how you know. The length consists of 𝟒𝟒 segments that each has a length of 𝟒𝟒 𝐜𝐜𝐜𝐜. 𝟒𝟒× 𝟒𝟒 𝐜𝐜𝐜𝐜= 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜. 𝑥𝑥 cm 4 cm 8 cm 𝑏𝑏 cm A STORY OF RATIOS 77 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers If the two large rectangles have equal lengths and widths, find the area of each rectangle. 𝟖𝟖 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜= 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐 Discuss with your partner how the formulas for the area of squares and rectangles can be used to evaluate area for a particular figure. Remember, a letter represents one number in an expression. When that number replaces the letter, the expression can be evaluated to one number. Exercise 2 (5 minutes) Ask students to complete the table for both rectangles in their student materials. Using a calculator is appropriate. Exercise 2 Length of Rectangle Width of Rectangle Rectangle’s Area Written as an Expression Rectangle’s Area Written as a Number 𝟕𝟕 units 𝟒𝟒 units 𝟕𝟕 units × 𝟒𝟒 units 𝟐𝟐𝟐𝟐 square units 𝟒𝟒𝟒𝟒 𝐦𝐦 𝟑𝟑𝟑𝟑 𝐦𝐦 𝟒𝟒𝟒𝟒 𝐦𝐦× 𝟑𝟑𝟑𝟑 𝐦𝐦 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 𝐦𝐦𝟐𝟐 𝟑𝟑𝟑𝟑 𝐦𝐦 𝟒𝟒𝟒𝟒 𝐦𝐦 A STORY OF RATIOS 78 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Example 3 (3 minutes) The formula 𝑉𝑉= 𝑙𝑙× 𝑤𝑤× ℎ is a quick way to determine the volume of right rectangular prisms. Take a look at the right rectangular prisms in your student materials. Example 3 What does the 𝒍𝒍 represent in the first diagram? The length of the rectangular prism What does the 𝒘𝒘 represent in the first diagram? The width of the rectangular prism What does the 𝒉𝒉 represent in the first diagram? The height of the rectangular prism Notice that the right rectangular prism in the second diagram is an exact copy of the first diagram. Since we know the formula to find the volume is 𝑽𝑽= 𝒍𝒍× 𝒘𝒘× 𝒉𝒉, what number can we substitute for the 𝒍𝒍 in the formula? Why? 𝟔𝟔, because the length of the second right rectangular prism is 𝟔𝟔 𝐜𝐜𝐜𝐜. What other number can we substitute for the 𝒍𝒍? No other number can replace the 𝒍𝒍. Only one number can replace one letter. What number can we substitute for the 𝒘𝒘 in the formula? Why? 𝟐𝟐, because the width of the second right rectangular prism is 𝟐𝟐 𝐜𝐜𝐜𝐜. What number can we substitute for the 𝒉𝒉 in the formula? 𝟖𝟖, because the height of the second right rectangular prism is 𝟖𝟖 𝐜𝐜𝐜𝐜. 𝒍𝒍 𝒘𝒘 𝒉𝒉 𝟔𝟔 𝐜𝐜𝐜𝐜 𝟐𝟐 𝐜𝐜𝐜𝐜 𝟖𝟖 𝐜𝐜𝐜𝐜 A STORY OF RATIOS 79 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Determine the volume of the second right rectangular prism by replacing the letters in the formula with their appropriate numbers. 𝑽𝑽= 𝒍𝒍× 𝒘𝒘× 𝒉𝒉; 𝑽𝑽= 𝟔𝟔 𝐜𝐜𝐜𝐜× 𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟖𝟖 𝐜𝐜𝐜𝐜= 𝟗𝟗𝟗𝟗 𝐜𝐜𝐦𝐦𝟑𝟑 Exercise 3 (5 minutes) Ask students to complete the table for both figures in their student materials. Using a calculator is appropriate. Exercise 3 Complete the table for both figures. Using a calculator is appropriate. Length of Rectangular Prism Width of Rectangular Prism Height of Rectangular Prism Rectangular Prism’s Volume Written as an Expression Rectangular Prism’s Volume Written as a Number 𝟏𝟏𝟏𝟏 units 𝟓𝟓 units 𝟏𝟏𝟏𝟏 units 𝟏𝟏𝟏𝟏 units × 𝟓𝟓 units × 𝟏𝟏𝟏𝟏 units 𝟗𝟗𝟗𝟗𝟗𝟗 cubic units 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 𝟒𝟒 𝐜𝐜𝐜𝐜 𝟕𝟕 𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟒𝟒 𝐜𝐜𝐜𝐜× 𝟕𝟕 𝐜𝐜𝐜𝐜 𝟔𝟔𝟔𝟔𝟔𝟔 𝐜𝐜𝐦𝐦𝟑𝟑 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝟓𝟓 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 𝟒𝟒 𝐜𝐜𝐜𝐜 𝟕𝟕 𝐜𝐜𝐜𝐜 A STORY OF RATIOS 80 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Closing (2 minutes) How many numbers are represented by one letter in an expression? One When that number replaces the letter, the expression can be evaluated to what? One number The key is to strongly link expressions back to computations with numbers. The description for expression given above is meant to work nicely with how students in Grade 6 and Grade 7 learn to manipulate expressions. In these grades, a lot of time is spent building expressions and evaluating expressions. Building and evaluating helps students see that expressions are really just a slight abstraction of arithmetic in elementary school. Building often occurs by thinking about examples of numerical expressions first and then replacing the numbers with letters in a numerical expression. The act of evaluating for students at this stage means they replace each of the variables with specific numbers and then compute to obtain a number. Exit Ticket (5 minutes) Lesson Summary VARIABLE (description): A variable is a symbol (such as a letter) that is a placeholder for a number. EXPRESSION (description): An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables. There are two ways to build expressions: 1. We can start out with a numerical expression, such as 𝟏𝟏 𝟑𝟑∙(𝟐𝟐+ 𝟒𝟒) + 𝟕𝟕, and replace some of the numbers with letters to get 𝟏𝟏 𝟑𝟑∙(𝒙𝒙+ 𝒚𝒚) + 𝒛𝒛. 2. We can build such expressions from scratch, as in 𝒙𝒙+ 𝒙𝒙(𝒚𝒚−𝒛𝒛), and note that if numbers were placed in the expression for the variables 𝒙𝒙, 𝒚𝒚, and 𝒛𝒛, the result would be a numerical expression. A STORY OF RATIOS 81 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Name Date Lesson 7: Replacing Letters with Numbers Exit Ticket 1. In the drawing below, what do the letters 𝑙𝑙 and 𝑤𝑤 represent? 2. What does the expression 𝑙𝑙+ 𝑤𝑤+ 𝑙𝑙+ 𝑤𝑤 represent? 3. What does the expression 𝑙𝑙∙𝑤𝑤 represent? 4. The rectangle below is congruent to the rectangle shown in Problem 1. Use this information to evaluate the expressions from Problems 2 and 3. A STORY OF RATIOS 82 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers Exit Ticket Sample Solutions 1. In the drawing below, what do the letters 𝒍𝒍 and 𝒘𝒘 represent? Length and width of the rectangle 2. What does the expression 𝒍𝒍+ 𝒘𝒘+ 𝒍𝒍+ 𝒘𝒘 represent? Perimeter of the rectangle, or the sum of the sides of the rectangle 3. What does the expression 𝒍𝒍∙𝒘𝒘 represent? Area of the rectangle 4. The rectangle below is congruent to the rectangle shown in Problem 1. Use this information to evaluate the expressions from Problems 2 and 3. 𝒍𝒍= 𝟓𝟓 and 𝒘𝒘= 𝟐𝟐 𝑷𝑷= 𝟏𝟏𝟏𝟏 units 𝑨𝑨= 𝟏𝟏𝟏𝟏 units2 Problem Set Sample Solutions 1. Replace the side length of this square with 𝟒𝟒 𝐢𝐢𝐢𝐢., and find the area. The student should draw a square, label the side 𝟒𝟒 𝐢𝐢𝐢𝐢., and calculate the area to be 𝟏𝟏𝟏𝟏 𝐢𝐢𝐧𝐧𝟐𝟐. 𝒔𝒔 A STORY OF RATIOS 83 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 7 Lesson 7: Replacing Letters with Numbers 2. Complete the table for each of the given figures. Length of Rectangle Width of Rectangle Rectangle’s Area Written as an Expression Rectangle’s Area Written as a Number 𝟑𝟑𝟑𝟑 𝐦𝐦 𝟐𝟐𝟐𝟐 𝐦𝐦 𝟑𝟑𝟑𝟑 𝐦𝐦× 𝟐𝟐𝟐𝟐 𝐦𝐦 𝟖𝟖𝟖𝟖𝟖𝟖 𝐦𝐦𝟐𝟐 𝟏𝟏𝟏𝟏 𝐲𝐲𝐲𝐲. 𝟑𝟑. 𝟓𝟓 𝐲𝐲𝐲𝐲. 𝟏𝟏𝟏𝟏 𝐲𝐲𝐲𝐲.× 𝟑𝟑. 𝟓𝟓 𝐲𝐲𝐲𝐲. 𝟒𝟒𝟒𝟒 𝐲𝐲𝐝𝐝𝟐𝟐 3. Find the perimeter of each quadrilateral in Problems 1 and 2. 𝑷𝑷= 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. 𝑷𝑷= 𝟏𝟏𝟏𝟏𝟏𝟏 𝐦𝐦 𝑷𝑷= 𝟑𝟑𝟑𝟑 𝐲𝐲𝐲𝐲. 4. Using the formula 𝑽𝑽= 𝒍𝒍× 𝒘𝒘× 𝒉𝒉, find the volume of a right rectangular prism when the length of the prism is 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜, the width is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜, and the height is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜. 𝑽𝑽= 𝒍𝒍× 𝒘𝒘× 𝒉𝒉; 𝑽𝑽= 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜= 𝟓𝟓, 𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐦𝐦𝟑𝟑 𝟐𝟐𝟐𝟐 𝐦𝐦 𝟑𝟑𝟑𝟑 𝐦𝐦 𝟏𝟏𝟏𝟏 𝐲𝐲𝐲𝐲. 𝟑𝟑. 𝟓𝟓 𝐲𝐲𝐲𝐲. A STORY OF RATIOS 84 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Lesson 8: Replacing Numbers with Letters Student Outcomes Students understand that a letter in an expression or an equation can represent a number. When that number is replaced with a letter, an expression or an equation is stated. Students discover the commutative properties of addition and multiplication, the additive identity property of zero, and the multiplicative identity property of one. They determine that 𝑔𝑔÷ 1 = 𝑔𝑔, 𝑔𝑔÷ 𝑔𝑔= 1, and 1 ÷ 𝑔𝑔= 1 𝑔𝑔. Classwork Fluency Exercise (10 minutes): Division of Fractions II Sprint: Refer to the Sprints and the Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening Exercise (5 minutes) Write this series of equations on the board: Opening Exercise 𝟒𝟒+ 𝟎𝟎= 𝟒𝟒 𝟒𝟒× 𝟏𝟏= 𝟒𝟒 𝟒𝟒÷ 𝟏𝟏= 𝟒𝟒 𝟒𝟒× 𝟎𝟎= 𝟎𝟎 𝟏𝟏÷ 𝟒𝟒= 𝟏𝟏 𝟒𝟒 Discussion (5 minutes) How many of these statements are true? All of them How many of those statements would be true if the number 𝟒𝟒 was replaced with the number 𝟕𝟕 in each of the number sentences? All of them Would the number sentences be true if we were to replace the number 𝟒𝟒 with any other number? MP.3 A STORY OF RATIOS 85 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Let students make conjectures about substitutions. What if we replaced the number 𝟒𝟒 with the number 𝟎𝟎? Would each of the number sentences be true? No. The first four are true, but the last one, dividing by zero, is not true. Division by zero is undefined. You cannot make zero groups of objects, and group size cannot be zero. It appears that we can replace the number 4 with any nonzero number, and each of the number sentences will be true. A letter in an expression can represent a number. When that number is replaced with a letter, an expression is stated. What if we replace the number 𝟒𝟒 with a letter 𝒈𝒈? Please write all 𝟒𝟒 expressions below, replacing each 𝟒𝟒 with a 𝒈𝒈. 𝒈𝒈+ 𝟎𝟎= 𝒈𝒈 𝒈𝒈× 𝟏𝟏= 𝒈𝒈 𝒈𝒈÷ 𝟏𝟏= 𝒈𝒈 𝒈𝒈× 𝟎𝟎= 𝟎𝟎 𝟏𝟏÷ 𝒈𝒈= 𝟏𝟏 𝒈𝒈 Are these all true (except for 𝒈𝒈= 𝟎𝟎) when dividing? Yes Let’s look at each of these a little closer and see if we can make some generalizations. Example 1 (5 minutes): Additive Identity Property of Zero Example 1: Additive Identity Property of Zero 𝒈𝒈+ 𝟎𝟎= 𝒈𝒈 Remember a letter in a mathematical expression represents a number. Can we replace 𝒈𝒈 with any number? Yes Choose a value for 𝒈𝒈, and replace 𝒈𝒈 with that number in the equation. What do you observe? The value of 𝒈𝒈 does not change when 𝟎𝟎 is added to 𝒈𝒈. Repeat this process several times, each time choosing a different number for 𝒈𝒈. MP.3 A STORY OF RATIOS 86 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Allow students to experiment for about a minute. Most quickly realize the additive identity property of zero: Any number added to zero equals itself. The number’s identity does not change. Will all values of 𝒈𝒈 result in a true number sentence? Yes Write the mathematical language for this property below: 𝒈𝒈+ 𝟎𝟎= 𝒈𝒈, additive identity property of zero. Any number added to zero equals itself. Example 2 (5 minutes): Multiplicative Identity Property of One Example 2: Multiplicative Identity Property of One 𝒈𝒈× 𝟏𝟏= 𝒈𝒈 Remember a letter in a mathematical expression represents a number. Can we replace 𝒈𝒈 with any number? Yes Choose a value for 𝒈𝒈, and replace 𝒈𝒈 with that number in the equation. What do you observe? The value of 𝒈𝒈 does not change when 𝒈𝒈 is multiplied by 𝟏𝟏. Allow students to experiment for about a minute with the next question. Most quickly realize the multiplicative identity property of one: Any number multiplied by 1 equals itself. The number’s identity does not change. Will all values of 𝒈𝒈 result in a true number sentence? Experiment with different values before making your claim. Yes Write the mathematical language for this property below: 𝒈𝒈× 𝟏𝟏= 𝒈𝒈, multiplicative identity property of one. Any number multiplied by one equals itself. Example 3 (6 minutes): Commutative Property of Addition and Multiplication Example 3: Commutative Property of Addition and Multiplication 𝟑𝟑+ 𝟒𝟒= 𝟒𝟒+ 𝟑𝟑 𝟑𝟑× 𝟒𝟒= 𝟒𝟒× 𝟑𝟑 Replace the 𝟑𝟑’s in these number sentences with the letter 𝒂𝒂. 𝒂𝒂+ 𝟒𝟒= 𝟒𝟒+ 𝒂𝒂 𝒂𝒂× 𝟒𝟒= 𝟒𝟒× 𝒂𝒂 A STORY OF RATIOS 87 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Choose a value for 𝒂𝒂, and replace 𝒂𝒂 with that number in each of the equations. What do you observe? The result is a true number sentence. Allow students to experiment for about a minute with the next question. Most quickly realize that the equations are examples of the commutative property of addition and commutative property of multiplication. These are sometimes called the “any-order properties.” Will all values of 𝒂𝒂 result in a true number sentence? Experiment with different values before making your claim. Yes, any number, even zero, can be used in place of the variable 𝒂𝒂. Now, write the equations again, this time replacing the number 𝟒𝟒 with a variable, 𝒃𝒃. 𝒂𝒂+ 𝒃𝒃= 𝒃𝒃+ 𝒂𝒂 𝒂𝒂× 𝒃𝒃= 𝒃𝒃× 𝒂𝒂 Will all values of 𝒂𝒂 and 𝒃𝒃 result in true number sentences for the first two equations? Experiment with different values before making your claim. Yes Write the mathematical language for this property below: 𝒂𝒂+ 𝒃𝒃= 𝒃𝒃+ 𝒂𝒂, commutative property of addition. Order does not matter when adding. 𝒂𝒂× 𝒃𝒃= 𝒃𝒃× 𝒂𝒂, commutative property of multiplication. Order does not matter when multiplying. Models are useful for making abstract ideas more concrete. A STORY OF RATIOS 88 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Example 4 (4 minutes) Display and discuss the models above as they relate to the commutative property of addition and the commutative property of multiplication. When finished, pose a new question: Will all values of 𝑎𝑎 and 𝑏𝑏 result in a true number sentence for the equation 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎= 𝑏𝑏× 𝑎𝑎? Experiment with different values before making your claim. Allow students to experiment for about a minute. They should discover that any value can be substituted for the variable 𝑎𝑎, but only 4 can be used for 𝑏𝑏, since there are exactly 4 copies of 𝑎𝑎 in the equation. Summarize your discoveries with a partner. In the equation 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎+ 𝑎𝑎= 𝑏𝑏× 𝑎𝑎, any value can be substituted for the variable 𝑎𝑎, but only 4 can be used for 𝑏𝑏, since there are exactly 4 copies of 𝑎𝑎 in the equation. Finally, consider the last equation, 𝑎𝑎÷ 𝑏𝑏= 𝑎𝑎 𝑏𝑏. Is this true for all values of 𝑎𝑎 and 𝑏𝑏? It is true for all values of 𝑎𝑎 and all values of 𝑏𝑏≠0. Example 4 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑+ 𝟑𝟑= 𝟒𝟒× 𝟑𝟑 𝟑𝟑÷ 𝟒𝟒= 𝟑𝟑 𝟒𝟒 Replace the 𝟑𝟑’s in these number sentences with the letter 𝒂𝒂. 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂= 𝟒𝟒× 𝒂𝒂 𝒂𝒂÷ 𝟒𝟒= 𝒂𝒂 𝟒𝟒 Choose a value for 𝒂𝒂, and replace 𝒂𝒂 with that number in each of the equations. What do you observe? The result is a true number sentence. Will all values of 𝒂𝒂 result in a true number sentence? Experiment with different values before making your claim. Yes, any number, even zero, can be used in place of the variable 𝒂𝒂. A STORY OF RATIOS 89 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Now, write the equations again, this time replacing the number 𝟒𝟒 with a variable, 𝒃𝒃. 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂= 𝒃𝒃× 𝒂𝒂 𝒂𝒂÷ 𝒃𝒃= 𝒂𝒂 𝒃𝒃, 𝒃𝒃≠𝟎𝟎 Will all values of 𝒂𝒂 and 𝒃𝒃 result in true number sentences for the equations? Experiment with different values before making your claim. In the equation 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂+ 𝒂𝒂= 𝒃𝒃× 𝒂𝒂, any value can be substituted for the variable 𝒂𝒂, but only 𝟒𝟒 can be used for 𝒃𝒃 since there are exactly 𝟒𝟒 copies of 𝒂𝒂 in the equation. It is true for all values of 𝒂𝒂 and all values of 𝒃𝒃≠𝟎𝟎. Closing (2 minutes) Tell your partner which of these properties of numbers is the easiest for you to remember. Allow sharing for a short time. Now, tell your partner which of these properties of numbers is the hardest for you to remember. Allow sharing for a short time. Although these properties might seem simple, we apply them in many different ways in mathematics. If you have a good grasp on them, you will recognize them and use them in many applications. With a partner, create two different division problems that support the following: 𝑔𝑔× 1 = 𝑔𝑔, and be ready to explain your reasoning. 𝑔𝑔÷ 𝑔𝑔= 1; 5 ÷ 5 = 1; 34 ÷ 34 = 1; 2 7 8 ÷ 2 7 8 = 1; and so on. Any nonzero number divided by itself equals 1. If a number 𝑔𝑔 is divided into 𝑔𝑔 equal parts, each part will have a size equal to one. If 𝑔𝑔 items are divided into groups of size 𝑔𝑔, there will be one group. What about any number divided by 1? What does this mean? 𝑔𝑔÷ 1 = 𝑔𝑔 If a number 𝑔𝑔 is divided into 1 part, then the size of that part will be 𝑔𝑔. Or, if 𝑔𝑔 items are divided into 1 group, there will be 𝑔𝑔 items in that group. Exit Ticket (5 minutes) A STORY OF RATIOS 90 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Name Date Lesson 8: Replacing Numbers with Letters Exit Ticket 1. State the commutative property of addition, and provide an example using two different numbers. 2. State the commutative property of multiplication, and provide an example using two different numbers. 3. State the additive property of zero, and provide an example using any other number. 4. State the multiplicative identity property of one, and provide an example using any other number. A STORY OF RATIOS 91 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Exit Ticket Sample Solutions 1. State the commutative property of addition, and provide an example using two different numbers. Any two different addends can be chosen, such as 𝟓𝟓+ 𝟔𝟔= 𝟔𝟔+ 𝟓𝟓. 2. State the commutative property of multiplication, and provide an example using two different numbers. Any two different factors can be chosen, such as 𝟒𝟒× 𝟗𝟗= 𝟗𝟗× 𝟒𝟒. 3. State the additive property of zero, and provide an example using any other number. Any nonzero addend can be chosen, such as 𝟑𝟑+ 𝟎𝟎= 𝟑𝟑. 4. State the multiplicative identity property of one, and provide an example using any other number. Any nonzero factor can be chosen, such as 𝟏𝟏𝟏𝟏× 𝟏𝟏= 𝟏𝟏𝟏𝟏. Problem Set Sample Solutions 1. State the commutative property of addition using the variables 𝒂𝒂 and 𝒃𝒃. 𝒂𝒂+ 𝒃𝒃= 𝒃𝒃+ 𝒂𝒂 2. State the commutative property of multiplication using the variables 𝒂𝒂 and 𝒃𝒃. 𝒂𝒂× 𝒃𝒃= 𝒃𝒃× 𝒂𝒂 3. State the additive property of zero using the variable 𝒃𝒃. 𝒃𝒃+ 𝟎𝟎= 𝒃𝒃 4. State the multiplicative identity property of one using the variable 𝒃𝒃. 𝒃𝒃× 𝟏𝟏= 𝒃𝒃 5. Demonstrate the property listed in the first column by filling in the third column of the table. Commutative Property of Addition 𝟐𝟐𝟐𝟐+ 𝒄𝒄= 𝒄𝒄+ 𝟐𝟐𝟐𝟐 Commutative Property of Multiplication 𝒍𝒍× 𝒘𝒘= 𝒘𝒘× 𝒍𝒍 Additive Property of Zero 𝒉𝒉+ 𝟎𝟎= 𝒉𝒉 Multiplicative Identity Property of One 𝒗𝒗× 𝟏𝟏= 𝒗𝒗 6. Why is there no commutative property for subtraction or division? Show examples. Answers will vary. Examples should show reasoning and proof that the commutative property does not work for subtraction and division. An example would be 𝟖𝟖÷ 𝟐𝟐 and 𝟐𝟐÷ 𝟖𝟖. 𝟖𝟖÷ 𝟐𝟐= 𝟒𝟒, but 𝟐𝟐÷ 𝟖𝟖= 𝟏𝟏 𝟒𝟒. A STORY OF RATIOS 92 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Division of Fractions II—Round 1 Directions: Determine the quotient of the fractions and simplify. 1. 4 10 ÷ 2 10 16. 3 1 8 ÷ 2 3 2. 9 12 ÷ 3 12 17. 1 5 6 ÷ 1 2 3. 6 10 ÷ 4 10 18. 5 8 ÷ 2 3 4 4. 2 8 ÷ 3 8 19. 1 3 ÷ 1 4 5 5. 2 7 ÷ 6 7 20. 3 4 ÷ 2 3 10 6. 11 9 ÷ 8 9 21. 2 1 5 ÷ 1 1 6 7. 5 13 ÷ 10 13 22. 2 4 9 ÷ 1 3 5 8. 7 8 ÷ 13 16 23. 1 2 9 ÷ 3 2 5 9. 3 5 ÷ 7 10 24. 2 2 3 ÷ 3 10. 9 30 ÷ 3 5 25. 1 3 4 ÷ 2 2 5 11. 1 3 ÷ 4 5 26. 4 ÷ 1 2 9 12. 2 5 ÷ 3 4 27. 3 1 5 ÷ 6 13. 3 4 ÷ 5 9 28. 2 5 6 ÷ 1 1 3 14. 4 5 ÷ 7 12 29. 10 2 3 ÷ 8 15. 3 8 ÷ 5 2 30. 15 ÷ 2 3 5 Number Correct: A STORY OF RATIOS 93 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Division of Fractions II—Round 1 [KEY] Directions: Determine the quotient of the fractions and simplify. 1. 4 10 ÷ 2 10 𝟒𝟒 𝟐𝟐= 𝟐𝟐 16. 3 1 8 ÷ 2 3 𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏= 𝟒𝟒𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 2. 9 12 ÷ 3 12 𝟗𝟗 𝟑𝟑= 𝟑𝟑 17. 1 5 6 ÷ 1 2 𝟐𝟐𝟐𝟐 𝟔𝟔= 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟐𝟐 𝟑𝟑 3. 6 10 ÷ 4 10 𝟔𝟔 𝟒𝟒= 𝟑𝟑 𝟐𝟐= 𝟏𝟏𝟏𝟏 𝟐𝟐 18. 5 8 ÷ 2 3 4 𝟐𝟐𝟐𝟐 𝟖𝟖𝟖𝟖= 𝟓𝟓 𝟐𝟐𝟐𝟐 4. 2 8 ÷ 3 8 𝟐𝟐 𝟑𝟑 19. 1 3 ÷ 1 4 5 𝟓𝟓 𝟐𝟐𝟐𝟐 5. 2 7 ÷ 6 7 𝟐𝟐 𝟔𝟔= 𝟏𝟏 𝟑𝟑 20. 3 4 ÷ 2 3 10 𝟑𝟑𝟑𝟑 𝟗𝟗𝟗𝟗= 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒 6. 11 9 ÷ 8 9 𝟏𝟏𝟏𝟏 𝟖𝟖= 𝟏𝟏𝟑𝟑 𝟖𝟖 21. 2 1 5 ÷ 1 1 6 𝟔𝟔𝟔𝟔 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 7. 5 13 ÷ 10 13 𝟓𝟓 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟐𝟐 22. 2 4 9 ÷ 1 3 5 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕𝟕𝟕= 𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 8. 7 8 ÷ 13 16 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 23. 1 2 9 ÷ 3 2 5 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 9. 3 5 ÷ 7 10 𝟔𝟔 𝟕𝟕 24. 2 2 3 ÷ 3 𝟖𝟖 𝟗𝟗 10. 9 30 ÷ 3 5 𝟗𝟗 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟐𝟐 25. 1 3 4 ÷ 2 2 5 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 11. 1 3 ÷ 4 5 𝟓𝟓 𝟏𝟏𝟏𝟏 26. 4 ÷ 1 2 9 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 12. 2 5 ÷ 3 4 𝟖𝟖 𝟏𝟏𝟏𝟏 27. 3 1 5 ÷ 6 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑= 𝟖𝟖 𝟏𝟏𝟏𝟏 13. 3 4 ÷ 5 9 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟕𝟕 𝟐𝟐𝟐𝟐 28. 2 5 6 ÷ 1 1 3 𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟑𝟑 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟖𝟖 14. 4 5 ÷ 7 12 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 29. 10 2 3 ÷ 8 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐= 𝟒𝟒 𝟑𝟑= 𝟏𝟏𝟏𝟏 𝟑𝟑 15. 3 8 ÷ 5 2 𝟔𝟔 𝟒𝟒𝟒𝟒= 𝟑𝟑 𝟐𝟐𝟐𝟐 30. 15 ÷ 2 3 5 𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏= 𝟓𝟓𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 94 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Division of Fractions II—Round 2 Directions: Determine the quotient of the fractions and simplify. 1. 10 2 ÷ 5 2 16. 5 8 ÷ 1 3 4 2. 6 5 ÷ 3 5 17. 1 4 ÷ 2 2 5 3. 10 7 ÷ 2 7 18. 2 3 5 ÷ 3 8 4. 3 8 ÷ 5 8 19. 1 3 5 ÷ 2 9 5. 1 4 ÷ 3 12 20. 4 ÷ 2 3 8 6. 7 5 ÷ 3 10 21. 1 1 2 ÷ 5 7. 8 15 ÷ 4 5 22. 3 1 3 ÷ 1 3 4 8. 5 6 ÷ 5 12 23. 2 2 5 ÷ 1 1 4 9. 3 5 ÷ 7 9 24. 3 1 2 ÷ 2 2 3 10. 3 10 ÷ 3 9 25. 1 4 5 ÷ 2 3 4 11. 3 4 ÷ 7 9 26. 3 1 6 ÷ 1 3 5 12. 7 10 ÷ 3 8 27. 3 3 5 ÷ 2 1 8 13. 4 ÷ 4 9 28. 5 ÷ 1 1 6 14. 5 8 ÷ 7 29. 3 3 4 ÷ 5 1 2 15. 9 ÷ 2 3 30. 4 2 3 ÷ 5 1 4 Number Correct: Improvement: A STORY OF RATIOS 95 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 8 Lesson 8: Replacing Numbers with Letters Division of Fractions II—Round 2 [KEY] Directions: Determine the quotient of the fractions and simplify. 1. 10 2 ÷ 5 2 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 16. 5 8 ÷ 1 3 4 𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓= 𝟓𝟓 𝟏𝟏𝟏𝟏 2. 6 5 ÷ 3 5 𝟔𝟔 𝟑𝟑= 𝟐𝟐 17. 1 4 ÷ 2 2 5 𝟓𝟓 𝟒𝟒𝟒𝟒 3. 10 7 ÷ 2 7 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟓𝟓 18. 2 3 5 ÷ 3 8 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏= 𝟔𝟔𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 4. 3 8 ÷ 5 8 𝟑𝟑 𝟓𝟓 19. 1 3 5 ÷ 2 9 𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏= 𝟕𝟕𝟐𝟐 𝟏𝟏𝟏𝟏= 𝟕𝟕𝟏𝟏 𝟓𝟓 5. 1 4 ÷ 3 12 𝟑𝟑 𝟑𝟑= 𝟏𝟏 20. 4 ÷ 2 3 8 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 6. 7 5 ÷ 3 10 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒𝟐𝟐 𝟑𝟑 21. 1 1 2 ÷ 5 𝟑𝟑 𝟏𝟏𝟏𝟏 7. 8 15 ÷ 4 5 𝟖𝟖 𝟏𝟏𝟏𝟏= 𝟐𝟐 𝟑𝟑 22. 3 1 3 ÷ 1 3 4 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 8. 5 6 ÷ 5 12 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 23. 2 2 5 ÷ 1 1 4 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 9. 3 5 ÷ 7 9 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 24. 3 1 2 ÷ 2 2 3 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟓𝟓 𝟏𝟏𝟏𝟏 10. 3 10 ÷ 3 9 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑= 𝟗𝟗 𝟏𝟏𝟏𝟏 25. 1 4 5 ÷ 2 3 4 𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 11. 3 4 ÷ 7 9 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 26. 3 1 6 ÷ 1 3 5 𝟗𝟗𝟗𝟗 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 12. 7 10 ÷ 3 8 𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑= 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 27. 3 3 5 ÷ 2 1 8 𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖= 𝟏𝟏𝟓𝟓𝟓𝟓 𝟖𝟖𝟖𝟖 13. 4 ÷ 4 9 𝟑𝟑𝟑𝟑 𝟒𝟒= 𝟗𝟗 28. 5 ÷ 1 1 6 𝟑𝟑𝟑𝟑 𝟕𝟕= 𝟒𝟒𝟐𝟐 𝟕𝟕 14. 5 8 ÷ 7 𝟓𝟓 𝟓𝟓𝟓𝟓 29. 3 3 4 ÷ 5 1 2 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 15. 9 ÷ 2 3 𝟐𝟐𝟐𝟐 𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐 30. 4 2 3 ÷ 5 1 4 𝟓𝟓𝟓𝟓 𝟔𝟔𝟔𝟔= 𝟖𝟖 𝟗𝟗 A STORY OF RATIOS 96 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 GRADE 6 • MODULE 4 Topic D: Expanding, Factoring, and Distributing Expressions 6 GRADE Mathematics Curriculum Topic D Expanding, Factoring, and Distributing Expressions 6.EE.A.2a, 6.EE.A.2b, 6.EE.A.3, 6.EE.A.4 Focus Standards: 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. a. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract 𝑦𝑦 from 5” as 5 −𝑦𝑦. b. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2(8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. 6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + 𝑥𝑥) to produce the equivalent expression 6 + 3𝑥𝑥; apply the distributive property to the expression 24𝑥𝑥+ 18𝑦𝑦 to produce the equivalent expression 6(4𝑥𝑥+ 3𝑦𝑦); apply properties of operations to 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 to produce the equivalent expression 3𝑦𝑦. 6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions 𝑦𝑦+ 𝑦𝑦+ 𝑦𝑦 and 3𝑦𝑦 are equivalent because they name the same number regardless of which number 𝑦𝑦 stands for. Instructional Days: 6 Lesson 9: Writing Addition and Subtraction Expressions (P)1 Lesson 10: Writing and Expanding Multiplication Expressions (P) Lesson 11: Factoring Expressions (P) Lesson 12: Distributing Expressions (P) Lessons 13–14: Writing Division Expressions (P, P) 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 97 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic D Topic D: Expanding, Factoring, and Distributing Expressions In Topic D, students formally utilize their understanding of expressions in order to expand, factor, and distribute. In Lesson 9, students write expressions that record addition and subtraction operations with numbers through the use of models. With a bar diagram, students understand that any number 𝑎𝑎 plus any number 𝑏𝑏 is the same as adding the numbers 𝑏𝑏+ 𝑎𝑎. Students also use bar diagrams to differentiate between the mathematical terms subtract and subtract from. For instance, when subtracting 𝑏𝑏 from 𝑎𝑎, they know they must first represent 𝑎𝑎 in order to take away 𝑏𝑏, leading to an understanding that the expression must be written 𝑎𝑎−𝑏𝑏. This concept deters students from writing the incorrect expression 𝑏𝑏−𝑎𝑎, which is a common misconception because the number 𝑏𝑏 is heard first in the expression “subtract 𝑏𝑏 from 𝑎𝑎.” Students continue to write expressions by combining operations with the use of parentheses. In Lesson 10, students identify parts of an expression using mathematical terms for multiplication. They view one or more parts of an expression as a single entity. They determine that through the use of models, when 𝑎𝑎 is represented 6 times, the expression is written as one entity: 6 × 𝑎𝑎, 6 ∙𝑎𝑎, or 6𝑎𝑎. In Lesson 11, students bring with them their previous knowledge of GCF and the distributive property from Module 2 to model and write expressions using the distributive property. They move from a factored form to an expanded form of an expression, while in Lesson 12, they move from an expanded form to a factored form. In Lesson 11, students are capable of moving from the expression 2𝑎𝑎+ 2𝑏𝑏 to 𝑎𝑎+ 𝑏𝑏 written twice as (𝑎𝑎+ 𝑏𝑏) + (𝑎𝑎+ 𝑏𝑏) and conclude that 2𝑎𝑎+ 2𝑏𝑏= 2(𝑎𝑎+ 𝑏𝑏). Conversely, students determine in Lesson 12 that (𝑎𝑎+ 𝑏𝑏) + (𝑎𝑎+ 𝑏𝑏) = 2𝑎𝑎+ 2𝑏𝑏 through the following model: Finally, in Lessons 13 and 14, students write division expressions in two forms: dividend ÷ divisor and dividend divisor , noting the relationship between the two. They determine from the model below that 1 ÷ 2 is the same as 1 2. They make an intuitive connection to expressions with letters and also determine that 𝑎𝑎÷ 2 is the same as 𝑎𝑎 2. 1 2 𝑎𝑎 2 A STORY OF RATIOS 98 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions Lesson 9: Writing Addition and Subtraction Expressions Student Outcomes Students write expressions that record addition and subtraction operations with numbers. Lesson Notes Individual white boards are recommended for this lesson. Classwork Example 1 (3 minutes) Example 1 Create a bar diagram to show 𝟑𝟑 plus 𝟓𝟓. How would this look if you were asked to show 𝟓𝟓 plus 𝟑𝟑? There would be 𝟓𝟓 tiles and then 𝟑𝟑 tiles. Are these two expressions equivalent? Yes. Both 𝟑𝟑+ 𝟓𝟓 and 𝟓𝟓+ 𝟑𝟑 have a sum of 𝟖𝟖. Example 2 (3 minutes) Example 2 How can we show a number increased by 𝟐𝟐? 𝒂𝒂+ 𝟐𝟐 or 𝟐𝟐+ 𝒂𝒂 Can you prove this using a model? If so, draw the model. Yes. I can use a bar diagram. 𝟑𝟑 + 𝟓𝟓 MP.2 𝒂𝒂 + 𝟐𝟐 𝟐𝟐+ 𝒂𝒂 A STORY OF RATIOS 99 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions 𝟏𝟏𝟏𝟏 𝒏𝒏 𝟑𝟑 Example 3 (3 minutes) Example 3 Write an expression to show the sum of 𝒎𝒎 and 𝒌𝒌. 𝒎𝒎+ 𝒌𝒌 or 𝒌𝒌+ 𝒎𝒎 Which property can be used in Examples 1–3 to show that both expressions given are equivalent? The commutative property of addition Example 4 (3 minutes) Example 4 How can we show 𝟏𝟏𝟏𝟏 minus 𝟔𝟔? Draw a bar diagram to model this expression. What expression would represent this model? 𝟏𝟏𝟏𝟏−𝟔𝟔 Could we also use 𝟔𝟔−𝟏𝟏𝟏𝟏? No. If we started with 𝟔𝟔 and tried to take 𝟏𝟏𝟏𝟏 away, the models would not match. Example 5 (3 minutes) Example 5 How can we write an expression to show 𝟑𝟑 less than a number? Start by drawing a diagram to model the subtraction. Are we taking away from the 𝟑𝟑 or the unknown number? We are taking 𝟑𝟑 away from the unknown number. We are starting with some number and then subtracting 3. MP.2 A STORY OF RATIOS 100 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions What expression would represent this model? The expression is 𝒏𝒏−𝟑𝟑. Example 6 (3 minutes) Example 6 How would we write an expression to show the number 𝒄𝒄 being subtracted from the sum of 𝒂𝒂 and 𝒃𝒃? Start by writing an expression for “the sum of 𝒂𝒂 and 𝒃𝒃.” 𝒂𝒂+ 𝒃𝒃 or 𝒃𝒃+ 𝒂𝒂 Now, show 𝒄𝒄 being subtracted from the sum. 𝒂𝒂+ 𝒃𝒃−𝒄𝒄 or 𝒃𝒃+ 𝒂𝒂−𝒄𝒄 Example 7 (3 minutes) Example 7 Write an expression to show 𝒄𝒄 minus the sum of 𝒂𝒂 and 𝒃𝒃. 𝒄𝒄−(𝒂𝒂+ 𝒃𝒃) Why are parentheses necessary in this example and not the others? Without the parentheses, only 𝒂𝒂 is being taken away from 𝒄𝒄, where the expression says that 𝒂𝒂+ 𝒃𝒃 should be taken away from 𝒄𝒄. Replace the variables with numbers to see if 𝒄𝒄−(𝒂𝒂+ 𝒃𝒃) is the same as 𝒄𝒄−𝒂𝒂+ 𝒃𝒃. If students do not see the necessity for the parentheses, have them replace the variables with numbers to see whether 𝑐𝑐−(𝑎𝑎+ 𝑏𝑏) is the same as 𝑐𝑐−𝑎𝑎+ 𝑏𝑏. Here is a sample of what they could try: 𝑎𝑎= 1, 𝑏𝑏= 2, 𝑐𝑐= 3 3 −(1 + 2) 3 −1 + 2 3 −3 or 2 + 2 0 4 Exercises (12 minutes) These questions can be done on the worksheet. However, if white boards, small chalkboards, or some other personal board is available, the teacher can give instant feedback as students show their boards after each question. MP.2 A STORY OF RATIOS 101 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions Exercises 1. Write an expression to show the sum of 𝟕𝟕 and 𝟏𝟏. 𝟓𝟓. 𝟕𝟕+ 𝟏𝟏. 𝟓𝟓 or 𝟏𝟏. 𝟓𝟓+ 𝟕𝟕 2. Write two expressions to show 𝒘𝒘 increased by 𝟒𝟒. Then, draw models to prove that both expressions represent the same thing. 𝒘𝒘+ 𝟒𝟒 and 𝟒𝟒+ 𝒘𝒘 3. Write an expression to show the sum of 𝒂𝒂, 𝒃𝒃, and 𝒄𝒄. Answers will vary. Below are possible answers. 𝒂𝒂+ 𝒃𝒃+ 𝒄𝒄 𝒃𝒃+ 𝒄𝒄+ 𝒂𝒂 𝒄𝒄+ 𝒃𝒃+ 𝒂𝒂 𝒂𝒂+ 𝒄𝒄+ 𝒃𝒃 𝒃𝒃+ 𝒂𝒂+ 𝒄𝒄 𝒄𝒄+ 𝒂𝒂+ 𝒃𝒃 4. Write an expression and a model showing 𝟑𝟑 less than 𝒑𝒑. 𝒑𝒑−𝟑𝟑 5. Write an expression to show the difference of 𝟑𝟑 and 𝒑𝒑. 𝟑𝟑−𝒑𝒑 6. Write an expression to show 𝟒𝟒 less than the sum of 𝒈𝒈 and 𝟓𝟓. 𝒈𝒈+ 𝟓𝟓−𝟒𝟒 or 𝟓𝟓+ 𝒈𝒈−𝟒𝟒 7. Write an expression to show 𝟒𝟒 decreased by the sum of 𝒈𝒈 and 𝟓𝟓. 𝟒𝟒−(𝒈𝒈+ 𝟓𝟓) or 𝟒𝟒−(𝟓𝟓+ 𝒈𝒈) 8. Should Exercises 6 and 7 have different expressions? Why or why not? The expressions are different because one includes the word “decreased by,” and the other has the words “less than.” The words “less than” give the amount that was taken away first, whereas the word “decreased by” gives us a starting amount and then the amount that was taken away. 𝒑𝒑 𝟑𝟑 𝒘𝒘 𝒘𝒘 𝟒𝟒 𝟒𝟒 A STORY OF RATIOS 102 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions Closing (7 minutes) Write the following in words. 𝑚𝑚+ 𝑘𝑘 Answers will vary; the sum of 𝑚𝑚 and 𝑘𝑘 𝑘𝑘+ 𝑚𝑚 Answers will vary; the sum of 𝑘𝑘 and 𝑚𝑚 𝑚𝑚−𝑘𝑘 Answers will vary; 𝑚𝑚 minus 𝑘𝑘 𝑘𝑘−𝑚𝑚 Answers will vary; 𝑘𝑘 minus 𝑚𝑚 Is 𝑚𝑚+ 𝑘𝑘 equivalent to 𝑘𝑘+ 𝑚𝑚? Is 𝑚𝑚−𝑘𝑘 equivalent to 𝑘𝑘−𝑚𝑚? Explain. 𝑚𝑚+ 𝑘𝑘 is equivalent to 𝑘𝑘+ 𝑚𝑚. Both of these expressions would evaluate to the same number regardless of the numbers substituted in for 𝑘𝑘 and 𝑚𝑚. However, 𝑚𝑚−𝑘𝑘 and 𝑘𝑘−𝑚𝑚 will not have the same result. I would be starting with a new total amount and taking away a different amount as well. The values of each expression would be different, so the expressions would not be equivalent. For example, 4 + 6 = 10, and 6 + 4 = 10. However, 6 −4 = 2, but 4 −6 ≠2. Exit Ticket (5 minutes) A STORY OF RATIOS 103 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions Name Date Lesson 9: Writing Addition and Subtraction Expressions Exit Ticket 1. Write an expression showing the sum of 8 and a number 𝑓𝑓. 2. Write an expression showing 5 less than the number 𝑘𝑘. 3. Write an expression showing the sum of a number ℎ and a number 𝑤𝑤 minus 11. A STORY OF RATIOS 104 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 9 Lesson 9: Writing Addition and Subtraction Expressions Exit Ticket Sample Solutions 1. Write an expression showing the sum of 𝟖𝟖 and a number 𝒇𝒇. 𝟖𝟖+ 𝒇𝒇 or 𝒇𝒇+ 𝟖𝟖 2. Write an expression showing 𝟓𝟓 less than the number 𝒌𝒌. 𝒌𝒌−𝟓𝟓 3. Write an expression showing the sum of a number 𝒉𝒉 and a number 𝒘𝒘 minus 𝟏𝟏𝟏𝟏. 𝒉𝒉+ 𝒘𝒘−𝟏𝟏𝟏𝟏 Problem Set Sample Solutions 1. Write two expressions to show a number increased by 𝟏𝟏𝟏𝟏. Then, draw models to prove that both expressions represent the same thing. 𝒂𝒂+ 𝟏𝟏𝟏𝟏 and 𝟏𝟏𝟏𝟏+ 𝒂𝒂 2. Write an expression to show the sum of 𝒙𝒙 and 𝒚𝒚. 𝒙𝒙+ 𝒚𝒚 or 𝒚𝒚+ 𝒙𝒙 3. Write an expression to show 𝒉𝒉 decreased by 𝟏𝟏𝟏𝟏. 𝒉𝒉−𝟏𝟏𝟏𝟏 4. Write an expression to show 𝒌𝒌 less than 𝟑𝟑. 𝟓𝟓. 𝟑𝟑. 𝟓𝟓−𝒌𝒌 5. Write an expression to show the sum of 𝒈𝒈 and 𝒉𝒉 reduced by 𝟏𝟏𝟏𝟏. 𝒈𝒈+ 𝒉𝒉−𝟏𝟏𝟏𝟏 6. Write an expression to show 𝟓𝟓 less than 𝒚𝒚, plus 𝒈𝒈. 𝒚𝒚−𝟓𝟓+ 𝒈𝒈 7. Write an expression to show 𝟓𝟓 less than the sum of 𝒚𝒚 and 𝒈𝒈. 𝒚𝒚+ 𝒈𝒈−𝟓𝟓 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝒂𝒂 𝒂𝒂 A STORY OF RATIOS 105 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions Lesson 10: Writing and Expanding Multiplication Expressions Student Outcomes Students identify parts of an expression using mathematical terms for multiplication. They view one or more parts of an expression as a single entity. Classwork Discussion (4 minutes) When we want to show multiplication of two numbers, like 5 and 7, we typically write 5 × 7, using the “×” to show the operation. When we start to use variables with multiplication, we can use other forms. 𝑎𝑎× 𝑏𝑏 𝑎𝑎∙𝑏𝑏 𝑎𝑎𝑎𝑎 (𝑎𝑎)(𝑏𝑏) Why might we want to use a form other than the × when variables are involved? The × can be confused for a variable instead of a symbol for an operation. Which of the three models can be used to show multiplication where there are no variables involved? 5 × 7, 5 ∙7, and (5)(7), but not 57 because it looks like the number fifty-seven and not five times seven. Example 1 (10 minutes) When writing expressions using the fewest number of symbols, we have to refrain from using the symbols ×, ∙, or ( ). We will also be using math terms to describe expressions and the parts of an expression. We will be using words like factor, product, quotient, coefficient, and term. A term is a part of an expression that can be added to or subtracted from the rest of the expression. In the expression 7𝑔𝑔+ 8ℎ+ 3, what are examples of terms? 7𝑔𝑔, 8ℎ, and 3 are all terms. A coefficient is a constant factor in a variable term. For example, in the term 4𝑚𝑚, 4 is the coefficient, and it is multiplied with 𝑚𝑚. MP.7 A STORY OF RATIOS 106 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions Example 1 Write each expression using the fewest number of symbols and characters. Use math terms to describe the expressions and parts of the expressions. a. 𝟔𝟔× 𝒃𝒃 𝟔𝟔𝟔𝟔; the 𝟔𝟔 is the coefficient and a factor, and the 𝒃𝒃 is the variable and a factor. We can call 𝟔𝟔𝟔𝟔 the product, and we can also call it a term. b. 𝟒𝟒∙𝟑𝟑∙𝒉𝒉 𝟏𝟏𝟏𝟏𝟏𝟏; the 𝟏𝟏𝟏𝟏 is the coefficient and a factor, and the 𝒉𝒉 is the variable and a factor. We can call 𝟏𝟏𝟏𝟏𝟏𝟏 the product, and we can also call it a term. c. 𝟐𝟐× 𝟐𝟐× 𝟐𝟐× 𝒂𝒂× 𝒃𝒃 𝟖𝟖𝟖𝟖𝟖𝟖; 𝟖𝟖 is the coefficient and a factor, 𝒂𝒂 and 𝒃𝒃 are both variables and factors, and 𝟖𝟖𝟖𝟖𝟖𝟖 is the product and also a term. Variables always follow the numbers and should be written in alphabetical order. Apply this knowledge to the examples below. d. 𝟓𝟓× 𝒎𝒎× 𝟑𝟑× 𝒑𝒑 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏 is the coefficient and factor, 𝒎𝒎 and 𝒑𝒑 are the variables and factors, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 is the product and also a term. If it is helpful, you can gather the numbers together and the variables together. You can do this because of the commutative property of multiplication. 5 × 3 × 𝑚𝑚× 𝑝𝑝 e. 𝟏𝟏× 𝒈𝒈× 𝒘𝒘 𝟏𝟏𝟏𝟏𝟏𝟏 or 𝒈𝒈𝒈𝒈; 𝒈𝒈 and 𝒘𝒘 are the variables and factors, 𝟏𝟏 is the coefficient and factor if it is included, and 𝒈𝒈𝒈𝒈 is the product and also a term. What happens when you multiply by 1? Multiplying by 1 is an example of the identity property. Any number times 1 is equal to that number. Therefore, we don’t always need to write the one because 1 × 𝑔𝑔𝑔𝑔= 𝑔𝑔𝑔𝑔. Example 2 (5 minutes) Example 2 To expand multiplication expressions, we will rewrite the expressions by including the “ ∙ ” back into the expressions. a. 𝟓𝟓𝒈𝒈 𝟓𝟓∙𝒈𝒈 MP.7 A STORY OF RATIOS 107 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions b. 𝟕𝟕𝒂𝒂𝒂𝒂𝒂𝒂 𝟕𝟕∙𝒂𝒂∙𝒃𝒃∙𝒄𝒄 c. 𝟏𝟏𝟏𝟏𝒈𝒈 𝟏𝟏𝟏𝟏∙𝒈𝒈 or 𝟐𝟐∙𝟐𝟐∙𝟑𝟑∙𝒈𝒈 d. 𝟑𝟑𝒉𝒉∙𝟖𝟖 𝟑𝟑∙𝒉𝒉∙𝟖𝟖 e. 𝟕𝟕𝒈𝒈∙𝟗𝟗𝒉𝒉 𝟕𝟕∙𝒈𝒈∙𝟗𝟗∙𝒉𝒉 or 𝟕𝟕∙𝒈𝒈∙𝟑𝟑∙𝟑𝟑∙𝒉𝒉 Example 3 (5 minutes) Example 3 a. Find the product of 𝟒𝟒𝟒𝟒∙𝟕𝟕𝟕𝟕. It may be easier to see how we will use the fewest number of symbols and characters by expanding the expression first. 𝟒𝟒∙𝒇𝒇∙𝟕𝟕∙𝒈𝒈 Now we can multiply the numbers and then multiply the variables. 𝟒𝟒∙𝟕𝟕∙𝒇𝒇∙𝒈𝒈 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 b. Multiply 𝟑𝟑𝒅𝒅𝒅𝒅∙𝟗𝟗𝒚𝒚𝒚𝒚. Let’s start again by expanding the expression. Then, we can rewrite the expression by multiplying the numbers and then multiplying the variables. 𝟑𝟑∙𝒅𝒅∙𝒆𝒆∙𝟗𝟗∙𝒚𝒚∙𝒛𝒛 𝟑𝟑∙𝟗𝟗∙𝒅𝒅∙𝒆𝒆∙𝒚𝒚∙𝒛𝒛 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 c. Double the product of 𝟔𝟔𝟔𝟔 and 𝟑𝟑𝟑𝟑𝟑𝟑. We can start by finding the product of 6𝑦𝑦 and 3𝑏𝑏𝑏𝑏. MP.7 A STORY OF RATIOS 108 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions 𝟔𝟔∙𝒚𝒚∙𝟑𝟑∙𝒃𝒃∙𝒄𝒄 𝟔𝟔∙𝟑𝟑∙𝒃𝒃∙𝒄𝒄∙𝒚𝒚 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 What does it mean to double something? It means to multiply by 2. 𝟐𝟐∙𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 Exercises (14 minutes) Students match expressions on a BINGO board. Some of the expressions are simplified, and some are expanded. To save time, provide students with a BINGO board with some of the squares already filled in. Have the remaining answers written on a smart board, a chalkboard, or an overhead projector so that students can randomly place them on the BINGO board. If there is not enough time for the BINGO game, these questions can be used on white boards, chalkboards, or some form of personal boards. Here are the clues to be given during the game, followed by the answers that are on the board. Questions/Clues Answers 1. 10𝑚𝑚 2 ∙5 ∙m 2. 8 ∙3 ∙𝑚𝑚 24𝑚𝑚 3. Has a coefficient of 11 11𝑚𝑚𝑚𝑚 4. 14𝑚𝑚𝑚𝑚 2 ∙7 ∙𝑚𝑚∙𝑝𝑝 5. (3𝑚𝑚)(9𝑝𝑝) 27𝑚𝑚𝑚𝑚 6. 11𝑚𝑚∙2𝑝𝑝 22𝑚𝑚𝑚𝑚 7. 36𝑚𝑚 2 ∙2 ∙3 ∙3 ∙𝑚𝑚 8. 2 ∙2 ∙2 ∙5 ∙𝑝𝑝 40𝑝𝑝 9. 7𝑚𝑚𝑚𝑚∙5𝑡𝑡 35𝑚𝑚𝑚𝑚𝑚𝑚 10. 18𝑝𝑝𝑝𝑝 2 ∙3 ∙3 ∙𝑝𝑝∙𝑡𝑡 11. 7 ∙2 ∙𝑡𝑡∙2 ∙𝑝𝑝 28𝑝𝑝𝑝𝑝 12. Has a coefficient of 5 5𝑚𝑚𝑚𝑚𝑚𝑚 13. 3 ∙3 ∙5 ∙𝑚𝑚∙𝑝𝑝 45𝑚𝑚𝑚𝑚 14. 5𝑚𝑚∙9𝑝𝑝𝑝𝑝 45𝑚𝑚𝑚𝑚𝑚𝑚 15. 10𝑚𝑚𝑚𝑚∙4𝑡𝑡 40𝑚𝑚𝑚𝑚𝑚𝑚 16. 1𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 17. 45𝑚𝑚𝑚𝑚 3 ∙3 ∙5 ∙𝑚𝑚∙𝑝𝑝 18. (4𝑚𝑚𝑚𝑚)(11) 44𝑚𝑚𝑚𝑚 19. 54𝑚𝑚𝑚𝑚𝑚𝑚 3 ∙3 ∙3 ∙2 ∙𝑚𝑚∙𝑝𝑝∙𝑡𝑡 20. Has a coefficient of 3 3𝑚𝑚 These answers have already been included on premade BINGO boards to save time. The other answers can be randomly placed in the remaining spaces. MP.7 A STORY OF RATIOS 109 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions 21. 2 ∙2 ∙2 ∙3 ∙𝑚𝑚∙𝑝𝑝 24𝑚𝑚𝑚𝑚 22. (5𝑚𝑚)(3𝑝𝑝)(2𝑡𝑡) 30𝑚𝑚𝑚𝑚𝑚𝑚 23. 13𝑚𝑚𝑚𝑚 (1𝑚𝑚𝑚𝑚)(13) 24. Has a coefficient of 2 2𝑝𝑝 Closing (3 minutes) What is the difference between standard form and expanded form? When we write an expression in standard form, we get rid of the operation symbol or symbols for multiplication, and we write the factors next to each other. Sometimes we might have to multiply numbers together before writing it next to the variable or variables. When we write an expression in expanded form, we write the expression as a product of the factors using the “ ∙ ” symbol for multiplication. How would you describe the following terms? 1. Factor A number or variable that is multiplied to get a product 2. Variable A letter used to represent a number 3. Product The solution when two factors are multiplied 4. Coefficient The numerical factor that multiplies the variable Note: Each summand of an expression in expanded form is called a term, and the number found by multiplying just the numbers in a term together is called the coefficient of the term. After the word term is defined, students can be shown what it means to “collect like terms” using the distributive property. Expressions in expanded form are analogous to polynomial expressions that are written as a sum of monomials. There are two reasons for introducing this term instead of the word polynomial: (1) In the Common Core State Standards, the word polynomial cannot be formally defined before high school, but the idea behind the word is needed much sooner. (2) The progressions are very clear about not asking problems that state, “Simplify.” However, they do describe standard form in the progressions, so students may be asked to put their answers in standard form. To get to standard form, students are asked to expand the expression and then collect like terms. Lesson Summary AN EXPRESSION IN EXPANDED FORM: An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form. A STORY OF RATIOS 110 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions AN EXPRESSION IN STANDARD FORM: An expression that is in expanded form where all like terms have been collected is said to be in standard form. Note: Students cannot be asked to “simplify,” but they can be asked to “put an expression in standard form” or “expand the expression and collect all like terms.” Exit Ticket (4 minutes) A STORY OF RATIOS 111 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions Name Date Lesson 10: Writing and Expanding Multiplication Expressions Exit Ticket 1. Rewrite the expression in standard form (use the fewest number of symbols and characters possible). a. 5𝑔𝑔∙7ℎ b. 3 ∙4 ∙5 ∙𝑚𝑚∙𝑛𝑛 2. Name the parts of the expression. Then, write it in expanded form. a. 14𝑏𝑏 b. 30𝑗𝑗𝑗𝑗 A STORY OF RATIOS 112 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions Exit Ticket Sample Solutions 1. Rewrite the expression in standard form (use the fewest number of symbols and characters possible). a. 𝟓𝟓𝒈𝒈∙𝟕𝟕𝒉𝒉 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 b. 𝟑𝟑∙𝟒𝟒∙𝟓𝟓∙𝒎𝒎∙𝒏𝒏 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔 2. Name the parts of the expression. Then, write it in expanded form. a. 𝟏𝟏𝟏𝟏𝒃𝒃 𝟏𝟏𝟏𝟏∙𝒃𝒃 or 𝟐𝟐∙𝟕𝟕∙𝒃𝒃 𝟏𝟏𝟏𝟏 is the coefficient, 𝒃𝒃 is the variable, and 𝟏𝟏𝟏𝟏𝟏𝟏 is a term and the product of 𝟏𝟏𝟏𝟏× 𝒃𝒃. b. 𝟑𝟑𝟑𝟑𝒋𝒋𝒋𝒋 𝟑𝟑𝟑𝟑∙𝒋𝒋∙𝒌𝒌 or 𝟐𝟐∙𝟑𝟑∙𝟓𝟓∙𝒋𝒋∙𝒌𝒌 𝟑𝟑𝟑𝟑 is the coefficient, 𝒋𝒋 and 𝒌𝒌 are the variables, and 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 is a term and the product of 𝟑𝟑𝟑𝟑∙𝒋𝒋∙𝒌𝒌. Problem Set Sample Solutions 1. Rewrite the expression in standard form (use the fewest number of symbols and characters possible). a. 𝟓𝟓∙𝒚𝒚 𝟓𝟓𝟓𝟓 b. 𝟕𝟕∙𝒅𝒅∙𝒆𝒆 𝟕𝟕𝟕𝟕𝟕𝟕 c. 𝟓𝟓∙𝟐𝟐∙𝟐𝟐∙𝒚𝒚∙𝒛𝒛 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 d. 𝟑𝟑∙𝟑𝟑∙𝟐𝟐∙𝟓𝟓∙𝒅𝒅 𝟗𝟗𝟗𝟗𝟗𝟗 2. Write the following expressions in expanded form. a. 𝟑𝟑𝒈𝒈 𝟑𝟑∙𝒈𝒈 b. 𝟏𝟏𝟏𝟏𝒎𝒎𝒎𝒎 𝟏𝟏𝟏𝟏∙𝒎𝒎∙𝒑𝒑 A STORY OF RATIOS 113 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions c. 𝟐𝟐𝟐𝟐𝒚𝒚𝒚𝒚 𝟐𝟐𝟐𝟐∙𝒚𝒚∙𝒛𝒛 or 𝟐𝟐∙𝟐𝟐∙𝟓𝟓∙𝒚𝒚∙𝒛𝒛 d. 𝟏𝟏𝟏𝟏𝒂𝒂𝒂𝒂𝒂𝒂 𝟏𝟏𝟏𝟏∙𝒂𝒂∙𝒃𝒃∙𝒄𝒄 or 𝟑𝟑∙𝟓𝟓∙𝒂𝒂∙𝒃𝒃∙𝒄𝒄 3. Find the product. a. 𝟓𝟓𝒅𝒅∙𝟕𝟕𝒈𝒈 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 b. 𝟏𝟏𝟏𝟏𝒂𝒂𝒂𝒂∙𝟑𝟑𝒄𝒄𝒄𝒄 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 A STORY OF RATIOS 114 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions A STORY OF RATIOS 115 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions 2 ● 5 ● 𝑚𝑚 35𝑚𝑚𝑚𝑚𝑚𝑚 45𝑚𝑚𝑚𝑚 40𝑝𝑝 24𝑚𝑚 2 ● 3 ● 3 ● 𝑝𝑝 ● 𝑡𝑡 2 ● 7 ● 𝑚𝑚 ● 𝑝𝑝 11𝑚𝑚𝑚𝑚 28𝑝𝑝𝑝𝑝 22𝑚𝑚𝑚𝑚 2 ● 2 ● 3 ● 3 ● 𝑚𝑚 27𝑚𝑚𝑚𝑚 5𝑚𝑚𝑚𝑚𝑚𝑚 45𝑚𝑚𝑚𝑚𝑚𝑚 A STORY OF RATIOS 116 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions 22𝑚𝑚𝑚𝑚 40𝑝𝑝 28𝑝𝑝𝑝𝑝 2 ● 5 ● 𝑚𝑚 2 ● 2 ● 3 ● 3 ● 𝑚𝑚 45𝑚𝑚𝑚𝑚 35𝑚𝑚𝑚𝑚𝑚𝑚 24𝑚𝑚 45𝑚𝑚𝑚𝑚𝑚𝑚 27𝑚𝑚𝑚𝑚 2 ● 7 ● 𝑚𝑚 ● 𝑝𝑝 5𝑚𝑚𝑚𝑚𝑚𝑚 11𝑚𝑚𝑚𝑚 2 ● 3 ● 3 ● 𝑝𝑝 ● 𝑡𝑡 A STORY OF RATIOS 117 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 10 Lesson 10: Writing and Expanding Multiplication Expressions 45𝑚𝑚𝑚𝑚 40𝑝𝑝 24𝑚𝑚 2 ● 3 ● 3 ● 𝑝𝑝 ●𝑡𝑡 5𝑚𝑚𝑚𝑚𝑚𝑚 22𝑚𝑚𝑚𝑚 11𝑚𝑚𝑚𝑚 45𝑚𝑚𝑚𝑚𝑚𝑚 2 ● 2 ●3 ● 3 ● 𝑚𝑚 27𝑚𝑚𝑚𝑚 2 ● 7 ● 𝑚𝑚 ● 𝑝𝑝 28𝑝𝑝𝑝𝑝 2 ● 5 ● 𝑚𝑚 35𝑚𝑚𝑚𝑚𝑚𝑚 A STORY OF RATIOS 118 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Lesson 11: Factoring Expressions Student Outcomes Students model and write equivalent expressions using the distributive property. They move from expanded form to factored form of an expression. Classwork Fluency Exercise (5 minutes): GCF Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Example 1 (8 minutes) Example 1 a. Use the model to answer the following questions. How many fives are in the model? 𝟐𝟐 How many threes are in the model? 𝟐𝟐 What does the expression represent in words? The sum of two groups of five and two groups of three What expression could we write to represent the model? 𝟐𝟐× 𝟓𝟓+ 𝟐𝟐× 𝟑𝟑 Scaffolding: For students struggling with variables, the concept can be further solidified by having them replace the variables with whole numbers to prove that the expressions are equivalent. 𝟐𝟐 × 𝟓𝟓 𝟐𝟐 × 𝟑𝟑 𝟓𝟓 𝟓𝟓 𝟑𝟑 𝟑𝟑 MP.7 A STORY OF RATIOS 119 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions 𝟓𝟓+ 𝟑𝟑 𝟓𝟓+ 𝟑𝟑 𝟓𝟓 𝟓𝟓 𝟑𝟑 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝒂𝒂 𝒂𝒂 𝒃𝒃 𝒃𝒃 b. Use the new model and the previous model to answer the next set of questions. How many fives are in the model? 𝟐𝟐 How many threes are in the model? 𝟐𝟐 What does the expression represent in words? Two groups of the sum of five and three What expression could we write to represent the model? (𝟓𝟓+ 𝟑𝟑) + (𝟓𝟓+ 𝟑𝟑) or 𝟐𝟐(𝟓𝟓+ 𝟑𝟑) c. Is the model in part (a) equivalent to the model in part (b)? Yes, because both expressions have two 𝟓𝟓's and two 𝟑𝟑's. Therefore, 𝟐𝟐× 𝟓𝟓+ 𝟐𝟐× 𝟑𝟑= 𝟐𝟐(𝟓𝟓+ 𝟑𝟑). d. What relationship do we see happening on either side of the equal sign? On the left-hand side, 𝟐𝟐 is being multiplied by 𝟓𝟓 and then by 𝟑𝟑 before adding the products together. On the right-hand side, the 𝟓𝟓 and 𝟑𝟑 are added first and then multiplied by 𝟐𝟐. e. In Grade 5 and in Module 2 of this year, you have used similar reasoning to solve problems. What is the name of the property that is used to say that 𝟐𝟐(𝟓𝟓+ 𝟑𝟑) is the same as 𝟐𝟐× 𝟓𝟓+ 𝟐𝟐× 𝟑𝟑? The name of the property is the distributive property. Example 2 (5 minutes) Example 2 Now we will take a look at an example with variables. Discuss the questions with your partner. What does the model represent in words? 𝒂𝒂 plus 𝒂𝒂 plus 𝒃𝒃 plus 𝒃𝒃, two 𝒂𝒂’s plus two 𝒃𝒃’s, two times 𝒂𝒂 plus two times 𝒃𝒃 MP.7 A STORY OF RATIOS 120 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions 𝒂𝒂+ 𝒃𝒃 𝒂𝒂+ 𝒃𝒃 𝒂𝒂 𝒂𝒂 𝒃𝒃 𝒃𝒃 What does 𝟐𝟐𝒂𝒂 mean? 𝟐𝟐𝟐𝟐 means that there are 𝟐𝟐 𝒂𝒂’s or 𝟐𝟐× 𝒂𝒂. How many 𝒂𝒂’s are in the model? 𝟐𝟐 How many 𝒃𝒃’s are in the model? 𝟐𝟐 What expression could we write to represent the model? 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 How many 𝒂𝒂’s are in the expression? 𝟐𝟐 How many 𝒃𝒃’s are in the expression? 𝟐𝟐 What expression could we write to represent the model? (𝒂𝒂+ 𝒃𝒃) + (𝒂𝒂+ 𝒃𝒃) = 𝟐𝟐(𝒂𝒂+ 𝒃𝒃) Are the two expressions equivalent? Yes. Both models include 𝟐𝟐 𝒂𝒂’s and 𝟐𝟐 𝒃𝒃’s. Therefore, 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐= 𝟐𝟐(𝒂𝒂+ 𝒃𝒃). Example 3 (8 minutes) Example 3 Use GCF and the distributive property to write equivalent expressions. 1. 𝟑𝟑𝒇𝒇+ 𝟑𝟑𝒈𝒈= 𝟑𝟑(𝒇𝒇+ 𝒈𝒈) What is the question asking us to do? We need to rewrite the expression as an equivalent expression in factored form, which means the expression is written as the product of factors. The number outside of the parentheses is the GCF. MP.7 A STORY OF RATIOS 121 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions How would Problem 1 look if we expanded each term? 𝟑𝟑∙𝒇𝒇+ 𝟑𝟑∙𝒈𝒈 What is the GCF in Problem 1? 𝟑𝟑 How can we use the GCF to rewrite this expression? 𝟑𝟑 goes on the outside, and 𝒇𝒇+ 𝒈𝒈 will go inside the parentheses. 𝟑𝟑(𝒇𝒇+ 𝒈𝒈) Let’s use the same ideas for Problem 2. Start by expanding the expression and naming the GCF. 2. 𝟔𝟔𝒙𝒙+ 𝟗𝟗𝒚𝒚= 𝟑𝟑(𝟐𝟐𝒙𝒙+ 𝟑𝟑𝒚𝒚) What is the question asking us to do? We need to rewrite the expression as an equivalent expression in factored form, which means the expression is written as the product of factors. The number outside of the parentheses is the GCF. How would Problem 2 look if we expanded each term? 𝟐𝟐∙𝟑𝟑∙𝒙𝒙+ 𝟑𝟑∙𝟑𝟑∙𝒚𝒚 What is the GCF in Problem 2? The GCF is 𝟑𝟑. How can we use the GCF to rewrite this expression? I will factor out the 𝟑𝟑 from both terms and place it in front of the parentheses. I will place what is left in the terms inside the parentheses: 𝟑𝟑(𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑). 3. 𝟑𝟑𝒄𝒄+ 𝟏𝟏𝟏𝟏𝒄𝒄= 𝒄𝒄(𝟑𝟑+ 𝟏𝟏𝟏𝟏) Is there a greatest common factor in Problem 3? Yes. When I expand, I can see that each term has a common factor 𝒄𝒄. 𝟑𝟑∙𝒄𝒄+ 𝟏𝟏𝟏𝟏∙𝒄𝒄 Rewrite the expression using the distributive property. 𝒄𝒄(𝟑𝟑+ 𝟏𝟏𝟏𝟏) 4. 𝟐𝟐𝟐𝟐𝒃𝒃+ 𝟖𝟖= 𝟖𝟖(𝟑𝟑𝒃𝒃+ 𝟏𝟏) MP.7 A STORY OF RATIOS 122 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Explain how you used GCF and the distributive property to rewrite the expression in Problem 4. I first expanded each term. I know that 𝟖𝟖 goes into 𝟐𝟐𝟐𝟐, so I used it in the expansion. 𝟐𝟐∙𝟐𝟐∙𝟐𝟐∙𝟑𝟑∙𝒃𝒃+ 𝟐𝟐∙𝟐𝟐∙𝟐𝟐 I determined that 𝟐𝟐∙𝟐𝟐∙𝟐𝟐, or 𝟖𝟖, is the common factor. So, on the outside of the parentheses I wrote 𝟖𝟖, and on the inside I wrote the leftover factor, 𝟑𝟑𝟑𝟑+ 𝟏𝟏. 𝟖𝟖(𝟑𝟑𝟑𝟑+ 𝟏𝟏) Why is there a 𝟏𝟏 in the parentheses? When I factor out a number, I am leaving behind the other factor that multiplies to make the original number. In this case, when I factor out an 𝟖𝟖 from 𝟖𝟖, I am left with a 𝟏𝟏 because 𝟖𝟖× 𝟏𝟏= 𝟖𝟖. How is this related to the first two examples? In the first two examples, we saw that we could rewrite the expressions by thinking about groups. We can either think of 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟖𝟖 as 𝟖𝟖 groups of 𝟑𝟑𝟑𝟑 and 𝟖𝟖 groups of 𝟏𝟏 or as 𝟖𝟖 groups of the sum of 𝟑𝟑𝟑𝟑+ 𝟏𝟏. This shows that 𝟖𝟖(𝟑𝟑𝟑𝟑) + 𝟖𝟖(𝟏𝟏) = 𝟖𝟖(𝟑𝟑𝟑𝟑+ 𝟏𝟏) is the same as 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟖𝟖. Exercises (12 minutes) If times allows, have students practice these questions on white boards or small personal boards. Exercises 1. Apply the distributive property to write equivalent expressions. a. 𝟕𝟕𝒙𝒙+ 𝟕𝟕𝒚𝒚 𝟕𝟕(𝒙𝒙+ 𝒚𝒚) b. 𝟏𝟏𝟏𝟏𝒈𝒈+ 𝟐𝟐𝟐𝟐𝒉𝒉 𝟓𝟓(𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒) c. 𝟏𝟏𝟏𝟏𝒎𝒎+ 𝟒𝟒𝟒𝟒𝒏𝒏 𝟔𝟔(𝟑𝟑𝟑𝟑+ 𝟕𝟕𝟕𝟕) d. 𝟑𝟑𝟑𝟑𝒂𝒂+ 𝟑𝟑𝟑𝟑𝒃𝒃 𝟑𝟑(𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏) e. 𝟏𝟏𝟏𝟏𝒇𝒇+ 𝟏𝟏𝟏𝟏𝒇𝒇 𝒇𝒇(𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏) f. 𝟏𝟏𝟏𝟏𝒉𝒉+ 𝟏𝟏𝟏𝟏𝒉𝒉 𝒉𝒉(𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏) g. 𝟓𝟓𝟓𝟓𝒎𝒎+ 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏(𝟓𝟓𝟓𝟓+ 𝟏𝟏) MP.7 A STORY OF RATIOS 123 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions h. 𝟕𝟕+ 𝟓𝟓𝟓𝟓𝒚𝒚 𝟕𝟕(𝟏𝟏+ 𝟖𝟖𝟖𝟖) 2. Evaluate each of the expressions below. a. 𝟔𝟔𝒙𝒙+ 𝟐𝟐𝟐𝟐𝒚𝒚 and 𝟑𝟑(𝟐𝟐𝒙𝒙+ 𝟕𝟕𝒚𝒚) 𝒙𝒙= 𝟑𝟑 and 𝒚𝒚= 𝟒𝟒 𝟔𝟔(𝟑𝟑) + 𝟐𝟐𝟐𝟐(𝟒𝟒) 𝟑𝟑(𝟐𝟐∙𝟑𝟑+ 𝟕𝟕∙𝟒𝟒) 𝟏𝟏𝟏𝟏+ 𝟖𝟖𝟖𝟖 𝟑𝟑(𝟔𝟔+ 𝟐𝟐𝟐𝟐) 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑(𝟑𝟑𝟑𝟑) 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 b. 𝟓𝟓𝒈𝒈+ 𝟕𝟕𝒈𝒈 and 𝒈𝒈(𝟓𝟓+ 𝟕𝟕) 𝒈𝒈= 𝟔𝟔 𝟓𝟓(𝟔𝟔) + 𝟕𝟕(𝟔𝟔) 𝟔𝟔(𝟓𝟓+ 𝟕𝟕) 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒 𝟔𝟔(𝟏𝟏𝟏𝟏) 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕 c. 𝟏𝟏𝟏𝟏𝒙𝒙+ 𝟐𝟐 and 𝟐𝟐(𝟕𝟕𝒙𝒙+ 𝟏𝟏) 𝒙𝒙= 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏(𝟏𝟏𝟏𝟏) + 𝟐𝟐 𝟐𝟐(𝟕𝟕∙𝟏𝟏𝟏𝟏+ 𝟏𝟏) 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟐𝟐 𝟐𝟐(𝟕𝟕𝟕𝟕+ 𝟏𝟏) 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐(𝟕𝟕𝟕𝟕) 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 d. Explain any patterns that you notice in the results to parts (a)–(c). Both expressions in parts (a)–(c) evaluated to the same number when the indicated value was substituted for the variable. This shows that the two expressions are equivalent for the given values. e. What would happen if other values were given for the variables? Because the two expressions in each part are equivalent, they evaluate to the same number, no matter what value is chosen for the variable. A STORY OF RATIOS 124 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Lesson Summary AN EXPRESSION IN FACTORED FORM: An expression that is a product of two or more expressions is said to be in factored form. Closing (3 minutes) Closing How can use you use your knowledge of GCF and the distributive property to write equivalent expressions? We can use our knowledge of GCF and the distributive property to change expressions from standard form to factored form. Find the missing value that makes the two expressions equivalent. 𝟒𝟒𝒙𝒙+ 𝟏𝟏𝟏𝟏𝒚𝒚 𝟒𝟒 (𝒙𝒙+ 𝟑𝟑𝒚𝒚) 𝟑𝟑𝟑𝟑𝒙𝒙+ 𝟓𝟓𝟓𝟓𝒚𝒚 𝟓𝟓 (𝟕𝟕𝒙𝒙+ 𝟏𝟏𝟏𝟏𝒚𝒚) 𝟏𝟏𝟏𝟏𝒙𝒙+ 𝟗𝟗𝒚𝒚 𝟗𝟗 (𝟐𝟐𝒙𝒙+ 𝒚𝒚) 𝟑𝟑𝟑𝟑𝒙𝒙+ 𝟖𝟖𝒚𝒚 𝟖𝟖 (𝟒𝟒𝒙𝒙+ 𝒚𝒚) 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙+ 𝟕𝟕𝟕𝟕𝟕𝟕𝒚𝒚 𝟏𝟏𝟏𝟏𝟏𝟏 (𝒙𝒙+ 𝟕𝟕𝒚𝒚) Explain how you determine the missing number. I would expand each term and determine the greatest common factor. The greatest common factor is the number that is placed on the blank line. Exit Ticket (4 minutes) A STORY OF RATIOS 125 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Name Date Lesson 11: Factoring Expressions Exit Ticket Use greatest common factor and the distributive property to write equivalent expressions in factored form. 1. 2𝑥𝑥+ 8𝑦𝑦 2. 13𝑎𝑎𝑎𝑎+ 15𝑎𝑎𝑎𝑎 3. 20𝑔𝑔+ 24ℎ A STORY OF RATIOS 126 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Exit Ticket Sample Solutions Use greatest common factor and the distributive property to write equivalent expressions in factored form. 1. 𝟐𝟐𝟐𝟐+ 𝟖𝟖𝟖𝟖 𝟐𝟐(𝒙𝒙+ 𝟒𝟒𝟒𝟒) 2. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒂𝒂𝒂𝒂(𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏) 3. 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒(𝟓𝟓𝟓𝟓+ 𝟔𝟔𝟔𝟔) Problem Set Sample Solutions 1. Use models to prove that 𝟑𝟑(𝒂𝒂+ 𝒃𝒃) is equivalent to 𝟑𝟑𝟑𝟑+ 𝟑𝟑𝟑𝟑. 2. Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expressions. a. 𝟒𝟒𝒅𝒅+ 𝟏𝟏𝟏𝟏𝒆𝒆 𝟒𝟒(𝒅𝒅+ 𝟑𝟑𝟑𝟑) or 𝟒𝟒(𝟏𝟏𝟏𝟏+ 𝟑𝟑𝟑𝟑) b. 𝟏𝟏𝟏𝟏𝒙𝒙+ 𝟑𝟑𝟑𝟑𝒚𝒚 𝟔𝟔(𝟑𝟑𝟑𝟑+ 𝟓𝟓𝟓𝟓) c. 𝟐𝟐𝟐𝟐𝒂𝒂+ 𝟐𝟐𝟐𝟐𝒚𝒚 𝟕𝟕(𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒) d. 𝟐𝟐𝟐𝟐𝒇𝒇+ 𝟓𝟓𝟓𝟓𝒈𝒈 𝟖𝟖(𝟑𝟑𝟑𝟑+ 𝟕𝟕𝟕𝟕) 𝒂𝒂+ 𝒃𝒃 𝒂𝒂+ 𝒃𝒃 𝒂𝒂 𝒃𝒃 𝒂𝒂 𝒂𝒂 𝒃𝒃 𝒃𝒃 𝒂𝒂+ 𝒃𝒃 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝒂𝒂 𝒃𝒃 𝒂𝒂 𝒂𝒂 𝒃𝒃 𝒃𝒃 A STORY OF RATIOS 127 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Greatest Common Factor—Round 1 Directions: Determine the greatest common factor of each pair of numbers. 1. GCF of 10 and 50 16. GCF of 45 and 72 2. GCF of 5 and 35 17. GCF of 28 and 48 3. GCF of 3 and 12 18. GCF of 44 and 77 4. GCF of 8 and 20 19. GCF of 39 and 66 5. GCF of 15 and 35 20. GCF of 64 and 88 6. GCF of 10 and 75 21. GCF of 42 and 56 7. GCF of 9 and 30 22. GCF of 28 and 42 8. GCF of 15 and 33 23. GCF of 13 and 91 9. GCF of 12 and 28 24. GCF of 16 and 84 10. GCF of 16 and 40 25. GCF of 36 and 99 11. GCF of 24 and 32 26. GCF of 39 and 65 12. GCF of 35 and 49 27. GCF of 27 and 87 13. GCF of 45 and 60 28. GCF of 28 and 70 14. GCF of 48 and 72 29. GCF of 26 and 91 15. GCF of 50 and 42 30. GCF of 34 and 51 Number Correct: A STORY OF RATIOS 128 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Greatest Common Factor—Round 1 [KEY] Directions: Determine the greatest common factor of each pair of numbers. 1. GCF of 10 and 50 𝟏𝟏𝟏𝟏 16. GCF of 45 and 72 𝟗𝟗 2. GCF of 5 and 35 𝟓𝟓 17. GCF of 28 and 48 𝟒𝟒 3. GCF of 3 and 12 𝟑𝟑 18. GCF of 44 and 77 𝟏𝟏𝟏𝟏 4. GCF of 8 and 20 𝟒𝟒 19. GCF of 39 and 66 𝟑𝟑 5. GCF of 15 and 35 𝟓𝟓 20. GCF of 64 and 88 𝟖𝟖 6. GCF of 10 and 75 𝟓𝟓 21. GCF of 42 and 56 𝟏𝟏𝟏𝟏 7. GCF of 9 and 30 𝟑𝟑 22. GCF of 28 and 42 𝟏𝟏𝟏𝟏 8. GCF of 15 and 33 𝟑𝟑 23. GCF of 13 and 91 𝟏𝟏𝟏𝟏 9. GCF of 12 and 28 𝟒𝟒 24. GCF of 16 and 84 𝟒𝟒 10. GCF of 16 and 40 𝟖𝟖 25. GCF of 36 and 99 𝟗𝟗 11. GCF of 24 and 32 𝟖𝟖 26. GCF of 39 and 65 𝟏𝟏𝟏𝟏 12. GCF of 35 and 49 𝟕𝟕 27. GCF of 27 and 87 𝟑𝟑 13. GCF of 45 and 60 𝟏𝟏𝟏𝟏 28. GCF of 28 and 70 𝟏𝟏𝟏𝟏 14. GCF of 48 and 72 𝟐𝟐𝟐𝟐 29. GCF of 26 and 91 𝟏𝟏𝟏𝟏 15. GCF of 50 and 42 𝟐𝟐 30. GCF of 34 and 51 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 129 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Greatest Common Factor—Round 2 Directions: Determine the greatest common factor of each pair of numbers. 1. GCF of 20 and 80 16. GCF of 33 and 99 2. GCF of 10 and 70 17. GCF of 38 and 76 3. GCF of 9 and 36 18. GCF of 26 and 65 4. GCF of 12 and 24 19. GCF of 39 and 48 5. GCF of 15 and 45 20. GCF of 72 and 88 6. GCF of 10 and 95 21. GCF of 21 and 56 7. GCF of 9 and 45 22. GCF of 28 and 52 8. GCF of 18 and 33 23. GCF of 51 and 68 9. GCF of 12 and 32 24. GCF of 48 and 84 10. GCF of 16 and 56 25. GCF of 21 and 63 11. GCF of 40 and 72 26. GCF of 64 and 80 12. GCF of 35 and 63 27. GCF of 36 and 90 13. GCF of 30 and 75 28. GCF of 28 and 98 14. GCF of 42 and 72 29. GCF of 39 and 91 15. GCF of 30 and 28 30. GCF of 38 and 95 Number Correct: Improvement: A STORY OF RATIOS 130 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 11 Lesson 11: Factoring Expressions Greatest Common Factor—Round 2 [KEY] Directions: Determine the greatest common factor of each pair of numbers. 1. GCF of 20 and 80 𝟐𝟐𝟐𝟐 16. GCF of 33 and 99 𝟑𝟑𝟑𝟑 2. GCF of 10 and 70 𝟏𝟏𝟏𝟏 17. GCF of 38 and 76 𝟑𝟑𝟑𝟑 3. GCF of 9 and 36 𝟗𝟗 18. GCF of 26 and 65 𝟏𝟏𝟏𝟏 4. GCF of 12 and 24 𝟏𝟏𝟏𝟏 19. GCF of 39 and 48 𝟑𝟑 5. GCF of 15 and 45 𝟏𝟏𝟏𝟏 20. GCF of 72 and 88 𝟖𝟖 6. GCF of 10 and 95 𝟓𝟓 21. GCF of 21 and 56 𝟕𝟕 7. GCF of 9 and 45 𝟗𝟗 22. GCF of 28 and 52 𝟒𝟒 8. GCF of 18 and 33 𝟑𝟑 23. GCF of 51 and 68 𝟏𝟏𝟏𝟏 9. GCF of 12 and 32 𝟒𝟒 24. GCF of 48 and 84 𝟏𝟏𝟏𝟏 10. GCF of 16 and 56 𝟖𝟖 25. GCF of 21 and 63 𝟐𝟐𝟐𝟐 11. GCF of 40 and 72 𝟖𝟖 26. GCF of 64 and 80 𝟏𝟏𝟏𝟏 12. GCF of 35 and 63 𝟕𝟕 27. GCF of 36 and 90 𝟏𝟏𝟏𝟏 13. GCF of 30 and 75 𝟏𝟏𝟏𝟏 28. GCF of 28 and 98 𝟏𝟏𝟏𝟏 14. GCF of 42 and 72 𝟔𝟔 29. GCF of 39 and 91 𝟏𝟏𝟏𝟏 15. GCF of 30 and 28 𝟐𝟐 30. GCF of 38 and 95 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 131 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions 𝒂𝒂 + 𝒃𝒃 𝒂𝒂 𝒃𝒃 𝒂𝒂 + 𝒃𝒃 𝒂𝒂 𝒃𝒃 Lesson 12: Distributing Expressions Student Outcomes Students model and write equivalent expressions using the distributive property. They move from the factored form to the expanded form of an expression. Classwork Opening Exercise (3 minutes) Opening Exercise a. Create a model to show 𝟐𝟐× 𝟓𝟓. b. Create a model to show 𝟐𝟐× 𝒃𝒃, or 𝟐𝟐𝒃𝒃. Example 1 (8 minutes) Example 1 Write an expression that is equivalent to 𝟐𝟐(𝒂𝒂+ 𝒃𝒃). In this example, we have been given the factored form of the expression. To answer this question, we can create a model to represent 2(𝑎𝑎+ 𝑏𝑏). Let’s start by creating a model to represent (𝑎𝑎+ 𝑏𝑏). Create a model to represent (𝒂𝒂+ 𝒃𝒃). The expression 𝟐𝟐(𝒂𝒂+ 𝒃𝒃) tells us that we have 𝟐𝟐 of the (𝒂𝒂+ 𝒃𝒃)’s. Create a model that shows 𝟐𝟐 groups of (𝒂𝒂+ 𝒃𝒃). 𝟓𝟓 𝟓𝟓 𝒃𝒃 𝒃𝒃 𝒂𝒂 + 𝒃𝒃 𝒂𝒂 𝒃𝒃 MP.7 A STORY OF RATIOS 132 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions How many 𝒂𝒂’s and how many 𝒃𝒃’s do you see in the diagram? There are 𝟐𝟐 𝒂𝒂’s and 𝟐𝟐 𝒃𝒃’s. How would the model look if we grouped together the 𝒂𝒂’s and then grouped together the 𝒃𝒃’s? What expression could we write to represent the new diagram? 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 This expression is written in expanded form. What conclusion can we draw from the models about equivalent expressions? 𝟐𝟐(𝒂𝒂+ 𝒃𝒃) = 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 To prove that these two forms are equivalent, let’s substitute some values for 𝑎𝑎 and 𝑏𝑏 and see what happens. Let 𝒂𝒂= 𝟑𝟑 and 𝒃𝒃= 𝟒𝟒. 𝟐𝟐(𝒂𝒂+ 𝒃𝒃) 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 𝟐𝟐(𝟑𝟑+ 𝟒𝟒) 𝟐𝟐(𝟑𝟑) + 𝟐𝟐(𝟒𝟒) 𝟐𝟐(𝟕𝟕) 𝟔𝟔+ 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 Note: If students do not believe yet that the two are equal, continue substituting more values for 𝑎𝑎 and 𝑏𝑏 until they are convinced. What happens when we double (𝒂𝒂+ 𝒃𝒃)? We double 𝒂𝒂, and we double 𝒃𝒃. Example 2 (5 minutes) Example 2 Write an expression that is equivalent to double (𝟑𝟑𝒙𝒙+ 𝟒𝟒𝒚𝒚). How can we rewrite double (𝟑𝟑𝒙𝒙+ 𝟒𝟒𝒚𝒚)? Double is the same as multiplying by two. 𝟐𝟐(𝟑𝟑𝒙𝒙+ 𝟒𝟒𝟒𝟒) or 𝟔𝟔𝟔𝟔+ 𝟖𝟖𝟖𝟖 𝟐𝟐𝟐𝟐 𝒂𝒂 𝒃𝒃 𝟐𝟐𝟐𝟐 𝒂𝒂 𝒃𝒃 MP.7 A STORY OF RATIOS 133 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions 𝟒𝟒𝟒𝟒 + 𝟓𝟓 𝒚𝒚 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟖𝟖𝟖𝟖 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 Is this expression in factored form, expanded form, or neither? The first expression is in factored form, and the second expression is in expanded form. Let’s start this problem the same way that we started the first example. What should we do? We can make a model of 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒. 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 How can we change the model to show 𝟐𝟐(𝟑𝟑𝒙𝒙+ 𝟒𝟒𝒚𝒚)? We can make two copies of the model. Are there terms that we can combine in this example? Yes. There are 𝟔𝟔 𝒙𝒙's and 𝟖𝟖 𝒚𝒚's. So, the model is showing 𝟔𝟔𝟔𝟔+ 𝟖𝟖𝟖𝟖. What is an equivalent expression that we can use to represent 𝟐𝟐(𝟑𝟑𝒙𝒙+ 𝟒𝟒𝒚𝒚)? 𝟐𝟐(𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒) = 𝟔𝟔𝟔𝟔+ 𝟖𝟖𝟖𝟖 This is the same as 𝟐𝟐(𝟑𝟑𝟑𝟑) + 𝟐𝟐(𝟒𝟒𝟒𝟒). Summarize how you would solve this question without the model. When there is a number outside the parentheses, I would multiply it by all the terms on the inside of the parentheses. Example 3 (3 minutes) Example 3 Write an expression in expanded form that is equivalent to the model below. What factored expression is represented in the model? 𝒚𝒚(𝟒𝟒𝟒𝟒+ 𝟓𝟓) MP.7 A STORY OF RATIOS 134 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions How can we rewrite this expression in expanded form? 𝒚𝒚(𝟒𝟒𝟒𝟒) + 𝒚𝒚(𝟓𝟓) 𝟒𝟒𝟒𝟒𝟒𝟒+ 𝟓𝟓𝟓𝟓 Example 4 (3 minutes) How can we use our work in the previous examples to write the following expression? Example 4 Write an expression in expanded form that is equivalent to 𝟑𝟑(𝟕𝟕𝒅𝒅+ 𝟒𝟒𝒆𝒆). We will multiply 𝟑𝟑× 𝟕𝟕𝟕𝟕 and 𝟑𝟑× 𝟒𝟒𝟒𝟒. We would get 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟏𝟏. So, 𝟑𝟑(𝟕𝟕𝟕𝟕+ 𝟒𝟒𝟒𝟒) = 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟏𝟏. Exercises (15 minutes) Exercises Create a model for each expression below. Then, write another equivalent expression using the distributive property. 1. 𝟑𝟑(𝒙𝒙+ 𝒚𝒚) 𝟑𝟑𝟑𝟑+ 𝟑𝟑𝟑𝟑 2. 𝟒𝟒(𝟐𝟐𝟐𝟐+ 𝒈𝒈) 𝟖𝟖𝟖𝟖+ 𝟒𝟒𝟒𝟒 𝒙𝒙+ 𝒚𝒚 𝒙𝒙+ 𝒚𝒚 𝒙𝒙 𝒚𝒚 𝒙𝒙 𝒙𝒙 𝒚𝒚 𝒚𝒚 𝒙𝒙+ 𝒚𝒚 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝒙𝒙 𝒚𝒚 𝒙𝒙 𝒙𝒙 𝒚𝒚 𝒚𝒚 𝟐𝟐𝟐𝟐+ 𝒈𝒈 𝟐𝟐𝟐𝟐+ 𝒈𝒈 𝟐𝟐𝟐𝟐 𝒈𝒈 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝒈𝒈 𝒈𝒈 𝟐𝟐𝟐𝟐+ 𝒈𝒈 𝟖𝟖𝟖𝟖 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝒈𝒈 𝟐𝟐𝟐𝟐+ 𝒈𝒈 𝟐𝟐𝟐𝟐 𝒈𝒈 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝒈𝒈 𝒈𝒈 𝒈𝒈 MP.7 A STORY OF RATIOS 135 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions Apply the distributive property to write equivalent expressions in expanded form. 3. 𝟖𝟖(𝒉𝒉+ 𝟑𝟑) 𝟖𝟖𝟖𝟖+ 𝟐𝟐𝟐𝟐 4. 𝟑𝟑(𝟐𝟐𝟐𝟐+ 𝟕𝟕) 𝟔𝟔𝟔𝟔+ 𝟐𝟐𝟐𝟐 5. 𝟓𝟓(𝟑𝟑𝟑𝟑+ 𝟗𝟗𝟗𝟗) 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟒𝟒𝟒𝟒𝟒𝟒 6. 𝟒𝟒(𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟑𝟑𝟑𝟑) 𝟒𝟒𝟒𝟒𝟒𝟒+ 𝟏𝟏𝟏𝟏𝟏𝟏 7. 𝟕𝟕𝟕𝟕𝟕𝟕+ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 8. 𝒂𝒂(𝟗𝟗𝟗𝟗+ 𝟏𝟏𝟏𝟏) 𝟗𝟗𝟗𝟗𝟗𝟗+ 𝟏𝟏𝟏𝟏𝟏𝟏 Closing (3 minutes) State what the expression 𝑎𝑎(𝑏𝑏+ 𝑐𝑐) represents. 𝑎𝑎 groups of the quantity 𝑏𝑏 plus 𝑐𝑐 Explain in your own words how to write an equivalent expression in expanded form when given an expression in the form of 𝑎𝑎(𝑏𝑏+ 𝑐𝑐). Then, create your own example to show off what you know. To write an equivalent expression, I would multiply 𝑎𝑎 times 𝑏𝑏 and 𝑎𝑎 times 𝑐𝑐. Then, I would add the two products together. Examples will vary. State what the equivalent expression in expanded form represents. 𝑎𝑎𝑎𝑎+ 𝑎𝑎𝑎𝑎 means 𝑎𝑎 groups of size 𝑏𝑏 plus 𝑎𝑎 groups of size 𝑐𝑐. Exit Ticket (5 minutes) 𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏 𝒋𝒋 A STORY OF RATIOS 136 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions Name Date Lesson 12: Distributing Expressions Exit Ticket Use the distributive property to write the following expressions in expanded form. 1. 2(𝑏𝑏+ 𝑐𝑐) 2. 5(7ℎ+ 3𝑚𝑚) 3. 𝑒𝑒(𝑓𝑓+ 𝑔𝑔) A STORY OF RATIOS 137 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions Exit Ticket Sample Solutions Use the distributive property to write the following expressions in expanded form. 1. 𝟐𝟐(𝒃𝒃+ 𝒄𝒄) 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 2. 𝟓𝟓(𝟕𝟕𝒉𝒉+ 𝟑𝟑𝒎𝒎) 𝟑𝟑𝟑𝟑𝟑𝟑+ 𝟏𝟏𝟏𝟏𝟏𝟏 3. 𝒆𝒆(𝒇𝒇+ 𝒈𝒈) 𝒆𝒆𝒆𝒆+ 𝒆𝒆𝒆𝒆 Problem Set Sample Solutions 1. Use the distributive property to write the following expressions in expanded form. a. 𝟒𝟒(𝒙𝒙+ 𝒚𝒚) 𝟒𝟒𝟒𝟒+ 𝟒𝟒𝟒𝟒 b. 𝟖𝟖(𝒂𝒂+ 𝟑𝟑𝟑𝟑) 𝟖𝟖𝟖𝟖+ 𝟐𝟐𝟐𝟐𝟐𝟐 c. 𝟑𝟑(𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟏𝟏) 𝟔𝟔𝟔𝟔+ 𝟑𝟑𝟑𝟑𝟑𝟑 d. 𝟗𝟗(𝟕𝟕𝟕𝟕+ 𝟔𝟔𝟔𝟔) 𝟔𝟔𝟔𝟔𝟔𝟔+ 𝟓𝟓𝟓𝟓𝟓𝟓 e. 𝒄𝒄(𝟑𝟑𝟑𝟑+ 𝒃𝒃) 𝟑𝟑𝟑𝟑𝟑𝟑+ 𝒃𝒃𝒃𝒃 f. 𝒚𝒚(𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟏𝟏) 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 A STORY OF RATIOS 138 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 12 Lesson 12: Distributing Expressions 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟔𝟔𝟔𝟔 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 2. Create a model to show that 𝟐𝟐(𝟐𝟐𝒙𝒙+ 𝟑𝟑𝒚𝒚) = 𝟒𝟒𝒙𝒙+ 𝟔𝟔𝒚𝒚. A STORY OF RATIOS 139 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions Lesson 13: Writing Division Expressions Student Outcomes Students write numerical expressions in two forms, “dividend ÷ divisor” and “ dividend divisor ”, and note the relationship between the two. Lesson Notes This is day one of a two-day lesson. Classwork Discussion (8 minutes) The discussion serves as a chance for students to show what they know about division and what division looks like. The discussion should conclude with the overall idea that writing 𝑎𝑎÷ 𝑏𝑏 as 𝑎𝑎 𝑏𝑏 is a strategic format when working algebraically. How can we write or show 8 divided by 2? (Students may be allowed to explain or even draw examples for the class to see.) Answers will vary. Students can draw models, arrays, use the division symbol, and some may even use a fraction. When working with algebraic expressions, are any of these expressions or models more efficient than others? Writing a fraction to show division is more efficient. Is 8 2 the same as 2 8? No, they are not the same. 8 2 = 4, while 2 8 = 1 4. How would we show 𝑎𝑎 divided by 𝑏𝑏 using a fraction? 𝑎𝑎 𝑏𝑏 Example 1 (5 minutes) Example 1 Write an expression showing 𝟏𝟏÷ 𝟐𝟐 without the use of the division symbol. MP.6 A STORY OF RATIOS 140 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions 𝟏𝟏 Let’s start by looking at a model of 1 ÷ 2. We can make a bar diagram. What can we determine from the model? 𝟏𝟏÷ 𝟐𝟐 is the same as 𝟏𝟏 𝟐𝟐. Example 2 (5 minutes) Example 2 Write an expression showing 𝒂𝒂÷ 𝟐𝟐 without the use of the division symbol. Here we have a variable being divided by 2. Let’s start by looking at a model of 𝑎𝑎÷ 2. We can make a bar diagram. What can we determine from the model? 𝒂𝒂÷ 𝟐𝟐 is the same as 𝒂𝒂 𝟐𝟐. When we write division expressions using the division symbol, we represent 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝÷ 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝. How would this look when we write division expressions using a fraction? 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 Example 3 (8 minutes) Example 3 a. Write an expression showing 𝒂𝒂÷ 𝒃𝒃 without the use of the division symbol. MP.6 𝒂𝒂 A STORY OF RATIOS 141 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions How can we use what we just learned in Examples 1 and 2 to help us with this example? The dividend is the numerator, and the divisor is the denominator. 𝒂𝒂 𝒃𝒃 b. Write an expression for 𝒈𝒈 divided by the quantity 𝒉𝒉 plus 𝟑𝟑. How would this look with the division symbol? 𝑔𝑔÷ (ℎ+ 3) Now, let’s rewrite this using a fraction. 𝒈𝒈 𝒉𝒉+ 𝟑𝟑 c. Write an expression for the quotient of the quantity 𝒎𝒎 reduced by 𝟑𝟑 and 𝟓𝟓. Let’s start again by writing this using a division symbol first. (𝑚𝑚−3) ÷ 5 Next, we will rewrite it using the fraction bar. 𝒎𝒎−𝟑𝟑 𝟓𝟓 Exercises (10 minutes) Have students use a white board or small board to practice the following questions. Exercises Write each expression two ways: using the division symbol and as a fraction. a. 𝟏𝟏𝟏𝟏 divided by 𝟒𝟒 𝟏𝟏𝟏𝟏÷ 𝟒𝟒 and 𝟏𝟏𝟏𝟏 𝟒𝟒 b. 𝟑𝟑 divided by 𝟓𝟓 𝟑𝟑÷ 𝟓𝟓 and 𝟑𝟑 𝟓𝟓 c. 𝒂𝒂 divided by 𝟒𝟒 𝒂𝒂÷ 𝟒𝟒 and 𝒂𝒂 𝟒𝟒 MP.6 A STORY OF RATIOS 142 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions d. The quotient of 𝟔𝟔 and 𝒎𝒎 𝟔𝟔÷ 𝒎𝒎 and 𝟔𝟔 𝒎𝒎 e. Seven divided by the quantity 𝒙𝒙 plus 𝒚𝒚 𝟕𝟕÷ (𝒙𝒙+ 𝒚𝒚) and 𝟕𝟕 𝒙𝒙 + 𝒚𝒚 f. 𝒚𝒚 divided by the quantity 𝒙𝒙 minus 𝟏𝟏𝟏𝟏 𝒚𝒚÷ (𝒙𝒙−𝟏𝟏𝟏𝟏) and 𝒚𝒚 𝒙𝒙 − 𝟏𝟏𝟏𝟏 g. The sum of the quantity 𝒉𝒉 and 𝟑𝟑 divided by 𝟒𝟒 (𝒉𝒉+ 𝟑𝟑) ÷ 𝟒𝟒 and 𝒉𝒉 + 𝟑𝟑 𝟒𝟒 h. The quotient of the quantity 𝒌𝒌 minus 𝟏𝟏𝟏𝟏 and 𝒎𝒎 (𝒌𝒌−𝟏𝟏𝟏𝟏) ÷ 𝒎𝒎 and 𝒌𝒌 − 𝟏𝟏𝟏𝟏 𝒎𝒎 Closing (4 minutes) Explain to your neighbor how you would rewrite any division problem using a fraction. The dividend would become the numerator, and the divisor would become the denominator. Exit Ticket (5 minutes) A STORY OF RATIOS 143 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions Name Date Lesson 13: Writing Division Expressions Exit Ticket Rewrite the expressions using the division symbol and as a fraction. 1. The quotient of 𝑚𝑚 and 7 2. Five divided by the sum of 𝑎𝑎 and 𝑏𝑏 3. The quotient of 𝑘𝑘 decreased by 4 and 9 A STORY OF RATIOS 144 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 13 Lesson 13: Writing Division Expressions Exit Ticket Sample Solutions Rewrite the expressions using the division symbol and as a fraction. 1. The quotient of 𝒎𝒎 and 𝟕𝟕 𝒎𝒎÷ 𝟕𝟕 and 𝒎𝒎 𝟕𝟕 2. Five divided by the sum of 𝒂𝒂 and 𝒃𝒃 𝟓𝟓÷ (𝒂𝒂+ 𝒃𝒃) and 𝟓𝟓 𝒂𝒂 + 𝒃𝒃 3. The quotient of 𝒌𝒌 decreased by 𝟒𝟒 and 𝟗𝟗 (𝒌𝒌−𝟒𝟒) ÷ 𝟗𝟗 and 𝒌𝒌 − 𝟒𝟒 𝟗𝟗 Problem Set Sample Solutions 1. Rewrite the expressions using the division symbol and as a fraction. a. Three divided by 𝟒𝟒 𝟑𝟑÷ 𝟒𝟒 and 𝟑𝟑 𝟒𝟒 b. The quotient of 𝒎𝒎 and 𝟏𝟏𝟏𝟏 𝒎𝒎÷ 𝟏𝟏𝟏𝟏 and 𝒎𝒎 𝟏𝟏𝟏𝟏 c. 𝟒𝟒 divided by the sum of 𝒉𝒉 and 𝟕𝟕 𝟒𝟒÷ (𝒉𝒉+ 𝟕𝟕) and 𝟒𝟒 𝒉𝒉 + 𝟕𝟕 d. The quantity 𝒙𝒙 minus 𝟑𝟑 divided by 𝒚𝒚 (𝒙𝒙−𝟑𝟑) ÷ 𝒚𝒚 and 𝒙𝒙 − 𝟑𝟑 𝒚𝒚 2. Draw a model to show that 𝒙𝒙÷ 𝟑𝟑 is the same as 𝒙𝒙 𝟑𝟑. 𝟏𝟏 𝟑𝟑𝒙𝒙 or 𝒙𝒙 𝟑𝟑 𝒙𝒙 A STORY OF RATIOS 145 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions The quotient of 𝟏𝟏𝟏𝟏 and 𝟑𝟑 𝟏𝟏𝟏𝟏÷ 𝟑𝟑 Equivalent Expressions 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟑𝟑 Lesson 14: Writing Division Expressions Student Outcomes Students write numerical expressions in two forms, “dividend ÷ divisor” and “ dividend divisor ”, and note the relationship between the two. Lesson Notes This is the second day of a two-day lesson. Classwork Fluency Exercise (5 Minutes): Long Division Algorithm RWBE: Refer to the Rapid White Board Exchanges sections in the Module Overview for directions on how to administer an RWBE. Example 1 (5 minutes) Example 1 Fill in the three remaining squares so that all the squares contain equivalent expressions. MP.2 A STORY OF RATIOS 146 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Example 2 (5 minutes) Example 2 Fill in a blank copy of the four boxes using the words dividend and divisor so that it is set up for any example. Exercises (20 minutes) Students work in pairs. Each pair is given a set of expressions to work on. There are several different versions that can be printed and used so that a variety of questions can be used throughout the classroom. Students fill in the four rectangles, one with the given information and three with equivalent expressions. Exercises Complete the missing spaces in each rectangle set. Set A Answers: 1. 𝟓𝟓 ÷ 𝒑𝒑 𝟓𝟓 divided by 𝒑𝒑, 𝟓𝟓 𝒑𝒑, 5 p 2. The quotient of 𝒈𝒈 and 𝒉𝒉 𝒈𝒈÷ 𝒉𝒉, 𝒈𝒈 𝒉𝒉, h g 3. 23 w 𝟐𝟐𝟐𝟐 divided by 𝒘𝒘, 𝟐𝟐𝟐𝟐÷ 𝒘𝒘, 𝟐𝟐𝟐𝟐 𝒘𝒘 4. 𝒚𝒚 𝒙𝒙+𝟖𝟖 𝒚𝒚 divided by the sum of 𝒙𝒙 and 𝟖𝟖, 𝒚𝒚÷ (𝒙𝒙+ 𝟖𝟖), 8 x y + 5. 𝟕𝟕 divided by the quantity 𝒂𝒂 minus 𝟔𝟔 𝟕𝟕÷ (𝒂𝒂−𝟔𝟔), 𝟕𝟕 𝒂𝒂−𝟔𝟔, 6 7 a − 6. 3 11 m + The sum of 𝒎𝒎 and 𝟏𝟏𝟏𝟏 divided by 𝟑𝟑, (𝒎𝒎+ 𝟏𝟏𝟏𝟏) ÷ 𝟑𝟑, 𝒎𝒎+𝟏𝟏𝟏𝟏 𝟑𝟑 Scaffolding: The sets of eight questions used in the Exercises can be tailored to fit the level at which students are working. A set for lower-level learners and/or a set for advanced learners may be written as needed. MP.2 The quotient of the dividend and the divisor 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝÷ 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 Equivalent Expressions divisor dividend dividend divisor A STORY OF RATIOS 147 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions 7. (𝒇𝒇 + 𝟐𝟐) ÷ 𝒈𝒈 The sum of 𝒇𝒇 and 𝟐𝟐 divided by 𝒈𝒈, 𝒇𝒇+𝟐𝟐 𝒈𝒈, 2 g f + 8. 𝒄𝒄−𝟗𝟗 𝒅𝒅+𝟑𝟑 The quotient of 𝒄𝒄 minus 𝟗𝟗 and 𝒅𝒅 plus 𝟑𝟑, (𝒄𝒄−𝟗𝟗) ÷ (𝒅𝒅+ 𝟑𝟑), 3 9 d c + − Set B Answers: 1. 𝒉𝒉 ÷ 𝟏𝟏𝟏𝟏 The quotient of 𝒉𝒉 and 𝟏𝟏𝟏𝟏, 11 h , 𝒉𝒉 𝟏𝟏𝟏𝟏 2. The quotient of 𝒎𝒎 and 𝒏𝒏 𝒎𝒎÷ 𝒏𝒏, n m , 𝒎𝒎 𝒏𝒏 3. 5 j The quotient of 𝒋𝒋 and 𝟓𝟓, 𝒋𝒋÷ 𝟓𝟓, 𝒋𝒋 𝟓𝟓 4. 𝒉𝒉 𝒎𝒎−𝟒𝟒 𝒉𝒉 divided by the quantity 𝒎𝒎 minus 𝟒𝟒, 𝒉𝒉÷ (𝒎𝒎−𝟒𝟒), 4 m h − 5. 𝒇𝒇 divided by the quantity 𝒈𝒈 minus 𝟏𝟏𝟏𝟏 𝒇𝒇÷ (𝒈𝒈−𝟏𝟏𝟏𝟏), 11 g f − , 𝒇𝒇 𝒈𝒈−𝟏𝟏𝟏𝟏 6. 18 5 a + The sum of 𝒂𝒂 and 𝟓𝟓 divided by 𝟏𝟏𝟏𝟏, (𝒂𝒂+ 𝟓𝟓) ÷ 𝟏𝟏𝟏𝟏, 𝒂𝒂+𝟓𝟓 𝟏𝟏𝟏𝟏 7. (𝒚𝒚−𝟑𝟑) ÷ 𝒙𝒙 The quantity 𝒚𝒚 minus 𝟑𝟑 divided by 𝒙𝒙, 3 x y − , 𝒚𝒚−𝟑𝟑 𝒙𝒙 8. 𝒈𝒈+𝟓𝟓 𝒉𝒉−𝟏𝟏𝟏𝟏 The quantity 𝒈𝒈 plus 𝟓𝟓 divided by the quantity 𝒉𝒉 minus 𝟏𝟏𝟏𝟏, (𝒈𝒈+ 𝟓𝟓) ÷ (𝒉𝒉−𝟏𝟏𝟏𝟏), 11 5 h g − + Set C Answers: 1. 𝟔𝟔÷ 𝒌𝒌 𝟔𝟔 divided by 𝒌𝒌, 6 k , 𝟔𝟔 𝒌𝒌 2. The quotient of 𝒋𝒋 and 𝒌𝒌 𝒋𝒋÷ 𝒌𝒌, k j , 𝒋𝒋 𝒌𝒌 3. 10 a 𝒂𝒂 divided by 𝟏𝟏𝟏𝟏, 𝒂𝒂÷ 𝟏𝟏𝟏𝟏, 𝒂𝒂 𝟏𝟏𝟏𝟏 4. 𝟏𝟏𝟏𝟏 𝒇𝒇−𝟐𝟐 𝟏𝟏𝟏𝟏 divided by the quantity 𝒇𝒇 minus 𝟐𝟐, 𝟏𝟏𝟏𝟏÷ (𝒇𝒇−𝟐𝟐), 2 15 f − 5. 𝟏𝟏𝟏𝟏 divided by the sum of 𝒉𝒉 and 𝟏𝟏 𝟏𝟏𝟏𝟏÷ (𝒉𝒉+ 𝟏𝟏), 1 13 h + , 𝟏𝟏𝟏𝟏 𝒉𝒉+𝟏𝟏 6. 3 18 c + The sum of 𝒄𝒄 plus 𝟏𝟏𝟏𝟏 divided by 𝟑𝟑, (𝒄𝒄+ 𝟏𝟏𝟏𝟏) ÷ 𝟑𝟑, 𝒄𝒄 + 𝟏𝟏𝟏𝟏 𝟑𝟑 7. (𝒉𝒉−𝟐𝟐) ÷ 𝒎𝒎 The quantity 𝒉𝒉 minus 𝟐𝟐 divided by 𝒎𝒎, 2 m h − , 𝒉𝒉−𝟐𝟐 𝒎𝒎 8. 𝟒𝟒 − 𝒎𝒎 𝒏𝒏+𝟏𝟏𝟏𝟏 The quantity 𝟒𝟒 minus 𝒎𝒎 divided by the sum of 𝒏𝒏 and 𝟏𝟏𝟏𝟏, (𝟒𝟒−𝒎𝒎) ÷ (𝒏𝒏+ 𝟏𝟏𝟏𝟏), 11 4 n m + − A STORY OF RATIOS 148 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Closing (7 minutes) Two pairs of students trade pages to check each other’s work. If all of the boxes are correct, students write a sentence that summarizes why the expressions are equivalent. If there are mistakes, students write sentences to explain how to correct them. Students evaluate some of the expressions. Many answers need to be written as fractions or decimals. Set A: 𝑝𝑝= 3, 𝑤𝑤= 5, 𝑎𝑎= 10 Set B: ℎ= 4, 𝑗𝑗= 8, 𝑎𝑎= 10 Set C: 𝑘𝑘= 2, 𝑎𝑎= 10, 𝑐𝑐= 6 Exit Ticket (3 minutes) A STORY OF RATIOS 149 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Name Date Lesson 14: Writing Division Expressions Exit Ticket 1. Write the division expression in words and as a fraction. (𝑔𝑔+ 12) ÷ ℎ 2. Write the following division expression using the division symbol and as a fraction: 𝑓𝑓 divided by the quantity ℎ minus 3. A STORY OF RATIOS 150 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Exit Ticket Sample Solutions 1. Write the division expression in words and as a fraction. (𝒈𝒈+ 𝟏𝟏𝟏𝟏) ÷ 𝒉𝒉 The sum of 𝒈𝒈 and 𝟏𝟏𝟏𝟏 divided by 𝒉𝒉, 𝒈𝒈+𝟏𝟏𝟏𝟏 𝒉𝒉 2. Write the following division expression using the division symbol and as a fraction: 𝒇𝒇 divided by the quantity 𝒉𝒉 minus 𝟑𝟑. 𝒇𝒇÷ (𝒉𝒉−𝟑𝟑) and 𝒇𝒇 𝒉𝒉−𝟑𝟑 Problem Set Sample Solutions Complete the missing spaces in each rectangle set. The quotient of 𝒉𝒉 and 𝟏𝟏𝟏𝟏, 16 h , 𝒉𝒉 𝟏𝟏𝟏𝟏 𝒎𝒎 divided by the quantity 𝒃𝒃 minus 𝟑𝟑𝟑𝟑, 33 b m − , 𝒎𝒎÷ (𝒃𝒃−𝟑𝟑𝟑𝟑) 7 x , 𝟕𝟕÷ 𝒙𝒙, 𝟕𝟕 𝒙𝒙 The sum of 𝒚𝒚 and 𝟏𝟏𝟏𝟏 divided by 𝟐𝟐, (𝒚𝒚+ 𝟏𝟏𝟏𝟏) ÷ 𝟐𝟐, 𝒚𝒚+𝟏𝟏𝟏𝟏 𝟐𝟐 𝒉𝒉 𝟏𝟏𝟏𝟏 𝒎𝒎 𝒃𝒃 − 𝟑𝟑𝟑𝟑 𝟕𝟕 divided by 𝒙𝒙 𝟐𝟐 𝒚𝒚+ 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 151 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Exercise Handout Set A 1. 5 ÷ 𝑝𝑝 2. The quotient of 𝑔𝑔 and ℎ 3. 23 w 4. 𝑦𝑦 𝑥𝑥+8 5. 7 divided by the quantity 𝑎𝑎 minus 6 6. 3 11 m + 7. (𝑓𝑓 + 2) ÷ 𝑔𝑔 8. 𝑐𝑐−9 𝑑𝑑+3 Set B 1. ℎ ÷ 11 2. The quotient of 𝑚𝑚 and 𝑛𝑛 3. 5 j 4. ℎ 𝑚𝑚−4 5. 𝑓𝑓 divided by the quantity 𝑔𝑔 minus 11 6. 18 5 a + 7. (𝑦𝑦−3) ÷ 𝑥𝑥 8. 𝑔𝑔+5 ℎ−11 Set C 1. 6 ÷ 𝑘𝑘 2. The quotient of 𝑗𝑗 and 𝑘𝑘 3. 10 a 4. 15 𝑓𝑓−2 5. 13 divided by the sum of ℎ and 1 6. 3 18 c + 7. (ℎ−2) ÷ 𝑚𝑚 8. 4 −𝑚𝑚 𝑛𝑛+11 A STORY OF RATIOS 152 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 14 Lesson 14: Writing Division Expressions Long Division Algorithm Progression of Exercises 1. 3,282 ÷ 6 𝟓𝟓𝟓𝟓𝟓𝟓 2. 2,712 ÷ 3 𝟗𝟗𝟗𝟗𝟗𝟗 3. 15,036 ÷ 7 𝟐𝟐, 𝟏𝟏𝟏𝟏𝟏𝟏 4. 1,788 ÷ 8 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟓𝟓 5. 5,736 ÷ 12 𝟒𝟒𝟒𝟒𝟒𝟒 6. 35,472 ÷ 16 𝟐𝟐, 𝟐𝟐𝟐𝟐𝟐𝟐 7. 13,384 ÷ 28 𝟒𝟒𝟒𝟒𝟒𝟒 8. 31,317 ÷ 39 𝟖𝟖𝟖𝟖𝟖𝟖 9. 1,113 ÷ 42 𝟐𝟐𝟐𝟐. 𝟓𝟓 10. 4,082 ÷ 52 𝟕𝟕𝟕𝟕. 𝟓𝟓 A STORY OF RATIOS 153 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Topic E: Expressing Operations in Algebraic Form GRADE 6 • MODULE 4 6 GRADE Mathematics Curriculum Topic E Expressing Operations in Algebraic Form 6.EE.A.2a, 6.EE.A.2b Focus Standards: 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. a. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract 𝑦𝑦 from 5” as 5 −𝑦𝑦. b. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2(8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. Instructional Days: 3 Lesson 15: Read Expressions in Which Letters Stand for Numbers (P)1 Lessons 16–17: Write Expressions in Which Letters Stand for Numbers (M, P) In Topic E, students express mathematical terms in algebraic form. They read and write expressions in which letters stand for numbers. In Lesson 15, students provide word descriptions for operations in an algebraic expression. Given the expression 4𝑏𝑏+ 𝑐𝑐, students assign the operation term product for multiplication and the term sum for addition. They verbalize the expression as “the sum of 𝑐𝑐 and the product of 4 and 𝑏𝑏.” However, in Lessons 16 and 17, students are given verbal expressions, and they write algebraic expressions to record operations with numbers and letters standing for numbers. Provided the verbal expression, “Devin quadrupled his money and deposited it with his mother’s,” students write the expression 4𝑎𝑎+ 𝑏𝑏, where 𝑎𝑎 represents the amount of money Devin originally had and 𝑏𝑏 represents the amount of money his mother has. Or, provided the verbal expression, “Crayons and markers were put together and distributed equally to six tables,” students create the algebraic expression 𝑎𝑎+𝑏𝑏 6 , where 𝑎𝑎 represents the number of crayons and 𝑏𝑏 represents the number of markers. Mastery of reading and writing expressions in this topic leads to a fluent transition in the next topic where students read, write, and evaluate expressions. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 154 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 15 Lesson 15: Read Expressions in Which Letters Stand for Numbers Lesson 15: Read Expressions in Which Letters Stand for Numbers Student Outcomes Students read expressions in which letters stand for numbers. They assign operation terms to operations when reading. Students identify parts of an algebraic expression using mathematical terms for all operations. Classwork Opening Exercise (10 minutes) Opening Exercise Complete the graphic organizer with mathematical words that indicate each operation. Some words may indicate more than one operation. Have different students share the vocabulary words they wrote in each category. If students are missing vocabulary words in their graphic organizers, have them add the new words. At the end of the Opening Exercise, every student should have the same lists of vocabulary words for each operation. Example 1 (13 minutes) Have students write down an expression using words. Encourage students to refer back to the graphic organizer created during the Opening Exercise. After providing students time to write each expression, have different students read each expression out loud. Each student should use different mathematical vocabulary. ADDITION SUBTRACTION MULTIPLICATION DIVISION SUM ADD MORE THAN TOTAL ALTOGETHER IN ALL INCREASED BY PLUS DIFFERENCE SUBTRACT FEWER THAN MINUS LESS THAN HOW MANY MORE LEFT DECREASED BY PRODUCT MULTIPLY TIMES EVERY DOUBLE, TRIPLE OF AS MUCH EACH QUOTIENT DIVIDE EACH PER SPLIT EXPONENTS POWER SQUARED CUBED A STORY OF RATIOS 155 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 15 Lesson 15: Read Expressions in Which Letters Stand for Numbers Example 1 Write an expression using words. a. 𝒂𝒂−𝒃𝒃 Possible answers: 𝒂𝒂 minus 𝒃𝒃; the difference of 𝒂𝒂 and 𝒃𝒃; 𝒂𝒂 decreased by 𝒃𝒃; 𝒃𝒃 subtracted from 𝒂𝒂 b. 𝒙𝒙𝒙𝒙 Possible answers: the product of 𝒙𝒙 and 𝒚𝒚; 𝒙𝒙 multiplied by 𝒚𝒚; 𝒙𝒙 times 𝒚𝒚 c. 𝟒𝟒𝒇𝒇+ 𝒑𝒑 Possible answers: 𝒑𝒑 added to the product of 𝟒𝟒 and f; 𝟒𝟒 times 𝒇𝒇 plus 𝒑𝒑; the sum of 𝟒𝟒 multiplied by 𝒇𝒇 and 𝒑𝒑 d. 𝒅𝒅−𝒃𝒃𝟑𝟑 Possible answers: 𝒅𝒅 minus 𝒃𝒃 cubed; the difference of 𝒅𝒅 and the quantity 𝒃𝒃 to the third power e. 𝟓𝟓(𝒖𝒖−𝟏𝟏𝟏𝟏) + 𝒉𝒉 Possible answers: Add 𝒉𝒉 to the product of 𝟓𝟓 and the difference of 𝒖𝒖 and 𝟏𝟏𝟏𝟏; 𝟓𝟓 times the quantity of 𝒖𝒖 minus 𝟏𝟏𝟏𝟏 added to 𝒉𝒉. f. 𝟑𝟑 𝒅𝒅+𝒇𝒇 Possible answers: Find the quotient of 𝟑𝟑 and the sum of 𝒅𝒅 and 𝒇𝒇; 𝟑𝟑 divided by the quantity 𝒅𝒅 plus 𝒇𝒇. Why is 3 divided by 𝑑𝑑 plus 𝑓𝑓 not a correct answer? Possible answer: 3 divided by 𝑑𝑑 plus 𝑓𝑓 would indicate that we divide 3 and 𝑑𝑑 first and then add 𝑓𝑓, but this is not what the expression is showing. Exercises (12 minutes) Students work with a partner to complete the following problems. Exercises Circle all the vocabulary words that could be used to describe the given expression. 1. 𝟔𝟔𝒉𝒉−𝟏𝟏𝟏𝟏 2. 𝟓𝟓𝒅𝒅 𝟔𝟔 Scaffolding: If students are using the vocabulary words well or finish early, ask students to write two different expressions for Exercises 1–4. ADDITION SUBTRACTION MULTIPLICATION DIVISION SUM DIFFERENCE PRODUCT QUOTIENT MP.6 A STORY OF RATIOS 156 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 15 Lesson 15: Read Expressions in Which Letters Stand for Numbers 3. 𝟓𝟓(𝟐𝟐+ 𝒅𝒅) −𝟖𝟖 4. 𝒂𝒂𝒂𝒂𝒂𝒂 Write an expression using vocabulary words to represent each given expression. 5. 𝟖𝟖−𝟐𝟐𝒈𝒈 Possible answers: 𝟖𝟖 minus the product of 𝟐𝟐 and 𝒈𝒈; 𝟐𝟐 times 𝒈𝒈 subtracted from 𝟖𝟖; 𝟖𝟖 decreased by 𝒈𝒈 doubled 6. 𝟏𝟏𝟏𝟏(𝒂𝒂+ 𝒄𝒄) Possible answers: 𝟏𝟏𝟏𝟏 times the quantity of 𝒂𝒂 increased by 𝒄𝒄; the product of 𝟏𝟏𝟏𝟏 and the sum of 𝒂𝒂 and 𝒄𝒄; 𝟏𝟏𝟏𝟏 multiplied by the total of 𝒂𝒂 and 𝒄𝒄 7. 𝒎𝒎+𝒏𝒏 𝟓𝟓 Possible answers: the sum of 𝒎𝒎 and 𝒏𝒏 divided by 𝟓𝟓; the quotient of the total of 𝒎𝒎 and 𝒏𝒏, and 𝟓𝟓; 𝒎𝒎 plus 𝒏𝒏 split into 𝟓𝟓 equal groups 8. 𝒃𝒃𝟑𝟑−𝟏𝟏𝟏𝟏 Possible answers: 𝒃𝒃 cube𝒅𝒅 minus 𝟏𝟏𝟏𝟏; 𝒃𝒃 to the third power decreased by 𝟏𝟏𝟏𝟏 9. 𝒇𝒇−𝒅𝒅 𝟐𝟐 Possible answers: 𝒇𝒇 minus the quotient of 𝒅𝒅 and 𝟐𝟐; 𝒅𝒅 split into 𝟐𝟐 groups and then subtracted from 𝒇𝒇; 𝒅𝒅 divided by 𝟐𝟐 less than 𝒇𝒇 10. 𝒖𝒖 𝒙𝒙 Possible answers: 𝒖𝒖 divided by 𝒙𝒙; the quotient of 𝒖𝒖 and 𝒙𝒙; 𝒖𝒖 divided into 𝒙𝒙 parts Closing (5 minutes) Peter says the expression 11 −3𝑐𝑐 is 3 times 𝑐𝑐 decreased by 11. Is he correct? Why or why not? Peter is not correct because the expression he wrote is in the wrong order. If Peter wanted to write a correct expression and use the same vocabulary words, he would have to write 11 decreased by 3 times 𝑐𝑐. Exit Ticket (5 minutes) ADD SUBTRACT MULTIPLY DIVIDE MORE THAN LESS THAN TIMES EACH A STORY OF RATIOS 157 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 15 Lesson 15: Read Expressions in Which Letters Stand for Numbers Name Date Lesson 15: Read Expressions in Which Letters Stand for Numbers Exit Ticket 1. Write two word expressions for each problem using different math vocabulary for each expression. a. 5𝑑𝑑−10 b. 𝑎𝑎 𝑏𝑏+2 2. List five different math vocabulary words that could be used to describe each given expression. a. 3(𝑑𝑑−2) + 10 b. 𝑎𝑎𝑎𝑎 𝑐𝑐 A STORY OF RATIOS 158 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 15 Lesson 15: Read Expressions in Which Letters Stand for Numbers Exit Ticket Sample Solutions 1. Write two word expressions for each problem using different math vocabulary for each expression. a. 𝟓𝟓𝟓𝟓−𝟏𝟏𝟏𝟏 Possible answers: the product of 𝟓𝟓 and 𝒅𝒅 minus 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 less than 𝟓𝟓 times 𝒅𝒅 b. 𝒂𝒂 𝒃𝒃+𝟐𝟐 Possible answers: the quotient of 𝒂𝒂 and the quantity of 𝒃𝒃 plus 𝟐𝟐, 𝒂𝒂 divided by the sum of 𝒃𝒃 and 𝟐𝟐 2. List five different math vocabulary words that could be used to describe each given expression. a. 𝟑𝟑(𝒅𝒅−𝟐𝟐) + 𝟏𝟏𝟏𝟏 Possible answers: difference, subtract, product, times, quantity, add, sum b. 𝒂𝒂𝒃𝒃 𝒄𝒄 Possible answers: quotient, divide, split, product, multiply, times, per, each Problem Set Sample Solutions 1. List five different vocabulary words that could be used to describe each given expression. a. 𝒂𝒂−𝒅𝒅+ 𝒄𝒄 Possible answers: sum, add, total, more than, increase, decrease, difference, subtract, less than b. 𝟐𝟐𝟐𝟐−𝟑𝟑𝟑𝟑 Possible answers: difference, subtract, fewer than, triple, times, product c. 𝒃𝒃 𝒅𝒅+𝟐𝟐 Possible answers: quotient, divide, split, per, sum, add, increase, more than 2. Write an expression using math vocabulary for each expression below. a. 𝟓𝟓𝟓𝟓−𝟏𝟏𝟏𝟏 Possible answers: the product of 𝟓𝟓 and 𝒃𝒃 minus 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏 less than 𝟓𝟓 times 𝒃𝒃 b. 𝒏𝒏 𝟐𝟐 Possible answers: the quotient of 𝒏𝒏 and 𝟐𝟐, 𝒏𝒏 split into 𝟐𝟐 equal groups c. 𝒂𝒂+ (𝒅𝒅−𝟔𝟔) Possible answers: 𝒂𝒂 plus the quantity 𝒅𝒅 minus 𝟔𝟔, 𝒂𝒂 increased by the difference of 𝒅𝒅 and 𝟔𝟔 d. 𝟏𝟏𝟏𝟏+ 𝟐𝟐𝒃𝒃 Possible answers: 𝟏𝟏𝟏𝟏 plus twice 𝒃𝒃, the total of 𝟏𝟏𝟏𝟏 and the product of 𝟐𝟐 and 𝒃𝒃 A STORY OF RATIOS 159 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Lesson 16: Write Expressions in Which Letters Stand for Numbers Student Outcomes Students write algebraic expressions that record all operations with numbers and letters standing for the numbers. Lesson Notes In general, key word readings should be avoided. However, at this initial phase, it is important for students to understand the direct relationship between words in a written phrase and their appearance in an algebraic expression. Classwork Opening Exercise (5 minutes) Students underline the key math vocabulary words in each statement. Opening Exercise Underline the key words in each statement. a. The sum of twice 𝒃𝒃 and 𝟓𝟓 The sum of twice 𝒃𝒃 and 𝟓𝟓 b. The quotient of 𝒄𝒄 and 𝒅𝒅 The quotient of 𝒄𝒄 and 𝒅𝒅 c. 𝒂𝒂 raised to the fifth power and then increased by the product of 𝟓𝟓 and 𝒄𝒄 𝒂𝒂 raised to the fifth power and then increased by the product of 𝟓𝟓 and 𝒄𝒄 d. The quantity of 𝒂𝒂 plus 𝒃𝒃 divided by 𝟒𝟒 The quantity of 𝒂𝒂 plus 𝒃𝒃 divided by 𝟒𝟒 e. 𝟏𝟏𝟏𝟏 less than the product of 𝟏𝟏𝟏𝟏 and 𝒄𝒄 𝟏𝟏𝟏𝟏 less than the product of 𝟏𝟏𝟏𝟏 and 𝒄𝒄 f. 𝟓𝟓 times 𝒅𝒅 and then increased by 𝟖𝟖 𝟓𝟓 times 𝒅𝒅 and then increased by 𝟖𝟖 A STORY OF RATIOS 160 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Mathematical Modeling Exercise 1 (10 minutes) Model how to change the expressions given in the Opening Exercise from words to variables and numbers. Mathematical Modeling Exercise 1 a. The sum of twice 𝒃𝒃 and 𝟓𝟓 Underline key words: the sum of twice 𝑏𝑏 and 5. Identify the operations each key word implies. “Sum” indicates addition, and “twice” indicates multiplication by 2. Write an expression. 𝟐𝟐𝟐𝟐+ 𝟓𝟓 b. The quotient of 𝒄𝒄 and 𝒅𝒅 Underline key words: the quotient of 𝑐𝑐 and 𝑑𝑑. Identify the operation the key word implies. “Quotient” implies division. Write an expression. 𝒄𝒄 𝒅𝒅 c. 𝒂𝒂 raised to the fifth power and then increased by the product of 𝟓𝟓 and 𝒄𝒄 Underline key words: 𝑎𝑎 raised to the fifth power and then increased by the product of 5 and 𝑐𝑐. Identify the operations each key word implies. “Power” indicates exponents, “increased” implies addition, and “product” implies multiplication. Write an expression. 𝒂𝒂𝟓𝟓+ 𝟓𝟓𝟓𝟓 d. The quantity of 𝒂𝒂 plus 𝒃𝒃 divided by 𝟒𝟒 Underline key words: the quantity of 𝑎𝑎 plus 𝑏𝑏 divided by 4. Identify the operations each key word implies. “Quantity” indicates parentheses, “plus” indicates addition, and “divided by” implies division. Write an expression. 𝒂𝒂+𝒃𝒃 𝟒𝟒 e. 𝟏𝟏𝟏𝟏 less than the product of 𝟏𝟏𝟏𝟏 and 𝒄𝒄 MP.6 A STORY OF RATIOS 161 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Underline key words: 10 less than the product of 15 and 𝑐𝑐. Identify the operations each key word implies. “Less than” indicates subtraction, and “product” implies multiplication. Write an expression. 𝟏𝟏𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 Would 10 −15𝑐𝑐 also be correct? Why or why not? This expression would not be correct. If the amount of money I have is 10 less than someone else, I would take the money the other person has and subtract the 10. f. 𝟓𝟓 times 𝒅𝒅 and then increased by 𝟖𝟖 Underline key words: 5 times 𝑑𝑑 and then increased by 8. Identify the operations each key word implies. “Times” indicates multiplication, and “increased” implies addition. Write an expression. 𝟓𝟓𝟓𝟓+ 𝟖𝟖 Mathematical Modeling Exercise 2 (10 minutes) Mathematical Modeling Exercise 2 Model how to change each real-world scenario to an expression using variables and numbers. Underline the text to show the key words before writing the expression. Marcus has 𝟒𝟒 more dollars than Yaseen. If 𝒚𝒚 is the amount of money Yaseen has, write an expression to show how much money Marcus has. Underline key words. Marcus has 4 more dollars than Yaseen. If Yaseen had $7, how much money would Marcus have? $11 How did you get that? Added 7 + 4 Write an expression using 𝑦𝑦 for the amount of money Yaseen has. 𝑦𝑦+ 4 Mario is missing half of his assignments. If 𝒂𝒂 represents the number of assignments, write an expression to show how many assignments Mario is missing. MP.6 MP.6 A STORY OF RATIOS 162 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Underline key words. Mario is missing half of his assignments. If Mario was assigned 10 assignments, how many is he missing? 5 How did you get that? 10 ÷ 2 Write an expression using 𝑎𝑎 for the number of assignments Mario was assigned. 𝑎𝑎 2 or 𝑎𝑎÷ 2 Kamilah’s weight has tripled since her first birthday. If 𝒘𝒘 represents the amount Kamilah weighed on her first birthday, write an expression to show how much Kamilah weighs now. Underline key words. Kamilah’s weight has tripled since her first birthday. If Kamilah weighed 20 pounds on her first birthday, how much does she weigh now? 60 pounds How did you get that? Multiplied 3 by 20 Write an expression using 𝑤𝑤 for Kamilah’s weight on her first birthday. 3𝑤𝑤 Nathan brings cupcakes to school and gives them to his five best friends, who share them equally. If 𝒄𝒄 represents the number of cupcakes Nathan brings to school, write an expression to show how many cupcakes each of his friends receive. Underline key words. Nathan brings cupcakes to school and gives them to his five best friends, who share them equally. If Nathan brings 15 cupcakes to school, how many will each friend receive? 3 How did you determine that? 15 ÷ 5 Write an expression using 𝑐𝑐 to represent the number of cupcakes Nathan brings to school. 𝑐𝑐 5 or 𝑐𝑐÷ 5 Mrs. Marcus combines her atlases and dictionaries and then divides them among 𝟏𝟏𝟏𝟏 different tables. If 𝒂𝒂 represents the number of atlases and 𝒅𝒅 represents the number of dictionaries Mrs. Marcus has, write an expression to show how many books would be on each table. Underline key words. Mrs. Marcus combines her atlases and dictionaries and then divides them among 10 different tables. MP.6 A STORY OF RATIOS 163 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers If Mrs. Marcus had 8 atlases and 12 dictionaries, how many books would be at each table? 2 How did you determine that? Added the atlases and dictionaries together and then divided by 10. Write an expression using 𝑎𝑎 for atlases and 𝑑𝑑 for dictionaries to represent how many books each table would receive. 𝑎𝑎+𝑑𝑑 10 or (𝑎𝑎+ 𝑑𝑑) ÷ 10 To improve in basketball, Ivan’s coach told him that he needs to take four times as many free throws and four times as many jump shots every day. If 𝒇𝒇 represents the number of free throws and 𝒋𝒋 represents the number of jump shots Ivan shoots daily, write an expression to show how many shots he will need to take in order to improve in basketball. Underline key words. To improve in basketball, Ivan needs to shoot 4 times more free throws and jump shots daily. If Ivan shoots 5 free throws and 10 jump shots, how many will he need to shoot in order to improve in basketball? 60 How did you determine that? Added the free throws and jump shots together and then multiplied by 4 Write an expression using 𝑓𝑓 for free throws and 𝑗𝑗 for jump shots to represent how many shots Ivan will have to take in order to improve in basketball. 4(𝑓𝑓+ 𝑗𝑗) or 4𝑓𝑓+ 4𝑗𝑗 Exercises (10 minutes) Have students work individually on the following exercises. Exercises Mark the text by underlining key words, and then write an expression using variables and/or numbers for each statement. 1. 𝒃𝒃 decreased by 𝒄𝒄 squared 𝒃𝒃 decreased by 𝒄𝒄 squared 𝒃𝒃−𝒄𝒄𝟐𝟐 2. 𝟐𝟐𝟐𝟐 divided by the product of 𝟐𝟐 and 𝒂𝒂 𝟐𝟐𝟐𝟐 divided by the product of 𝟐𝟐 and 𝒂𝒂 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 or 𝟐𝟐𝟐𝟐÷ (𝟐𝟐𝟐𝟐) MP.6 A STORY OF RATIOS 164 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers 3. 𝟏𝟏𝟏𝟏𝟏𝟏 decreased by the quantity of 𝟔𝟔 plus 𝒃𝒃 𝟏𝟏𝟏𝟏𝟏𝟏 decreased by the quantity of 𝟔𝟔 plus 𝒃𝒃 𝟏𝟏𝟏𝟏𝟏𝟏−(𝟔𝟔+ 𝒃𝒃) 4. The sum of twice 𝒄𝒄 and 𝟏𝟏𝟏𝟏 The sum of twice 𝒄𝒄 and 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏 5. Marlo had $𝟑𝟑𝟑𝟑 but then spent $𝒎𝒎 Mario had $𝟑𝟑𝟑𝟑 but then spent $𝒎𝒎. 𝟑𝟑𝟑𝟑−𝒎𝒎 6. Samantha saved her money and was able to quadruple the original amount, 𝒎𝒎. Samantha saved her money and was able to quadruple the original amount, 𝒎𝒎. 𝟒𝟒𝟒𝟒 7. Veronica increased her grade, 𝒈𝒈, by 𝟒𝟒 points and then doubled it. Veronica increased her grade, 𝒈𝒈, by 𝟒𝟒 points and then doubled it. 𝟐𝟐(𝒈𝒈+ 𝟒𝟒) 8. Adbell had 𝒎𝒎 pieces of candy and ate 𝟓𝟓 of them. Then, he split the remaining candy equally among 𝟒𝟒 friends. Adbell had 𝒎𝒎 pieces of candy and ate 𝟓𝟓 of them. Then, he split the remaining candy equally among 𝟒𝟒 friends. 𝒎𝒎−𝟓𝟓 𝟒𝟒 or (𝒎𝒎−𝟓𝟓) ÷ 𝟒𝟒 9. To find out how much paint is needed, Mr. Jones must square the side length, 𝒔𝒔, of the gate and then subtract 𝟏𝟏𝟏𝟏. To find out how much paint is needed, Mr. Jones must square the side length, 𝒔𝒔, of the gate and then subtract 𝟏𝟏𝟏𝟏. 𝒔𝒔𝟐𝟐−𝟏𝟏𝟏𝟏 10. Luis brought 𝒙𝒙 cans of cola to the party, Faith brought 𝒅𝒅 cans of cola, and De’Shawn brought 𝒉𝒉 cans of cola. How many cans of cola did they bring altogether? Luis brought 𝒙𝒙 cans of cola to the party, Faith brought 𝒅𝒅 cans of cola, and De’Shawn brought 𝒉𝒉 cans of cola. How many cans of cola did they bring altogether? 𝒙𝒙+ 𝒅𝒅+ 𝒉𝒉 A STORY OF RATIOS 165 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Closing (5 minutes) How is writing expressions with variables and numbers similar to writing expressions using words? Possible answers: The same vocabulary words can be used; identifying parts of the expression before writing the expression is helpful. How is writing expressions with variables and numbers different than writing expressions using words? Possible answers: When an expression with words is provided, it is possible that it might be represented mathematically in more than one way. However, when an algebraic expression is written, there can only be one correct answer. Exit Ticket (5 minutes) A STORY OF RATIOS 166 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Name Date Lesson 16: Write Expressions in Which Letters Stand for Numbers Exit Ticket Mark the text by underlining key words, and then write an expression using variables and/or numbers for each of the statements below. 1. Omaya picked 𝑥𝑥 amount of apples, took a break, and then picked 𝑣𝑣 more. Write the expression that models the total number of apples Omaya picked. 2. A number ℎ is tripled and then decreased by 8. 3. Sidney brought 𝑠𝑠 carrots to school and combined them with Jenan’s 𝑗𝑗 carrots. She then splits them equally among 8 friends. 4. 15 less than the quotient of 𝑒𝑒 and 𝑑𝑑 5. Marissa’s hair was 10 inches long, and then she cut ℎ inches. A STORY OF RATIOS 167 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers Exit Ticket Sample Solutions Mark the text by underlining key words, and then write an expression using variables and/or numbers for each of the statements below. 1. Omaya picked 𝒙𝒙 amount of apples, took a break, and then picked 𝒗𝒗 more. Write the expression that models the total number of apples Omaya picked. Omaya picked 𝒙𝒙 amount of apples, took a break, and then picked 𝒗𝒗 more. 𝒙𝒙+ 𝒗𝒗 2. A number 𝒉𝒉 is tripled and then decreased by 𝟖𝟖. A number 𝒉𝒉 is tripled and then decreased by 𝟖𝟖. 𝟑𝟑𝟑𝟑−𝟖𝟖 3. Sidney brought 𝒔𝒔 carrots to school and combined them with Jenan’s 𝒋𝒋 carrots. She then split them equally among 𝟖𝟖 friends. Sidney brought 𝒔𝒔 carrots to school and combined them with Jenan’s 𝒋𝒋 carrots. She then split them equally among 𝟖𝟖 friends. 𝒔𝒔+𝒋𝒋 𝟖𝟖 or (𝒔𝒔+ 𝒋𝒋) ÷ 𝟖𝟖 4. 𝟏𝟏𝟏𝟏 less than the quotient of 𝒆𝒆 and 𝒅𝒅 𝟏𝟏𝟏𝟏 less than the quotient of 𝒆𝒆 and 𝒅𝒅 𝒆𝒆 𝒅𝒅−𝟏𝟏𝟏𝟏 or 𝒆𝒆÷ 𝒅𝒅−𝟏𝟏𝟏𝟏 5. Marissa’s hair was 𝟏𝟏𝟏𝟏 inches long, and then she cut 𝒉𝒉 inches. Marissa’s hair was 𝟏𝟏𝟏𝟏 inches long, and then she cut 𝒉𝒉 inches. 𝟏𝟏𝟏𝟏−𝒉𝒉 Problem Set Sample Solutions Mark the text by underlining key words, and then write an expression using variables and numbers for each of the statements below. 1. Justin can type 𝒘𝒘 words per minute. Melvin can type 𝟒𝟒 times as many words as Justin. Write an expression that represents the rate at which Melvin can type. Justin can type 𝒘𝒘 words per minute. Melvin can type 𝟒𝟒 times as many words as Justin. Write an expression that represents the rate at which Melvin can type. 𝟒𝟒𝟒𝟒 A STORY OF RATIOS 168 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 16 Lesson 16: Write Expressions in Which Letters Stand for Numbers 2. Yohanna swam 𝒚𝒚 yards yesterday. Sheylin swam 𝟓𝟓 yards less than half the amount of yards as Yohanna. Write an expression that represents the number of yards Sheylin swam yesterday. Yohanna swam 𝒚𝒚 yards yesterday. Sheylin swam 𝟓𝟓 yards less than half the amount of yards as Yohanna. Write an expression that represents the number of yards Sheylin swam yesterday. 𝒚𝒚 𝟐𝟐−𝟓𝟓 or 𝒚𝒚÷ 𝟐𝟐−𝟓𝟓 or 𝟏𝟏 𝟐𝟐𝒚𝒚−𝟓𝟓 3. A number 𝒅𝒅 is decreased by 𝟓𝟓 and then doubled. A number 𝒅𝒅 is decreased by 𝟓𝟓 and then doubled. 𝟐𝟐(𝒅𝒅−𝟓𝟓) 4. Nahom had 𝒏𝒏 baseball cards, and Semir had 𝒔𝒔 baseball cards. They combined their baseball cards and then sold 𝟏𝟏𝟏𝟏 of them. Nahom had 𝒏𝒏 baseball cards, and Semir had 𝒔𝒔 baseball cards. They combined their baseball cards and then sold 𝟏𝟏𝟏𝟏 of them. 𝒏𝒏+ 𝒔𝒔−𝟏𝟏𝟏𝟏 5. The sum of 𝟐𝟐𝟐𝟐 and 𝒉𝒉 is divided by 𝒇𝒇 cubed. The sum of 𝟐𝟐𝟐𝟐 and 𝒉𝒉 is divided by 𝒇𝒇 cubed. 𝟐𝟐𝟐𝟐+𝒉𝒉 𝒇𝒇𝟑𝟑 or (𝟐𝟐𝟐𝟐+ 𝒉𝒉) ÷ 𝒇𝒇𝟑𝟑 A STORY OF RATIOS 169 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Lesson 17: Write Expressions in Which Letters Stand for Numbers Student Outcomes Students write algebraic expressions that record all operations with numbers and/or letters standing for the numbers. Lesson Notes Large paper is needed to complete this lesson. Classwork Fluency Exercise (5 minutes): Addition of Decimals Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening (5 minutes) Discuss the Exit Ticket from Lesson 16. Students continue to work on writing expressions, so discuss any common mistakes from the previous lesson. Exercises (25 minutes) Students work in groups of two or three to complete the stations. At each station, students write down the problem and the expression with variables and/or numbers. Encourage students to underline key words in each problem. Exercises Station One: 1. The sum of 𝒂𝒂 and 𝒃𝒃 𝒂𝒂+ 𝒃𝒃 2. Five more than twice a number 𝒄𝒄 𝟓𝟓+ 𝟐𝟐𝟐𝟐 or 𝟐𝟐𝟐𝟐+ 𝟓𝟓 3. Martha bought 𝒅𝒅 number of apples and then ate 𝟔𝟔 of them. 𝒅𝒅−𝟔𝟔 Scaffolding: If students struggled during Lesson 16, complete some examples with students before moving into the exercises. A STORY OF RATIOS 170 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Station Two: 1. 𝟏𝟏𝟏𝟏 decreased by 𝒑𝒑 𝟏𝟏𝟏𝟏−𝒑𝒑 2. The total of 𝒅𝒅 and 𝒇𝒇, divided by 𝟖𝟖 𝒅𝒅+𝒇𝒇 𝟖𝟖 or (𝒅𝒅+ 𝒇𝒇) ÷ 𝟖𝟖 3. Rashod scored 𝟔𝟔 less than 𝟑𝟑 times as many baskets as Mike. Mike scored 𝒃𝒃 baskets. 𝟑𝟑𝟑𝟑−𝟔𝟔 Station Three: 1. The quotient of 𝒄𝒄 and 𝟔𝟔 𝒄𝒄 𝟔𝟔 2. Triple the sum of 𝒙𝒙 and 𝟏𝟏𝟏𝟏 𝟑𝟑(𝒙𝒙+ 𝟏𝟏𝟏𝟏) 3. Gabrielle had 𝒃𝒃 buttons but then lost 𝟔𝟔. Gabrielle took the remaining buttons and split them equally among her 𝟓𝟓 friends. 𝒃𝒃−𝟔𝟔 𝟓𝟓 or (𝒃𝒃−𝟔𝟔) ÷ 𝟓𝟓 Station Four: 1. 𝒅𝒅 doubled 𝟐𝟐𝟐𝟐 2. Three more than 𝟒𝟒 times a number 𝒙𝒙 𝟒𝟒𝟒𝟒+ 𝟑𝟑 or 𝟑𝟑+ 𝟒𝟒𝟒𝟒 3. Mali has 𝒄𝒄 pieces of candy. She doubles the amount of candy she has and then gives away 𝟏𝟏𝟏𝟏 pieces. 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 Station Five: 1. 𝒇𝒇 cubed 𝒇𝒇𝟑𝟑 2. The quantity of 𝟒𝟒 increased by 𝒂𝒂, and then the sum is divided by 𝟗𝟗. 𝟒𝟒+𝒂𝒂 𝟗𝟗 or (𝟒𝟒+ 𝒂𝒂) ÷ 𝟗𝟗 3. Tai earned 𝟒𝟒 points fewer than double Oden’s points. Oden earned 𝒑𝒑 points. 𝟐𝟐𝟐𝟐−𝟒𝟒 A STORY OF RATIOS 171 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Station Six: 1. The difference between 𝒅𝒅 and 𝟖𝟖 𝒅𝒅−𝟖𝟖 2. 𝟔𝟔 less than the sum of 𝒅𝒅 and 𝟗𝟗 (𝒅𝒅+ 𝟗𝟗) −𝟔𝟔 3. Adalyn has 𝒙𝒙 pants and 𝒔𝒔 shirts. She combined them and sold half of them. How many items did Adalyn sell? 𝒙𝒙+𝒔𝒔 𝟐𝟐 or 𝟏𝟏 𝟐𝟐(𝒙𝒙+ 𝒔𝒔) When students reach the final station, they complete the station on larger paper. Students should put all of their work on the top half of the paper. After all students have completed every station, they travel through the stations again to look at the answers provided on the larger paper. Students compare their answers with the answers at the stations and leave feedback on the bottom half of the paper. This may be positive feedback (“I agree with all of your answers” or “Great job”) or critiques (“I think your subtraction is in the incorrect order” or “Why did you write your answer in that order?”). Closing (5 minutes) Discuss feedback that was left on the larger sheets of paper. This answers any questions and provides an opportunity to discuss common mistakes. Is it possible to have more than one correct answer? Why or why not? When writing some of the expressions, it is possible to have more than one correct answer. For example, when writing an expression with addition, the order can be different. Also, we learned how to write division expressions in more than one way. Exit Ticket (5 minutes) MP.3 A STORY OF RATIOS 172 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Name Date Lesson 17: Write Expressions in Which Letters Stand for Numbers Exit Ticket Write an expression using letters and/or numbers for each problem below. 1. 𝑑𝑑 squared 2. A number 𝑥𝑥 increased by 6, and then the sum is doubled. 3. The total of ℎ and 𝑏𝑏 is split into 5 equal groups. 4. Jazmin has increased her $45 by 𝑚𝑚 dollars and then spends a third of the entire amount. 5. Bill has 𝑑𝑑 more than 3 times the number of baseball cards as Frank. Frank has 𝑓𝑓 baseball cards. A STORY OF RATIOS 173 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Exit Ticket Sample Solutions Write an expression using letters and/or numbers for each problem below. 1. 𝒅𝒅 squared 𝒅𝒅𝟐𝟐 2. A number 𝒙𝒙 increased by 𝟔𝟔, and then the sum is doubled. 𝟐𝟐(𝒙𝒙+ 𝟔𝟔) 3. The total of 𝒉𝒉 and 𝒃𝒃 is split into 𝟓𝟓 equal groups. 𝒉𝒉+𝒃𝒃 𝟓𝟓 or (𝒉𝒉+ 𝒃𝒃) ÷ 𝟓𝟓 4. Jazmin has increased her $𝟒𝟒𝟒𝟒 by 𝒎𝒎 dollars and then spends a third of the entire amount. 𝟒𝟒𝟒𝟒+ 𝒎𝒎 𝟑𝟑 or 𝟏𝟏 𝟑𝟑(𝟒𝟒𝟒𝟒+ 𝒎𝒎) 5. Bill has 𝒅𝒅 more than 𝟑𝟑 times the number of baseball cards as Frank. Frank has 𝒇𝒇 baseball cards. 𝟑𝟑𝟑𝟑+ 𝒅𝒅 or 𝒅𝒅+ 𝟑𝟑𝟑𝟑 Problem Set Sample Solutions Write an expression using letters and/or numbers for each problem below. 1. 𝟒𝟒 less than the quantity of 𝟖𝟖 times 𝒏𝒏 𝟖𝟖𝟖𝟖−𝟒𝟒 2. 𝟔𝟔 times the sum of 𝒚𝒚 and 𝟏𝟏𝟏𝟏 𝟔𝟔(𝒚𝒚+ 𝟏𝟏𝟏𝟏) 3. The square of 𝒎𝒎 reduced by 𝟒𝟒𝟒𝟒 𝒎𝒎𝟐𝟐−𝟒𝟒𝟒𝟒 4. The quotient when the quantity of 𝟏𝟏𝟏𝟏 plus 𝒑𝒑 is divided by 𝟖𝟖 𝟏𝟏𝟏𝟏+𝒑𝒑 𝟖𝟖 or (𝟏𝟏𝟏𝟏+ 𝒑𝒑) ÷ 𝟖𝟖 5. Jim earned 𝒋𝒋 in tips, and Steve earned 𝒔𝒔 in tips. They combine their tips and then split them equally. 𝒋𝒋+𝒔𝒔 𝟐𝟐 or (𝒋𝒋+ 𝒔𝒔) ÷ 𝟐𝟐 A STORY OF RATIOS 174 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers 6. Owen has 𝒄𝒄 collector cards. He quadruples the number of cards he has and then combines them with Ian, who has 𝒊𝒊 collector cards. 𝟒𝟒𝟒𝟒+ 𝒊𝒊 7. Rae runs 𝟒𝟒 times as many miles as Madison and Aaliyah combined. Madison runs 𝒎𝒎 miles, and Aaliyah runs 𝒂𝒂 miles. 𝟒𝟒(𝒎𝒎+ 𝒂𝒂) 8. By using coupons, Mary Jo is able to decrease the retail price of her groceries, 𝒈𝒈, by $𝟏𝟏𝟏𝟏𝟏𝟏. 𝒈𝒈−𝟏𝟏𝟏𝟏𝟏𝟏 9. To calculate the area of a triangle, you find the product of the base and height and then divide by 𝟐𝟐. 𝒃𝒃𝒃𝒃 𝟐𝟐 or 𝒃𝒃𝒃𝒃÷ 𝟐𝟐 10. The temperature today was 𝟏𝟏𝟏𝟏 degrees colder than twice yesterday’s temperature, 𝒕𝒕. 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 A STORY OF RATIOS 175 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Addition of Decimals I—Round 1 Directions: Evaluate each expression. 1. 5.1 + 6 23. 3.6 + 2.1 2. 5.1 + 0.6 24. 3.6 + 0.21 3. 5.1 + 0.06 25. 3.6 + 0.021 4. 5.1 + 0.006 26. 0.36 + 0.021 5. 5.1 + 0.0006 27. 0.036 + 0.021 6. 3 + 2.4 28. 1.4 + 42 7. 0.3 + 2.4 29. 1.4 + 4.2 8. 0.03 + 2.4 30. 1.4 + 0.42 9. 0.003 + 2.4 31. 1.4 + 0.042 10. 0.0003 + 2.4 32. 0.14 + 0.042 11. 24 + 0.3 33. 0.014 + 0.042 12. 2 + 0.3 34. 0.8 + 2 13. 0.2 + 0.03 35. 0.8 + 0.2 14. 0.02 + 0.3 36. 0.08 + 0.02 15. 0.2 + 3 37. 0.008 + 0.002 16. 2 + 0.03 38. 6 + 0.4 17. 5 + 0.4 39. 0.6 + 0.4 18. 0.5 + 0.04 40. 0.06 + 0.04 19. 0.05 + 0.4 41. 0.006 + 0.004 20. 0.5 + 4 42. 0.1 + 9 21. 5 + 0.04 43. 0.1 + 0.9 22. 0.5 + 0.4 44. 0.01 + 0.09 Number Correct: Number Correct: A STORY OF RATIOS 176 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Addition of Decimals I—Round 1 [KEY] Directions: Evaluate each expression. 1. 5.1 + 6 𝟏𝟏𝟏𝟏. 𝟏𝟏 23. 3.6 + 2.1 𝟓𝟓. 𝟕𝟕 2. 5.1 + 0.6 𝟓𝟓. 𝟕𝟕 24. 3.6 + 0.21 𝟑𝟑. 𝟖𝟖𝟖𝟖 3. 5.1 + 0.06 𝟓𝟓. 𝟏𝟏𝟏𝟏 25. 3.6 + 0.021 𝟑𝟑. 𝟔𝟔𝟔𝟔𝟔𝟔 4. 5.1 + 0.006 𝟓𝟓. 𝟏𝟏𝟏𝟏𝟏𝟏 26. 0.36 + 0.021 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 5. 5.1 + 0.0006 𝟓𝟓. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 27. 0.036 + 0.021 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 6. 3 + 2.4 𝟓𝟓. 𝟒𝟒 28. 1.4 + 42 𝟒𝟒𝟒𝟒. 𝟒𝟒 7. 0.3 + 2.4 𝟐𝟐. 𝟕𝟕 29. 1.4 + 4.2 𝟓𝟓. 𝟔𝟔 8. 0.03 + 2.4 𝟐𝟐. 𝟒𝟒𝟒𝟒 30. 1.4 + 0.42 𝟏𝟏. 𝟖𝟖𝟖𝟖 9. 0.003 + 2.4 𝟐𝟐. 𝟒𝟒𝟒𝟒𝟒𝟒 31. 1.4 + 0.042 𝟏𝟏. 𝟒𝟒𝟒𝟒𝟒𝟒 10. 0.0003 + 2.4 𝟐𝟐. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 32. 0.14 + 0.042 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏 11. 24 + 0.3 𝟐𝟐𝟐𝟐. 𝟑𝟑 33. 0.014 + 0.042 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 12. 2 + 0.3 𝟐𝟐. 𝟑𝟑 34. 0.8 + 2 𝟐𝟐. 𝟖𝟖 13. 0.2 + 0.03 𝟎𝟎. 𝟐𝟐𝟐𝟐 35. 0.8 + 0.2 𝟏𝟏 14. 0.02 + 0.3 𝟎𝟎. 𝟑𝟑𝟑𝟑 36. 0.08 + 0.02 𝟎𝟎. 𝟏𝟏 15. 0.2 + 3 𝟑𝟑. 𝟐𝟐 37. 0.008 + 0.002 𝟎𝟎. 𝟎𝟎𝟎𝟎 16. 2 + 0.03 𝟐𝟐. 𝟎𝟎𝟎𝟎 38. 6 + 0.4 𝟔𝟔. 𝟒𝟒 17. 5 + 0.4 𝟓𝟓. 𝟒𝟒 39. 0.6 + 0.4 𝟏𝟏 18. 0.5 + 0.04 𝟎𝟎. 𝟓𝟓𝟓𝟓 40. 0.06 + 0.04 𝟎𝟎. 𝟏𝟏 19. 0.05 + 0.4 𝟎𝟎. 𝟒𝟒𝟒𝟒 41. 0.006 + 0.004 𝟎𝟎. 𝟎𝟎𝟎𝟎 20. 0.5 + 4 𝟒𝟒. 𝟓𝟓 42. 0.1 + 9 𝟗𝟗. 𝟏𝟏 21. 5 + 0.04 𝟓𝟓. 𝟎𝟎𝟎𝟎 43. 0.1 + 0.9 𝟏𝟏 22. 0.5 + 0.4 𝟎𝟎. 𝟗𝟗 44. 0.01 + 0.09 𝟎𝟎. 𝟏𝟏 A STORY OF RATIOS 177 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Addition of Decimals I—Round 2 Directions: Evaluate each expression. 1. 3.2 + 5 23. 4.2 + 5.5 2. 3.2 + 0.5 24. 4.2 + 0.55 3. 3.2 + 0.05 25. 4.2 + 0.055 4. 3.2 + 0.005 26. 0.42 + 0.055 5. 3.2 + 0.0005 27. 0.042 + 0.055 6. 4 + 5.3 28. 2.7 + 12 7. 0.4 + 5.3 29. 2.7 + 1.2 8. 0.04 + 5.3 30. 2.7 + 0.12 9. 0.004 + 5.3 31. 2.7 + 0.012 10. 0.0004 + 5.3 32. 0.27 + 0.012 11. 4 + 0.53 33. 0.027 + 0.012 12. 6 + 0.2 34. 0.7 + 3 13. 0.6 + 0.02 35. 0.7 + 0.3 14. 0.06 + 0.2 36. 0.07 + 0.03 15. 0.6 + 2 37. 0.007 + 0.003 16. 2 + 0.06 38. 5 + 0.5 17. 1 + 0.7 39. 0.5 + 0.5 18. 0.1 + 0.07 40. 0.05 + 0.05 19. 0.01 + 0.7 41. 0.005 + 0.005 20. 0.1 + 7 42. 0.2 + 8 21. 1 + 0.07 43. 0.2 + 0.8 22. 0.1 + 0.7 44. 0.02 + 0.08 Number Correct: Improvement: A STORY OF RATIOS 178 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 17 Lesson 17: Write Expressions in Which Letters Stand for Numbers Addition of Decimals I—Round 2 [KEY] Directions: Evaluate each expression. 1. 3.2 + 5 𝟖𝟖. 𝟐𝟐 23. 4.2 + 5.5 𝟗𝟗. 𝟕𝟕 2. 3.2 + 0.5 𝟑𝟑. 𝟕𝟕 24. 4.2 + 0.55 𝟒𝟒. 𝟕𝟕𝟕𝟕 3. 3.2 + 0.05 𝟑𝟑. 𝟐𝟐𝟐𝟐 25. 4.2 + 0.055 𝟒𝟒. 𝟐𝟐𝟐𝟐𝟐𝟐 4. 3.2 + 0.005 𝟑𝟑. 𝟐𝟐𝟐𝟐𝟐𝟐 26. 0.42 + 0.055 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 5. 3.2 + 0.0005 𝟑𝟑. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 27. 0.042 + 0.055 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 6. 4 + 5.3 𝟗𝟗. 𝟑𝟑 28. 2.7 + 12 𝟏𝟏𝟏𝟏. 𝟕𝟕 7. 0.4 + 5.3 𝟓𝟓. 𝟕𝟕 29. 2.7 + 1.2 𝟑𝟑. 𝟗𝟗 8. 0.04 + 5.3 𝟓𝟓. 𝟑𝟑𝟑𝟑 30. 2.7 + 0.12 𝟐𝟐. 𝟖𝟖𝟖𝟖 9. 0.004 + 5.3 𝟓𝟓. 𝟑𝟑𝟑𝟑𝟑𝟑 31. 2.7 + 0.012 𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕 10. 0.0004 + 5.3 𝟓𝟓. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 32. 0.27 + 0.012 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐 11. 4 + 0.53 𝟒𝟒. 𝟓𝟓𝟓𝟓 33. 0.027 + 0.012 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 12. 6 + 0.2 𝟔𝟔. 𝟐𝟐 34. 0.7 + 3 𝟑𝟑. 𝟕𝟕 13. 0.6 + 0.02 𝟎𝟎. 𝟔𝟔𝟔𝟔 35. 0.7 + 0.3 𝟏𝟏 14. 0.06 + 0.2 𝟎𝟎. 𝟐𝟐𝟐𝟐 36. 0.07 + 0.03 𝟎𝟎. 𝟏𝟏 15. 0.6 + 2 𝟐𝟐. 𝟔𝟔 37. 0.007 + 0.003 𝟎𝟎. 𝟎𝟎𝟎𝟎 16. 2 + 0.06 𝟐𝟐. 𝟎𝟎𝟎𝟎 38. 5 + 0.5 𝟓𝟓. 𝟓𝟓 17. 1 + 0.7 𝟏𝟏. 𝟕𝟕 39. 0.5 + 0.5 𝟏𝟏 18. 0.1 + 0.07 𝟎𝟎. 𝟏𝟏𝟏𝟏 40. 0.05 + 0.05 𝟎𝟎. 𝟏𝟏 19. 0.01 + 0.7 𝟎𝟎. 𝟕𝟕𝟕𝟕 41. 0.005 + 0.005 𝟎𝟎. 𝟎𝟎𝟎𝟎 20. 0.1 + 7 𝟕𝟕. 𝟏𝟏 42. 0.2 + 8 𝟖𝟖. 𝟐𝟐 21. 1 + 0.07 𝟏𝟏. 𝟎𝟎𝟎𝟎 43. 0.2 + 0.8 𝟏𝟏 22. 0.1 + 0.7 𝟎𝟎. 𝟖𝟖 44. 0.02 + 0.08 𝟎𝟎. 𝟏𝟏 A STORY OF RATIOS 179 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task Name Date 1. Yolanda is planning out her vegetable garden. She decides that her garden will be square. Below are possible sizes of the garden she will create. a. Complete the table by continuing the pattern. Side Length 𝟏𝟏 foot 𝟐𝟐 feet 𝟑𝟑 feet 𝟒𝟒 feet 𝟓𝟓 feet 𝒙𝒙 feet Notation 12 = 1 ∙1 = 1 Formula 𝐴𝐴= 𝑙𝑙∙𝑤𝑤 𝐴𝐴= 1 ft ∙1 ft 𝐴𝐴= 12 ft2 𝐴𝐴= 1 ft2 Representation b. Yolanda decides the length of her square vegetable garden will be 17 ft. She calculates that the area of the garden is 34 ft2. Determine if Yolanda’s calculation is correct. Explain. A STORY OF RATIOS 180 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 2. Yolanda creates garden cubes to plant flowers. She will fill the cubes with soil and needs to know the amount of soil that will fill each garden cube. The volume of a cube is determined by the following formula: 𝑉𝑉= 𝑠𝑠3, where 𝑠𝑠 represents the side length. a. Represent the volume, in cubic inches, of the garden cube above using a numerical expression. b. Evaluate the expression to determine the volume of the garden cube and the amount of soil, in cubic inches, she will need for each cube. 3. Explain why ቀ1 2ቁ 4 = 1 16. 32 inches A STORY OF RATIOS 181 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 4. Yolanda is building a patio in her backyard. She is interested in using both brick and wood for the flooring of the patio. Below is the plan she has created for the patio. All measurements are in feet. a. Create an expression to represent the area of the patio. b. Yolanda’s husband develops another plan for the patio because he prefers the patio to be much wider than Yolanda’s plan. Determine the length of the brick section and the length of the wood section. Then, use the dimensions to write an expression that represents the area of the entire patio. 5. The landscaper hired for Yolanda’s lawn suggests a patio that has the same measure of wood as it has brick. a. Express the perimeter of the patio in terms of 𝑥𝑥, first using addition and then using multiplication. b. Use substitution to determine if your expressions are equivalent. Explain. brick wood 𝟏𝟏𝟏𝟏. 𝟓𝟓 𝟑𝟑 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒𝟒𝟒 𝟗𝟗𝟗𝟗 𝟐𝟐𝟐𝟐 A STORY OF RATIOS 182 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 6. Elena and Jorge have similar problems and find the same answer. Each determines that the solution to the problem is 24. Elena: (14 + 42) ÷ 7 + 42 Jorge: 14 + (42 ÷ 7) + 42 a. Evaluate each expression to determine if both Elena and Jorge are correct. b. Why would each find the solution of 24? What mistakes were made, if any? 7. Jackson gave Lena this expression to evaluate: 14(8 + 12). Lena said that to evaluate the expression was simple; just multiply the factors 14 and 20. Jackson told Lena she was wrong. He solved it by finding the product of 14 and 8 and then adding that to the product of 14 and 12. a. Evaluate the expression using each student’s method. Lena’s Method Jackson’s Method b. Who was right in this discussion? Why? A STORY OF RATIOS 183 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task A Progression Toward Mastery Assessment Task Item STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem. STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem. STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, OR an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem. STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem. 1 a 6.EE.A.1 6.EE.A.2 Student completes the table with fewer than 10 cells correct. An entire row may be incorrect, indicating that incomplete understanding is still present. Student completes the table with at least 10 of the 15 cells correct. One or more of the three rows in the last column are incorrect, indicating the student does not understand the general form of the notation, formula, and representation. Student completes the table with at least 13 of the 15 cells correct. All three rows in the last column are correct, indicating the student understands the general form of the notation, formula, and representation. Student completes the table without error. All notations, formulas, and representations are correct. Please note student exemplar below. b 6.EE.A.1 6.EE.A.2 Student states that Yolanda was correct in her calculation or states that Yolanda was incorrect but offers no explanation. Student states that Yolanda was incorrect but offers an incomplete analysis of the error. Student states that Yolanda was incorrect and that Yolanda calculated 17 · 2. Student states that Yolanda was incorrect and that Yolanda found 17 · 2, the base times the exponent. AND Student calculates the correct area: 172 ft2 = 289 ft2. A STORY OF RATIOS 184 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 2 a 6.EE.A.2 Student does not write a numerical expression or writes an expression unrelated to the problem. Student writes a numerical expression that relates the volume and side length, but the student makes an error, such as using 2 as an exponent. Student writes an equation that correctly represents the data, 𝑉𝑉= 323 or 𝑉𝑉= 32 ∙32 ∙32, instead of a numerical expression. Student correctly writes the numerical expression for the volume of the cube: 323, or 32 ∙32 ∙ 32. b 6.EE.A.1 Student does not attempt to evaluate the expression or has no expression from part (a) to evaluate. Student attempts to evaluate the expression but makes an arithmetic error. Student correctly evaluates the expression and finds 𝑉𝑉= 32768. The unit of volume, in3, is missing. OR Student correctly evaluates an incorrect expression from part (a). Student correctly evaluates the expression and uses the correct unit. Student gives the answer 𝑉𝑉= 32,768 in3. 3 6.EE.A.1 Student does not demonstrate understanding of exponential notation. One example would be adding 1 2 four times. Student makes a common error, such as ቀ1 2ቁ 4 = 4 2 or ቀ1 2ቁ 4 = 1 8 . Student shows that ቀ1 2ቁ 4 = 1 2 ∙1 2 ∙1 2 ∙1 2 but makes an arithmetic error and arrives at an answer other than 1 16. Student shows that ቀ1 2ቁ 4 = 1 2 ∙1 2 ∙1 2 ∙1 2 = 1 16. 4 a 6.EE.A.3 Student does not write an expression or does not indicate an understanding of 𝐴𝐴= 𝑙𝑙∙𝑤𝑤. Student writes an expression relating the width (12.5) to only one part of the length (3 or 𝑥𝑥). Student writes the expression incorrectly, without parentheses: 12.5 ⋅3 + 𝑥𝑥, but includes each term needed to find the area. Student writes the correct expression: 12.5(3 + 𝑥𝑥) or 37.5 + 12.5𝑥𝑥. b 6.EE.A.3 Student does not write an expression or does not indicate an understanding of 𝐴𝐴= 𝑙𝑙∙𝑤𝑤. Student writes an expression using the width, 24 feet, but does not calculate the length, 2𝑥𝑥+ 4, correctly. Student writes the correct expression, 24(2𝑥𝑥+ 4). Student writes the correct expression, 24(2𝑥𝑥+ 4), and identifies the width, 24 feet, and the length, 2𝑥𝑥+ 4 feet. 5 a 6.EE.A.3 Student does not express the perimeter of the figure in terms of 𝑥𝑥, using neither addition nor multiplication. Student expresses the perimeter of the figure in terms of 𝑥𝑥, using either addition or multiplication but not both. Student expresses the perimeter of the figure in terms of 𝑥𝑥, using addition and multiplication but makes an error in one of the expressions. Student expresses the perimeter of the figure as: 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥+ 2𝑥𝑥+ 2𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥+ 𝑥𝑥 (or uses any other order of addends that is equivalent) and writes the expression 10𝑥𝑥. A STORY OF RATIOS 185 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task b 6.EE.A.4 Student states that the expressions are not equivalent, but student does not use substitution and offers no explanation. Student states that the expressions are equivalent, but student does not use substitution and offers no explanation. Student substitutes a value for 𝑥𝑥 in both equations but makes one or more arithmetic mistakes and claims that the two expressions are not equivalent. Student substitutes any value for 𝑥𝑥 into both the addition and multiplication expression, calculates them accurately, and finds them equivalent. 6 a 6.EE.A.1 6.EE.A.2 Student evaluates both expressions incorrectly. Errors are both in order of operations and arithmetic. Student evaluates one expression correctly and one incorrectly. Errors are due to lack of application of order of operation rules. Student follows the correct order of operations on both expressions but fails to compute the exponents correctly on one or both expressions. Student evaluates both expressions accurately, applying the rules of order of operations correctly. Elena’s answer is 24, and Jorge’s answer is 36. b 6.EE.A.1 6.EE.A.2 Student offers no credible reason why both Elena and Jorge would arrive at the answer, 24. Jorge’s mistakes are not identified. Student shows partial understanding of order of operations but is unable to find or describe Jorge’s mistake. Student may have an incomplete understanding of exponents. Student finds that Elena followed the order of operation rules correctly. Jorge’s mistake is noted, but it is not described in detail. Student finds that Elena followed the order of operation rules correctly. Also, Jorge’s mistake is identified: Jorge did not evaluate the operation inside the parentheses first. Instead, he added 14 + 42 first, arriving at a sum of 56. He then divided 56 by 7 to get 8, added 42 to 8, and arrived at a final answer of 24. 7 a 6.EE.A.3 Student evaluates neither Lena’s nor Jackson’s methods accurately. Student evaluates both expressions using the same method. Student evaluates either Lena’s or Jackson’s methods accurately. The other is evaluated inaccurately. Student evaluates both Lena’s and Jackson’s methods accurately. A STORY OF RATIOS 186 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task b 6.EE.A.2 6.EE.A.3 Student claims both Lena and Jackson are incorrect. Evidence is missing or lacking. Student chooses either Lena or Jackson as being correct, implying that the other is wrong. Evidence is not fully articulated. Student indicates that methods used by both Lena and Jackson are correct. Both methods are described. No mention of the distributive property is made. Student indicates that methods used by both Lena and Jackson are correct. Student claims Lena followed the order of operations by adding 8 + 12 first because they were contained in parentheses, and then Lena multiplied the sum (20) by 14 to arrive at a product of 280. Student also identifies Jackson’s method as an application of the distributive property. Partial products of 14(8) = 112 and 14(12) = 168 are found first and then added to arrive at 280. A STORY OF RATIOS 187 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task Name Date 1. Yolanda is planning out her vegetable garden. She decides that her garden will be square. Below are possible sizes of the garden she will create. a. Complete the table by continuing the pattern. Side Length 𝟏𝟏 foot 𝟐𝟐 feet 𝟑𝟑 feet 𝟒𝟒 feet 𝟓𝟓 feet 𝒙𝒙 feet Notation 12 = 1 ∙1 = 1 Formula 𝐴𝐴 = 𝑙𝑙∙𝑤𝑤 𝐴𝐴 = 1 ft ∙1 ft 𝐴𝐴= 12 ft2 𝐴𝐴 = 1 ft2 Representation b. Yolanda decides the length of her square vegetable garden will be 17 ft. She calculates that the area of the garden is 34 ft2. Determine if Yolanda’s calculation is correct. Explain. A STORY OF RATIOS 188 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 2. Yolanda creates garden cubes to plant flowers. She will fill the cubes with soil and needs to know the amount of soil that will fill each garden cube. The volume of a cube is determined by the following formula: 𝑉𝑉= 𝑠𝑠3, where 𝑠𝑠 equals the side length. a. Represent the volume, in cubic inches, of the garden cube above using a numerical expression. b. Evaluate the expression to determine the volume of the garden cube and the amount of soil, in cubic inches, she will need for each cube. 3. Explain why ቀ1 2ቁ 4 = 1 16. 32 inches A STORY OF RATIOS 189 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 4. Yolanda is building a patio in her backyard. She is interested in using both brick and wood for the flooring of the patio. Below is the plan she has created for the patio. All measurements are in feet. a. Create an expression to represent the area of the patio. b. Yolanda’s husband develops another plan for the patio because he prefers the patio to be much wider than Yolanda’s plan. Determine the length of the brick section and the length of the wood section. Then, use the dimensions to write an expression that represents the area of the entire patio. 5. The landscaper hired for Yolanda’s lawn suggests a patio that has the same measure of wood as it has brick. a. Express the perimeter of the patio in terms of 𝑥𝑥, first using addition and then using multiplication. b. Use substitution to determine if your expressions are equivalent. Explain. brick wood 12.5 3 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 𝑥𝑥 2𝑥𝑥 2𝑥𝑥 48𝑥𝑥 96 24 A STORY OF RATIOS 190 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Module 4: Expressions and Equations 6•4 Mid-Module Assessment Task 6. Elena and Jorge have similar problems and find the same answer. Each determines that the solution to the problem is 24. Elena: (14 + 42) ÷ 7 + 42 Jorge: 14 + (42 ÷ 7) + 42 a. Evaluate each expression to determine if both Elena and Jorge are correct. b. Why would each find the solution of 24? What mistakes were made, if any? 7. Jackson gave Lena this expression to evaluate: 14(8 + 12). Lena said that to evaluate the expression was simple; just multiply the factors 14 and 20. Jackson told Lena she was wrong. He solved it by finding the product of 14 and 8 and then adding that to the product of 14 and 12. a. Evaluate the expression using each student’s method. Lena’s Method Jackson’s Method b. Who was right in this discussion? Why? A STORY OF RATIOS 191 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 GRADE 6 • MODULE 4 Topic F: Writing and Evaluating Expressions and Formulas 6 GRADE Mathematics Curriculum Topic F Writing and Evaluating Expressions and Formulas 6.EE.A.2a, 6.EE.A.2c, 6.EE.B.6 Focus Standards: 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. a. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract 𝑦𝑦 from 5” as 5 −𝑦𝑦. b. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas 𝑉𝑉= 𝑠𝑠3 and 𝐴𝐴= 6𝑠𝑠2 to find the volume and surface area of a cube with sides of length 𝑠𝑠= 1/2. 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specific set. Instructional Days: 5 Lesson 18: Writing and Evaluating Expressions—Addition and Subtraction (P)1 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions (P) Lesson 20: Writing and Evaluating Expressions—Multiplication and Division (P) Lesson 21: Writing and Evaluating Expressions—Multiplication and Addition (M) Lesson 22: Writing and Evaluating Expressions—Exponents (P) 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 192 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic F Topic F: Writing and Evaluating Expressions and Formulas In Topic F, students demonstrate their knowledge of expressions from previous topics in order to write and evaluate expressions and formulas. Students bridge their understanding of reading and writing expressions to substituting values in order to evaluate expressions. In Lesson 18, students use variables to write expressions involving addition and subtraction from real-world problems. They evaluate those expressions when they are given the value of the variable. For example, given the problem “Quentin has two more dollars than his sister, Juanita,” students determine the variable to represent the unknown. In this case, students let 𝑥𝑥 represent Juanita’s money, in dollars. Since Quentin has two more dollars than Juanita, students represent his quantity as 𝑥𝑥+ 2. Now students can substitute given values for the variable to determine the amount of money Quentin and Juanita each have. If Juanita has fourteen dollars, students substitute the 𝑥𝑥 with the amount, 14, and evaluate the expression: 𝑥𝑥+ 2. 14 + 2 16 Here, students determine that the amount of money Quentin has is 16 dollars because 16 is two more than the 14 dollars Juanita has. In Lesson 19, students develop expressions involving addition and subtraction from real-world problems. They use tables to organize the information provided and evaluate expressions for given values. They continue to Lesson 20 where they develop expressions again, this time focusing on multiplication and division from real-world problems. Students bridge their study of the relationships between operations from Topic A to further develop and evaluate expressions in Lesson 21, focusing on multiplication and addition in real-world contexts. Building from their previous experiences in this topic, students create formulas in Lesson 22 by setting expressions equal to another variable. Students assume, for example, that there are 𝑝𝑝 peanuts in a bag. There are three bags and four extra peanuts altogether. Students express the total number of peanuts in terms of 𝑝𝑝: 3𝑝𝑝+ 4. Students let 𝑡𝑡 be the total number of peanuts and determine a formula that expresses the relationship between the number of peanuts in a bag and the total number of peanuts, 𝑡𝑡= 3𝑝𝑝+ 4. From there, students are provided a value for 𝑝𝑝, which they substitute into the formula: If 𝑝𝑝= 10, then 3(10) + 4 = 30 + 4 = 34, and they determine that there are 34 peanuts. In the final lesson of the topic, students evaluate formulas involving exponents for given values in real-world problems. A STORY OF RATIOS 193 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Student Outcomes Students use variables to write expressions involving addition and subtraction from real-world problems. Students evaluate these expressions when given the value of the variable. Lesson Notes When students write expressions, make sure they are as specific as possible. Students should understand the importance of specifying units when defining letters. For example, students should say, “Let 𝐾𝐾 represent Karolyn’s weight in pounds” instead of “Let 𝐾𝐾 represent Karolyn’s weight” because weight is not a number until it is specified by pounds, ounces, grams, and so on. They also must be taught that it is inaccurate to define 𝐾𝐾 as Karolyn because Karolyn is not a number. Students conclude that in word problems, each letter represents a number, and its meaning must be clearly stated. Classwork Opening Exercise (4 minutes) Opening Exercise How can we show a number increased by 𝟐𝟐? 𝒂𝒂+ 𝟐𝟐 or 𝟐𝟐+ 𝒂𝒂 Can you prove this using a model? Yes. I can use a tape diagram. Discussion (5 minutes) In this lesson, you connect real-world problems to addition and subtraction expressions. What story problem could you make up to go along with the expression 𝑎𝑎+ 2? Allow a few moments for students to form realistic scenarios. As students share these, critique them. Answers will vary. Ronnie has some apples, but Gayle has two more apples than Ronnie. How many apples does Gayle have? 𝒂𝒂 𝟐𝟐 𝒂𝒂 𝟐𝟐 or A STORY OF RATIOS 194 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Example 1 (1 minute): The Importance of Being Specific in Naming Variables Example 1: The Importance of Being Specific in Naming Variables When naming variables in expressions, it is important to be very clear about what they represent. The units of measure must be included if something is measured. Exercises 1–2 (5 minutes) Ask students to read the variables listed in the table and correct them for specificity. Exercises 1–2 1. Read the variable in the table, and improve the description given, making it more specific. Answers may vary because students may choose a different unit. Variable Incomplete Description Complete Description with Units Joshua’s speed (𝑱𝑱) Let 𝑱𝑱 represent Joshua’s speed. Let 𝑱𝑱 represent Joshua’s speed in meters per second. Rufus’s height (𝑹𝑹) Let 𝑹𝑹 represent Rufus’s height. Let 𝑹𝑹 represent Rufus’s height in centimeters. Milk sold (𝑴𝑴) Let 𝑴𝑴 represent the amount of milk sold. Let 𝑴𝑴 represent the amount of milk sold in gallons. Colleen’s time in the 𝟒𝟒𝟒𝟒-meter hurdles (𝑪𝑪) Let 𝑪𝑪 represent Colleen’s time. Let 𝑪𝑪 represent Colleen’s time in seconds. Sean’s age (𝑺𝑺) Let 𝑺𝑺 represent Sean’s age. Let 𝑺𝑺 represent Sean’s age in years. Again, when naming variables in expressions, it is important to be very clear about what they represent. When a variable represents a quantity of items, this too must be specified. Review the concept of speed from the above table. Recall from Module 1 that speed is a rate. Emphasize that there are two different units needed to express speed (meters per second in the above example). 2. Read each variable in the table, and improve the description given, making it more specific. Variable Incomplete Description Complete Description with Units Karolyn’s CDs (𝑲𝑲) Let 𝑲𝑲 represent Karolyn’s CDs. Let 𝑲𝑲 represent the number of CDs Karolyn has. Joshua’s merit badges (𝑱𝑱) Let 𝑱𝑱 represent Joshua’s merit badges. Let 𝑱𝑱 represent the number of merit badges Joshua has earned. Rufus’s trading cards (𝑹𝑹) Let 𝑹𝑹 representRufus’s trading cards. Let 𝑹𝑹 represent the number of trading cards in Rufus’s collection. Milk money (𝑴𝑴) Let 𝑴𝑴 represent the amount of milk money. Let 𝑴𝑴 represent the amount of milk money collected in dollars. A STORY OF RATIOS 195 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Example 2 (17 minutes): Writing and Evaluating Addition and Subtraction Expressions Read the following story descriptions, and write an addition or subtraction expression for each one in the table. Example 2: Writing and Evaluating Addition and Subtraction Expressions Read each story problem. Identify the unknown quantity, and write the addition or subtraction expression that is described. Finally, evaluate your expression using the information given in column four. Story Problem Description with Units Expression Evaluate the Expression If: Show Your Work and Evaluate Gregg has two more dollars than his brother Jeff. Write an expression for the amount of money Gregg has. Let 𝒋𝒋 represent Jeff’s money in dollars. 𝒋𝒋+ 𝟐𝟐 Jeff has $𝟏𝟏𝟏𝟏. 𝒋𝒋+ 𝟐𝟐 𝟏𝟏𝟏𝟏+ 𝟐𝟐 𝟏𝟏𝟏𝟏 Gregg has $𝟏𝟏𝟏𝟏. Gregg has two more dollars than his brother Jeff. Write an expression for the amount of money Jeff has. Let 𝒈𝒈 represent Gregg’s money in dollars. 𝒈𝒈−𝟐𝟐 Gregg has $𝟏𝟏𝟏𝟏. 𝒈𝒈−𝟐𝟐 𝟏𝟏𝟏𝟏−𝟐𝟐 𝟏𝟏𝟏𝟏 Jeff has $𝟏𝟏𝟏𝟏. Abby read 𝟖𝟖 more books than Kristen in the first marking period. Write an expression for the number of books Abby read. Let 𝒌𝒌 represent the number of books Kristen read in the first marking period. 𝒌𝒌+ 𝟖𝟖 Kristen read 𝟗𝟗 books in the first marking period. 𝒌𝒌+ 𝟖𝟖 𝟗𝟗+ 𝟖𝟖 𝟏𝟏𝟏𝟏 Abby read 𝟏𝟏𝟏𝟏 books in the first marking period. Abby read 𝟔𝟔 more books than Kristen in the second marking period. Write an expression for the number of books Kristen read. Let 𝒂𝒂 represent the number of books Abby read in the second marking period. 𝒂𝒂−𝟔𝟔 Abby read 𝟐𝟐𝟐𝟐 books in the second marking period. 𝒂𝒂−𝟔𝟔 𝟐𝟐𝟐𝟐−𝟔𝟔 𝟏𝟏𝟏𝟏 Kristen read 𝟏𝟏𝟏𝟏 books in the second marking period. Daryl has been teaching for one year longer than Julie. Write an expression for the number of years that Daryl has been teaching. Let 𝒋𝒋 represent the number of years Julie has been teaching. 𝒋𝒋+ 𝟏𝟏 Julie has been teaching for 𝟐𝟐𝟐𝟐 years. 𝒋𝒋+ 𝟏𝟏 𝟐𝟐𝟐𝟐+ 𝟏𝟏 𝟐𝟐𝟐𝟐 Daryl has been teaching for 𝟐𝟐𝟐𝟐 years. Ian scored 𝟒𝟒 fewer goals than Julia in the first half of the season. Write an expression for the number of goals Ian scored. Let 𝒋𝒋 represent the number of goals scored by Julia. 𝒋𝒋−𝟒𝟒 Julia scored 𝟏𝟏𝟏𝟏 goals. 𝒋𝒋−𝟒𝟒 𝟏𝟏𝟏𝟏−𝟒𝟒 𝟗𝟗 Ian scored 𝟗𝟗 goals in the first half of the season. Ian scored 𝟑𝟑 fewer goals than Julia in the second half of the season. Write an expression for the number of goals Julia scored. Let 𝒊𝒊 represent the number of goals scored by Ian. 𝒊𝒊+ 𝟑𝟑 Ian scored 𝟖𝟖 goals. 𝒊𝒊+ 𝟑𝟑 𝟖𝟖+ 𝟑𝟑 𝟏𝟏𝟏𝟏 Julia scored 𝟏𝟏𝟏𝟏 goals in the second half of the season. Johann visited Niagara Falls 𝟑𝟑 times fewer than Arthur. Write an expression for the number of times Johann visited Niagara Falls. Let 𝒇𝒇 represent the number of times Arthur visited Niagara Falls. 𝒇𝒇−𝟑𝟑 Arthur visited Niagara Falls 𝟓𝟓 times. 𝒇𝒇−𝟑𝟑 𝟓𝟓−𝟑𝟑 𝟐𝟐 Johann visited Niagara Falls twice. A STORY OF RATIOS 196 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Closing (5 minutes) Why is it important to describe the variable in an expression? The biggest reason to define the variable is to know what the expression represents. If something is measured, include units. If something is counted, include that it is a number of items. How do you determine if an expression will be an addition expression or a subtraction expression? In the first problem in the table on the previous page, if we define 𝑥𝑥 as the amount of money that Jeff has, then we would write an expression for the amount of money that Gregg has as 𝑥𝑥+ 2. However, if we define the variable to be the amount of money that Gregg has, then we would write an expression to represent the amount of money that Jeff has as 𝑥𝑥−2. Since the story problem represents a relationship between two quantities, both expressions are equally relevant. Exit Ticket (5 minutes) A STORY OF RATIOS 197 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Name Date Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Exit Ticket Kathleen lost a tooth today. Now she has lost 4 more than her sister Cara lost. 1. Write an expression to represent the number of teeth Cara has lost. Let 𝐾𝐾 represent the number of teeth Kathleen lost. Expression: 2. Write an expression to represent the number of teeth Kathleen has lost. Let 𝐶𝐶 represent the number of teeth Cara lost. Expression: 3. If Cara lost 3 teeth, how many teeth has Kathleen lost? A STORY OF RATIOS 198 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 18 Lesson 18: Writing and Evaluating Expressions―Addition and Subtraction Exit Ticket Sample Solutions Kathleen lost a tooth today. Now she has lost 𝟒𝟒 more than her sister Cara lost. 1. Write an expression to represent the number of teeth Cara has lost. Let 𝑲𝑲 represent the number of teeth Kathleen lost. Expression: 𝑲𝑲−𝟒𝟒 2. Write an expression to represent the number of teeth Kathleen lost. Let 𝑪𝑪 represent the number of teeth Cara lost. Expression: 𝑪𝑪+ 𝟒𝟒 3. If Cara lost 𝟑𝟑 teeth, how many teeth has Kathleen lost? 𝑪𝑪+ 𝟒𝟒; 𝟑𝟑+ 𝟒𝟒; Kathleen has lost 𝟕𝟕 teeth. Problem Set Sample Solutions 1. Read each story problem. Identify the unknown quantity, and write the addition or subtraction expression that is described. Finally, evaluate your expression using the information given in column four. Sample answers are shown. An additional expression can be written for each. Story Problem Description with Units Expression Evaluate the Expression If: Show Your Work and Evaluate Sammy has two more baseballs than his brother Ethan. Let 𝒆𝒆 represent the number of balls Ethan has. 𝒆𝒆+ 𝟐𝟐 Ethan has 𝟕𝟕 baseballs. 𝒆𝒆+ 𝟐𝟐 𝟕𝟕+ 𝟐𝟐 𝟗𝟗 Sammy has 𝟗𝟗 baseballs. Ella wrote 𝟖𝟖 more stories than Anna in the fifth grade. Let 𝒔𝒔 represent the number of stories Anna wrote in the fifth grade. 𝒔𝒔+ 𝟖𝟖 Anna wrote 𝟏𝟏𝟏𝟏 stories in the fifth grade. 𝒔𝒔+ 𝟖𝟖 𝟏𝟏𝟏𝟏+ 𝟖𝟖 𝟏𝟏𝟏𝟏 Ella wrote 𝟏𝟏𝟏𝟏 stories in the fifth grade. Lisa has been dancing for 𝟑𝟑 more years than Danika. Let 𝒚𝒚 represent the number of years Danika has been dancing. 𝒚𝒚+ 𝟑𝟑 Danika has been dancing for 𝟔𝟔 years. 𝒚𝒚+ 𝟑𝟑 𝟔𝟔+ 𝟑𝟑 𝟗𝟗 Lisa has been dancing for 𝟗𝟗 years. The New York Rangers scored 𝟐𝟐 fewer goals than the Buffalo Sabres last night. Let 𝒈𝒈 represent the number of goals scored by the Rangers. 𝒈𝒈+ 𝟐𝟐 The Rangers scored 𝟑𝟑 goals last night. 𝒈𝒈+ 𝟐𝟐 𝟑𝟑+ 𝟐𝟐 𝟓𝟓 The Buffalo Sabres scored 𝟓𝟓 goals last night. George has gone camping 𝟑𝟑 times fewer than Dave. Let 𝒄𝒄 represent the number of times George has gone camping. 𝒄𝒄+ 𝟑𝟑 George has gone camping 𝟖𝟖 times. 𝒄𝒄+ 𝟑𝟑 𝟖𝟖+ 𝟑𝟑 𝟏𝟏𝟏𝟏 Dave has gone camping 𝟏𝟏𝟏𝟏 times. 2. If George went camping 𝟏𝟏𝟏𝟏 times, how could you figure out how many times Dave went camping? Adding 𝟑𝟑 to George’s camping trip total (𝟏𝟏𝟏𝟏) would yield an answer of 𝟏𝟏𝟏𝟏 trips for Dave. A STORY OF RATIOS 199 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Student Outcomes Students develop expressions involving addition and subtraction from real-world problems. Students evaluate these expressions for given values. Lesson Notes In this lesson, students begin by filling in data tables to help them organize data and see patterns; then, they move to drawing their own tables. Encourage students to label the columns completely. Classwork Fluency Exercise (5 minutes): Subtraction of Decimals Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening Exercise (3 minutes) Opening Exercise My older sister is exactly two years older than I am. Sharing a birthday is both fun and annoying. Every year on our birthday, we have a party, which is fun, but she always brags that she is two years older than I am, which is annoying. Shown below is a table of our ages, starting when I was born: My Age (in years) My Sister’s Age (in years) 𝟎𝟎 𝟐𝟐 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟒𝟒 𝟑𝟑 𝟓𝟓 𝟒𝟒 𝟔𝟔 Make sure students understand the context of the story problem. It should be clear that the day I was born was my sister’s second birthday. My first birthday was her third birthday; my second birthday was her fourth birthday; and so on. Scaffolding: Some students benefit from having blank tables prepared ahead of time. A STORY OF RATIOS 200 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Discussion (5 minutes) Today in class, we will use data tables. They help us organize data and see patterns. We can use variables to make generalizations about the patterns we see. a. Looking at the table, what patterns do you see? Tell a partner. My sister’s age is always two years more than my age. b. On the day I turned 𝟖𝟖 years old, how old was my sister? 𝟏𝟏𝟏𝟏 years old c. How do you know? Since my sister’s age is always two years more than my age, we just add 𝟐𝟐 to my age. 𝟖𝟖+ 𝟐𝟐= 𝟏𝟏𝟏𝟏 d. On the day I turned 𝟏𝟏𝟏𝟏 years old, how old was my sister? 𝟏𝟏𝟏𝟏 years old e. How do you know? Since my sister’s age is always two years more than my age, we just add 𝟐𝟐 to my age. 𝟏𝟏𝟏𝟏+ 𝟐𝟐= 𝟏𝟏𝟏𝟏 f. Do we need to extend the table to calculate these answers? No; the pattern is to add 𝟐𝟐 to your age to calculate your sister’s age. Example 1 (5 minutes) Example 1 My Age (in years) My Sister’s Age (in years) 𝟎𝟎 𝟐𝟐 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟒𝟒 𝟑𝟑 𝟓𝟓 𝟒𝟒 𝟔𝟔 𝒀𝒀 𝒀𝒀+ 𝟐𝟐 a. What if you don’t know how old I am? Let’s use a variable for my age. Let 𝒀𝒀= my age in years. Can you develop an expression to describe how old my sister is? Your sister is 𝒀𝒀+ 𝟐𝟐 years old. b. Please add that to the last row of the table. My age is 𝑌𝑌 years. My sister is 𝑌𝑌+ 2 years old. So, no matter what my age is (or was), my sister’s age in years will always be two years greater than mine. Scaffolding: A number line in the classroom can provide an additional reference for students. A cardboard sheet with two windows cut to reveal 𝑌𝑌 and 𝑌𝑌+ 2 works with Examples 1 and 2. MP.6 A STORY OF RATIOS 201 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Example 2 (5 minutes) Example 2 My Age (in years) My Sister’s Age (in years) 𝟎𝟎 𝟐𝟐 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟒𝟒 𝟑𝟑 𝟓𝟓 𝟒𝟒 𝟔𝟔 𝑮𝑮−𝟐𝟐 𝑮𝑮 a. How old was I when my sister was 𝟔𝟔 years old? 𝟒𝟒 years old b. How old was I when my sister was 𝟏𝟏𝟏𝟏 years old? 𝟏𝟏𝟏𝟏 years old c. How do you know? My age is always 𝟐𝟐 years less than my sister’s age. d. Look at the table in Example 2. If you know my sister’s age, can you determine my age? We can subtract two from your sister’s age, and that will equal your age. e. If we use the variable 𝑮𝑮 for my sister’s age in years, what expression would describe my age in years? 𝑮𝑮−𝟐𝟐 f. Fill in the last row of the table with the expressions. My age is 𝑮𝑮−𝟐𝟐 years. My sister is 𝑮𝑮 years old. g. With a partner, calculate how old I was when my sister was 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐, and 𝟐𝟐𝟐𝟐 years old. You were 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐, and 𝟐𝟐𝟐𝟐 years old, respectively. A STORY OF RATIOS 202 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Exercises (15 minutes) Exercises 1. Noah and Carter are collecting box tops for their school. They each bring in 𝟏𝟏 box top per day starting on the first day of school. However, Carter had a head start because his aunt sent him 𝟏𝟏𝟏𝟏 box tops before school began. Noah’s grandma saved 𝟏𝟏𝟏𝟏 box tops, and Noah added those on his first day. a. Fill in the missing values that indicate the total number of box tops each boy brought to school. School Day Number of Box Tops Noah Has Number of Box Tops Carter Has 𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 b. If we let 𝑫𝑫 be the number of days since the new school year began, on day 𝑫𝑫 of school, how many box tops will Noah have brought to school? 𝑫𝑫+ 𝟏𝟏𝟏𝟏 box tops c. On day 𝑫𝑫 of school, how many box tops will Carter have brought to school? 𝑫𝑫+ 𝟏𝟏𝟏𝟏 box tops d. On day 𝟏𝟏𝟏𝟏 of school, how many box tops will Noah have brought to school? 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐; On day 𝟏𝟏𝟏𝟏, Noah would have brought in 𝟐𝟐𝟐𝟐 box tops. e. On day 𝟏𝟏𝟏𝟏 of school, how many box tops will Carter have brought to school? 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐; On day 𝟏𝟏𝟏𝟏, Carter would have brought in 𝟐𝟐𝟐𝟐 box tops. 2. Each week the Primary School recycles 𝟐𝟐𝟐𝟐𝟐𝟐 pounds of paper. The Intermediate School also recycles the same amount but had another 𝟑𝟑𝟑𝟑𝟑𝟑 pounds left over from summer school. The Intermediate School custodian added this extra 𝟑𝟑𝟑𝟑𝟑𝟑 pounds to the first recycle week. a. Number the weeks, and record the amount of paper recycled by both schools. Week Total Amount of Paper Recycled by the Primary School This School Year in Pounds Total Amount of Paper Recycled by the Intermediate School This School Year in Pounds 𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓𝟓𝟓 𝟐𝟐 𝟒𝟒𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕𝟕𝟕 𝟑𝟑 𝟔𝟔𝟔𝟔𝟔𝟔 𝟗𝟗𝟗𝟗𝟗𝟗 𝟒𝟒 𝟖𝟖𝟖𝟖𝟖𝟖 𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟏𝟏, 𝟑𝟑𝟑𝟑𝟑𝟑 b. If this trend continues, what will be the total amount collected for each school on Week 𝟏𝟏𝟏𝟏? The Primary School will have collected 𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 pounds. The Intermediate School will have collected 𝟐𝟐, 𝟑𝟑𝟑𝟑𝟑𝟑 pounds. A STORY OF RATIOS 203 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions 3. Shelly and Kristen share a birthday, but Shelly is 𝟓𝟓 years older. a. Make a table showing their ages every year, beginning when Kristen was born. Kristen’s Age (in years) Shelly’s Age (in years) 𝟎𝟎 𝟓𝟓 𝟏𝟏 𝟔𝟔 𝟐𝟐 𝟕𝟕 𝟑𝟑 𝟖𝟖 b. If Kristen is 𝟏𝟏𝟏𝟏 years old, how old is Shelly? If Kristen is 𝟏𝟏𝟏𝟏 years old, Shelly is 𝟐𝟐𝟐𝟐 years old. c. If Kristen is 𝑲𝑲 years old, how old is Shelly? If Kristen is 𝑲𝑲 years old, Shelly is 𝑲𝑲+ 𝟓𝟓 years old. d. If Shelly is 𝑺𝑺 years old, how old is Kristen? If Shelly is 𝑺𝑺 years old, Kristen is 𝑺𝑺−𝟓𝟓 years old. Closing (2 minutes) Why were we able to write these expressions? There was a relationship between the two quantities that we could identify. What is important to remember about labeling columns in a table? The label should be complete, with units, so the reader understands precisely what is meant. How are addition and subtraction expressions related to one another? They are inverse operations. One undoes the other. Exit Ticket (5 minutes) A STORY OF RATIOS 204 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Name Date Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Exit Ticket Jenna and Allie work together at a piano factory. They both were hired on January 3, but Jenna was hired in 2005, and Allie was hired in 2009. a. Fill in the table below to summarize the two workers’ experience totals. Year Allie’s Years of Experience Jenna’s Years of Experience 2010 2011 2012 2013 2014 b. If both workers continue working at the piano factory, when Allie has 𝐴𝐴 years of experience on the job, how many years of experience will Jenna have on the job? c. If both workers continue working at the piano factory, when Allie has 20 years of experience on the job, how many years of experience will Jenna have on the job? A STORY OF RATIOS 205 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Exit Ticket Sample Solutions Jenna and Allie work together at a piano factory. They both were hired on January 3, but Jenna was hired in 2005, and Allie was hired in 2009. a. Fill in the table below to summarize the two workers’ experience totals. Year Allie’s Years of Experience Jenna’s Years of Experience 2010 𝟏𝟏 𝟓𝟓 2011 𝟐𝟐 𝟔𝟔 2012 𝟑𝟑 𝟕𝟕 2013 𝟒𝟒 𝟖𝟖 2014 𝟓𝟓 𝟗𝟗 b. If both workers continue working at the piano factory, when Allie has 𝑨𝑨 years of experience on the job, how many years of experience will Jenna have on the job? Jenna will have been on the job for 𝑨𝑨+ 𝟒𝟒 years. c. If both workers continue working at the piano factory, when Allie has 𝟐𝟐𝟐𝟐 years of experience on the job, how many years of experience will Jenna have on the job? 𝟐𝟐𝟐𝟐+ 𝟒𝟒= 𝟐𝟐𝟐𝟐 Jenna will have been on the job for 𝟐𝟐𝟐𝟐 years. Problem Set Sample Solutions 1. Suellen and Tara are in sixth grade, and both take dance lessons at Twinkle Toes Dance Studio. This is Suellen’s first year, while this is Tara’s fifth year of dance lessons. Both girls plan to continue taking lessons throughout high school. a. Complete the table showing the number of years the girls will have danced at the studio. Grade Suellen’s Years of Experience Dancing Tara’s Years of Experience Dancing Sixth 𝟏𝟏 𝟓𝟓 Seventh 𝟐𝟐 𝟔𝟔 Eighth 𝟑𝟑 𝟕𝟕 Ninth 𝟒𝟒 𝟖𝟖 Tenth 𝟓𝟓 𝟗𝟗 Eleventh 𝟔𝟔 𝟏𝟏𝟏𝟏 Twelfth 𝟕𝟕 𝟏𝟏𝟏𝟏 b. If Suellen has been taking dance lessons for 𝒀𝒀 years, how many years has Tara been taking lessons? Tara has been taking dance lessons for 𝒀𝒀+ 𝟒𝟒 years. A STORY OF RATIOS 206 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions 2. Daejoy and Damian collect fossils. Before they went on a fossil-hunting trip, Daejoy had 𝟐𝟐𝟐𝟐 fossils in her collection, and Damian had 𝟏𝟏𝟏𝟏 fossils in his collection. On a 𝟏𝟏𝟏𝟏-day fossil-hunting trip, they each collected 𝟐𝟐 new fossils each day. a. Make a table showing how many fossils each person had in their collection at the end of each day. Day Number of Fossils in Daejoy’s Collection Number of Fossils in Damian’s Collection 𝟏𝟏 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟒𝟒 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟔𝟔 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟕𝟕 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟖𝟖 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟗𝟗 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 b. If this pattern of fossil finding continues, how many fossils does Damian have when Daejoy has 𝑭𝑭 fossils? When Daejoy has 𝑭𝑭 fossils, Damian has 𝑭𝑭−𝟗𝟗 fossils. c. If this pattern of fossil finding continues, how many fossils does Damian have when Daejoy has 𝟓𝟓𝟓𝟓 fossils? 𝟓𝟓𝟓𝟓−𝟗𝟗= 𝟒𝟒𝟒𝟒 When Daejoy has 𝟓𝟓𝟓𝟓 fossils, Damian has 𝟒𝟒𝟒𝟒 fossils. 3. A train consists of three types of cars: box cars, an engine, and a caboose. The relationship among the types of cars is demonstrated in the table below. Number of Box Cars Number of Cars in the Train 𝟎𝟎 𝟐𝟐 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 a. Tom wrote an expression for the relationship depicted in the table as 𝑩𝑩+ 𝟐𝟐. Theresa wrote an expression for the same relationship as 𝑪𝑪−𝟐𝟐. Is it possible to have two different expressions to represent one relationship? Explain. Both expressions can represent the same relationship, depending on the point of view. The expression 𝑩𝑩+ 𝟐𝟐 represents the number of box cars plus an engine and a caboose. The expression 𝑪𝑪−𝟐𝟐 represents the whole car length of the train, less the engine and caboose. b. What do you think the variable in each student’s expression represents? How would you define them? The variable 𝑪𝑪 would represent the total cars in the train. The variable 𝑩𝑩 would represent the number of box cars. A STORY OF RATIOS 207 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions 4. David was 𝟑𝟑 when Marieka was born. Complete the table. Marieka’s Age in Years David’s Age in Years 𝟓𝟓 𝟖𝟖 𝟔𝟔 𝟗𝟗 𝟕𝟕 𝟏𝟏𝟏𝟏 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝑴𝑴 𝑴𝑴+ 𝟑𝟑 𝑫𝑫−𝟑𝟑 𝑫𝑫 5. Caitlin and Michael are playing a card game. In the first round, Caitlin scored 𝟐𝟐𝟐𝟐𝟐𝟐 points, and Michael scored 𝟏𝟏𝟏𝟏𝟏𝟏 points. In each of the next few rounds, they each scored 𝟓𝟓𝟓𝟓 points. Their score sheet is below. Caitlin’s Points Michael’s Points 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 a. If this trend continues, how many points will Michael have when Caitlin has 𝟔𝟔𝟔𝟔𝟔𝟔 points? 𝟔𝟔𝟔𝟔𝟔𝟔−𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓𝟓𝟓 Michael will have 𝟓𝟓𝟓𝟓𝟓𝟓 points. b. If this trend continues, how many points will Michael have when Caitlin has 𝑪𝑪 points? Michael will have 𝑪𝑪−𝟐𝟐𝟐𝟐 points. c. If this trend continues, how many points will Caitlin have when Michael has 𝟗𝟗𝟗𝟗𝟗𝟗 points? 𝟗𝟗𝟗𝟗𝟗𝟗+ 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 Caitlin will have 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 points. d. If this trend continues, how many points will Caitlin have when Michael has 𝑴𝑴 points? Caitlin will have 𝑴𝑴+ 𝟐𝟐𝟐𝟐 points. 6. The high school marching band has 𝟏𝟏𝟏𝟏 drummers this year. The band director insists that there are to be 𝟓𝟓 more trumpet players than drummers at all times. a. How many trumpet players are in the marching band this year? 𝟏𝟏𝟏𝟏+ 𝟓𝟓= 𝟐𝟐𝟐𝟐 . There are 𝟐𝟐𝟐𝟐 trumpet players this year. b. Write an expression that describes the relationship of the number of trumpet players (𝑻𝑻) and the number of drummers (𝑫𝑫). 𝑻𝑻= 𝑫𝑫+ 𝟓𝟓 or 𝑫𝑫= 𝑻𝑻−𝟓𝟓 c. If there are only 𝟏𝟏𝟏𝟏 trumpet players interested in joining the marching band next year, how many drummers will the band director want in the band? 𝟏𝟏𝟏𝟏−𝟓𝟓= 𝟗𝟗. The band director will want 𝟗𝟗 drummers. A STORY OF RATIOS 208 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Subtraction of Decimals—Round 1 Directions: Evaluate each expression. 1. 55 −50 23. 9.9 −5 2. 55 −5 24. 9.9 −0.5 3. 5.5 −5 25. 0.99 −0.5 4. 5.5 −0.5 26. 0.99 −0.05 5. 88 −80 27. 4.7 −2 6. 88 −8 28. 4.7 −0.2 7. 8.8 −8 29. 0.47 −0.2 8. 8.8 −0.8 30. 0.47 −0.02 9. 33 −30 31. 8.4 −1 10. 33 −3 32. 8.4 −0.1 11. 3.3 −3 33. 0.84 −0.1 12. 1 −0.3 34. 7.2 −5 13. 1 −0.03 35. 7.2 −0.5 14. 1 −0.003 36. 0.72 −0.5 15. 0.1 −0.03 37. 0.72 −0.05 16. 4 −0.8 38. 8.6 −7 17. 4 −0.08 39. 8.6 −0.7 18. 4 −0.008 40. 0.86 −0.7 19. 0.4 −0.08 41. 0.86 −0.07 20. 9 −0.4 42. 5.1 −4 21. 9 −0.04 43. 5.1 −0.4 22. 9 −0.004 44. 0.51 −0.4 Number Correct: Number Correct: A STORY OF RATIOS 209 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Subtraction of Decimals—Round 1 [KEY] Directions: Evaluate each expression. 1. 55 −50 𝟓𝟓 23. 9.9 −5 𝟒𝟒. 𝟗𝟗 2. 55 −5 𝟓𝟓𝟓𝟓 24. 9.9 −0.5 𝟗𝟗. 𝟒𝟒 3. 5.5 −5 𝟎𝟎. 𝟓𝟓 25. 0.99 −0.5 𝟎𝟎. 𝟒𝟒𝟒𝟒 4. 5.5 −0.5 𝟓𝟓 26. 0.99 −0.05 𝟎𝟎. 𝟗𝟗𝟗𝟗 5. 88 −80 𝟖𝟖 27. 4.7 −2 𝟐𝟐. 𝟕𝟕 6. 88 −8 𝟖𝟖𝟖𝟖 28. 4.7 −0.2 𝟒𝟒. 𝟓𝟓 7. 8.8 −8 𝟎𝟎. 𝟖𝟖 29. 0.47 −0.2 𝟎𝟎. 𝟐𝟐𝟐𝟐 8. 8.8 −0.8 𝟖𝟖 30. 0.47 −0.02 𝟎𝟎. 𝟒𝟒𝟒𝟒 9. 33 −30 𝟑𝟑 31. 8.4 −1 𝟕𝟕. 𝟒𝟒 10. 33 −3 𝟑𝟑𝟑𝟑 32. 8.4 −0.1 𝟖𝟖. 𝟑𝟑 11. 3.3 −3 𝟎𝟎. 𝟑𝟑 33. 0.84 −0.1 𝟎𝟎. 𝟕𝟕𝟕𝟕 12. 1 −0.3 𝟎𝟎. 𝟕𝟕 34. 7.2 −5 𝟐𝟐. 𝟐𝟐 13. 1 −0.03 𝟎𝟎. 𝟗𝟗𝟗𝟗 35. 7.2 −0.5 𝟔𝟔. 𝟕𝟕 14. 1 −0.003 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗 36. 0.72 −0.5 𝟎𝟎. 𝟐𝟐𝟐𝟐 15. 0.1 −0.03 𝟎𝟎. 𝟎𝟎𝟎𝟎 37. 0.72 −0.05 𝟎𝟎. 𝟔𝟔𝟔𝟔 16. 4 −0.8 𝟑𝟑. 𝟐𝟐 38. 8.6 −7 𝟏𝟏. 𝟔𝟔 17. 4 −0.08 𝟑𝟑. 𝟗𝟗𝟗𝟗 39. 8.6 −0.7 𝟕𝟕. 𝟗𝟗 18. 4 −0.008 𝟑𝟑. 𝟗𝟗𝟗𝟗𝟗𝟗 40. 0.86 −0.7 𝟎𝟎. 𝟏𝟏𝟏𝟏 19. 0.4 −0.08 𝟎𝟎. 𝟑𝟑𝟑𝟑 41. 0.86 −0.07 𝟎𝟎. 𝟕𝟕𝟕𝟕 20. 9 −0.4 𝟖𝟖. 𝟔𝟔 42. 5.1 −4 𝟏𝟏. 𝟏𝟏 21. 9 −0.04 𝟖𝟖. 𝟗𝟗𝟗𝟗 43. 5.1 −0.4 𝟒𝟒. 𝟕𝟕 22. 9 −0.004 𝟖𝟖. 𝟗𝟗𝟗𝟗𝟗𝟗 44. 0.51 −0.4 𝟎𝟎. 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 210 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Subtraction of Decimals—Round 2 Directions: Evaluate each expression. 1. 66 −60 23. 6.8 −4 2. 66 −6 24. 6.8 −0.4 3. 6.6 −6 25. 0.68 −0.4 4. 6.6 −0.6 26. 0.68 −0.04 5. 99 −90 27. 7.3 −1 6. 99 −9 28. 7.3 −0.1 7. 9.9 −9 29. 0.73 −0.1 8. 9.9 −0.9 30. 0.73 −0.01 9. 22 −20 31. 9.5 −2 10. 22 −2 32. 9.5 −0.2 11. 2.2 −2 33. 0.95 −0.2 12. 3 −0.4 34. 8.3 −5 13. 3 −0.04 35. 8.3 −0.5 14. 3 −0.004 36. 0.83 −0.5 15. 0.3 −0.04 37. 0.83 −0.05 16. 8 −0.2 38. 7.2 −4 17. 8 −0.02 39. 7.2 −0.4 18. 8 −0.002 40. 0.72 −0.4 19. 0.8 −0.02 41. 0.72 −0.04 20. 5 −0.1 42. 9.3 −7 21. 5 −0.01 43. 9.3 −0.7 22. 5 −0.001 44. 0.93 −0.7 Number Correct: Improvement: A STORY OF RATIOS 211 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 19 Lesson 19: Substituting to Evaluate Addition and Subtraction Expressions Subtraction of Decimals—Round 2 [KEY] Directions: Evaluate each expression. 1. 66 −60 𝟔𝟔 23. 6.8 −4 𝟐𝟐. 𝟖𝟖 2. 66 −6 𝟔𝟔𝟔𝟔 24. 6.8 −0.4 𝟔𝟔. 𝟒𝟒 3. 6.6 −6 𝟎𝟎. 𝟔𝟔 25. 0.68 −0.4 𝟎𝟎. 𝟐𝟐𝟐𝟐 4. 6.6 −0.6 𝟔𝟔 26. 0.68 −0.04 𝟎𝟎. 𝟔𝟔𝟔𝟔 5. 99 −90 𝟗𝟗 27. 7.3 −1 𝟔𝟔. 𝟑𝟑 6. 99 −9 𝟗𝟗𝟗𝟗 28. 7.3 −0.1 𝟕𝟕. 𝟐𝟐 7. 9.9 −9 𝟎𝟎. 𝟗𝟗 29. 0.73 −0.1 𝟎𝟎. 𝟔𝟔𝟔𝟔 8. 9.9 −0.9 𝟗𝟗 30. 0.73 −0.01 𝟎𝟎. 𝟕𝟕𝟕𝟕 9. 22 −20 𝟐𝟐 31. 9.5 −2 𝟕𝟕. 𝟓𝟓 10. 22 −2 𝟐𝟐𝟐𝟐 32. 9.5 −0.2 𝟗𝟗. 𝟑𝟑 11. 2.2 −2 𝟎𝟎. 𝟐𝟐 33. 0.95 −0.2 𝟎𝟎. 𝟕𝟕𝟕𝟕 12. 3 −0.4 𝟐𝟐. 𝟔𝟔 34. 8.3 −5 𝟑𝟑. 𝟑𝟑 13. 3 −0.04 𝟐𝟐. 𝟗𝟗𝟗𝟗 35. 8.3 −0.5 𝟕𝟕. 𝟖𝟖 14. 3 −0.004 𝟐𝟐. 𝟗𝟗𝟗𝟗𝟗𝟗 36. 0.83 −0.5 𝟎𝟎. 𝟑𝟑𝟑𝟑 15. 0.3 −0.04 𝟎𝟎. 𝟐𝟐𝟐𝟐 37. 0.83 −0.05 𝟎𝟎. 𝟕𝟕𝟕𝟕 16. 8 −0.2 𝟕𝟕. 𝟖𝟖 38. 7.2 −4 𝟑𝟑. 𝟐𝟐 17. 8 −0.02 𝟕𝟕. 𝟗𝟗𝟗𝟗 39. 7.2 −0.4 𝟔𝟔. 𝟖𝟖 18. 8 −0.002 𝟕𝟕. 𝟗𝟗𝟗𝟗𝟗𝟗 40. 0.72 −0.4 𝟎𝟎. 𝟑𝟑𝟑𝟑 19. 0.8 −0.02 𝟎𝟎. 𝟕𝟕𝟕𝟕 41. 0.72 −0.04 𝟎𝟎. 𝟔𝟔𝟔𝟔 20. 5 −0.1 𝟒𝟒. 𝟗𝟗 42. 9.3 −7 𝟐𝟐. 𝟑𝟑 21. 5 −0.01 𝟒𝟒. 𝟗𝟗𝟗𝟗 43. 9.3 −0.7 𝟖𝟖. 𝟔𝟔 22. 5 −0.001 𝟒𝟒. 𝟗𝟗𝟗𝟗𝟗𝟗 44. 0.93 −0.7 𝟎𝟎. 𝟐𝟐𝟐𝟐 A STORY OF RATIOS 212 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Lesson 20: Writing and Evaluating Expressions— Multiplication and Division Student Outcomes Students develop expressions involving multiplication and division from real-world problems. Students evaluate these expressions for given values. Lesson Notes This lesson builds on Lessons 18 and 19, extending the concepts using multiplication and division expressions. Classwork Opening (3 minutes) Take time to make sure the answers to the Problem Set from the previous lesson are clear. The labels on the tables should be complete. Discussion (3 minutes) In the previous lessons, we created expressions that used addition and subtraction to describe the relationship between two quantities. How did using tables help your understanding? Answers will vary. Patterns were easy to see. Looking down the columns revealed a number pattern. Looking across the rows revealed a constant difference between columns. In this lesson, we are going to develop expressions involving multiplication and division, much like the last lesson. We also evaluate these expressions for given values. Example 1 (10 minutes) The farmers’ market is selling bags of apples. In every bag, there are 3 apples. If I buy one bag, how many apples will I have? Three If I buy two bags, how many apples will I have? Since 2 ∙3 = 6, you will have 6 apples. If I buy three bags, how many apples will I have? Since 3 ∙3 = 9, you will have 9 apples. Fill in the table for a purchase of 4 bags of apples. Check your answer with a partner. Scaffolding: Having interlocking cubes ready in groups of three makes a concrete visual for students to see and hold for Example 1. Put these in clear plastic bags, if desired. A STORY OF RATIOS 213 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Example 1 The farmers’ market is selling bags of apples. In every bag, there are 𝟑𝟑 apples. a. Complete the table. Number of Bags Total Number of Apples 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟔𝟔 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝑩𝑩 𝟑𝟑𝟑𝟑 What if I bought some other number of bags? If I told you how many bags, could you calculate the number of apples I would have altogether? Yes. Multiply the number of bags by 3 to find the total number of apples. What if I bought 𝐵𝐵 bags of apples? Can you write an expression in the table that describes the total number of apples I have purchased? 3𝐵𝐵 or 3(𝐵𝐵) or 3 · 𝐵𝐵 Take a moment to review the different notations used for multiplication. Students should be comfortable reading and writing the expressions in all three forms. What if the market had 25 bags of apples to sell? How many apples is that in all? If 𝐵𝐵= 25, then 3𝐵𝐵= 3 ∙25 = 75. The market had 75 apples to sell. b. What if the market had 𝟐𝟐𝟐𝟐 bags of apples to sell? How many apples is that in all? If 𝑩𝑩= 𝟐𝟐𝟐𝟐, then 𝟑𝟑𝟑𝟑= 𝟑𝟑∙𝟐𝟐𝟐𝟐= 𝟕𝟕𝟕𝟕. The market had 𝟕𝟕𝟕𝟕 apples to sell. c. If a truck arrived that had some number, 𝒂𝒂, more apples on it, then how many bags would the clerks use to bag up the apples? 𝒂𝒂÷ 𝟑𝟑 bags are needed. If there are 𝟏𝟏 or 𝟐𝟐 apples left over, an extra bag will be needed (although not full). d. If a truck arrived that had 𝟔𝟔𝟔𝟔𝟔𝟔 apples on it, how many bags would the clerks use to bag up the apples? 𝟔𝟔𝟔𝟔𝟔𝟔 𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚÷ 𝟑𝟑𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 𝐛𝐛𝐛𝐛𝐛𝐛 = 𝟐𝟐𝟐𝟐𝟐𝟐 𝐛𝐛𝐛𝐛𝐛𝐛𝐛𝐛 e. How is part (d) different from part (b)? Part (d) gives the number of apples and asks for the number of bags. Therefore, we needed to divide the number of apples by 𝟑𝟑. Part (b) gives the number of bags and asks for the number of apples. Therefore, we needed to multiply the number of bags by 𝟑𝟑. A STORY OF RATIOS 214 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Exercise 1 (5 minutes) Students work on Exercise 1 independently. Exercises 1–3 1. In New York State, there is a five-cent deposit on all carbonated beverage cans and bottles. When you return the empty can or bottle, you get the five cents back. a. Complete the table. Number of Containers Returned Refund in Dollars 𝟏𝟏 𝟎𝟎. 𝟎𝟎𝟎𝟎 𝟐𝟐 𝟎𝟎. 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟎𝟎. 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟎𝟎. 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟎𝟎. 𝟓𝟓𝟓𝟓 𝟓𝟓𝟓𝟓 𝟐𝟐. 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓. 𝟎𝟎𝟎𝟎 𝑪𝑪 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 b. If we let 𝑪𝑪 represent the number of cans, what is the expression that shows how much money is returned? 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 c. Use the expression to find out how much money Brett would receive if he returned 𝟐𝟐𝟐𝟐𝟐𝟐 cans. If 𝑪𝑪= 𝟐𝟐𝟐𝟐𝟐𝟐, then 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎= 𝟎𝟎. 𝟎𝟎𝟎𝟎∙𝟐𝟐𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏. Brett would receive $𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏 if he returned 𝟐𝟐𝟐𝟐𝟐𝟐 cans. d. If Gavin needs to earn $𝟒𝟒. 𝟓𝟓𝟓𝟓 for returning cans, how many cans does he need to collect and return? 𝟒𝟒. 𝟓𝟓𝟓𝟓÷ 𝟎𝟎. 𝟎𝟎𝟎𝟎= 𝟗𝟗𝟗𝟗. Gavin needs to collect and return 𝟗𝟗𝟗𝟗 cans. e. How is part (d) different from part (c)? Part (d) gives the amount of money and asks for the number of cans. Therefore, we needed to divide the amount of money by 𝟎𝟎. 𝟎𝟎𝟎𝟎. Part (c) gives the number of cans and asks for the amount of money. Therefore, we needed to multiply the number of cans by 𝟎𝟎. 𝟎𝟎𝟎𝟎. Discuss the similarities and differences between Example 1 and Exercise 1. In both problems, the second quantity is a multiple of the first. Multiplication by the constant term is used to show the relationship between the quantities in the first column and the quantities in the second column. Division is used to show the relationship between the quantities in the second column and the quantities in the first column. A STORY OF RATIOS 215 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Exercise 2 (10 minutes) Students work on Exercise 2 independently. 2. The fare for a subway or a local bus ride is $𝟐𝟐. 𝟓𝟓𝟓𝟓. a. Complete the table. Number of Rides Cost of Rides in Dollars 𝟏𝟏 𝟐𝟐. 𝟓𝟓𝟓𝟓 𝟐𝟐 𝟓𝟓. 𝟎𝟎𝟎𝟎 𝟑𝟑 𝟕𝟕. 𝟓𝟓𝟓𝟓 𝟒𝟒 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟓𝟓 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 𝟑𝟑𝟑𝟑 𝟕𝟕𝟕𝟕. 𝟎𝟎𝟎𝟎 𝑹𝑹 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓 or 𝟐𝟐. 𝟓𝟓𝟓𝟓 b. If we let 𝑹𝑹 represent the number of rides, what is the expression that shows the cost of the rides? 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓 or 𝟐𝟐. 𝟓𝟓𝟓𝟓 c. Use the expression to find out how much money 𝟔𝟔𝟔𝟔 rides would cost. If 𝑹𝑹= 𝟔𝟔𝟔𝟔, then 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓= 𝟐𝟐. 𝟓𝟓𝟓𝟓∙𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎. Sixty rides would cost $𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎. d. If a commuter spends $𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 on subway or bus rides, how many trips did the commuter take? 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎÷ 𝟐𝟐. 𝟓𝟓𝟓𝟓= 𝟕𝟕𝟕𝟕. The commuter took 𝟕𝟕𝟕𝟕 trips. e. How is part (d) different from part (c)? Part (d) gives the amount of money and asks for the number of rides. Therefore, we needed to divide the amount of money by the cost of each ride ($𝟐𝟐. 𝟓𝟓𝟓𝟓). Part (c) gives the number of rides and asks for the amount of money. Therefore, we needed to multiply the number of rides by $𝟐𝟐. 𝟓𝟓𝟓𝟓. A STORY OF RATIOS 216 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Exercise 3 (10 minutes): Challenge Problem Challenge Problem 3. A pendulum swings though a certain number of cycles in a given time. Owen made a pendulum that swings 𝟏𝟏𝟏𝟏 times every 𝟏𝟏𝟏𝟏 seconds. a. Construct a table showing the number of cycles through which a pendulum swings. Include data for up to one minute. Use the last row for 𝑪𝑪 cycles, and write an expression for the time it takes for the pendulum to make 𝑪𝑪 cycles. Number of Cycles Time in Seconds 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝑪𝑪 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 b. Owen and his pendulum team set their pendulum in motion and counted 𝟏𝟏𝟏𝟏 cycles. What was the elapsed time? 𝑪𝑪= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐. The elapsed time is 𝟐𝟐𝟐𝟐 seconds. c. Write an expression for the number of cycles a pendulum swings in 𝑺𝑺 seconds. 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝑺𝑺 or 𝟒𝟒 𝟓𝟓𝑺𝑺 or 𝟎𝟎. 𝟖𝟖∙𝑺𝑺 d. In a different experiment, Owen and his pendulum team counted the cycles of the pendulum for 𝟑𝟑𝟑𝟑 seconds. How many cycles did they count? 𝑺𝑺= 𝟑𝟑𝟑𝟑; 𝟎𝟎. 𝟖𝟖 ∙𝟑𝟑𝟑𝟑= 𝟐𝟐𝟐𝟐. They counted 𝟐𝟐𝟐𝟐 cycles. Closing (2 minutes) In Example 1, we looked at the relationship between the number of bags purchased at the farmers’ market and the total number of apples purchased. We created two different expressions: 3𝐵𝐵 and 𝑎𝑎÷ 3. What does each variable represent, and why did we multiply by 3 in the first expression and divide by 3 in the second? The variable 𝐵𝐵 represented the number of bags. We had to multiply by 3 because we were given the number of bags, and there were 3 apples packaged in each bag. The variable 𝑎𝑎 represented the number of apples. We divided by 3 because we were given the number of apples and need to determine the number of bags needed. What would the expressions be if the farmers’ market sold bags that contained 5 apples in a bag instead of 3? 5𝐵𝐵 and 𝑎𝑎÷ 5, respectively Exit Ticket (3 minutes) A STORY OF RATIOS 217 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Name Date Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Exit Ticket Anna charges $8.50 per hour to babysit. Complete the table, and answer the questions below. Number of Hours Amount Anna Charges in Dollars 1 2 5 8 𝐻𝐻 a. Write an expression describing her earnings for working 𝐻𝐻 hours. b. How much will she earn if she works for 3 1 2 hours? c. How long will it take Anna to earn $51.00? A STORY OF RATIOS 218 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division Exit Ticket Sample Solutions 1. Anna charges $𝟖𝟖. 𝟓𝟓𝟓𝟓 per hour to babysit. Complete the table, and answer the questions below. Number of Hours Amount Anna Charges in Dollars 𝟏𝟏 𝟖𝟖. 𝟓𝟓𝟓𝟓 𝟐𝟐 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟓𝟓 𝟒𝟒𝟒𝟒. 𝟓𝟓𝟓𝟓 𝟖𝟖 𝟔𝟔𝟔𝟔 𝑯𝑯 𝟖𝟖. 𝟓𝟓𝟓𝟓𝟓𝟓 or 𝟖𝟖. 𝟓𝟓𝟓𝟓 a. Write an expression describing her earnings for working 𝑯𝑯 hours. 𝟖𝟖. 𝟓𝟓𝟓𝟓𝟓𝟓 or 𝟖𝟖. 𝟓𝟓𝟓𝟓 b. How much will she earn if she works for 𝟑𝟑𝟏𝟏 𝟐𝟐 hours? If 𝑯𝑯= 𝟑𝟑. 𝟓𝟓, then 𝟖𝟖. 𝟓𝟓𝟓𝟓= 𝟖𝟖. 𝟓𝟓∙𝟑𝟑. 𝟓𝟓= 𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕. She will earn $𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕. c. How long will it take Anna to earn $𝟓𝟓𝟓𝟓. 𝟎𝟎𝟎𝟎? 𝟓𝟓𝟓𝟓÷ 𝟖𝟖. 𝟓𝟓= 𝟔𝟔. It will take Anna 𝟔𝟔 hours to earn $𝟓𝟓𝟓𝟓. 𝟎𝟎𝟎𝟎. Problem Set Sample Solutions 1. A radio station plays 𝟏𝟏𝟏𝟏 songs each hour. They never stop for commercials, news, weather, or traffic reports. a. Write an expression describing how many songs are played by the radio station in 𝑯𝑯 hours. 𝟏𝟏𝟏𝟏𝟏𝟏 b. How many songs will be played in an entire day (𝟐𝟐𝟐𝟐 hours)? 𝟏𝟏𝟏𝟏∙𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐𝟐𝟐. There will be 𝟐𝟐𝟐𝟐𝟐𝟐 songs played. c. How long does it take the radio station to play 𝟔𝟔𝟔𝟔 consecutive songs? 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 ÷ 𝟏𝟏𝟏𝟏𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡= 𝟓𝟓 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡 2. A ski area has a high-speed lift that can move 𝟐𝟐, 𝟒𝟒𝟒𝟒𝟒𝟒 skiers to the top of the mountain each hour. a. Write an expression describing how many skiers can be lifted in 𝑯𝑯 hours. 𝟐𝟐, 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 b. How many skiers can be moved to the top of the mountain in 𝟏𝟏𝟏𝟏 hours? 𝟏𝟏𝟏𝟏∙𝟐𝟐, 𝟒𝟒𝟒𝟒𝟒𝟒= 𝟑𝟑𝟑𝟑, 𝟔𝟔𝟔𝟔𝟔𝟔. 𝟑𝟑𝟑𝟑, 𝟔𝟔𝟔𝟔𝟔𝟔 skiers can be moved. A STORY OF RATIOS 219 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division c. How long will it take to move 𝟑𝟑, 𝟔𝟔𝟎𝟎𝟎𝟎 skiers to the top of the mountain? 𝟑𝟑, 𝟔𝟔𝟔𝟔𝟔𝟔÷ 𝟐𝟐, 𝟒𝟒𝟒𝟒𝟒𝟒= 𝟏𝟏. 𝟓𝟓. It will take an hour and a half to move 𝟑𝟑, 𝟔𝟔𝟔𝟔𝟔𝟔 skiers to the top of the mountain. 3. Polly writes a magazine column, for which she earns $𝟑𝟑𝟑𝟑 per hour. Create a table of values that shows the relationship between the number of hours that Polly works, 𝑯𝑯, and the amount of money Polly earns in dollars, 𝑬𝑬. Answers will vary. Sample answers are shown. Hours Polly Works (𝑯𝑯) Polly’s Earnings in Dollars (𝑬𝑬) 𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐 𝟕𝟕𝟕𝟕 𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 a. If you know how many hours Polly works, can you determine how much money she earned? Write the corresponding expression. Multiplying the number of hours that Polly works by her rate ($𝟑𝟑𝟑𝟑 per hour) will calculate her pay. 𝟑𝟑𝟑𝟑𝟑𝟑 is the expression for her pay in dollars. b. Use your expression to determine how much Polly earned after working for 𝟑𝟑𝟏𝟏 𝟐𝟐 hours. 𝟑𝟑𝟑𝟑𝟑𝟑= 𝟑𝟑𝟑𝟑∙𝟑𝟑. 𝟓𝟓= 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓. Polly makes $𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 for working 𝟑𝟑𝟏𝟏 𝟐𝟐 hours. c. If you know how much money Polly earned, can you determine how long she worked? Write the corresponding expression. Dividing Polly’s pay by 𝟑𝟑𝟑𝟑 will calculate the number of hours she worked. 𝑬𝑬÷ 𝟑𝟑𝟑𝟑 is the expression for the number of hours she worked. d. Use your expression to determine how long Polly worked if she earned $𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓÷ 𝟑𝟑𝟑𝟑= 𝟏𝟏. 𝟓𝟓; Polly worked an hour and a half for $𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓. 4. Mitchell delivers newspapers after school, for which he earns $𝟎𝟎. 𝟎𝟎𝟎𝟎 per paper. Create a table of values that shows the relationship between the number of papers that Mitchell delivers, 𝑷𝑷, and the amount of money Mitchell earns in dollars, 𝑬𝑬. Answers will vary. Sample answers are shown. Number of Papers Delivered (𝑷𝑷) Mitchell’s Earnings in Dollars (𝑬𝑬) 𝟏𝟏 𝟎𝟎. 𝟎𝟎𝟎𝟎 𝟏𝟏𝟏𝟏 𝟎𝟎. 𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗. 𝟎𝟎𝟎𝟎 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟗𝟗𝟗𝟗. 𝟎𝟎𝟎𝟎 a. If you know how many papers Mitchell delivered, can you determine how much money he earned? Write the corresponding expression. Multiplying the number of papers that Mitchell delivers by his rate ($𝟎𝟎. 𝟎𝟎𝟎𝟎 per paper) will calculate his pay. 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 is the expression for his pay in dollars. A STORY OF RATIOS 220 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 20 Lesson 20: Writing and Evaluating Expressions—Multiplication and Division b. Use your expression to determine how much Mitchell earned by delivering 𝟑𝟑𝟑𝟑𝟑𝟑 newspapers. 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎= 𝟎𝟎. 𝟎𝟎𝟎𝟎∙𝟑𝟑𝟑𝟑𝟑𝟑= 𝟐𝟐𝟐𝟐. Mitchell earned $𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 for delivering 𝟑𝟑𝟑𝟑𝟑𝟑 newspapers. c. If you know how much money Mitchell earned, can you determine how many papers he delivered? Write the corresponding expression. Dividing Mitchell’s pay by $𝟎𝟎. 𝟎𝟎𝟎𝟎 will calculate the number of papers he delivered. 𝑬𝑬÷ 𝟎𝟎. 𝟎𝟎𝟎𝟎 is the expression for the number of papers he delivered. d. Use your expression to determine how many papers Mitchell delivered if he earned $𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓 last week. 𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓÷ 𝟎𝟎. 𝟎𝟎𝟎𝟎= 𝟔𝟔𝟔𝟔𝟔𝟔; therefore, Mitchell delivered 𝟔𝟔𝟔𝟔𝟔𝟔 newspapers last week. 5. Randy is an art dealer who sells reproductions of famous paintings. Copies of the Mona Lisa sell for $𝟒𝟒𝟒𝟒𝟒𝟒. a. Last year Randy sold $𝟗𝟗, 𝟗𝟗𝟗𝟗𝟗𝟗 worth of Mona Lisa reproductions. How many did he sell? 𝟗𝟗, 𝟗𝟗𝟗𝟗𝟗𝟗÷ 𝟒𝟒𝟒𝟒𝟒𝟒= 𝟐𝟐𝟐𝟐. He sold 𝟐𝟐𝟐𝟐 copies of the painting. b. If Randy wants to increase his sales to at least $𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 this year, how many copies will he need to sell (without changing the price per painting)? 𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎÷ 𝟒𝟒𝟒𝟒𝟒𝟒 is about 𝟑𝟑𝟑𝟑. 𝟔𝟔. He will have to sell 𝟑𝟑𝟑𝟑 paintings in order to increase his sales to at least $𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎. A STORY OF RATIOS 221 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Student Outcomes Students develop formulas involving multiplication and addition from real-world problems. Students evaluate these formulas for given values. Lesson Notes This lesson begins with students making a model of a real-world problem they most likely have already encountered: moving enough tables together so that a large group of people can sit together. After the problem is posed, students use square tiles to model the problem. Using this data and looking for patterns, they make generalizations about the expression that describes the problem. It is necessary to prepare bags of five square tiles before class and to group students thoughtfully. Classwork Opening (2 minutes) Move students into groups of two or three, and distribute the bags of tiles. Mathematical Modeling Exercise (15 minutes) Today, we will model a problem that happens in restaurants every day: moving tables together so that everyone in a group can sit together. Use the square tiles to represent square tables. One person can sit along each edge of a table side, no crowding. Our first goal is to find how many people can sit at tables made of various numbers of square tables pushed together end to end. How many chairs can fit around one square table? What is the perimeter of the square if the edge length is one yard? Record the results in your table. 4; 4 yards If two square tables are pushed together to form a longer rectangular table, how many chairs will fit around the new table? What is the perimeter of the rectangle? Record the results in your table. 6; 6 yards Make sure that each student can connect the square model on the desk to the picture on the classwork sheet. If there are twice as many square tables in the new rectangular table, why can’t twice as many chairs fit around it? No chairs will fit right where the tables come together. A STORY OF RATIOS 222 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Make a record of the number of chairs that will fit around longer rectangular tables when 3, 4, and 5 square tables are pushed together to form long rectangular tables. Mathematical Modeling Exercise The Italian Villa Restaurant has square tables that the servers can push together to accommodate the customers. Only one chair fits along the side of the square table. Make a model of each situation to determine how many seats will fit around various rectangular tables. Number of Square Tables Number of Seats at the Table 𝟏𝟏 𝟒𝟒 𝟐𝟐 𝟔𝟔 𝟑𝟑 𝟖𝟖 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒𝟒𝟒 𝑻𝑻 𝟐𝟐𝟐𝟐+ 𝟐𝟐 or 𝟐𝟐(𝑻𝑻+ 𝟏𝟏) Are there any other ways to think about solutions to this problem? Regardless of the number of tables, there is one chair on each end, and each table has two chairs opposite one another. It is impractical to make a model of pushing 𝟓𝟓𝟓𝟓 tables together to make a long rectangle. If we did have a rectangle that long, how many chairs would fit on the long sides of the table? 𝟓𝟓𝟓𝟓 on each side, for a total of 𝟏𝟏𝟏𝟏𝟏𝟏 How many chairs fit on the ends of the long table? 𝟐𝟐 chairs, one on each end How many chairs fit in all? Record it on your table. 𝟏𝟏𝟏𝟏𝟏𝟏 chairs in all Work with your group to determine how many chairs would fit around a very long rectangular table if 𝟐𝟐𝟐𝟐𝟐𝟐 square tables were pushed together. 𝟐𝟐𝟐𝟐𝟐𝟐 chairs on each side, totaling 𝟒𝟒𝟒𝟒𝟒𝟒, plus one on each end; grand total 𝟒𝟒𝟒𝟒𝟒𝟒 If we let 𝑻𝑻 represent the number of square tables that make one long rectangular table, what is the expression for the number of chairs that will fit around it? 𝟐𝟐𝟐𝟐+ 𝟐𝟐 A STORY OF RATIOS 223 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Example (13 minutes) Example Look at Example 1 with your group. Determine the cost for various numbers of pizzas, and also determine the expression that describes the cost of having 𝑷𝑷 pizzas delivered. a. Pizza Queen has a special offer on lunch pizzas: $𝟒𝟒. 𝟎𝟎𝟎𝟎 each. They charge $𝟐𝟐. 𝟎𝟎𝟎𝟎 to deliver, regardless of how many pizzas are ordered. Determine the cost for various numbers of pizzas, and also determine the expression that describes the cost of having 𝑷𝑷 pizzas delivered. Number of Pizzas Delivered Total Cost in Dollars 𝟏𝟏 𝟔𝟔 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒 𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐𝟐𝟐 𝑷𝑷 𝟒𝟒𝟒𝟒+ 𝟐𝟐 Allow the groups to discover patterns and share them. What mathematical operations did you need to perform to find the total cost? Multiplication and addition. We multiplied the number of pizzas by $𝟒𝟒 and then added the $𝟐𝟐 delivery fee. Suppose our principal wanted to buy a pizza for everyone in our class. Determine how much this would cost. Answers will vary depending on the number of students in your class. b. If the booster club had $𝟒𝟒𝟒𝟒𝟒𝟒 to spend on pizza, what is the greatest number of pizzas they could order? Students can use the “guess and check” method for answering this question. A scaffold question might be, “Could they order 100 pizzas at this price?” The greatest number of pizzas they could order would be 𝟗𝟗𝟗𝟗. The pizzas themselves would cost 𝟗𝟗𝟗𝟗× $𝟒𝟒= $𝟑𝟑𝟑𝟑𝟑𝟑, and then add $𝟐𝟐. 𝟎𝟎𝟎𝟎 for delivery. The total bill is $𝟑𝟑𝟑𝟑𝟑𝟑. A STORY OF RATIOS 224 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition c. If the pizza price was raised to $𝟓𝟓. 𝟎𝟎𝟎𝟎 and the delivery price was raised to $𝟑𝟑. 𝟎𝟎𝟎𝟎, create a table that shows the total cost (pizza plus delivery) of 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒, and 𝟓𝟓 pizzas. Include the expression that describes the new cost of ordering 𝑷𝑷 pizzas. Number of Pizzas Delivered Total Cost in Dollars 𝟏𝟏 𝟖𝟖 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟐𝟐𝟐𝟐 𝟓𝟓 𝟐𝟐𝟐𝟐 𝑷𝑷 𝟓𝟓𝟓𝟓+ 𝟑𝟑 Closing (8 minutes) Some mathematical expressions use both multiplication and addition. With your partner, make up a new example of a problem that uses both multiplication and addition. Allow a short time for groups to make up a situation. Share these as a group. Ensure that there is both a coefficient and a constant in each problem. Naming these terms is not important for this lesson. Exit Ticket (7 minutes) A STORY OF RATIOS 225 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Name Date Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Exit Ticket Krystal Klear Cell Phone Company charges $5.00 per month for service. The company also charges $0.10 for each text message sent. a. Complete the table below to calculate the monthly charges for various numbers of text messages sent. Number of Text Messages Sent (𝑻𝑻) Total Monthly Bill in Dollars 0 10 20 30 𝑇𝑇 b. If Suzannah’s budget limit is $10 per month, how many text messages can she send in one month? A STORY OF RATIOS 226 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition Exit Ticket Sample Solutions Krystal Klear Cell Phone Company charges $𝟓𝟓. 𝟎𝟎𝟎𝟎 per month for service. The company also charges $𝟎𝟎. 𝟏𝟏𝟏𝟏 for each text message sent. a. Complete the table below to calculate the monthly charges for various numbers of text messages sent. Number of Text Messages Sent (𝑻𝑻) Total Monthly Bill in Dollars 𝟎𝟎 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟔𝟔 𝟐𝟐𝟐𝟐 𝟕𝟕 𝟑𝟑𝟑𝟑 𝟖𝟖 𝑻𝑻 𝟎𝟎. 𝟏𝟏𝟏𝟏+ 𝟓𝟓 b. If Suzannah’s budget limit is $𝟏𝟏𝟏𝟏 per month, how many text messages can she send in one month? Suzannah can send 𝟓𝟓𝟓𝟓 text messages in one month for $𝟏𝟏𝟏𝟏. Problem Set Sample Solutions 1. Compact discs (CDs) cost $𝟏𝟏𝟏𝟏 each at the Music Emporium. The company charges $𝟒𝟒. 𝟓𝟓𝟓𝟓 for shipping and handling, regardless of how many compact discs are purchased. a. Create a table of values that shows the relationship between the number of compact discs that Mickey buys, 𝑫𝑫, and the amount of money Mickey spends, 𝑪𝑪, in dollars. Number of CDs Mickey Buys (𝑫𝑫) Total Cost in Dollars (𝑪𝑪) 𝟏𝟏 $𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟐𝟐 $𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓 𝟑𝟑 $𝟒𝟒𝟒𝟒. 𝟓𝟓𝟓𝟓 b. If you know how many CDs Mickey orders, can you determine how much money he spends? Write the corresponding expression. 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟒𝟒. 𝟓𝟓 c. Use your expression to determine how much Mickey spent buying 𝟖𝟖 CDs. 𝟖𝟖(𝟏𝟏𝟏𝟏) + 𝟒𝟒. 𝟓𝟓𝟓𝟓= 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓. Mickey spent $𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓. 2. Mr. Gee’s class orders paperback books from a book club. The books cost $𝟐𝟐. 𝟗𝟗𝟗𝟗 each. Shipping charges are set at $𝟒𝟒. 𝟎𝟎𝟎𝟎, regardless of the number of books purchased. a. Create a table of values that shows the relationship between the number of books that Mr. Gee’s class buys, 𝑩𝑩, and the amount of money they spend, 𝑪𝑪, in dollars. Number of Books Ordered(𝑩𝑩) Amount of Money Spent in Dollars (𝑪𝑪) 𝟏𝟏 𝟔𝟔. 𝟗𝟗𝟗𝟗 𝟐𝟐 𝟗𝟗. 𝟗𝟗𝟗𝟗 𝟑𝟑 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 A STORY OF RATIOS 227 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition b. If you know how many books Mr. Gee’s class orders, can you determine how much money they spend? Write the corresponding expression. 𝟐𝟐. 𝟗𝟗𝟗𝟗𝟗𝟗+ 𝟒𝟒 c. Use your expression to determine how much Mr. Gee’s class spent buying 𝟐𝟐𝟐𝟐 books. 𝟐𝟐𝟐𝟐(𝟐𝟐. 𝟗𝟗𝟗𝟗) + 𝟒𝟒= 𝟕𝟕𝟕𝟕. Mr. Gee’s class spent $𝟕𝟕𝟕𝟕. 𝟖𝟖𝟖𝟖. 3. Sarah is saving money to take a trip to Oregon. She received $𝟒𝟒𝟒𝟒𝟒𝟒 in graduation gifts and saves $𝟏𝟏𝟏𝟏𝟏𝟏 per week working. a. Write an expression that shows how much money Sarah has after working 𝑾𝑾 weeks. 𝟒𝟒𝟒𝟒𝟒𝟒+ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 b. Create a table that shows the relationship between the amount of money Sarah has (𝑴𝑴) and the number of weeks she works (𝑾𝑾). Amount of Money Sarah Has (𝑴𝑴) Number of Weeks Worked (𝑾𝑾) 𝟓𝟓𝟓𝟓𝟓𝟓 𝟏𝟏 𝟔𝟔𝟔𝟔𝟔𝟔 𝟐𝟐 𝟖𝟖𝟖𝟖𝟖𝟖 𝟑𝟑 𝟗𝟗𝟗𝟗𝟗𝟗 𝟒𝟒 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟓𝟓 𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔 𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐 𝟕𝟕 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 𝟖𝟖 c. The trip will cost $𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐. How many weeks will Sarah have to work to earn enough for the trip? Sarah will have to work 𝟕𝟕 weeks to earn enough for the trip. 4. Mr. Gee’s language arts class keeps track of how many words per minute are read aloud by each of the students. They collect this oral reading fluency data each month. Below is the data they collected for one student in the first four months of school. a. Assume this increase in oral reading fluency continues throughout the rest of the school year. Complete the table to project the reading rate for this student for the rest of the year. Month Number of Words Read Aloud in One Minute September 𝟏𝟏𝟏𝟏𝟏𝟏 October 𝟏𝟏𝟏𝟏𝟏𝟏 November 𝟏𝟏𝟏𝟏𝟏𝟏 December 𝟏𝟏𝟏𝟏𝟏𝟏 January 𝟏𝟏𝟏𝟏𝟏𝟏 February 𝟏𝟏𝟏𝟏𝟏𝟏 March 𝟏𝟏𝟏𝟏𝟏𝟏 April 𝟏𝟏𝟏𝟏𝟏𝟏 May 𝟏𝟏𝟏𝟏𝟏𝟏 June 𝟏𝟏𝟏𝟏𝟏𝟏 A STORY OF RATIOS 228 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 21 Lesson 21: Writing and Evaluating Expressions―Multiplication and Addition b. If this increase in oral reading fluency continues throughout the rest of the school year, when would this student achieve the goal of reading 𝟏𝟏𝟏𝟏𝟏𝟏 words per minute? The student will meet the goal in May. c. The expression for this student’s oral reading fluency is 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟓𝟓𝟓𝟓, where 𝒎𝒎 represents the number of months during the school year. Use this expression to determine how many words per minute the student would read after 𝟏𝟏𝟏𝟏 months of instruction. The student would read 𝟏𝟏𝟏𝟏𝟏𝟏 words per minute: 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟓𝟓× 𝟏𝟏𝟏𝟏. 5. When corn seeds germinate, they tend to grow 𝟓𝟓 inches in the first week and then 𝟑𝟑 inches per week for the remainder of the season. The relationship between the height (𝑯𝑯) and the number of weeks since germination (𝑾𝑾) is shown below. a. Complete the missing values in the table. Number of Weeks Since Germination (𝑾𝑾) Height of Corn Plant (𝑯𝑯) 𝟏𝟏 𝟓𝟓 𝟐𝟐 𝟖𝟖 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟔𝟔 𝟐𝟐𝟐𝟐 b. The expression for this height is 𝟐𝟐+ 𝟑𝟑𝟑𝟑. How tall will the corn plant be after 𝟏𝟏𝟏𝟏 weeks of growth? 𝟐𝟐+ 𝟑𝟑(𝟏𝟏𝟏𝟏) = 𝟒𝟒𝟒𝟒. The plant will be 𝟒𝟒𝟒𝟒 inches tall. 6. The Honeymoon Charter Fishing Boat Company only allows newlywed couples on their sunrise trips. There is a captain, a first mate, and a deck hand manning the boat on these trips. a. Write an expression that shows the number of people on the boat when there are 𝑪𝑪 couples booked for the trip. 𝟑𝟑+ 𝟐𝟐𝟐𝟐 b. If the boat can hold a maximum of 𝟐𝟐𝟐𝟐 people, how many couples can go on the sunrise fishing trip? Eight couples (𝟏𝟏𝟏𝟏 passengers) can fit along with the 𝟑𝟑 crew members, totaling 𝟏𝟏𝟏𝟏 people on the boat. A ninth couple would overload the boat. A STORY OF RATIOS 229 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Lesson 22: Writing and Evaluating Expressions—Exponents Student Outcomes Students evaluate and write formulas involving exponents for given values in real-world problems. Lesson Notes Exponents are used in calculations of both area and volume. Other examples of exponential applications involve bacterial growth (powers of 2) and compound interest. Students need a full-size sheet of paper (8 1 2 × 11 inches) for the first example. Teachers should try the folding activity ahead of time to anticipate outcomes. If time permits at the end of the lesson, a larger sheet of paper can be used to experiment further. Classwork Fluency Exercise (10 minutes): Multiplication of Decimals RWBE: Refer to the Rapid White Board Exchanges sections in the Module Overview for directions on how to administer an RWBE. Example 1 (5 minutes): Folding Paper Ask students to predict how many times they can fold a piece of paper in half. Allow a short discussion before allowing students to try it. Predict how many times you can fold a piece of paper in half. The folds must be as close to a half as possible. Record your prediction in Exercise 1. Students repeatedly fold a piece of paper until it is impossible, about seven folds. Remind students they must fold the paper the same way each time. Fold the paper once. Record the number of layers of paper that result in the table in Exercise 2. 2 Fold again. Record the number of layers of paper that result. 4 Ensure that students see that doubling the two sheets results in four sheets. At this stage, the layers can easily be counted. During subsequent stages, it is impractical to do so. Focus the count on the corner that has four loose pieces. Fold again. Count and record the number of layers you have now. 8 The number of layers is doubling from one stage to the next; so, the pattern is modeled by multiplying by 2, not adding 2. It is critical that students find that there are eight layers here, not six. Scaffolding: Some students benefit from unfolding and counting rectangles on the paper throughout Example 1. This provides a concrete representation of the exponential relationship at the heart of this lesson. A STORY OF RATIOS 230 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Continue folding and recording the number of layers you make. Use a calculator if desired. Record your answers as both numbers in standard form and exponential form, as powers of 2. Exercises (5 minutes) Exercises 1. Predict how many times you can fold a piece of paper in half. My prediction: 2. Before any folding (zero folds), there is only one layer of paper. This is recorded in the first row of the table. Fold your paper in half. Record the number of layers of paper that result. Continue as long as possible. Number of Folds Number of Paper Layers That Result Number of Paper Layers Written as a Power of 𝟐𝟐 𝟎𝟎 𝟏𝟏 𝟐𝟐𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟐𝟐𝟏𝟏 𝟐𝟐 𝟒𝟒 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟖𝟖 𝟐𝟐𝟑𝟑 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟐𝟐𝟒𝟒 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟐𝟐𝟓𝟓 𝟔𝟔 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔 𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟕𝟕 𝟖𝟖 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟖𝟖 a. Are you able to continue folding the paper indefinitely? Why or why not? No. The stack got too thick on one corner because it kept doubling each time. b. How could you use a calculator to find the next number in the series? I could multiply the number by 𝟐𝟐 to find the number of layers after another fold. c. What is the relationship between the number of folds and the number of layers? As the number of folds increases by one, the number of layers doubles. d. How is this relationship represented in exponential form of the numerical expression? I could use 𝟐𝟐 as a base and the number of folds as the exponent. e. If you fold a paper 𝒇𝒇 times, write an expression to show the number of paper layers. There would be 𝟐𝟐𝒇𝒇 layers of paper. 3. If the paper were to be cut instead of folded, the height of the stack would double at each successive stage, and it would be possible to continue. a. Write an expression that describes how many layers of paper result from 𝟏𝟏𝟏𝟏 cuts. 𝟐𝟐𝟏𝟏𝟏𝟏 b. Evaluate this expression by writing it in standard form. 𝟐𝟐𝟏𝟏𝟏𝟏= 𝟔𝟔𝟔𝟔, 𝟓𝟓𝟓𝟓𝟓𝟓 A STORY OF RATIOS 231 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Example 2 (10 minutes): Bacterial Infection Modeling of exponents in real life leads to our next example of the power of doubling. Think about the last time you had a cut or a wound that became infected. What caused the infection? Bacteria growing in the wound When colonies of certain types of bacteria are allowed to grow unchecked, serious illness can result. Example 2: Bacterial Infection Bacteria are microscopic single-celled organisms that reproduce in a couple of different ways, one of which is called binary fission. In binary fission, a bacterium increases its size until it is large enough to split into two parts that are identical. These two grow until they are both large enough to split into two individual bacteria. This continues as long as growing conditions are favorable. a. Record the number of bacteria that result from each generation. Generation Number of Bacteria Number of Bacteria Written as a Power of 𝟐𝟐 𝟏𝟏 𝟐𝟐 𝟐𝟐𝟏𝟏 𝟐𝟐 𝟒𝟒 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟖𝟖 𝟐𝟐𝟑𝟑 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟐𝟐𝟒𝟒 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟐𝟐𝟓𝟓 𝟔𝟔 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔 𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟕𝟕 𝟖𝟖 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟖𝟖 𝟗𝟗 𝟓𝟓𝟓𝟓𝟓𝟓 𝟐𝟐𝟗𝟗 𝟏𝟏𝟏𝟏 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟐𝟐𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟐𝟐𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟒𝟒, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟐𝟐𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟖𝟖, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐𝟏𝟏𝟏𝟏 b. How many generations would it take until there were over one million bacteria present? 𝟐𝟐𝟐𝟐 generations will produce more than one million bacteria. 𝟐𝟐𝟐𝟐𝟐𝟐= 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟓𝟓𝟓𝟓𝟓𝟓 c. Under the right growing conditions, many bacteria can reproduce every 𝟏𝟏𝟏𝟏 minutes. Under these conditions, how long would it take for one bacterium to reproduce itself into more than one million bacteria? It would take 𝟐𝟐𝟐𝟐 fifteen-minute periods, or 𝟓𝟓 hours. d. Write an expression for how many bacteria would be present after 𝒈𝒈 generations. There will be 𝟐𝟐𝒈𝒈 bacteria present after 𝒈𝒈 generations. A STORY OF RATIOS 232 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Example 3 (10 minutes): Volume of a Rectangular Solid Exponents are used when we calculate the volume of rectangular solids. Example 3: Volume of a Rectangular Solid This box has a width, 𝒘𝒘. The height of the box, 𝒉𝒉, is twice the width. The length of the box, 𝒍𝒍, is three times the width. That is, the width, height, and length of a rectangular prism are in the ratio of 𝟏𝟏: 𝟐𝟐: 𝟑𝟑. For rectangular solids like this, the volume is calculated by multiplying length times width times height. 𝑽𝑽= 𝒍𝒍· 𝒘𝒘· 𝒉𝒉 𝑽𝑽= 𝟑𝟑𝟑𝟑· 𝒘𝒘· 𝟐𝟐𝟐𝟐 𝑽𝑽= 𝟑𝟑· 𝟐𝟐· 𝒘𝒘· 𝒘𝒘· 𝒘𝒘 𝑽𝑽= 𝟔𝟔𝒘𝒘𝟑𝟑 Follow the above example to calculate the volume of these rectangular solids, given the width, 𝒘𝒘. Width in Centimeters (𝐜𝐜𝐜𝐜) Volume in Cubic Centimeters (𝐜𝐜𝐦𝐦𝟑𝟑) 𝟏𝟏 𝟏𝟏 𝐜𝐜𝐜𝐜× 𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟑𝟑 𝐜𝐜𝐜𝐜= 𝟔𝟔 𝐜𝐜𝐦𝐦𝟑𝟑 𝟐𝟐 𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟒𝟒 𝐜𝐜𝐜𝐜× 𝟔𝟔 𝐜𝐜𝐜𝐜= 𝟒𝟒𝟒𝟒 𝐜𝐜𝐦𝐦𝟑𝟑 𝟑𝟑 𝟑𝟑 𝐜𝐜𝐜𝐜× 𝟔𝟔 𝐜𝐜𝐜𝐜× 𝟗𝟗 𝐜𝐜𝐜𝐜= 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟑𝟑 𝟒𝟒 𝟒𝟒 𝐜𝐜𝐜𝐜× 𝟖𝟖 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜= 𝟑𝟑𝟑𝟑𝟑𝟑 𝐜𝐜𝐦𝐦𝟑𝟑 𝒘𝒘 𝒘𝒘 𝐜𝐜𝐜𝐜× 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜× 𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜= 𝟔𝟔 𝒘𝒘𝟑𝟑 𝐜𝐜𝐦𝐦𝟑𝟑 Closing (2 minutes) Why is 53 different from 5 × 3? 53 means 5 × 5 × 5. Five is the factor that will be multiplied by itself 3 times. That equals 125. On the other hand, 5 × 3 means 5 + 5 + 5. Five is the addend that will be added to itself 3 times. This equals 15. Exit Ticket (3 minutes) 𝒘𝒘 𝒉𝒉= 𝟐𝟐𝟐𝟐 𝒍𝒍= 𝟑𝟑𝟑𝟑 MP.3 A STORY OF RATIOS 233 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Name Date Lesson 22: Writing and Evaluating Expressions—Exponents Exit Ticket 1. Naomi’s allowance is $2.00 per week. If she convinces her parents to double her allowance each week for two months, what will her weekly allowance be at the end of the second month (week 8)? Week Number Allowance 1 $2.00 2 3 4 5 6 7 8 𝑤𝑤 2. Write the expression that describes Naomi’s allowance during week 𝑤𝑤 in dollars. A STORY OF RATIOS 234 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Exit Ticket Sample Solutions 1. Naomi’s allowance is $𝟐𝟐. 𝟎𝟎𝟎𝟎 per week. If she convinces her parents to double her allowance each week for two months, what will her weekly allowance be at the end of the second month (week 𝟖𝟖)? Week Number Allowance 𝟏𝟏 $𝟐𝟐. 𝟎𝟎𝟎𝟎 𝟐𝟐 $𝟒𝟒. 𝟎𝟎𝟎𝟎 𝟑𝟑 $𝟖𝟖. 𝟎𝟎𝟎𝟎 𝟒𝟒 $𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟓𝟓 $𝟑𝟑𝟑𝟑. 𝟎𝟎𝟎𝟎 𝟔𝟔 $𝟔𝟔𝟔𝟔. 𝟎𝟎𝟎𝟎 𝟕𝟕 $𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟖𝟖 $𝟐𝟐𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 𝒘𝒘 $𝟐𝟐𝒘𝒘 2. Write the expression that describes Naomi’s allowance during week 𝒘𝒘 in dollars. $𝟐𝟐𝒘𝒘 A STORY OF RATIOS 235 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Problem Set Sample Solutions 1. A checkerboard has 𝟔𝟔𝟔𝟔 squares on it. a. If one grain of rice is put on the first square, 𝟐𝟐 grains of rice on the second square, 𝟒𝟒 grains of rice on the third square, 𝟖𝟖 grains of rice on the fourth square, and so on (doubling each time), complete the table to show how many grains of rice are on each square. Write your answers in exponential form on the table below. Checkerboard Square Grains of Rice Checkerboard Square Grains of Rice Checkerboard Square Grains of Rice Checkerboard Square Grains of Rice 𝟏𝟏 𝟐𝟐𝟎𝟎 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟐𝟐 𝟐𝟐𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟒𝟒𝟒𝟒 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟒𝟒 𝟐𝟐𝟑𝟑 𝟐𝟐𝟐𝟐 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟓𝟓 𝟐𝟐𝟒𝟒 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟔𝟔 𝟐𝟐𝟓𝟓 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟕𝟕 𝟐𝟐𝟔𝟔 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟖𝟖 𝟐𝟐𝟕𝟕 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟑𝟑𝟑𝟑 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟗𝟗 𝟐𝟐𝟖𝟖 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟗𝟗 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟓𝟓𝟓𝟓 𝟐𝟐𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟐𝟐𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 𝟐𝟐𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔 𝟐𝟐𝟔𝟔𝟔𝟔 b. How many grains of rice would be on the last square? Represent your answer in exponential form and standard form. Use the table above to help solve the problem. There would be 𝟐𝟐𝟔𝟔𝟔𝟔or 𝟗𝟗 , 𝟐𝟐𝟐𝟐𝟐𝟐 , 𝟑𝟑𝟑𝟑𝟑𝟑 , 𝟎𝟎𝟎𝟎𝟎𝟎 , 𝟖𝟖𝟖𝟖𝟖𝟖 , 𝟕𝟕𝟕𝟕𝟕𝟕 , 𝟖𝟖𝟖𝟖𝟖𝟖 grains of rice. c. Would it have been easier to write your answer to part (b) in exponential form or standard form? Answers will vary. Exponential form is more concise: 𝟐𝟐𝟔𝟔𝟔𝟔. Standard form is longer and more complicated to calculate: 𝟗𝟗, 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟑𝟑𝟑𝟑𝟑𝟑, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟖𝟖𝟖𝟖𝟖𝟖, 𝟕𝟕𝟕𝟕𝟕𝟕, 𝟖𝟖𝟖𝟖𝟖𝟖. (In word form: nine quintillion, two hundred twenty-three quadrillion, three hundred seventy-two trillion, thirty-six billion, eight hundred fifty-four million, seven hundred seventy-five thousand, eight hundred eight.) A STORY OF RATIOS 236 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents 2. If an amount of money is invested at an annual interest rate of 𝟔𝟔%, it doubles every 𝟏𝟏𝟏𝟏 years. If Alejandra invests $𝟓𝟓𝟓𝟓𝟓𝟓, how long will it take for her investment to reach $𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 (assuming she does not contribute any additional funds)? It will take 𝟐𝟐𝟐𝟐 years. After 𝟏𝟏𝟏𝟏 years, Alejandra will have doubled her money and will have $𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎. If she waits an additional 𝟏𝟏𝟏𝟏 years, she will have $𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎. 3. The athletics director at Peter’s school has created a phone tree that is used to notify team players in the event a game has to be canceled or rescheduled. The phone tree is initiated when the director calls two captains. During the second stage of the phone tree, the captains each call two players. During the third stage of the phone tree, these players each call two other players. The phone tree continues until all players have been notified. If there are 𝟓𝟓𝟓𝟓 players on the teams, how many stages will it take to notify all of the players? It will take five stages. After the first stage, two players have been called, and 𝟒𝟒𝟒𝟒 will not have been called. After the second stage, four more players will have been called, for a total of six; 𝟒𝟒𝟒𝟒 players will remain uncalled. After the third stage, 𝟐𝟐𝟑𝟑 players (eight) more will have been called, totaling 𝟏𝟏𝟏𝟏; 𝟑𝟑𝟑𝟑 remain uncalled. After the fourth stage, 𝟐𝟐𝟒𝟒 more players (𝟏𝟏𝟏𝟏) will have gotten a call, for a total of 𝟑𝟑𝟑𝟑 players notified. Twenty remain uncalled at this stage. The fifth round of calls will cover all of them because 𝟐𝟐𝟓𝟓 includes 𝟑𝟑𝟑𝟑 more players. A STORY OF RATIOS 237 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 22 Lesson 22: Writing and Evaluating Expressions—Exponents Multiplication of Decimals Progression of Exercises 1. 0.5 × 0.5 = 𝟎𝟎. 𝟐𝟐𝟐𝟐 2. 0.6 × 0.6 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 3. 0.7 × 0.7 = 𝟎𝟎. 𝟒𝟒𝟒𝟒 4. 0.5 × 0.6 = 𝟎𝟎. 𝟑𝟑 5. 1.5 × 1.5 = 𝟐𝟐. 𝟐𝟐𝟐𝟐 6. 2.5 × 2.5 = 𝟔𝟔. 𝟐𝟐𝟐𝟐 7. 0.25 × 0.25 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 8. 0.1 × 0.1 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 9. 0.1 × 123.4 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 10. 0.01 × 123.4 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 A STORY OF RATIOS 238 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Topic G: Solving Equations GRADE 6 • MODULE 4 6 GRADE Mathematics Curriculum Topic G Solving Equations 6.EE.B.5, 6.EE.B.6, 6.EE.B.7 Focus Standards: 6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form 𝑥𝑥+ 𝑝𝑝= 𝑞𝑞 and 𝑝𝑝𝑝𝑝= 𝑞𝑞 for cases in which 𝑝𝑝, 𝑞𝑞, and 𝑥𝑥 are all nonnegative rational numbers. Instructional Days: 7 Lessons 23–24: True and False Number Sentences (P, P)1 Lesson 25: Finding Solutions to Make Equations True (P) Lesson 26: One-Step Equations—Addition and Subtraction (M) Lesson 27: One-Step Equations—Multiplication and Division (E) Lesson 28: Two-Step Problems—All Operations (M) Lesson 29: Multi-Step Problems—All Operations (P) In Topic G, students move from identifying true and false number sentences to making true number sentences false and false number sentences true. In Lesson 23, students explain what equality and inequality symbols represent. They determine if a number sentence is true or false based on the equality or inequality symbol. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 239 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic G Topic G: Solving Equations Symbol Meaning Example = Is equal to 1.8 + 3 = 4.8 ≠ Is not equal to 6 ÷ 1 2 ≠3 > Is greater than 1 > 0.9 < Is less than 1 4 < 1 2 In Lesson 24, students move to identifying a value or a set of values that makes number sentences true. They identify values that make a true sentence false. For example, students substitute 4 for the variable in 𝑥𝑥+ 12 = 14 to determine if the sentence is true or false. They note that when 4 is substituted for 𝑥𝑥, the sum of 𝑥𝑥+ 12 is 16, which makes the sentence false because 16 ≠14. They change course in the lesson to find what they can do to make the sentence true. They ask themselves, “What number must we add to 12 to find the sum of 14?” By substituting 2 for 𝑥𝑥, the sentence becomes true because 𝑥𝑥+ 12 = 14, 2 + 12 = 14, and 14 = 14. They bridge this discovery to Lesson 25 where students understand that the solution of an equation is the value or values of the variable that make the equation true. Students begin solving equations in Lesson 26. They use bar models or tape diagrams to depict an equation and apply previously learned properties of equality for addition and subtraction to solve the equation. Students check to determine if their solutions make the equation true. Given the equation 1 + 𝑎𝑎= 6, students represent the equation with the following model: Students recognize that the solution can also be found using properties of operations. They make connections to the model and determine that 1 + 𝑎𝑎−1 = 6 −1 and, ultimately, that 𝑎𝑎= 5. Students represent two-step and multi-step equations involving all operations with bar models or tape diagrams while continuing to apply properties of operations and the order of operations to solve equations in the remaining lessons in this topic. 6 1 5 6 1 𝑎𝑎 5 𝑎𝑎 A STORY OF RATIOS 240 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Lesson 23: True and False Number Sentences Student Outcomes Students explain what the equality and inequality symbols including =, <, >, ≤, and ≥ represent. They determine if a number sentence is true or false based on the given symbol. Lesson Notes For students to be prepared to solve equations later in this module, it is important that they understand truth values in number sentences and in equations. In the next three lessons, students learn to differentiate between number sentences and generic equations. Later, in Lesson 26, students learn why number sentences play a fundamental role in understanding both equations with unknown numbers and solution sets of equations that contain variables. Number sentences are special among types of equations because they have truth values. A number sentence is the most concrete version of an equation. It has the very important property that it is always true or always false, and it is this property that distinguishes it from a generic equation. Examples include 2 + 3 = 5 (true) and 2 + 2 = 5 (false). The property guarantees the ability to check whether or not a number is a solution to an equation with a variable. Just substitute the number into the variable; then, check to see if the resulting number sentence is either true or false. If the number sentence is true, the number is a solution. For that reason, number sentences are the first and most important type of equation that students need to understand. Lesson 23 begins by first determining what is true and false and then moving to true and false number sentences using equality and inequality symbols. Classwork Opening (4 minutes) Discuss the meaning of true and false by calling on students to tell if the following statements are true or false. Conclude with a discussion of what makes a number sentence true or false. Earth orbits the sun. True George Washington was the first president of the United States. True There are 25 hours in a day. False 3 + 3 = 6 True 2 + 2 = 5 False Why is 2 + 2 = 5 a false number sentence? Answers will vary but should include the idea that the expressions on both sides of the equal sign should evaluate to the same number; so, for this number sentence to be true, either the first or second addend should be three, resulting in the sum of five. A STORY OF RATIOS 241 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Opening Exercise/Discussion (8 minutes) Discuss what each symbol below stands for, and provide students with an example. Students can complete the table in their student materials. Have one student come up to the front of the room and stand against the wall or board. Mark the student’s height with a piece of masking tape. (It is important to pick a student who is about average height.) Measure the height of the tape mark using a tape measure, and record it as a mixed number in feet, along with the student’s name, on the piece of masking tape. Next, start the following table on the board/screen: Opening Exercise Determine what each symbol stands for, and provide an example. Symbol What the Symbol Stands For Example = is equal to 𝟒𝟒𝟕𝟕 𝟖𝟖= 𝟒𝟒. 𝟖𝟖𝟖𝟖𝟖𝟖 What is another example of a number sentence that includes an equal symbol? Answers will vary. Ask more than one student. The student’s height is the height marked by the tape on the wall. Have students stand next to the marked height. Discuss how their heights compare to the height of the tape. Are there other students in the room who have the same height? > is greater than 𝟓𝟓𝟏𝟏 𝟒𝟒> 𝟒𝟒𝟕𝟕 𝟖𝟖 Use the student’s height measurement in the example (the example uses a student 4 7 8 ft. in height). What is another example of a number sentence that includes a greater than symbol? Answers will vary. Ask more than one student. Have students taller than the tape on the wall stand near the tape. Discuss how more than one student has a height that is greater than the tape, so there could be more than one number inserted into the inequality: > 4 7 8. < is less than 𝟒𝟒𝟏𝟏 𝟐𝟐< 𝟒𝟒𝟕𝟕 𝟖𝟖 What is another example of a number sentence that includes a less than symbol? Answers will vary. Ask more than one student. Have students shorter than the tape on the wall stand near the tape. Discuss how more than one student has a height that is less than the tape, so there could be more than one number inserted into the inequality: < 4 7 8. A STORY OF RATIOS 242 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences ≤ is less than or equal to 𝟒𝟒𝟕𝟕 𝟖𝟖≤𝟒𝟒𝟕𝟕 𝟖𝟖 What is another example of a number sentence that includes a less than or equal to symbol? Answers will vary. Ask more than one student. Ask students who are the exact height as the tape and students who are shorter than the tape to stand near the tape. Discuss how this symbol is different from the previous symbol. ≥ is greater than or equal to 𝟓𝟓𝟏𝟏 𝟒𝟒≥𝟒𝟒𝟕𝟕 𝟖𝟖 What is another example of a number sentence that includes a greater than or equal to symbol? Answers will vary. Ask more than one student. Which students would stand near the tape to demonstrate this symbol? Students who are the same height as or taller than the tape Example 1 (5 minutes) Display each of the equations and inequalities one by one for students to review and determine whether they result in a true or a false number sentence. Example 1 For each equation or inequality your teacher displays, write the equation or inequality, and then substitute 𝟑𝟑 for every 𝒙𝒙. Determine if the equation or inequality results in a true number sentence or a false number sentence. Display 5 + 𝑥𝑥= 8. Substitute 3 for 𝑥𝑥 and evaluate. Does this result in a true number sentence or a false number sentence? True Why is the number sentence a true number sentence? Each expression on either side of the equal sign evaluates to 8. 8 = 8 Display 5𝑥𝑥= 8. Substitute 3 for 𝑥𝑥 and evaluate. Does this result in a true number sentence or a false number sentence? False Why is the number sentence a false number sentence? Five times three equals fifteen. Fifteen does not equal eight, so the number sentence 5(3) = 8 is false. Display 5 + 𝑥𝑥> 8. Substitute 3 for 𝑥𝑥 and evaluate. Does this result in a true number sentence or a false number sentence? False A STORY OF RATIOS 243 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Why is the number sentence a false number sentence? Each expression on either side of the inequality sign evaluates to 8. However, the inequality sign states that eight is greater than eight, which is not true. We have already shown that 8 = 8. Display 5𝑥𝑥> 8. Substitute 3 for 𝑥𝑥 and evaluate. Does this result in a true number sentence or a false number sentence? True Why is the number sentence a true number sentence? When three is substituted, the left side of the inequality evaluates to fifteen. Fifteen is greater than eight, so the number sentence is true. Display 5 + 𝑥𝑥≥8. Substitute 3 for 𝑥𝑥 and evaluate. Does this result in a true number sentence or a false number sentence? True Why is the number sentence a true number sentence? Each expression on either side of the inequality sign evaluates to 8. Because the inequality sign states that the expression on the left side can be greater than or equal to the expression on the right side, the number sentence is true because we have already shown that 8 = 8. Can you find a number other than three that we can substitute for 𝑥𝑥 that will result in a false number sentence? Answers will vary, but any number less than three will result in a false number sentence. Exercises (13 minutes) Students work on the following exercises independently. Note that students are writing complete sentences to describe what happens when a variable is substituted with a number and whether it turns the equation into a true number sentence or a false number sentence. Exercises Substitute the indicated value into the variable, and state (in a complete sentence) whether the resulting number sentence is true or false. If true, find a value that would result in a false number sentence. If false, find a value that would result in a true number sentence. 1. 𝟒𝟒+ 𝒙𝒙= 𝟏𝟏𝟏𝟏. Substitute 𝟖𝟖 for 𝒙𝒙. When 𝟖𝟖 is substituted for 𝒙𝒙, the number sentence is true. Answers will vary on values to make the sentence false; any number other than 𝟖𝟖 will make the sentence false. 2. 𝟑𝟑𝟑𝟑> 𝟏𝟏𝟏𝟏. Substitute 𝟒𝟒𝟏𝟏 𝟐𝟐 for 𝒈𝒈. When 𝟒𝟒𝟏𝟏 𝟐𝟐 is substituted for 𝒈𝒈, the number sentence is false. Answers will vary on values that make the sentence true; any number greater than 𝟓𝟓 will make the sentence true. MP.6 A STORY OF RATIOS 244 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences 3. 𝒇𝒇 𝟒𝟒< 𝟐𝟐. Substitute 𝟖𝟖 for 𝒇𝒇. When 𝟖𝟖 is substituted for 𝒇𝒇, the number sentence is false. Answers will vary on values to make the sentence true; any number less than 𝟖𝟖 will make the sentence true. 4. 𝟏𝟏𝟏𝟏. 𝟐𝟐≤𝒉𝒉−𝟏𝟏𝟏𝟏. 𝟑𝟑. Substitute 𝟐𝟐𝟐𝟐. 𝟖𝟖 for 𝒉𝒉. When 𝟐𝟐𝟐𝟐. 𝟖𝟖 is substituted for 𝒉𝒉, the number sentence is true. Answers will vary on values to make the sentence false; any number less than 𝟐𝟐𝟐𝟐. 𝟓𝟓 will make the sentence false. 5. 𝟒𝟒= 𝟖𝟖 𝒉𝒉. Substitute 𝟔𝟔 for 𝒉𝒉. When 𝟔𝟔 is substituted for 𝒉𝒉, the number sentence is false. 𝟐𝟐 is the only value that will make the sentence true. 6. 𝟑𝟑> 𝒌𝒌+ 𝟏𝟏 𝟒𝟒. Substitute 𝟏𝟏𝟏𝟏 𝟐𝟐 for 𝒌𝒌. When 𝟏𝟏𝟏𝟏 𝟐𝟐 is substituted for 𝒌𝒌, the number sentence is true. Answers will vary on values to make the sentence false; the number 𝟐𝟐𝟑𝟑 𝟒𝟒 or any number greater than 𝟐𝟐𝟑𝟑 𝟒𝟒 will make the sentence false. 7. 𝟒𝟒. 𝟓𝟓−𝒅𝒅> 𝟐𝟐. 𝟓𝟓. Substitute 𝟐𝟐. 𝟓𝟓 for 𝒅𝒅. When 𝟐𝟐. 𝟓𝟓 is substituted for 𝒅𝒅, the number sentence is false. Answers will vary on values to make the sentence true; any number less than 𝟐𝟐 will make the number sentence true. 8. 𝟖𝟖≥𝟑𝟑𝟑𝟑𝟑𝟑. Substitute 𝟏𝟏 𝟐𝟐 for 𝒑𝒑. When 𝟏𝟏 𝟐𝟐 is substituted for 𝒑𝒑, the number sentence is false. Answers will vary on values to make the sentence true; the number 𝟏𝟏 𝟒𝟒 or any number less than 𝟏𝟏 𝟒𝟒 will make the sentence true. 9. 𝒘𝒘 𝟐𝟐< 𝟑𝟑𝟑𝟑. Substitute 𝟏𝟏𝟏𝟏 for 𝒘𝒘. When 𝟏𝟏𝟏𝟏 is substituted for 𝒑𝒑, the number sentence is true. Answers will vary on values to make the sentence false; the number 𝟔𝟔𝟔𝟔 or any other number greater than 𝟔𝟔𝟔𝟔 will make the sentence false. 10. 𝟏𝟏𝟏𝟏≤𝟑𝟑𝟑𝟑−𝒃𝒃. Substitute 𝟏𝟏𝟏𝟏 for 𝒃𝒃. When 𝟏𝟏𝟏𝟏 is substituted for 𝒃𝒃, the number sentence is true. Answers will vary on values to make the sentence false; any number greater than 𝟏𝟏𝟏𝟏 will make the sentence false. MP.6 A STORY OF RATIOS 245 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Closing (10 minutes) Take 3 minutes to discuss the answers from the exercises. From the exercises, continue the discussion from the lesson. Let’s take a look at Exercise 1: 4 + 𝑥𝑥= 12. We substituted 8 for 𝑥𝑥. What did we determine? When we substituted 8 for 𝑥𝑥, the number sentence was true. And then we tried to find values to substitute for 𝑥𝑥 to make the number sentence false. What number did you substitute for 𝑥𝑥 to make this number sentence false? Elicit responses from the class. Answers will vary, but collect all that make the number sentence false, and record them on the board. Elicit responses for the next set of questions: Did anyone substitute with zero? A thousand? A trillion? How about a fraction? A decimal? Answers will vary. If all of these responses result in a false number sentence, what can you conclude? Only one number can be substituted to make the two expressions equal, and that number is 8. Look at all of the numbers that will make this number sentence false, and then look at the one number that will make this number sentence true. Why do you think the number 8 is important compared to all the others that make the number sentence false? Elicit various responses. The goal is for students to understand that since all numbers other than 8 result in false statements, those numbers do not contribute valuable information about the equation in the same way that the number 8 does. What about inequalities? Let’s take another look at Exercise 2: 3𝑔𝑔> 15. We substituted 4 1 2 for 𝑔𝑔 and determined that after we evaluated the inequality, it created a false number sentence because 13 1 2 is not greater than 15. What number did you substitute for 𝑔𝑔 to make this number sentence true? Elicit responses from the class. Answers will vary, but collect all that make the number sentence true, and record them on the board. Elicit responses for the next set of questions: What about 14, 16, 18, or 20? 100? 200? How about 5.1? 5.01? 5.001? 5.0000000000000001? Answers will vary. What can you conclude about the substituted numbers that will make this number sentence true? To make this number sentence true, any number greater than five can be substituted for the variable, whether it is a whole number, fraction, or decimal. Which substituted numbers made this number sentence false? Answers will vary but must be five or any number less than five. Visualize a number line in your mind. If we can only substitute numbers greater than five on the number line to make this number sentence true, what would that number line look like? Answers will vary. The goal is for students to visualize that only part of the number line works for the inequality in order to create a true number sentence, while the other part does not work and makes the number sentence false. A STORY OF RATIOS 246 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Lesson Summary NUMBER SENTENCE: A number sentence is a statement of equality (or inequality) between two numerical expressions. TRUTH VALUES OF A NUMBER SENTENCE: A number sentence is said to be true if both numerical expressions evaluate to the same number; it is said to be false otherwise. True and false are called truth values. Number sentences that are inequalities also have truth values. For example, 𝟑𝟑< 𝟒𝟒, 𝟔𝟔+ 𝟖𝟖> 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏, and (𝟏𝟏𝟏𝟏+ 𝟑𝟑)𝟐𝟐< 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎−𝟑𝟑𝟑𝟑 are all true number sentences, while the sentence 𝟗𝟗> 𝟑𝟑(𝟒𝟒) is false. Exit Ticket (5 minutes) A STORY OF RATIOS 247 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Name Date Lesson 23: True and False Number Sentences Exit Ticket Substitute the value for the variable, and state in a complete sentence whether the resulting number sentence is true or false. If true, find a value that would result in a false number sentence. If false, find a value that would result in a true number sentence. 1. 15𝑎𝑎≥75. Substitute 5 for 𝑎𝑎. 2. 23 + 𝑏𝑏= 30. Substitute 10 for 𝑏𝑏. 3. 20 > 86 −ℎ. Substitute 46 for ℎ. 4. 32 ≥8𝑚𝑚. Substitute 5 for 𝑚𝑚. A STORY OF RATIOS 248 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences Exit Ticket Sample Solutions Substitute the value for the variable, and state in a complete sentence whether the resulting number sentence is true or false. If true, find a value that would result in a false number sentence. If false, find a value that would result in a true number sentence. 1. 𝟏𝟏𝟏𝟏𝟏𝟏≥𝟕𝟕𝟕𝟕. Substitute 𝟓𝟓 for 𝒂𝒂. When 𝟓𝟓 is substituted in for 𝒂𝒂, the number sentence is true. Answers will vary, but any value for 𝒂𝒂 less than 𝟓𝟓 will result in a false number sentence. 2. 𝟐𝟐𝟐𝟐+ 𝒃𝒃= 𝟑𝟑𝟑𝟑. Substitute 𝟏𝟏𝟏𝟏 for 𝒃𝒃. When 𝟏𝟏𝟏𝟏 is substituted in for 𝒃𝒃, the number sentence is false. The only value for 𝒃𝒃 that will result in a true number sentence is 𝟕𝟕. 3. 𝟐𝟐𝟐𝟐> 𝟖𝟖𝟖𝟖−𝒉𝒉. Substitute 𝟒𝟒𝟒𝟒 for 𝒉𝒉. When 𝟒𝟒𝟒𝟒 is substituted in for 𝒉𝒉, the number sentence will be false. Answers will vary, but any value for 𝒉𝒉 greater than 𝟔𝟔𝟔𝟔 will result in a true number sentence. 4. 𝟑𝟑𝟑𝟑≥𝟖𝟖𝟖𝟖. Substitute 𝟓𝟓 for 𝒎𝒎. When 𝟓𝟓 is substituted in for 𝒎𝒎, the number sentence is false. Answers will vary, but the value of 𝟒𝟒 and any value less than 𝟒𝟒 for 𝒎𝒎 will result in a true number sentence. Problem Set Sample Solutions Substitute the value for the variable, and state (in a complete sentence) whether the resulting number sentence is true or false. If true, find a value that would result in a false number sentence. If false, find a value that would result in a true number sentence. 1. 𝟑𝟑𝟓𝟓 𝟔𝟔= 𝟏𝟏𝟐𝟐 𝟑𝟑+ 𝒉𝒉. Substitute 𝟐𝟐𝟏𝟏 𝟔𝟔 for 𝒉𝒉. When 𝟐𝟐𝟏𝟏 𝟔𝟔 is substituted in for 𝒉𝒉, the number sentence is true. Answers will vary, but any value for 𝒉𝒉 other than 𝟐𝟐𝟏𝟏 𝟔𝟔 will result in a false number sentence. 2. 𝟑𝟑𝟑𝟑> 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. Substitute 𝟏𝟏 𝟒𝟒 for 𝒈𝒈. When 𝟏𝟏 𝟒𝟒 is substituted in for 𝒈𝒈, the number sentence is false. Answers will vary, but any value for 𝒈𝒈 less than 𝟏𝟏 𝟒𝟒 will result in a true number sentence. 3. 𝒇𝒇 𝟒𝟒≤𝟑𝟑. Substitute 𝟏𝟏𝟏𝟏 for 𝒇𝒇. When 𝟏𝟏𝟏𝟏 is substituted in for 𝒇𝒇, the number sentence is true. Answers will vary, but any value for 𝒇𝒇 greater than 𝟏𝟏𝟏𝟏 will result in a false number sentence. A STORY OF RATIOS 249 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 23 Lesson 23: True and False Number Sentences 4. 𝟏𝟏𝟏𝟏𝟏𝟏−𝟗𝟗𝟗𝟗≥𝒓𝒓. Substitute 𝟐𝟐𝟐𝟐 for 𝒓𝒓. When 𝟐𝟐𝟐𝟐 is substituted in for 𝒓𝒓, the number sentence is true. Answers will vary, but any value for 𝒓𝒓 greater than 𝟐𝟐𝟐𝟐 will result in a false number sentence. 5. 𝟓𝟓𝟓𝟓 𝒒𝒒= 𝟔𝟔. Substitute 𝟏𝟏𝟏𝟏 for 𝒒𝒒. When 𝟏𝟏𝟏𝟏 is substituted in for 𝒒𝒒, the number sentence is false. The number 𝟗𝟗 is the only value for 𝒒𝒒 that will result in a true number sentence. Create a number sentence using the given variable and symbol. The number sentence you write must be true for the given value of the variable. 6. Variable: 𝒅𝒅 Symbol: ≥ The sentence is true when 𝟓𝟓 is substituted for 𝒅𝒅. 7. Variable: 𝒚𝒚 Symbol: ≠ The sentence is true when 𝟏𝟏𝟏𝟏 is substituted for 𝒚𝒚. 8. Variable: 𝒌𝒌 Symbol: < The sentence is true when 𝟖𝟖 is substituted for 𝒌𝒌. 9. Variable: 𝒂𝒂 Symbol: ≤ The sentence is true when 𝟗𝟗 is substituted for 𝒂𝒂. Answers will vary for Problems 6–9. A STORY OF RATIOS 250 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Lesson 24: True and False Number Sentences Student Outcomes Students identify values for the variables in equations and inequalities that result in true number sentences. Students identify values for the variables in equations and inequalities that result in false number sentences. Lesson Notes Beginning in the previous lesson and continuing here, the language used in answering questions has been carefully chosen. Responses have been purposefully elicited from students in the form of numbers, quantities, or sentences. Soon, students see that another way to report an answer to an equation or inequality is another equation or inequality. For example, the solution to 3𝑥𝑥≥15 can be reported as 𝑥𝑥≥5. During this lesson, students discover that solutions and solution sets can be represented by a sentence description, leading to (or followed by) the use of equations or inequalities. This discussion provides students with knowledge to systemically solve and check one-step equations later in the module. For example, in this lesson, students transition from “The inequality is true for any value of 𝑥𝑥 that is greater than or equal to five,” to “The inequality is true when 𝑥𝑥≥5.” This transition is preparing students to understand why they rewrite complicated-looking equations and inequalities as simpler ones (such as 𝑥𝑥= 5 or 𝑥𝑥≥5) to describe solutions. This is an important goal in the solution process. The ≠ symbol has purposefully been omitted in these lessons because it does not satisfy all of the properties listed in Tables 4 and 5 of the Common Core State Standards. However, it is a symbol that is useful and easy to understand. Its absence from the lessons does not mean that it cannot be used in class, nor should it be forgotten. Classwork Opening Exercise (3 minutes) Opening Exercise State whether each number sentence is true or false. If the number sentence is false, explain why. a. 𝟒𝟒+ 𝟓𝟓> 𝟗𝟗 False. 𝟒𝟒+ 𝟓𝟓 is not greater than 𝟗𝟗. b. 𝟑𝟑∙𝟔𝟔= 𝟏𝟏𝟏𝟏 True c. 𝟑𝟑𝟑𝟑> 𝟔𝟔𝟔𝟔 𝟒𝟒 True A STORY OF RATIOS 251 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 d. 𝟕𝟕𝟕𝟕−𝟏𝟏𝟏𝟏< 𝟔𝟔𝟔𝟔 True e. 𝟐𝟐𝟐𝟐≥𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 False. 𝟐𝟐𝟐𝟐 is not greater than or equal to 𝟐𝟐𝟐𝟐. Students share their answers and defend their decisions for each problem. Example 1 (10 minutes) The teacher leads the following discussion after students complete the table below. Have students work on the first two columns alone or in groups of two, writing true or false if the number substituted for 𝑔𝑔 results in a true or false number sentence. Example 1 Write true or false if the number substituted for 𝒈𝒈 results in a true or false number sentence. Substitute 𝒈𝒈 with 𝟒𝟒𝒈𝒈= 𝟑𝟑𝟑𝟑 𝒈𝒈= 𝟖𝟖 𝟑𝟑𝒈𝒈≥𝟑𝟑𝟑𝟑 𝒈𝒈≥𝟏𝟏𝟏𝟏 𝒈𝒈 𝟐𝟐> 𝟐𝟐 𝒈𝒈> 𝟒𝟒 𝟑𝟑𝟑𝟑≥𝟑𝟑𝟑𝟑−𝒈𝒈 𝒈𝒈≥𝟖𝟖 𝟖𝟖 True True False False True True True True 𝟒𝟒 False False False False False False False False 𝟐𝟐 False False False False False False False False 𝟎𝟎 False False False False False False False False 𝟏𝟏𝟏𝟏 False False True True True True True True Let’s look at 4𝑔𝑔= 32 and 𝑔𝑔= 8. What do you notice happens when 8 is substituted for 𝑔𝑔 in both of the equations? 8 makes both of the equations result in true number sentences. What do you notice about the substitutions with 4, 2, 0, and 10? Each of those substituted values makes the equations result in false number sentences. Why do you think that happened? Because they are both equations, we expect that only one number can be substituted for 𝑔𝑔 to result in a true number sentence. In this case, 8 is the only number that can be substituted to make both equations true. How are 4𝑔𝑔= 32 and 𝑔𝑔= 8 related? You can get from 4𝑔𝑔= 32 to 𝑔𝑔= 8 by dividing both sides of 4𝑔𝑔= 32 by 4. You can get from 𝑔𝑔= 8 to 4𝑔𝑔= 32 by multiplying both sides of 𝑔𝑔= 8 by 4. In which equation is it easier to observe the value of 𝑔𝑔 that makes the number sentence true? The second. It is certainly easier to recognize the value in the second equation. Let’s look at the next set of inequalities: 3𝑔𝑔≥30 and 𝑔𝑔≥10. (Let students fill out the table for these two columns.) What do you notice happens when 10 is substituted for 𝑔𝑔 in both of the inequalities? 10 makes both of the inequalities result in true number sentences. A STORY OF RATIOS 252 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Let’s substitute some numbers into each inequality to test. For the second inequality, as long as the number is greater than or equal to 10, the inequality will result in a true number sentence. Let’s read the inequality aloud together. Chorally: 𝑔𝑔 is greater than or equal to 10. Let’s try 11. 11 is greater than or equal to 10. Substitute 11 for 𝑔𝑔 in 3𝑔𝑔≥30. Does this result in a true number sentence? Yes How are 3𝑔𝑔≥30 and 𝑔𝑔≥10 related? You can get from 3𝑔𝑔≥30 to 𝑔𝑔≥10 by dividing both sides of 3𝑔𝑔≥30 by 3. You can get from 𝑔𝑔≥10 to 3𝑔𝑔≥30 by multiplying both sides of 𝑔𝑔≥10 by 3. In which inequality is it easier to observe the value of 𝑔𝑔 that makes the number sentence true? The second, which is similar to the first example Continue testing the substitutions, and continue the discussion for the remaining sets of inequalities (but do not ask how the last two inequalities are related). The goal is to have students discover that for each set of equations and inequalities, the second in the set represents a much clearer way to represent the solutions. Point out to students that the second equation or inequality plays an important role in the next few lessons. Please note that it is not necessary that students fully understand a process for solving equations and inequalities from these examples. Example 2 (10 minutes) Guide students in how to use mental math to answer the questions. Students do not know how to solve one-step equations using a formal method; therefore, they need guidance in solving these problems. Please note that the second problem includes the use of subtraction to get a negative value. While operations with integers is a Grade 7 topic, this example should be accessible using a visual model. Example 2 State when the following equations/inequalities will be true and when they will be false. a. 𝒓𝒓+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 Can you think of a number that will make this equation true? Yes. Substituting 10 for 𝑟𝑟 will make a true number sentence. Is 10 the only number that results in a true number sentence? Why or why not? Yes. There is only one value that, if substituted, will result in a true number sentence. There is only one number that can be added to 15 to get exactly 25. What will make the number sentence false? Any number that is not 10 will result in a false number sentence. If we look back to the original questions, how can we state when the equation will be true? False? The equation is true when the value substituted for 𝑟𝑟 is 10 and false when the value of 𝑟𝑟 is any other number. b. 𝟔𝟔−𝒅𝒅> 𝟎𝟎 MP.6 A STORY OF RATIOS 253 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 2 3 4 5 6 7 1 0 If we wanted 6 −𝑑𝑑 to equ𝑎𝑎𝑎𝑎 0, what would the value of 𝑑𝑑 have to be? Why? The value of 𝑑𝑑 would have to be 6 because 6 −6 = 0. Will substituting 6 for 𝑑𝑑 result in a true number sentence? Why or why not? If 𝑑𝑑 has a value of 6, then the resulting number sentence would not be true because the left side has to be greater than 0, not equal to 0. How about substituting 5 for 𝑑𝑑? 4? 3? 2? Yes. Substituting any of these numbers for 𝑑𝑑 into the inequality results in true number sentences. What values can we substitute for 𝑑𝑑 in order for the resulting number sentence to be true? The inequality is true for any value of 𝑑𝑑 that is less than 6. What values for 𝑑𝑑 would make the resulting number sentence false? The inequality is false for any value of 𝑑𝑑 that is greater than or equal to 6. Let’s take a look at a number line and see why these statements make sense. Display a number line on the board. Label the number line as shown below. Let’s begin at 6. If I were to subtract 1 from 6, where would that place be on the number line? 5 So, if we substitute 1 for 𝑑𝑑, then 6 −1 = 5, and the resulting number sentence is true. How about if I subtracted 2 from 6? Would our number sentence be true for the value 2? Yes What if I subtracted 6 from the 6 on the number line? Where would that be on the number line? 0 So, if we substitute 6 for 𝑑𝑑, will the resulting number sentence be true or false? False Let’s try one more. We have already determined that any number greater than or equal to 6 will result in a false number sentence. Let’s try a number greater than 6. Let’s try the number 7. Start with the 6 on the number line. If we were to subtract 7, in which direction on the number line would we move? To the left And how many times will we move to the left? 7 Model beginning at 6 on the number line, and move a finger, or draw the unit skips, while continually moving to the left on the number line 7 times. So, it seems we have ended up at a place to the left of 0. What number is represented by this position? −1 2 3 4 5 6 7 1 0 MP.6 A STORY OF RATIOS 254 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Label the number line with −1. Using our knowledge of ordering rational numbers, is −1 greater than or less than 0? Less than So, we have shown that the inequality is true for any value of 𝑑𝑑 that is less than 6 (𝑑𝑑< 6) and is false when the value of 𝑑𝑑 is greater than or equal to 6 (𝑑𝑑≥6). Continue to discuss how to answer each question below with students. As students gain more confidence, have them try to solve the problems individually; discuss the answers when students are finished. c. 𝟏𝟏 𝟐𝟐𝒇𝒇= 𝟏𝟏𝟏𝟏 The equation is true when the value substituted for 𝒇𝒇 is 𝟑𝟑𝟑𝟑 (𝒇𝒇= 𝟑𝟑𝟑𝟑) and false when the value of 𝒇𝒇 is any other number (𝒇𝒇≠𝟑𝟑𝟑𝟑). d. 𝒚𝒚 𝟑𝟑< 𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒚𝒚 that is less than 𝟑𝟑𝟑𝟑 (𝒚𝒚< 𝟑𝟑𝟑𝟑) and false when the value of 𝒚𝒚 is greater than or equal to 𝟑𝟑𝟑𝟑 (𝒚𝒚≥𝟑𝟑𝟑𝟑). e. 𝟕𝟕𝒈𝒈≥𝟒𝟒𝟒𝟒 The inequality is true for any value of 𝒈𝒈 that is greater than or equal to 𝟔𝟔 (𝒈𝒈≥𝟔𝟔) and false when the value of 𝒈𝒈 is less than (𝒈𝒈< 𝟔𝟔). f. 𝒂𝒂−𝟖𝟖≤𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒂𝒂 that is less than or equal to 𝟐𝟐𝟐𝟐 (𝒂𝒂≤𝟐𝟐𝟐𝟐) and false when the value of 𝒂𝒂 is greater than 𝟐𝟐𝟐𝟐 (𝒂𝒂> 𝟐𝟐𝟐𝟐). Exercises (10 minutes) Students complete the following problems in pairs. Exercises Complete the following problems in pairs. State when the following equations and inequalities will be true and when they will be false. 1. 𝟏𝟏𝟏𝟏𝒄𝒄> 𝟒𝟒𝟒𝟒 The inequality is true for any value of 𝒄𝒄 that is greater than 𝟑𝟑 (𝒄𝒄> 𝟑𝟑) and false when the value of 𝒄𝒄 is less than or equal to (𝒄𝒄≤𝟑𝟑). 2. 𝟐𝟐𝟐𝟐= 𝒅𝒅−𝟏𝟏𝟏𝟏 The equation is true when the value of 𝒅𝒅 is 𝟑𝟑𝟑𝟑 (𝒅𝒅= 𝟑𝟑𝟑𝟑) and false when the value of 𝒅𝒅 is any other number (𝒅𝒅≠𝟑𝟑𝟑𝟑). 3. 𝟓𝟓𝟓𝟓≥𝟐𝟐𝒆𝒆 The inequality is true for any value of 𝒆𝒆 that is less than or equal to 𝟐𝟐𝟐𝟐 (𝒆𝒆≤𝟐𝟐𝟐𝟐) and false when the value of 𝒆𝒆 is greater than 𝟖𝟖 (𝒆𝒆> 𝟐𝟐𝟐𝟐). MP.6 A STORY OF RATIOS 255 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 4. 𝒉𝒉 𝟓𝟓≥𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒉𝒉 that is greater than or equal to 𝟔𝟔𝟔𝟔 (𝒉𝒉≥𝟔𝟔𝟔𝟔) and false when the value of 𝒉𝒉 is less than (𝒉𝒉< 𝟔𝟔𝟔𝟔). 5. 𝟒𝟒𝟒𝟒> 𝒉𝒉+ 𝟐𝟐𝟐𝟐 The inequality is true for any value of 𝒉𝒉 that is less than 𝟏𝟏𝟏𝟏 (𝒉𝒉< 𝟏𝟏𝟏𝟏) and false when the value of 𝒉𝒉 is greater than or equal to 𝟏𝟏𝟏𝟏 (𝒉𝒉≥𝟏𝟏𝟏𝟏). 6. 𝟒𝟒𝒂𝒂≤𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒂𝒂 that is less than or equal to 𝟒𝟒 (𝒂𝒂≤𝟒𝟒) and false when the value of 𝒂𝒂 is greater than (𝒂𝒂> 𝟒𝟒). 7. 𝟑𝟑𝒙𝒙= 𝟐𝟐𝟐𝟐 The equation is true when the value of 𝒙𝒙 is 𝟖𝟖 (𝒙𝒙= 𝟖𝟖) and false when the value of 𝒙𝒙 is any other number (𝒙𝒙≠𝟖𝟖). Identify all equality and inequality signs that can be placed into the blank to make a true number sentence. 8. 𝟏𝟏𝟏𝟏+ 𝟗𝟗 𝟐𝟐𝟐𝟐 = or ≥ or ≤ 9. 𝟖𝟖∙𝟕𝟕 𝟓𝟓𝟓𝟓 > or ≥ 10. 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟏𝟏𝟏𝟏 < or ≤ 11. 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏∙𝟐𝟐 = or ≥ or ≤ 12. 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐. 𝟓𝟓−𝟔𝟔 < or ≤ MP.6 A STORY OF RATIOS 256 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Closing (7 minutes) For the past two lessons, we have been using sentences to describe when values substituted for variables in equations and inequalities result in true number sentences or false number sentences. Let’s take a look at an example from each of the past two lessons. Display the following equation on the board: 5 + 𝑥𝑥= 8. Substituting 3 for 𝑥𝑥 in the equation results in a true number sentence: 5 + 3 = 8. Let’s evaluate to be sure. What is the sum of 5 + 3? 8 Does 8 = 8? Yes So, when we substitute 3 for 𝑥𝑥, the equation results in a true number sentence. Let’s try to substitute 3 for 𝑥𝑥 in 𝑥𝑥= 3. Display 𝑥𝑥= 3 on the board. If we substituted 3 for 𝑥𝑥, what would our number sentence look like? 3 = 3 Is this a true number sentence? Yes Previously, we described the values of 𝑥𝑥 that would make the equation 5 + 𝑥𝑥= 8 true in a sentence. Display on the board: The equation is true when the value of 𝑥𝑥 is 3. This is the same sentence we would write for the equation 𝑥𝑥= 3. Therefore, we can shorten this sentence and, instead, say: The equation is true when 𝑥𝑥= 3. Display on the board: The equation is true when 𝑥𝑥= 3. Let’s look at an inequality from today: Display 4𝑎𝑎≤16 on the board. What numbers did we determine would make this inequality result in a true number sentence? We determined that any number less than or equal to 4 would result in a true number sentence. Write this statement on the board: The inequality is true for any value of 𝑎𝑎 that is less than or equal to 4. Is there any way we can abbreviate or shorten this statement using symbols instead of words? Display 𝑎𝑎≤4 on the board. Let’s read this aloud: 𝑎𝑎 is less than or equal to four. We can use this inequality to rewrite the sentence. Display on the board: The inequality is true when 𝑎𝑎≤4. Either sentence is a correct way to state the values that make 4𝑎𝑎≤16 true. Exit Ticket (5 minutes) A STORY OF RATIOS 257 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Name Date Lesson 24: True and False Number Sentences Exit Ticket State when the following equations and inequalities will be true and when they will be false. 1. 5𝑔𝑔> 45 2. 14 = 5 + 𝑘𝑘 3. 26 −𝑤𝑤< 12 4. 32 ≤𝑎𝑎+ 8 5. 2 ∙ℎ≤16 A STORY OF RATIOS 258 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 Exit Ticket Sample Solutions State when the following equations and inequalities will be true and when they will be false. 1. 𝟓𝟓𝒈𝒈> 𝟒𝟒𝟒𝟒 The inequality is true for any value of 𝒈𝒈 that is greater than 𝟗𝟗 and false when the value of 𝒈𝒈 is less than or equal to 𝟗𝟗. OR The inequality is true when 𝒈𝒈> 𝟗𝟗 and false when 𝒈𝒈≤𝟗𝟗. 2. 𝟏𝟏𝟒𝟒= 𝟓𝟓+ 𝒌𝒌 The equation is true when the value of 𝒌𝒌 is 𝟗𝟗 and false when the value of 𝒌𝒌 is any other number. OR The equation is true when 𝒌𝒌= 𝟗𝟗 and false when 𝒌𝒌≠𝟗𝟗. 3. 𝟐𝟐𝟐𝟐−𝒘𝒘< 𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒘𝒘 that is greater than 𝟏𝟏𝟏𝟏 and false when the value of 𝒘𝒘 is less than or equal to 𝟏𝟏𝟏𝟏. OR The inequality is true when 𝒘𝒘> 𝟏𝟏𝟏𝟏 and false when 𝒘𝒘≤𝟏𝟏𝟏𝟏. 4. 𝟑𝟑𝟑𝟑≤𝒂𝒂+ 𝟖𝟖 The inequality is true for any value of 𝒂𝒂 that is greater than or equal to 𝟐𝟐𝟐𝟐 and false when the value of 𝒂𝒂 is less than 𝟐𝟐𝟐𝟐. OR The inequality is true when 𝒂𝒂≥𝟐𝟐𝟐𝟐 and false when 𝒂𝒂< 𝟐𝟐𝟐𝟐. 5. 𝟐𝟐∙𝒉𝒉≤𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒉𝒉 that is less than or equal to 𝟖𝟖 and false when the value of 𝒉𝒉 is greater than 𝟖𝟖. OR The inequality is true when 𝒉𝒉≤𝟖𝟖 and false when 𝒉𝒉> 𝟖𝟖. Problem Set Sample Solutions State when the following equations and inequalities will be true and when they will be false. 1. 𝟑𝟑𝟑𝟑= 𝟗𝟗𝒌𝒌 The equation is true when the value of 𝒌𝒌 is 𝟒𝟒 and false when the value of 𝒌𝒌 is any number other than 𝟒𝟒. OR The equation is true when 𝒌𝒌= 𝟒𝟒 and false when 𝒌𝒌≠𝟒𝟒. 2. 𝟔𝟔𝟔𝟔> 𝒇𝒇−𝟏𝟏𝟏𝟏 The inequality is true for any value of 𝒇𝒇 that is less than 𝟖𝟖𝟖𝟖 and false when the value of 𝒇𝒇 is greater than or equal to 𝟖𝟖𝟖𝟖. OR The inequality is true when 𝒇𝒇< 𝟖𝟖𝟖𝟖 and false when 𝒇𝒇≥𝟖𝟖𝟖𝟖. A STORY OF RATIOS 259 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 24: True and False Number Sentences 6•4 Lesson 24 3. 𝒗𝒗 𝟗𝟗= 𝟑𝟑 The equation is true when the value of 𝒗𝒗 is 𝟐𝟐𝟐𝟐 and false when the value of 𝒗𝒗 is any number other than 𝟐𝟐𝟐𝟐. OR The equation is true when 𝒗𝒗= 𝟐𝟐𝟐𝟐 and false when 𝒗𝒗≠𝟐𝟐𝟐𝟐. 4. 𝟏𝟏𝟏𝟏+ 𝒃𝒃> 𝟒𝟒𝟒𝟒 The inequality is true for any value of 𝒃𝒃 that is greater than 𝟑𝟑𝟑𝟑 and false when the value of 𝒃𝒃 is less than or equal to 𝟑𝟑𝟑𝟑. OR The inequality is true when 𝒃𝒃> 𝟑𝟑𝟑𝟑 and false when 𝒃𝒃≤𝟑𝟑𝟑𝟑. 5. 𝒅𝒅−𝟖𝟖≥𝟑𝟑𝟑𝟑 The inequality is true for any value of 𝒅𝒅 that is greater than or equal to 𝟒𝟒𝟒𝟒 and false when the value of 𝒅𝒅 is less than 𝟒𝟒𝟒𝟒. OR The inequality is true when 𝒅𝒅≥𝟒𝟒𝟒𝟒 and false when 𝒅𝒅< 𝟒𝟒𝟒𝟒. 6. 𝟑𝟑𝟑𝟑𝒇𝒇< 𝟔𝟔𝟔𝟔 The inequality is true for any value of 𝒇𝒇 that is less than 𝟐𝟐 and false when the value of 𝒇𝒇 is greater than or equal to 𝟐𝟐. OR The inequality is true when 𝒇𝒇< 𝟐𝟐 and false when 𝒇𝒇≥𝟐𝟐. 7. 𝟏𝟏𝟏𝟏−𝒉𝒉≤𝟕𝟕 The inequality is true for any value of 𝒉𝒉 that is greater than or equal to 𝟑𝟑 and false when the value of 𝒉𝒉 is less than 𝟑𝟑. OR The inequality is true when 𝒉𝒉≥𝟑𝟑 and false when 𝒉𝒉< 𝟑𝟑. 8. 𝟒𝟒𝟒𝟒+ 𝟖𝟖≥𝒈𝒈 The inequality is true for any value of 𝒈𝒈 that is less than or equal to 𝟓𝟓𝟓𝟓 and false when the value of 𝒈𝒈 is greater than 𝟓𝟓𝟓𝟓. OR The inequality is true when 𝒈𝒈≤𝟓𝟓𝟓𝟓 and false when 𝒈𝒈> 𝟓𝟓𝟓𝟓. 9. 𝒎𝒎 𝟑𝟑= 𝟏𝟏𝟏𝟏 The equation is true when the value of 𝒎𝒎 is 𝟒𝟒𝟒𝟒 and false when the value of 𝒎𝒎 is any number other than 𝟒𝟒𝟒𝟒. OR The equation is true when 𝒎𝒎= 𝟒𝟒𝟒𝟒 and false when 𝒎𝒎≠𝟒𝟒𝟒𝟒. A STORY OF RATIOS 260 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Lesson 25: Finding Solutions to Make Equations True Student Outcomes Students learn the definition of solution in the context of placing a value into a variable to see if that value makes the equation true. Lesson Notes In previous lessons, students used sentences and symbols to describe the values that, when substituted for the variable in an equation, resulted in a true number sentence. In this lesson, students make the transition from their previous learning (e.g., substituting numbers into equations, writing complete sentences to describe when an equation results in a true number sentence, and using symbols to reduce the wordiness of a description) to today’s lesson where they identify the value that makes an equation true as a solution. As they did in previous lessons, students test for solutions by substituting numbers into equations and by checking whether the resulting number sentence is true. They have already seen how equations like 𝑥𝑥= 3 relate to the original equation and that it is valuable to find ways to simplify equations until they are in the form of 𝑥𝑥= “a number.” In the next lesson, students begin to learn the formal process of “solving an equation,” that is, the process of transforming the original equation to an equation of the form 𝑥𝑥= “a number,” where it is easy to identify the solution. Materials: Students complete a matching game that needs to be cut out and prepared before the class period begins. Ideally, there should be 20 sets prepared, each in a separate bag, so that students may work in pairs. Specific directions for the game are below. Classwork Fluency Exercise (5 minutes): Division of Fractions Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening Exercise (5 minutes) Opening Exercise Identify a value for the variable that would make each equation or inequality into a true number sentence. Is this the only possible answer? State when the equation or inequality is true using equality and inequality symbols. a. 𝟑𝟑+ 𝒈𝒈= 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 is the only value of 𝒈𝒈 that will make the equation true. The equation is true when 𝒈𝒈= 𝟏𝟏𝟐𝟐. b. 𝟑𝟑𝟑𝟑> 𝟐𝟐𝟐𝟐 Answers will vary. There is more than one value of 𝒅𝒅 that will make the inequality true. The inequality is true when 𝒅𝒅< 𝟏𝟏𝟏𝟏. A STORY OF RATIOS 261 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True c. 𝟏𝟏𝟏𝟏 𝒇𝒇< 𝟓𝟓 Answers will vary. There is more than one value of 𝒇𝒇 that will make the inequality true. The inequality is true when 𝒇𝒇> 𝟑𝟑. d. 𝟒𝟒𝟒𝟒≤𝟓𝟓𝟓𝟓−𝒎𝒎 Answers will vary. There is more than one value of 𝒎𝒎 that will make the inequality true. The inequality is true when 𝒎𝒎≤𝟖𝟖. Example (5 minutes) Example Each of the following numbers, if substituted for the variable, makes one of the equations below into a true number sentence. Match the number to that equation: 𝟑𝟑, 𝟔𝟔, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟒𝟒𝟒𝟒. a. 𝒏𝒏+ 𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑 𝟔𝟔 b. 𝒏𝒏−𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒 c. 𝟏𝟏𝟏𝟏𝒏𝒏= 𝟓𝟓𝟓𝟓 𝟑𝟑 d. 𝟒𝟒𝟐𝟐= 𝒏𝒏 𝟏𝟏𝟏𝟏 e. 𝒏𝒏 𝟑𝟑= 𝟓𝟓 𝟏𝟏𝟏𝟏 Discussion (2 minutes) In most of the equations we have looked at so far, the numbers we used to substitute in for the variable have resulted in true number sentences. A number or value for the variable that results in a true number sentence is special and is called a solution to the equation. In the example above, 6 is a solution to 𝑛𝑛+ 26 = 32, 44 is a solution to 𝑛𝑛−12 = 32, and so on. A STORY OF RATIOS 262 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Exercises (15 minutes) Students work with a partner to match the equation with its solution. Please note that below are the answers for the activity. The actual game cut-out pieces are at the end of the lesson. 𝑎𝑎+ 14 = 36 22 33 = 𝑏𝑏 27 𝑐𝑐 5 = 3 15 𝑑𝑑−10 = 32 42 24 = 𝑒𝑒+ 11 13 32 = 4 ∙𝑓𝑓 8 9 = 45 𝑔𝑔 5 43 = ℎ−17 60 1.5 + 0.5 = 𝑗𝑗 2 9 ∙1 3 = 𝑘𝑘 3 𝑚𝑚= 56 8 7 𝑛𝑛= 35.5 −9.5 26 𝑝𝑝+ 13 3 4 = 32 3 4 19 4 = 1 4 𝑞𝑞 16 63 𝑟𝑟= 7 9 99 −𝑢𝑢= 45 54 Closing (8 minutes) Let’s look at the equation 8𝑛𝑛= 72. We know that 9 is a value that we can substitute for 𝑛𝑛 that results in a true number sentence. In previous lessons, we described this solution as “The equation is true when the value of 𝑛𝑛 is 9” and noted that the equation is false when any number other than 9 is substituted for 𝑛𝑛, or when 𝑛𝑛≠ 9. Therefore, there is only one solution to 8𝑛𝑛= 72, and it is 9. We also saw that both statements (i.e., the numbers that make the equation true and the numbers that make it false) can be summarized with one sentence, “The equation is true when 𝑛𝑛= 9,” because the values that make 𝑛𝑛= 9 true or false are the same as the values that make 8𝑛𝑛= 72 true or false. Thus, we can represent the solution as “The solution is 9,” or 𝑛𝑛= 9. The next lesson shows the process for transforming an equation like 8𝑛𝑛= 72 until it is in the form 𝑥𝑥= 9. You have been doing this process for many years in tape diagrams and unknown angle problems, but now we describe explicitly the steps you were following. Note that the domain of the variable is just the set of numbers from which we are looking for solutions. For example, sometimes we only want to consider integers as solutions. In those cases, the domain of the variable would be the set of integer numbers. MP.3 A STORY OF RATIOS 263 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Lesson Summary VARIABLE: A variable is a symbol (such as a letter) that is a placeholder for a number. A variable is a placeholder for “a number” that does not “vary.” EXPRESSION: An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables. EQUATION: An equation is a statement of equality between two expressions. If 𝑨𝑨 and 𝑩𝑩 are two expressions in the variable 𝒙𝒙, then 𝑨𝑨= 𝑩𝑩 is an equation in the variable 𝒙𝒙. Teacher notes: A common description of a variable in the U.S. is “a quantity that varies.” Ask yourselves, how can a quantity vary? A less vague description of a variable is “a placeholder for a number”; this is better because it denotes a single, non-varying number. The upside of the description of variable (and this is a point that must be made explicit to students) is that it is the user of the variable who controls what number to insert into the placeholder. Hence, it is the student who has the power to change or vary the number as he so desires. The power to vary rests with the student, not with the variable itself! Exit Ticket (5 minutes) A STORY OF RATIOS 264 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Name Date Lesson 25: Finding Solutions to Make Equations True Exit Ticket Find the solution to each equation. 1. 7𝑓𝑓= 49 2. 1 = 𝑟𝑟 12 3. 1.5 = 𝑑𝑑+ 0.8 4. 92 = ℎ 5. 𝑞𝑞= 45 −19 6. 40 = 1 2 𝑝𝑝 A STORY OF RATIOS 265 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Exit Ticket Sample Solutions Find the solution to each equation. 1. 𝟕𝟕𝟕𝟕= 𝟒𝟒𝟒𝟒 𝒇𝒇= 𝟕𝟕 2. 𝟏𝟏= 𝒓𝒓 𝟏𝟏𝟏𝟏 𝒓𝒓= 𝟏𝟏𝟏𝟏 3. 𝟏𝟏. 𝟓𝟓= 𝒅𝒅+ 𝟎𝟎. 𝟖𝟖 𝒅𝒅= 𝟎𝟎. 𝟕𝟕 4. 𝟗𝟗𝟐𝟐= 𝒉𝒉 𝒉𝒉= 𝟖𝟖𝟖𝟖 5. 𝒒𝒒= 𝟒𝟒𝟒𝟒−𝟏𝟏𝟏𝟏 𝒒𝒒= 𝟐𝟐𝟐𝟐 6. 𝟒𝟒𝟒𝟒= 𝟏𝟏 𝟐𝟐𝒑𝒑 𝒑𝒑= 𝟖𝟖𝟖𝟖 Problem Set Sample Solutions Find the solution to each equation. 1. 𝟒𝟒𝟑𝟑= 𝒚𝒚 𝒚𝒚= 𝟔𝟔𝟔𝟔 2. 𝟖𝟖𝟖𝟖= 𝟐𝟐𝟐𝟐 𝒂𝒂= 𝟑𝟑 3. 𝟑𝟑𝟑𝟑= 𝒈𝒈−𝟒𝟒 𝒈𝒈= 𝟑𝟑𝟑𝟑 4. 𝟓𝟓𝟓𝟓= 𝒋𝒋+ 𝟐𝟐𝟐𝟐 𝒋𝒋= 𝟐𝟐𝟐𝟐 5. 𝟒𝟒𝟒𝟒 𝒓𝒓= 𝟏𝟏𝟏𝟏 𝒓𝒓= 𝟒𝟒 A STORY OF RATIOS 266 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True 6. 𝒌𝒌= 𝟏𝟏𝟏𝟏−𝟗𝟗 𝒌𝒌= 𝟔𝟔 7. 𝒙𝒙∙𝟏𝟏 𝟓𝟓= 𝟔𝟔𝟔𝟔 𝒙𝒙= 𝟑𝟑𝟑𝟑𝟑𝟑 8. 𝒎𝒎+ 𝟑𝟑. 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏. 𝟖𝟖 𝒎𝒎= 𝟗𝟗. 𝟑𝟑𝟑𝟑 9. 𝒂𝒂= 𝟏𝟏𝟓𝟓 𝒂𝒂= 𝟏𝟏 A STORY OF RATIOS 267 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True 𝑎𝑎+ 14 = 36 22 33 = 𝑏𝑏 27 𝑐𝑐 5 = 3 15 𝑑𝑑−10 = 32 42 24 = 𝑒𝑒+ 11 13 32 = 4 ∙𝑓𝑓 8 9 = 45 𝑔𝑔 5 43 = ℎ−17 A STORY OF RATIOS 268 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True 60 1.5 + 0.5 = 𝑗𝑗 2 9 ∙1 3 = 𝑘𝑘 3 𝑚𝑚= 56 8 7 𝑛𝑛= 35.5 −9.5 26 𝑝𝑝+ 13 3 4 = 32 3 4 19 4 = 1 4 𝑞𝑞 16 63 𝑟𝑟= 7 9 A STORY OF RATIOS 269 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True 99 −𝑢𝑢= 45 54 A STORY OF RATIOS 270 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Division of Fractions—Round 1 Directions: Evaluate each expression and simplify. 1. 9 ones ÷ 3 ones 23. 6 10 ÷ 4 10 2. 9 ÷ 3 24. 6 10 ÷ 2 5 = 6 10 ÷ 10 3. 9 tens ÷ 3 tens 25. 10 12 ÷ 5 12 4. 90 ÷ 30 26. 5 6 ÷ 5 12 = 12 ÷ 5 12 5. 9 hundreds ÷ 3 hundreds 27. 10 12 ÷ 3 12 6. 900 ÷ 300 28. 10 12 ÷ 1 4 = 10 12 ÷ 12 7. 9 halves ÷ 3 halves 29. 5 6 ÷ 3 12 = 12 ÷ 3 12 8. 9 2 ÷ 3 2 30. 5 10 ÷ 2 10 9. 9 fourths ÷ 3 fourths 31. 5 10 ÷ 1 5 = 5 10 ÷ 10 10. 9 4 ÷ 3 4 32. 1 2 ÷ 2 10 = 10 ÷ 2 10 11. 9 8 ÷ 3 8 33. 1 2 ÷ 2 4 12. 2 3 ÷ 1 3 34. 3 4 ÷ 2 8 13. 1 3 ÷ 2 3 35. 1 2 ÷ 3 8 14. 6 7 ÷ 2 7 36. 1 2 ÷ 1 5 = 10 ÷ 10 15. 5 7 ÷ 2 7 37. 2 4 ÷ 1 3 16. 3 7 ÷ 4 7 38. 1 4 ÷ 4 6 17. 6 10 ÷ 2 10 39. 3 4 ÷ 2 6 18. 6 10 ÷ 4 10 40. 5 6 ÷ 1 4 19. 6 10 ÷ 8 10 41. 2 9 ÷ 5 6 20. 7 12 ÷ 2 12 42. 5 9 ÷ 1 6 21. 6 12 ÷ 9 12 43. 1 2 ÷ 1 7 22. 4 12 ÷ 11 12 44. 5 7 ÷ 1 2 Number Correct: A STORY OF RATIOS 271 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Division of Fractions—Round 1 [KEY] Directions: Evaluate each expression and simplify. 1. 9 ones ÷ 3 ones 𝟗𝟗 𝟑𝟑= 𝟑𝟑 23. 6 10 ÷ 4 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 2. 9 ÷ 3 𝟗𝟗 𝟑𝟑= 𝟑𝟑 24. 6 10 ÷ 2 5 = 6 10 ÷ 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 3. 9 tens ÷ 3 tens 𝟗𝟗 𝟑𝟑= 𝟑𝟑 25. 10 12 ÷ 5 12 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 4. 90 ÷ 30 𝟗𝟗 𝟑𝟑= 𝟑𝟑 26. 5 6 ÷ 5 12 = 12 ÷ 5 12 𝟏𝟏𝟏𝟏 𝟓𝟓= 𝟐𝟐 5. 9 hundreds ÷ 3 hundreds 𝟗𝟗 𝟑𝟑= 𝟑𝟑 27. 10 12 ÷ 3 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 6. 900 ÷ 300 𝟗𝟗 𝟑𝟑= 𝟑𝟑 28. 10 12 ÷ 1 4 = 10 12 ÷ 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 7. 9 halves ÷ 3 halves 𝟗𝟗 𝟑𝟑= 𝟑𝟑 29. 5 6 ÷ 3 12 = 12 ÷ 3 12 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 8. 9 2 ÷ 3 2 𝟗𝟗 𝟑𝟑= 𝟑𝟑 30. 5 10 ÷ 2 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 9. 9 fourths ÷ 3 fourths 𝟗𝟗 𝟑𝟑= 𝟑𝟑 31. 5 10 ÷ 1 5 = 5 10 ÷ 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 10. 9 4 ÷ 3 4 𝟗𝟗 𝟑𝟑= 𝟑𝟑 32. 1 2 ÷ 2 10 = 10 ÷ 2 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 11. 9 8 ÷ 3 8 𝟗𝟗 𝟑𝟑= 𝟑𝟑 33. 1 2 ÷ 2 4 𝟐𝟐 𝟐𝟐= 𝟏𝟏 12. 2 3 ÷ 1 3 𝟐𝟐 𝟏𝟏= 𝟐𝟐 34. 3 4 ÷ 2 8 𝟑𝟑 13. 1 3 ÷ 2 3 𝟏𝟏 𝟐𝟐 35. 1 2 ÷ 3 8 𝟒𝟒 𝟑𝟑= 𝟏𝟏𝟏𝟏 𝟑𝟑 14. 6 7 ÷ 2 7 𝟔𝟔 𝟐𝟐= 𝟑𝟑 36. 1 2 ÷ 1 5 = 10 ÷ 10 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 15. 5 7 ÷ 2 7 𝟓𝟓 𝟐𝟐= 𝟐𝟐𝟏𝟏 𝟐𝟐 37. 2 4 ÷ 1 3 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 16. 3 7 ÷ 4 7 𝟑𝟑 𝟒𝟒 38. 1 4 ÷ 4 6 𝟑𝟑 𝟖𝟖 17. 6 10 ÷ 2 10 𝟔𝟔 𝟐𝟐= 𝟑𝟑 39. 3 4 ÷ 2 6 𝟗𝟗 𝟒𝟒= 𝟐𝟐𝟏𝟏 𝟒𝟒 18. 6 10 ÷ 4 10 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 40. 5 6 ÷ 1 4 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟑𝟑𝟏𝟏 𝟑𝟑 19. 6 10 ÷ 8 10 𝟔𝟔 𝟖𝟖= 𝟑𝟑 𝟒𝟒 41. 2 9 ÷ 5 6 𝟒𝟒 𝟏𝟏𝟏𝟏 20. 7 12 ÷ 2 12 𝟕𝟕 𝟐𝟐= 𝟑𝟑𝟏𝟏 𝟐𝟐 42. 5 9 ÷ 1 6 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟓𝟓 21. 6 12 ÷ 9 12 𝟔𝟔 𝟗𝟗= 𝟐𝟐 𝟑𝟑 43. 1 2 ÷ 1 7 𝟕𝟕 𝟐𝟐= 𝟑𝟑𝟏𝟏 𝟐𝟐 22. 4 12 ÷ 11 12 𝟒𝟒 𝟏𝟏𝟏𝟏 44. 5 7 ÷ 1 2 𝟏𝟏𝟏𝟏 𝟕𝟕= 𝟏𝟏𝟑𝟑 𝟕𝟕 A STORY OF RATIOS 272 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Number Correct: Improvement: Division of Fractions—Round 2 Directions: Evaluate each expression and simplify. 1. 12 ones ÷ 2 ones 23. 6 12 ÷ 4 12 2. 12 ÷ 2 24. 6 12 ÷ 2 6 = 6 12 ÷ 12 3. 12 tens ÷ 2 tens 25. 8 14 ÷ 7 14 4. 120 ÷ 20 26. 8 14 ÷ 1 2 = 8 14 ÷ 14 5. 12 hundreds ÷ 2 hundreds 27. 11 14 ÷ 2 14 6. 1,200 ÷ 200 28. 11 14 ÷ 1 7 = 11 14 ÷ 14 7. 12 halves ÷ 2 halves 29. 1 7 ÷ 6 14 = 14 ÷ 6 14 8. 12 2 ÷ 2 2 30. 7 18 ÷ 3 18 9. 12 fourths ÷ 3 fourths 31. 7 18 ÷ 1 6 = 7 18 ÷ 18 10. 12 4 ÷ 3 4 32. 1 3 ÷ 12 18 = 18 ÷ 12 18 11. 12 8 ÷ 3 8 33. 1 6 ÷ 4 18 12. 2 4 ÷ 1 4 34. 4 12 ÷ 8 6 13. 1 4 ÷ 2 4 35. 1 3 ÷ 3 15 14. 4 5 ÷ 2 5 36. 2 6 ÷ 1 9 = 18 ÷ 18 15. 2 5 ÷ 4 5 37. 1 6 ÷ 4 9 16. 3 5 ÷ 4 5 38. 2 3 ÷ 3 4 17. 6 8 ÷ 2 8 39. 1 3 ÷ 3 5 18. 6 8 ÷ 4 8 40. 1 7 ÷ 1 2 19. 6 8 ÷ 5 8 41. 5 6 ÷ 2 9 20. 6 10 ÷ 2 10 42. 5 9 ÷ 2 6 21. 7 10 ÷ 8 10 43. 5 6 ÷ 4 9 22. 4 10 ÷ 7 10 44. 1 2 ÷ 4 5 A STORY OF RATIOS 273 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 25 Lesson 25: Finding Solutions to Make Equations True Division of Fractions—Round 2 [KEY] Directions: Evaluate each expression and simplify. 1. 12 ones ÷ 2 ones 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 23. 6 12 ÷ 4 12 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 2. 12 ÷ 2 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 24. 6 12 ÷ 2 6 = 6 12 ÷ 12 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 3. 12 tens ÷ 2 tens 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 25. 8 14 ÷ 7 14 𝟖𝟖 𝟕𝟕= 𝟏𝟏𝟏𝟏 𝟕𝟕 4. 120 ÷ 20 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 26. 8 14 ÷ 1 2 = 8 14 ÷ 14 𝟖𝟖 𝟕𝟕= 𝟏𝟏𝟏𝟏 𝟕𝟕 5. 12 hundreds ÷ 2 hundreds 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 27. 11 14 ÷ 2 14 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟓𝟓𝟏𝟏 𝟐𝟐 6. 1,200 ÷ 200 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 28. 11 14 ÷ 1 7 = 11 14 ÷ 14 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟓𝟓𝟏𝟏 𝟐𝟐 7. 12 halves ÷ 2 halves 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 29. 1 7 ÷ 6 14 = 14 ÷ 6 14 𝟐𝟐 𝟔𝟔= 𝟏𝟏 𝟑𝟑 8. 12 2 ÷ 2 2 𝟏𝟏𝟏𝟏 𝟐𝟐= 𝟔𝟔 30. 7 18 ÷ 3 18 𝟕𝟕 𝟑𝟑= 𝟐𝟐𝟏𝟏 𝟑𝟑 9. 12 fourths ÷ 3 fourths 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 31. 7 18 ÷ 1 6 = 7 18 ÷ 18 𝟕𝟕 𝟑𝟑= 𝟐𝟐𝟏𝟏 𝟑𝟑 10. 12 4 ÷ 3 4 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 32. 1 3 ÷ 12 18 = 18 ÷ 12 18 𝟔𝟔 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟐𝟐 11. 12 8 ÷ 3 8 𝟏𝟏𝟏𝟏 𝟑𝟑= 𝟒𝟒 33. 1 6 ÷ 4 18 𝟑𝟑 𝟒𝟒 12. 2 4 ÷ 1 4 𝟐𝟐 𝟏𝟏= 𝟐𝟐 34. 4 12 ÷ 8 6 𝟒𝟒 𝟏𝟏𝟏𝟏= 𝟏𝟏 𝟒𝟒 13. 1 4 ÷ 2 4 𝟏𝟏 𝟐𝟐 35. 1 3 ÷ 3 15 𝟓𝟓 𝟑𝟑= 𝟏𝟏𝟐𝟐 𝟑𝟑 14. 4 5 ÷ 2 5 𝟒𝟒 𝟐𝟐= 𝟐𝟐 36. 2 6 ÷ 1 9 = 18 ÷ 18 𝟔𝟔 𝟐𝟐= 𝟑𝟑 15. 2 5 ÷ 4 5 𝟐𝟐 𝟒𝟒= 𝟏𝟏 𝟐𝟐 37. 1 6 ÷ 4 9 𝟑𝟑 𝟖𝟖 16. 3 5 ÷ 4 5 𝟑𝟑 𝟒𝟒 38. 2 3 ÷ 3 4 𝟖𝟖 𝟗𝟗 17. 6 8 ÷ 2 8 𝟔𝟔 𝟐𝟐= 𝟑𝟑 39. 1 3 ÷ 3 5 𝟓𝟓 𝟗𝟗 18. 6 8 ÷ 4 8 𝟔𝟔 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝟐𝟐 40. 1 7 ÷ 1 2 𝟐𝟐 𝟕𝟕 19. 6 8 ÷ 5 8 𝟔𝟔 𝟓𝟓= 𝟏𝟏𝟏𝟏 𝟓𝟓 41. 5 6 ÷ 2 9 𝟏𝟏𝟏𝟏 𝟒𝟒= 𝟑𝟑𝟑𝟑 𝟒𝟒 20. 6 10 ÷ 2 10 𝟔𝟔 𝟐𝟐= 𝟑𝟑 42. 5 9 ÷ 2 6 𝟏𝟏𝟏𝟏 𝟔𝟔= 𝟏𝟏𝟐𝟐 𝟑𝟑 21. 7 10 ÷ 8 10 𝟕𝟕 𝟖𝟖 43. 5 6 ÷ 4 9 𝟏𝟏𝟏𝟏 𝟖𝟖= 𝟏𝟏𝟕𝟕 𝟖𝟖 22. 4 10 ÷ 7 10 𝟒𝟒 𝟕𝟕 44. 1 2 ÷ 4 5 𝟓𝟓 𝟖𝟖 A STORY OF RATIOS 274 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 8 Lesson 26: One-Step Equations―Addition and Subtraction Student Outcomes Students solve one-step equations by relating an equation to a diagram. Students check to determine if their solutions make the equations true. Lesson Notes This lesson serves as a means for students to solve one-step equations through the use of tape diagrams. Through the construction of tape diagrams, students create algebraic equations and solve for one variable. In this lesson, students continue their study of the properties of operations and identity and develop intuition of the properties of equality. This lesson continues the informal study of the properties of equality students have practiced since Grade 1 and also serves as a springboard to the formal study, use, and application of the properties of equality seen in Grade 7. While students intuitively use the properties of equality, understand that diagrams are driving the meaning behind the content of this lesson. This lesson purposefully omits focus on the actual properties of equality, which is reserved for Grade 7. Students relate an equation directly to diagrams and verbalize what they do with diagrams to construct and solve algebraic equations. Classwork Opening (3 minutes) In order for students to learn how to solve multi-step equations (in future grades), they must first learn how to solve basic equations. Although a majority of students have the ability to find the solutions to the equations using mental math, it is crucial that they understand the importance of knowing and understanding the process for solving equations so they can apply it to more complex equations in the future. Mathematical Modeling Exercise (8 minutes) Model the example to show students how to use tape diagrams to calculate solutions to one-step equations. Calculate the solution: 𝑎𝑎+ 2 = 8 Draw two tape diagrams that are the same length. Label the first tape diagram 8. MP.3 & MP.4 A STORY OF RATIOS 275 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 2 6 2 8 𝑎𝑎 2 𝑎𝑎 2 6 2 Represent 2 on the second tape diagram. What must the remaining section of the tape diagram represent? How do you know? The remaining part of the tape diagram represents 6 because the entire tape diagram is 8, and we know one section is 2. Therefore, we can compute the difference, 8 −2, to determine the remaining part. Label your tape diagram. Draw another set of tape diagrams to represent the given equation: 𝑎𝑎+ 2 = 8. Because both of the following tape diagrams represent the same value, what would the value of 𝑎𝑎 be? Explain. Since both of the tape diagrams represent the same value, both parts that have 𝑎𝑎 and 6 must represent the same value. Therefore, 𝑎𝑎 must have a value of 6. Using this knowledge, try to show or explain how to solve equations without tape diagrams. What actually happened when constructing the tape diagrams? Guide and promote this discussion with students: The first set of tape diagrams shows that the quantity of 6 + 2 is equal to 8. To write this algebraically, we can use the equal sign. 6 + 2 = 8 The second set of tape diagrams shows two things: first, that 𝑎𝑎+ 2 is equal to 8, and also that 𝑎𝑎+ 2 = 8 is equal to 6 + 2 = 8. We found that the only number that 𝑎𝑎 can represent in the equation is 6. Therefore, when 𝑎𝑎+ 2 = 8, the only solution for 𝑎𝑎 is 6. In previous lessons, we discussed identity properties. How can we explain why 𝑎𝑎+ 2 −2 = 𝑎𝑎 using the identity properties? We know that when we add a number and then subtract the same number, the result is the original number. Previously, we demonstrated this identity with 𝑎𝑎+ 𝑏𝑏−𝑏𝑏= 𝑎𝑎. How can we check our answer? Substitute 6 in for 𝑎𝑎 to determine if the number sentence is true. 6 + 2 = 8 is a true number sentence because 6 + 2 −2 = 8 −2, resulting in 6 = 6. So, our answer is correct. MP.3 & MP.4 A STORY OF RATIOS 276 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Exercise 1 (8 minutes) Students work with partners to complete the following problems. They show how to solve each equation using tape diagrams and algebraically. Then, students use substitution to check their answers after each problem. Exercise 1 Solve each equation. Use both tape diagrams and algebraic methods for each problem. Use substitution to check your answers. a. 𝒃𝒃+ 𝟗𝟗= 𝟏𝟏𝟏𝟏 Algebraically: 𝒃𝒃+ 𝟗𝟗= 𝟏𝟏𝟏𝟏 𝒃𝒃+ 𝟗𝟗−𝟗𝟗= 𝟏𝟏𝟏𝟏−𝟗𝟗 𝒃𝒃= 𝟔𝟔 Check: 𝟔𝟔+ 𝟗𝟗−𝟗𝟗= 𝟏𝟏𝟏𝟏−𝟗𝟗; 𝟔𝟔= 𝟔𝟔. This is a true number sentence, so 𝟔𝟔 is the correct solution. 𝟔𝟔 𝟗𝟗 𝟏𝟏𝟏𝟏 𝒃𝒃 𝟗𝟗 𝟏𝟏𝟏𝟏 𝒃𝒃 𝟗𝟗 𝟔𝟔 𝟗𝟗 𝟔𝟔 𝒃𝒃 A STORY OF RATIOS 277 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 𝟖𝟖 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟖𝟖 𝒄𝒄 𝟏𝟏𝟏𝟏 𝟖𝟖 𝒄𝒄 𝟖𝟖 𝟒𝟒 𝒄𝒄 𝟒𝟒 b. 𝟏𝟏𝟏𝟏= 𝟖𝟖+ 𝒄𝒄 Algebraically: 𝟏𝟏𝟏𝟏= 𝟖𝟖+ 𝒄𝒄 𝟏𝟏𝟏𝟏−𝟖𝟖= 𝟖𝟖+ 𝒄𝒄−𝟖𝟖 𝟒𝟒= 𝒄𝒄 Check: 𝟏𝟏𝟏𝟏−𝟖𝟖= 𝟖𝟖+ 𝟒𝟒−𝟖𝟖; 𝟒𝟒= 𝟒𝟒. This is a true number sentence, so 𝟒𝟒 is the correct solution. A STORY OF RATIOS 278 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 𝟓𝟓 𝒅𝒅−𝟓𝟓= 𝟕𝟕 𝒅𝒅 𝟓𝟓 𝟕𝟕 𝟏𝟏𝟏𝟏 𝒅𝒅 Exercise 2 (8 minutes) Students use the knowledge gained in the first part of the lesson to determine how to solve an equation with subtraction. Exercise 2 Given the equation 𝒅𝒅−𝟓𝟓= 𝟕𝟕: a. Demonstrate how to solve the equation using tape diagrams. b. Demonstrate how to solve the equation algebraically. 𝒅𝒅−𝟓𝟓= 𝟕𝟕 𝒅𝒅−𝟓𝟓+ 𝟓𝟓= 𝟕𝟕+ 𝟓𝟓 𝒅𝒅= 𝟏𝟏𝟏𝟏 c. Check your answer. 𝟏𝟏𝟏𝟏−𝟓𝟓+ 𝟓𝟓= 𝟕𝟕+ 𝟓𝟓; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This is a true number sentence, so our solution is correct. Provide students time to work and then provide some examples that show how to solve the equations using both methods. At this time, remind students of the identity with subtraction to explain why 𝑑𝑑−5 + 5 = 𝑑𝑑. MP.4 A STORY OF RATIOS 279 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Exercise 3 (8 minutes) Students solve each problem using the method of their choice, but they must show their work. Have students check their answers. Exercise 3 Solve each problem, and show your work. You may choose which method (tape diagrams or algebraically) you prefer. Check your answers after solving each problem. a. 𝒆𝒆+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 Algebraically: 𝒆𝒆+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒆𝒆+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 𝒆𝒆= 𝟖𝟖 Check: 𝟖𝟖+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏; 𝟖𝟖= 𝟖𝟖. This is a true number sentence, so our answer is correct. 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒆𝒆 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒆𝒆 𝟏𝟏𝟏𝟏 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟖𝟖 𝒆𝒆 A STORY OF RATIOS 280 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 b. 𝒇𝒇−𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏 Algebraically: 𝒇𝒇−𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏 𝒇𝒇−𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 𝒇𝒇= 𝟐𝟐𝟐𝟐 Check: 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏; 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐. This is a true number sentence, so our solution is correct. c. 𝒈𝒈−𝟖𝟖= 𝟗𝟗 Algebraically: 𝒈𝒈−𝟖𝟖= 𝟗𝟗 𝒈𝒈−𝟖𝟖+ 𝟖𝟖= 𝟗𝟗+ 𝟖𝟖 𝒈𝒈= 𝟏𝟏𝟏𝟏 Check: 𝟏𝟏𝟏𝟏−𝟖𝟖+ 𝟖𝟖= 𝟗𝟗+ 𝟖𝟖; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so our solution is correct. 𝒇𝒇 𝟏𝟏𝟏𝟏 𝒇𝒇−𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏 𝒇𝒇 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝒈𝒈 𝟖𝟖 𝒈𝒈−𝟖𝟖= 𝟗𝟗 𝒈𝒈 𝟏𝟏𝟏𝟏 𝟖𝟖 𝟗𝟗 A STORY OF RATIOS 281 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Closing (5 minutes) John checked his answer and found that it was incorrect. John’s work is below. What did he do incorrectly? ℎ+ 10 = 25 ℎ+ 10 + 10 = 25 + 10 ℎ= 35 John should have subtracted 10 on each side of the equation instead of adding because ℎ+ 10 + 10 ≠ℎ. Use a tape diagram to show why John’s method does not lead to the correct answer. 𝟐𝟐𝟐𝟐 𝒉𝒉 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝒉𝒉 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 When John added 10 to both sides of the equation, the equation would change to ℎ+ 20 = 35. Therefore, the value of ℎ cannot equal 35. Why do you do the inverse operation to calculate the solution of the equation? Include a tape diagram as part of your explanation. When you do the inverse operation, the result is zero. Using the identity property, we know any number added to zero is the original number. 𝒉𝒉 𝟏𝟏𝟏𝟏 This tape diagram demonstrates ℎ+ 10; however, we want to know the value of just ℎ. Therefore, we would subtract 10 from this tape diagram. 𝒉𝒉 Therefore, ℎ+ 10 −10 = ℎ. Exit Ticket (5 minutes) A STORY OF RATIOS 282 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Name Date Lesson 26: One-Step Equations—Addition and Subtraction Exit Ticket 1. If you know the answer, state it. Then, use a tape diagram to demonstrate why this is the correct answer. If you do not know the answer, find the solution using a tape diagram. 𝑗𝑗+ 12 = 25 2. Find the solution to the equation algebraically. Check your answer. 𝑘𝑘−16 = 4 A STORY OF RATIOS 283 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Exit Ticket Sample Solutions 1. If you know the answer, state it. Then, use a tape diagram to demonstrate why this is the correct answer. If you do not know the answer, find the solution using a tape diagram. 𝒋𝒋+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒋𝒋 is equal to 𝟏𝟏𝟏𝟏; 𝒋𝒋= 𝟏𝟏𝟏𝟏. Check: 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐. This is a true number sentence, so the solution is correct. 2. Find the solution to the equation algebraically. Check your answer. 𝒌𝒌−𝟏𝟏𝟏𝟏= 𝟒𝟒 𝒌𝒌−𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟒𝟒+ 𝟏𝟏𝟏𝟏 𝒌𝒌= 𝟐𝟐𝟐𝟐 Check: 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏= 𝟒𝟒; 𝟒𝟒= 𝟒𝟒. This is a true number sentence, so the solution is correct. 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒋𝒋 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒋𝒋 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝒋𝒋 A STORY OF RATIOS 284 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 Problem Set Sample Solutions 1. Find the solution to the equation below using tape diagrams. Check your answer. 𝒎𝒎−𝟕𝟕= 𝟏𝟏𝟏𝟏 𝒎𝒎 is equal to 𝟐𝟐𝟐𝟐; 𝒎𝒎= 𝟐𝟐𝟐𝟐. Check: 𝟐𝟐𝟐𝟐−𝟕𝟕= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so the solution is correct. 2. Find the solution of the equation below algebraically. Check your answer. 𝒏𝒏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒏𝒏+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 𝒏𝒏= 𝟏𝟏𝟏𝟏 Check: 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐. This number sentence is true, so the solution is correct. 3. Find the solution of the equation below using tape diagrams. Check your answer. 𝒑𝒑+ 𝟖𝟖= 𝟏𝟏𝟏𝟏 𝐩𝐩= 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟖𝟖 𝟏𝟏𝟏𝟏 𝒑𝒑 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟖𝟖 𝒑𝒑 𝟖𝟖 𝒑𝒑 𝟏𝟏𝟏𝟏 Check: 𝟏𝟏𝟏𝟏+ 𝟖𝟖= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so the solution is correct. 𝒎𝒎 𝟕𝟕 𝒎𝒎−𝟕𝟕= 𝟏𝟏𝟏𝟏 𝒎𝒎 𝟐𝟐𝟐𝟐 𝟕𝟕 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 285 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 4. Find the solution to the equation algebraically. Check your answer. 𝒈𝒈−𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏 𝒈𝒈−𝟔𝟔𝟔𝟔+ 𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏+ 𝟔𝟔𝟔𝟔 𝒈𝒈= 𝟕𝟕𝟕𝟕 Check: 𝟕𝟕𝟕𝟕−𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so the solution is correct. 5. Find the solution to the equation using the method of your choice. Check your answer. 𝒎𝒎+ 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐𝟐𝟐 Tape Diagrams: 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝒎𝒎 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎 Algebraically: 𝒎𝒎+ 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐𝟐𝟐 𝒎𝒎+ 𝟏𝟏𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎= 𝟏𝟏𝟏𝟏𝟏𝟏 Check: 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐𝟐𝟐. This number sentence is true, so the solution is correct. 6. Identify the mistake in the problem below. Then, correct the mistake. 𝒑𝒑−𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑 𝒑𝒑−𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑−𝟐𝟐𝟐𝟐 𝒑𝒑= 𝟏𝟏𝟏𝟏 The mistake is subtracting rather than adding 𝟐𝟐𝟐𝟐. This is incorrect because 𝒑𝒑−𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐 would not equal 𝒑𝒑. 𝒑𝒑−𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑 𝒑𝒑−𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑+ 𝟐𝟐𝟐𝟐 𝒑𝒑= 𝟓𝟓𝟓𝟓 A STORY OF RATIOS 286 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Lesson 26: One-Step Equations―Addition and Subtraction 6•4 Lesson 26 7. Identify the mistake in the problem below. Then, correct the mistake. 𝒒𝒒+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒒𝒒+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐+ 𝟏𝟏𝟏𝟏 𝒒𝒒= 𝟒𝟒𝟒𝟒 The mistake is adding 𝟏𝟏𝟏𝟏 on the right side of the equation instead of subtracting it from both sides. 𝒒𝒒+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒒𝒒+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐−𝟏𝟏𝟏𝟏 𝒒𝒒= 𝟒𝟒 8. Match the equation with the correct solution on the right. 𝒓𝒓+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐 𝒓𝒓= 𝟏𝟏𝟏𝟏 𝒓𝒓−𝟏𝟏𝟏𝟏= 𝟓𝟓 𝒓𝒓= 𝟐𝟐𝟐𝟐 𝒓𝒓−𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏 𝒓𝒓= 𝟏𝟏𝟏𝟏 𝒓𝒓+ 𝟓𝟓= 𝟏𝟏𝟏𝟏 𝒓𝒓= 𝟑𝟑𝟑𝟑 A STORY OF RATIOS 287 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝒛𝒛 𝟑𝟑 𝟑𝟑 𝟑𝟑 Lesson 27: One-Step Equations―Multiplication and Division Student Outcomes Students solve one-step equations by relating an equation to a diagram. Students check to determine if their solutions make the equations true. Lesson Notes This lesson teaches students to solve one-step equations using tape diagrams. Through the construction of tape diagrams, students create algebraic equations and solve for one variable. This lesson not only allows students to continue studying the properties of operations and identity but also allows students to develop intuition of the properties of equality. This lesson continues the informal study of the properties of equality students have practiced since Grade 1 and also serves as a springboard to the formal study, use, and application of the properties of equality seen in Grade 7. Understand that, while students intuitively use the properties of equality, diagrams are the focus of this lesson. This lesson purposefully omits focusing on the actual properties of equality, which are covered in Grade 7. Students relate an equation directly to diagrams and verbalize what they do with diagrams to construct and solve algebraic equations. Poster paper is needed for this lesson. Posters need to be prepared ahead of time, one set of questions per poster. Classwork Example 1 (5 minutes) Example 1 Solve 𝟑𝟑𝟑𝟑= 𝟗𝟗 using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation. If 𝟗𝟗 had to be split into three groups, how big would each group be? 𝟑𝟑 Demonstrate the value of 𝒛𝒛 using tape diagrams. 𝟗𝟗 𝒛𝒛 𝒛𝒛 𝒛𝒛 A STORY OF RATIOS 288 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝒚𝒚 𝒚𝒚÷ 𝟒𝟒 𝒚𝒚÷ 𝟒𝟒 𝒚𝒚÷ 𝟒𝟒 𝒚𝒚÷ 𝟒𝟒 How can we demonstrate this algebraically? We know we have to split 𝟗𝟗 into three equal groups, so we have to divide by 𝟑𝟑 to show this algebraically. 𝟑𝟑𝟑𝟑÷ 𝟑𝟑= 𝟗𝟗÷ 𝟑𝟑 How does this get us the value of 𝒛𝒛? The left side of the equation will equal 𝒛𝒛 because we know the identity property, where 𝒂𝒂∙𝒃𝒃÷ 𝒃𝒃= 𝒂𝒂, so we can use this identity here. The right side of the equation will be 𝟑𝟑 because 𝟗𝟗÷ 𝟑𝟑= 𝟑𝟑. Therefore, the value of 𝒛𝒛 is 𝟑𝟑. How can we check our answer? We can substitute the value of 𝒛𝒛 into the original equation to see if the number sentence is true. 𝟑𝟑(𝟑𝟑) = 𝟗𝟗; 𝟗𝟗= 𝟗𝟗. This number sentence is true, so our answer is correct. Example 2 (5 minutes) Example 2 Solve 𝒚𝒚 𝟒𝟒= 𝟐𝟐 using tape diagrams and algebraically. Then, check your answer. First, draw two tape diagrams, one to represent each side of the equation. If the first tape diagram shows the size of 𝒚𝒚÷ 𝟒𝟒, how can we draw a tape diagram to represent 𝒚𝒚? The tape diagram to represent 𝒚𝒚 should be four sections of the size 𝒚𝒚÷ 𝟒𝟒. Draw this tape diagram. What value does each 𝒚𝒚÷ 𝟒𝟒 section represent? How do you know? Each 𝒚𝒚÷ 𝟒𝟒 section represents a value of 𝟐𝟐. We know this from our original tape diagram. How can you use a tape diagram to show the value of 𝒚𝒚? Draw four equal sections of 𝟐𝟐, which will give 𝒚𝒚 the value of 𝟖𝟖. 𝒚𝒚÷ 𝟒𝟒 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 A STORY OF RATIOS 289 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division How can we demonstrate this algebraically? 𝒚𝒚 𝟒𝟒∙𝟒𝟒= 𝟐𝟐∙𝟒𝟒. Because we multiplied the number of sections in the original equation by 𝟒𝟒, we know the identity 𝒂𝒂 𝒃𝒃∙𝒃𝒃= 𝒂𝒂 can be used here. How does this help us find the value of 𝒚𝒚? The left side of the equation will equal 𝒚𝒚, and the right side will equal 𝟖𝟖. Therefore, the value of 𝒚𝒚 is 𝟖𝟖. How can we check our answer? Substitute 𝟖𝟖 into the equation for 𝒚𝒚, and then check to see if the number sentence is true. 𝟖𝟖 𝟒𝟒= 𝟐𝟐. This is a true number sentence, so 𝟖𝟖 is the correct answer. Exploratory Challenge (15 minutes) Each group (two or three) of students receives one set of problems. Have students solve both problems on poster paper with tape diagrams and algebraically. Students should also check their answers on the poster paper. More than one group may have each set of problems. Set 1 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 2𝑎𝑎= 16 Tape Diagrams: Algebraically: 2𝑎𝑎= 16 2𝑎𝑎÷ 2 = 16 ÷ 2 𝑎𝑎= 8 Check: 2 ∙8 = 16; 16 = 16. This is a true number sentence, so 8 is the correct solution. Scaffolding: If students are struggling, model one set of problems before continuing with the Exploratory Challenge. 𝟏𝟏𝟏𝟏 𝟖𝟖 𝒂𝒂 𝒂𝒂 𝟖𝟖 𝒂𝒂 MP.1 A STORY OF RATIOS 290 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝟏𝟏𝟏𝟏 𝒃𝒃 𝒃𝒃÷ 𝟑𝟑 𝟒𝟒 𝒃𝒃÷ 𝟑𝟑 𝒃𝒃÷ 𝟑𝟑 𝒃𝒃÷ 𝟑𝟑 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝒃𝒃 2. 𝑏𝑏 3 = 4 Tape Diagrams: Algebraically: 𝑏𝑏 3 = 4 𝑏𝑏 3 ∙3 = 4 ∙3 𝑏𝑏= 12 Check: 12 3 = 4; 4 = 4. This number sentence is true, so 12 is the correct solution. Set 2 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 4 ∙𝑐𝑐= 24 Tape Diagrams: 𝒄𝒄 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟐𝟐𝟐𝟐 𝒄𝒄 𝒄𝒄 𝒄𝒄 𝒄𝒄 MP.1 A STORY OF RATIOS 291 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Algebraically: 4 ∙𝑐𝑐= 24 4 ∙𝑐𝑐÷ 4 = 24 ÷ 4 𝑐𝑐= 6 Check: 4 ∙6 = 24; 24 = 24. This number sentence is true, so 6 is the correct solution. 2. 𝑑𝑑 7 = 1 Tape Diagrams: Algebraically: 𝑑𝑑 7 = 1 𝑑𝑑 7 ∙7 = 1 ∙7 𝑑𝑑= 7 Check: 7 7 = 1; 1 = 1. This number sentence is true, so 7 is the correct solution. 𝟕𝟕 𝒅𝒅 𝒅𝒅÷ 𝟕𝟕 𝟏𝟏 𝒅𝒅 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝒅𝒅÷ 𝟕𝟕 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 MP.1 A STORY OF RATIOS 292 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝟑𝟑𝟑𝟑 𝒇𝒇 𝒇𝒇÷ 𝟑𝟑 𝟏𝟏𝟏𝟏 𝒇𝒇÷ 𝟑𝟑 𝒇𝒇÷ 𝟑𝟑 𝒇𝒇÷ 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝒇𝒇 Set 3 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 5𝑒𝑒= 45 Tape Diagrams: Algebraically: 5𝑒𝑒= 45 5𝑒𝑒÷ 5 = 45 ÷ 5 𝑒𝑒= 9 Check: 5(9) = 45; 45 = 45. This number sentence is true, so 9 is the correct solution. 2. 𝑓𝑓 3 = 10 Tape Diagrams: Algebraically: 𝑓𝑓 3 = 10 𝑓𝑓 3 ∙3 = 10 ∙3 𝑓𝑓= 30 Check: 30 3 = 10; 10 = 10. This number sentence is true, so 30 is the correct solution. 𝟒𝟒𝟒𝟒 𝒆𝒆 𝒆𝒆 𝒆𝒆 𝒆𝒆 𝒆𝒆 𝒆𝒆 𝟗𝟗 𝟗𝟗 𝟗𝟗 𝟗𝟗 𝟗𝟗 MP.1 A STORY OF RATIOS 293 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Set 4 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 9 ∙𝑔𝑔= 54 Tape Diagrams: Algebraically: 9 ∙𝑔𝑔= 54 9 ∙𝑔𝑔÷ 9 = 54 ÷ 9 𝑔𝑔= 6 Check: 9 ∙6 = 54; 54 = 54. This number sentence is true, so 6 is the correct solution. 2. 2 = ℎ 7 Tape Diagrams: 𝟓𝟓𝟓𝟓 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝒈𝒈 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝒈𝒈 𝟏𝟏𝟏𝟏 𝒉𝒉 𝒉𝒉÷ 𝟕𝟕 𝟐𝟐 𝒉𝒉 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝒉𝒉÷ 𝟕𝟕 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟐𝟐 MP.1 A STORY OF RATIOS 294 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Algebraically: 2 = ℎ 7 2 ∙7 = ℎ 7 ∙7 14 = ℎ Check: 2 = 14 7 ; 2 = 2. This number sentence is true, so 14 is the correct solution. Set 5 On poster paper, solve each problem below algebraically and using tape diagrams. Check each answer to show that you solved the equation correctly (algebraic and tape diagram sample responses are below). 1. 50 = 10𝑗𝑗 Tape Diagrams: Algebraically: 50 = 10𝑗𝑗 50 ÷ 10 = 10𝑗𝑗÷ 10 5 = 𝑗𝑗 Check: 50 = 10(5); 50 = 50. This number sentence is true, so 5 is the correct solution. 𝟓𝟓𝟓𝟓 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝒋𝒋 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝟓𝟓 𝒋𝒋 𝟓𝟓 MP.1 A STORY OF RATIOS 295 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝒌𝒌 𝒌𝒌÷ 𝟖𝟖 𝟑𝟑 𝟐𝟐𝟐𝟐 𝒌𝒌 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝒌𝒌÷ 𝟖𝟖 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑 2. 𝑘𝑘 8 = 3 Tape Diagrams: Algebraically: 𝑘𝑘 8 = 3 𝑘𝑘 8 ∙8 = 3 ∙8 𝑘𝑘= 24 Check: 24 8 = 3; 3 = 3. This number sentence is true, so 24 is the correct solution. Hang completed posters around the room. Students walk around to examine other groups’ posters. Students may either write on a piece of paper, write on Post-it notes, or write on the posters any questions or comments they may have. Answer students’ questions after providing time for students to examine posters. MP.3 MP.1 A STORY OF RATIOS 296 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 𝟐𝟐𝟐𝟐 𝒎𝒎 𝒎𝒎 𝒎𝒎 𝒎𝒎 𝟕𝟕 𝟕𝟕 𝟕𝟕 Exercises (10 minutes) Students complete the following problems individually. Remind students to check their solutions. Exercises 1. Use tape diagrams to solve the following problem: 𝟑𝟑𝟑𝟑= 𝟐𝟐𝟐𝟐. Check: 𝟑𝟑(𝟕𝟕) = 𝟐𝟐𝟐𝟐; 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐. This number sentence is true, so 𝟕𝟕 is the correct solution. 2. Solve the following problem algebraically: 𝟏𝟏𝟏𝟏= 𝒏𝒏 𝟓𝟓. 𝟏𝟏𝟏𝟏= 𝒏𝒏 𝟓𝟓 𝟏𝟏𝟏𝟏∙𝟓𝟓= 𝒏𝒏 𝟓𝟓∙𝟓𝟓 𝟕𝟕𝟕𝟕= 𝒏𝒏 Check: 𝟏𝟏𝟏𝟏= 𝟕𝟕𝟕𝟕 𝟓𝟓; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so 𝟕𝟕𝟕𝟕 is the correct solution. 3. Calculate the solution of the equation using the method of your choice: 𝟒𝟒𝒑𝒑= 𝟑𝟑𝟑𝟑. Tape Diagrams: Algebraically: 𝟒𝟒𝟒𝟒= 𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒÷ 𝟒𝟒= 𝟑𝟑𝟑𝟑÷ 𝟒𝟒 𝒑𝒑= 𝟗𝟗 Check: 𝟒𝟒(𝟗𝟗) = 𝟑𝟑𝟑𝟑; 𝟑𝟑𝟑𝟑= 𝟑𝟑𝟑𝟑. This number sentence is true, so 𝟗𝟗 is the correct solution. 𝟑𝟑𝟑𝟑 𝒑𝒑 𝒑𝒑 𝒑𝒑 𝒑𝒑 𝒑𝒑 𝟗𝟗 𝟗𝟗 𝟗𝟗 𝟗𝟗 A STORY OF RATIOS 297 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 4. Examine the tape diagram below, and write an equation it represents. Then, calculate the solution to the equation using the method of your choice. 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕 or 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕 Tape Diagram: Algebraically: 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕÷ 𝟕𝟕= 𝟕𝟕𝟕𝟕÷ 𝟕𝟕 𝒒𝒒= 𝟏𝟏𝟏𝟏 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕÷ 𝟕𝟕= 𝟕𝟕𝟕𝟕÷ 𝟕𝟕 𝒒𝒒= 𝟏𝟏𝟏𝟏 Check: 𝟕𝟕(𝟏𝟏𝟏𝟏) = 𝟕𝟕𝟕𝟕, 𝟕𝟕𝟕𝟕= 𝟕𝟕(𝟏𝟏𝟏𝟏); 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕. This number sentence is true, so 𝟏𝟏𝟏𝟏 is the correct answer. 5. Write a multiplication equation that has a solution of 𝟏𝟏𝟏𝟏. Use tape diagrams to prove that your equation has a solution of 𝟏𝟏𝟏𝟏. Answers will vary. 6. Write a division equation that has a solution of 𝟏𝟏𝟏𝟏. Prove that your equation has a solution of 𝟏𝟏𝟏𝟏 using algebraic methods. Answers will vary. Closing (5 minutes) How is solving addition and subtraction equations similar to and different from solving multiplication and division equations? Solving addition and subtraction equations is similar to solving multiplication and division equations because identities are used for all of these equations. Solving addition and subtraction equations is different from solving multiplication and division equations because they require different identities. What do you know about the pattern in the operations you used to solve the equations today? We used inverse operations to solve the equations today. Division was used to solve multiplication equations, and multiplication was used to solve division equations. Exit Ticket (5 minutes) 𝟕𝟕𝟕𝟕 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝒒𝒒 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 298 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Name Date Lesson 27: One-Step Equations—Multiplication and Division Exit Ticket Calculate the solution to each equation below using the indicated method. Remember to check your answers. 1. Use tape diagrams to find the solution of 𝑟𝑟 10 = 4. 2. Find the solution of 64 = 16𝑢𝑢 algebraically. A STORY OF RATIOS 299 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 3. Use the method of your choice to find the solution of 12 = 3𝑣𝑣. A STORY OF RATIOS 300 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Exit Ticket Sample Solutions Calculate the solution to each equation below using the indicated method. Remember to check your answers. 1. Use tape diagrams to find the solution of 𝒓𝒓 𝟏𝟏𝟏𝟏= 𝟒𝟒. Check: 𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏= 𝟒𝟒; 𝟒𝟒= 𝟒𝟒. This number sentence is true, so 𝟒𝟒𝟒𝟒 is the correct solution. 2. Find the solution of 𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏𝟏𝟏 algebraically. 𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔÷ 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏𝟏𝟏÷ 𝟏𝟏𝟏𝟏 𝟒𝟒= 𝒖𝒖 Check: 𝟔𝟔𝟔𝟔= 𝟏𝟏𝟏𝟏(𝟒𝟒); 𝟔𝟔𝟔𝟔= 𝟔𝟔𝟔𝟔. This number sentence is true, so 𝟒𝟒 is the correct solution. 3. Use the method of your choice to find the solution of 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑. Tape Diagrams: Algebraically: 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏÷ 𝟑𝟑= 𝟑𝟑𝟑𝟑÷ 𝟑𝟑 𝟒𝟒= 𝒗𝒗 Check: 𝟏𝟏𝟏𝟏= 𝟑𝟑(𝟒𝟒); 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so 𝟒𝟒 is the correct solution. 𝟒𝟒𝟒𝟒 𝒓𝒓 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝟒𝟒 𝒓𝒓 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝒓𝒓÷ 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟒𝟒 𝟏𝟏𝟏𝟏 𝒗𝒗 𝒗𝒗 𝒗𝒗 𝒗𝒗 𝟒𝟒 𝟒𝟒 𝟒𝟒 A STORY OF RATIOS 301 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division Problem Set Sample Solutions 1. Use tape diagrams to calculate the solution of 𝟑𝟑𝟑𝟑= 𝟓𝟓𝟓𝟓. Then, check your answer. Check: 𝟑𝟑𝟑𝟑= 𝟓𝟓(𝟔𝟔); 𝟑𝟑𝟑𝟑= 𝟑𝟑𝟑𝟑. This number sentence is true, so 𝟔𝟔 is the correct solution. 2. Solve 𝟏𝟏𝟏𝟏= 𝒙𝒙 𝟒𝟒 algebraically. Then, check your answer. 𝟏𝟏𝟏𝟏= 𝒙𝒙 𝟒𝟒 𝟏𝟏𝟏𝟏∙𝟒𝟒= 𝒙𝒙 𝟒𝟒∙𝟒𝟒 𝟒𝟒𝟒𝟒= 𝒙𝒙 Check: 𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒 𝟒𝟒; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so 𝟒𝟒𝟒𝟒 is the correct solution. 3. Use tape diagrams to calculate the solution of 𝒚𝒚 𝟓𝟓= 𝟏𝟏𝟏𝟏. Then, check your answer. 𝒚𝒚÷ 𝟓𝟓 𝟏𝟏𝟏𝟏 𝒚𝒚 𝒚𝒚÷ 𝟓𝟓 𝒚𝒚÷ 𝟓𝟓 𝒚𝒚÷ 𝟓𝟓 𝒚𝒚÷ 𝟓𝟓 𝒚𝒚÷ 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝒚𝒚 𝟕𝟕𝟕𝟕 Check: 𝟕𝟕𝟕𝟕 𝟓𝟓= 𝟏𝟏𝟏𝟏; 𝟏𝟏𝟏𝟏= 𝟏𝟏𝟏𝟏. This number sentence is true, so 𝟕𝟕𝟕𝟕 is the correct solution. 𝟑𝟑𝟑𝟑 𝒘𝒘 𝒘𝒘 𝒘𝒘 𝒘𝒘 𝒘𝒘 𝒘𝒘 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 𝟔𝟔 A STORY OF RATIOS 302 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 27 Lesson 27: One-Step Equations―Multiplication and Division 4. Solve 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟕𝟕𝟕𝟕 algebraically. Then, check your answer. 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟕𝟕𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏÷ 𝟏𝟏𝟏𝟏= 𝟕𝟕𝟕𝟕÷ 𝟏𝟏𝟏𝟏 𝒛𝒛= 𝟒𝟒 Check: 𝟏𝟏𝟏𝟏(𝟒𝟒) = 𝟕𝟕𝟕𝟕; 𝟕𝟕𝟕𝟕= 𝟕𝟕𝟕𝟕. This number sentence is true, so 𝟒𝟒 is the correct solution. 5. Write a division equation that has a solution of 𝟖𝟖. Prove that your solution is correct by using tape diagrams. Answers will vary. 6. Write a multiplication equation that has a solution of 𝟖𝟖. Solve the equation algebraically to prove that your solution is correct. Answers will vary. 7. When solving equations algebraically, Meghan and Meredith each got a different solution. Who is correct? Why did the other person not get the correct answer? Meghan Meredith 𝒚𝒚 𝟐𝟐= 𝟒𝟒 𝒚𝒚 𝟐𝟐= 𝟒𝟒 𝒚𝒚 𝟐𝟐∙𝟐𝟐= 𝟒𝟒∙𝟐𝟐 𝒚𝒚 𝟐𝟐÷ 𝟐𝟐= 𝟒𝟒÷ 𝟐𝟐 𝒚𝒚= 𝟖𝟖 𝒚𝒚= 𝟐𝟐 Meghan is correct. Meredith divided by 𝟐𝟐 to solve the equation, which is not correct because she would end up with 𝒚𝒚 𝟒𝟒= 𝟐𝟐. To solve a division equation, Meredith must multiply by 𝟐𝟐 to end up with 𝒚𝒚 because the identity states 𝒚𝒚÷ 𝟐𝟐∙𝟐𝟐= 𝒚𝒚. A STORY OF RATIOS 303 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝟏𝟏𝟐𝟐𝟐𝟐 𝒋𝒋 𝟐𝟐𝟐𝟐 𝒅𝒅 Lesson 28: Two-Step Problems—All Operations Student Outcomes Students calculate the solutions of two-step equations by using their knowledge of order of operations and the properties of equality for addition, subtraction, multiplication, and division. Students employ tape diagrams to determine their answers. Students check to determine if their solutions make the equations true. Classwork Fluency Exercise (5 minutes): Addition of Decimals Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Mathematical Modeling Exercise (6 minutes) Model the problems while students follow along. Mathematical Modeling Exercise Juan has gained 𝟐𝟐𝟐𝟐 𝐥𝐥𝐥𝐥. since last year. He now weighs 𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥. Rashod is 𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥. heavier than Diego. If Rashod and Juan weighed the same amount last year, how much does Diego weigh? Let 𝒋𝒋 represent Juan’s weight last year in pounds, and let 𝒅𝒅 represent Diego’s weight in pounds. Draw a tape diagram to represent Juan’s weight. Draw a tape diagram to represent Rashod’s weight. Draw a tape diagram to represent Diego’s weight. 𝒅𝒅 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 304 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations What would combining all three tape diagrams look like? Write an equation to represent Juan’s tape diagram. 𝒋𝒋+ 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 Write an equation to represent Rashod’s tape diagram. 𝒅𝒅+ 𝟏𝟏𝟏𝟏+ 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 How can we use the final tape diagram or the equations above to answer the question presented? By combining 𝟏𝟏𝟏𝟏 and 𝟐𝟐𝟐𝟐 from Rashod’s equation, we can use our knowledge of addition identities to determine Diego’s weight. The final tape diagram can be used to write a third equation, 𝒅𝒅+ 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏. We can use our knowledge of addition identities to determine Diego’s weight. Calculate Diego’s weight. 𝒅𝒅+ 𝟑𝟑𝟑𝟑−𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏−𝟑𝟑𝟑𝟑 𝒅𝒅= 𝟖𝟖𝟖𝟖 We can use identities to defend our thought that 𝒅𝒅+ 𝟑𝟑𝟑𝟑−𝟑𝟑𝟑𝟑= 𝒅𝒅. Does your answer make sense? Yes. If Diego weighs 𝟖𝟖𝟖𝟖 𝐥𝐥𝐥𝐥., and Rashod weighs 𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥. more than Diego, then Rashod weighs 𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥., which is what Juan weighed before he gained 𝟐𝟐𝟐𝟐 𝐥𝐥𝐥𝐥. A STORY OF RATIOS 305 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Example 1 (5 minutes) Assist students in solving the problem by providing step-by-step guidance. Example 1 Marissa has twice as much money as Frank. Christina has $𝟐𝟐𝟐𝟐 more than Marissa. If Christina has $𝟏𝟏𝟏𝟏𝟏𝟏, how much money does Frank have? Let 𝒇𝒇 represent the amount of money Frank has in dollars and 𝒎𝒎 represent the amount of money Marissa has in dollars. Draw a tape diagram to represent the amount of money Frank has. Draw a tape diagram to represent the amount of money Marissa has. Draw a tape diagram to represent the amount of money Christina has. Which tape diagram provides enough information to determine the value of the variable 𝒎𝒎? The tape diagram that represents the amount of money Christina has. Write and solve the equation. 𝒎𝒎+ 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎+ 𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏−𝟐𝟐𝟐𝟐 𝒎𝒎= 𝟖𝟖𝟖𝟖 The identities we have discussed throughout the module solidify that 𝒎𝒎+ 𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐= 𝒎𝒎. What does the 𝟖𝟖𝟖𝟖 represent? 𝟖𝟖𝟖𝟖 is the amount of money, in dollars, that Marissa has. Now that we know Marissa has $𝟖𝟖𝟖𝟖, how can we use this information to find out how much money Frank has? We can write an equation to represent Marissa’s tape diagram since we now know the length is 𝟖𝟖𝟖𝟖. Write an equation. 𝟐𝟐𝟐𝟐= 𝟖𝟖𝟖𝟖 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎 𝟐𝟐𝟐𝟐 𝒇𝒇 𝒇𝒇 𝒇𝒇 A STORY OF RATIOS 306 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝟓𝟓𝟓𝟓 𝒓𝒓 𝟏𝟏𝟏𝟏 Solve the equation. 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝟖𝟖𝟖𝟖÷ 𝟐𝟐 𝒇𝒇= 𝟒𝟒𝟒𝟒 Once again, the identities we have used throughout the module can solidify that 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝒇𝒇. What does the 𝟒𝟒𝟒𝟒 represent? The 𝟒𝟒𝟒𝟒 represents the amount of money Frank has in dollars. Does 𝟒𝟒𝟒𝟒 make sense in the problem? Yes, because if Frank has $𝟒𝟒𝟒𝟒, then Marissa has twice this, which is $𝟖𝟖𝟖𝟖. Then, Christina has $𝟏𝟏𝟏𝟏𝟏𝟏 because she has $𝟐𝟐𝟐𝟐 more than Marissa, which is what the problem stated. Exercises (20 minutes; 5 minutes per station) Students work in small groups to complete the following stations. Station One: Use tape diagrams to solve the problem. Raeana is twice as old as Madeline, and Laura is 𝟏𝟏𝟏𝟏 years older than Raeana. If Laura is 𝟓𝟓𝟓𝟓 years old, how old is Madeline? Let 𝒎𝒎 represent Madeline’s age in years, and let 𝒓𝒓 represent Raeana’s age in years. Raeana’s Tape Diagram: Madeline’s Tape Diagram: Laura’s Tape Diagram: Equation for Laura’s Tape Diagram: 𝒓𝒓+ 𝟏𝟏𝟏𝟏= 𝟓𝟓𝟓𝟓 𝒓𝒓+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟓𝟓𝟓𝟓−𝟏𝟏𝟏𝟏 𝒓𝒓= 𝟒𝟒𝟒𝟒 We now know that Raeana is 𝟒𝟒𝟒𝟒 years old, and we can use this and Raeana’s tape diagram to determine the age of Madeline. 𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝟒𝟒𝟒𝟒÷ 𝟐𝟐 𝒎𝒎= 𝟐𝟐𝟐𝟐 Therefore, Madeline is 𝟐𝟐𝟐𝟐 years old. 𝒎𝒎 𝒎𝒎 𝒎𝒎 MP.1 A STORY OF RATIOS 307 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝟗𝟗𝟗𝟗 𝒕𝒕 𝒕𝒕 𝒕𝒕 𝟏𝟏𝟏𝟏𝟏𝟏 𝒓𝒓 𝟒𝟒𝟒𝟒 Station Two: Use tape diagrams to solve the problem. Carli has 𝟗𝟗𝟗𝟗 apps on her phone. Braylen has half the amount of apps as Theiss. If Carli has three times the amount of apps as Theiss, how many apps does Braylen have? Let 𝒃𝒃 represent the number of Braylen’s apps and 𝒕𝒕 represent the number of Theiss’s apps. Theiss’s Tape Diagram: Braylen’s Tape Diagram: Carli’s Tape Diagram: Equation for Carli’s Tape Diagram: We now know that Theiss has 𝟑𝟑𝟑𝟑 apps on his phone. We can use this information to write an equation for Braylen’s tape diagram and determine how many apps are on Braylen’s phone. 𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝟑𝟑𝟑𝟑÷ 𝟐𝟐 𝒃𝒃= 𝟏𝟏𝟏𝟏 Therefore, Braylen has 𝟏𝟏𝟏𝟏 apps on his phone. Station Three: Use tape diagrams to solve the problem. Reggie ran for 𝟏𝟏𝟏𝟏𝟏𝟏 yards during the last football game, which is 𝟒𝟒𝟒𝟒 more yards than his previous personal best. Monte ran 𝟓𝟓𝟓𝟓 more yards than Adrian during the same game. If Monte ran the same amount of yards Reggie ran in one game for his previous personal best, how many yards did Adrian run? Let 𝒓𝒓 represent the number of yards Reggie ran during his previous personal best and 𝒂𝒂 represent the number of yards Adrian ran. Reggie’s Tape Diagram: Monte’s Tape Diagram: Adrian’s Tape Diagram: 𝒃𝒃 𝒃𝒃 𝒕𝒕 𝒂𝒂 𝟓𝟓𝟓𝟓 MP.1 𝒂𝒂 A STORY OF RATIOS 308 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝒇𝒇 𝒇𝒇 𝒇𝒇 𝒇𝒇 𝟔𝟔𝟔𝟔 Combining all 𝟑𝟑 tape diagrams: Equation for Reggie’s Tape Diagram: 𝒓𝒓+ 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏𝟏𝟏 Equation for Monte’s Tape Diagram: 𝒂𝒂+ 𝟓𝟓𝟓𝟓+ 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏𝟏𝟏 𝒂𝒂+ 𝟗𝟗𝟗𝟗= 𝟏𝟏𝟏𝟏𝟏𝟏 𝒂𝒂+ 𝟗𝟗𝟗𝟗−𝟗𝟗𝟗𝟗= 𝟏𝟏𝟏𝟏𝟏𝟏−𝟗𝟗𝟗𝟗 𝒂𝒂= 𝟗𝟗𝟗𝟗 Therefore, Adrian ran 𝟗𝟗𝟗𝟗 yards during the football game. Station Four: Use tape diagrams to solve the problem. Lance rides his bike downhill at a pace of 𝟔𝟔𝟔𝟔 miles per hour. When Lance is riding uphill, he rides 𝟖𝟖 miles per hour slower than on flat roads. If Lance’s downhill speed is 𝟒𝟒 times faster than his flat-road speed, how fast does he travel uphill? Let 𝒇𝒇 represent Lance’s pace on flat roads in miles per hour and 𝒖𝒖 represent Lance’s pace uphill in miles per hour. Tape Diagram for Uphill Pace: Tape Diagram for Downhill Pace: Equation for Downhill Pace: 𝟒𝟒𝟒𝟒= 𝟔𝟔𝟔𝟔 𝟒𝟒𝟒𝟒÷ 𝟒𝟒= 𝟔𝟔𝟔𝟔÷ 𝟒𝟒 𝒇𝒇= 𝟏𝟏𝟏𝟏 Equation for Uphill Pace: 𝒖𝒖+ 𝟖𝟖= 𝟏𝟏𝟏𝟏 𝒖𝒖+ 𝟖𝟖−𝟖𝟖= 𝟏𝟏𝟏𝟏−𝟖𝟖 𝒖𝒖= 𝟕𝟕 Therefore, Lance travels at a pace of 𝟕𝟕 miles per hour uphill. Reggie Monte 𝒂𝒂 𝟓𝟓𝟓𝟓 𝒂𝒂 𝟏𝟏𝟏𝟏𝟏𝟏 𝒓𝒓 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 𝟗𝟗𝟗𝟗 𝒇𝒇 𝒖𝒖 𝟖𝟖 MP.1 A STORY OF RATIOS 309 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Closing (4 minutes) Use this time to go over the solutions to the stations and answer student questions. How did the tape diagrams help you create the expressions and equations that you used to solve the problems? Answers will vary. Exit Ticket (5 minutes) A STORY OF RATIOS 310 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Name Date Lesson 28: Two-Step Problems―All Operations Exit Ticket Use tape diagrams and equations to solve the problem with visual models and algebraic methods. Alyssa is twice as old as Brittany, and Jazmyn is 15 years older than Alyssa. If Jazmyn is 35 years old, how old is Brittany? Let 𝑎𝑎 represent Alyssa’s age in years and 𝑏𝑏 represent Brittany’s age in years. A STORY OF RATIOS 311 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝒂𝒂 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 Exit Ticket Sample Solutions Use tape diagrams and equations to solve the problem with visual models and algebraic methods. Alyssa is twice as old as Brittany, and Jazmyn is 𝟏𝟏𝟏𝟏 years older than Alyssa. If Jazmyn is 𝟑𝟑𝟑𝟑 years old, how old is Brittany? Let 𝒂𝒂 represent Alyssa’s age in years and 𝒃𝒃 represent Brittany’s age in years. Brittany’s Tape Diagram: Alyssa’s Tape Diagram: Jazmyn’s Tape Diagram: Equation for Jazmyn’s Tape Diagram: 𝒂𝒂+ 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 𝒂𝒂+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑−𝟏𝟏𝟏𝟏 𝒂𝒂= 𝟐𝟐𝟐𝟐 Now that we know Alyssa is 𝟐𝟐𝟐𝟐 years old, we can use this information and Alyssa’s tape diagram to determine Brittany’s age. 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝟐𝟐𝟐𝟐÷ 𝟐𝟐 𝒃𝒃= 𝟏𝟏𝟏𝟏 Therefore, Brittany is 𝟏𝟏𝟏𝟏 years old. 𝒃𝒃 𝒃𝒃 𝒃𝒃 A STORY OF RATIOS 312 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝒄𝒄 𝟏𝟏𝟏𝟏 𝒄𝒄 𝟓𝟓𝟓𝟓 𝒅𝒅 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 Problem Set Sample Solutions Use tape diagrams to solve each problem. 1. Dwayne scored 𝟓𝟓𝟓𝟓 points in the last basketball game, which is 𝟏𝟏𝟏𝟏 points more than his previous personal best. Lebron scored 𝟏𝟏𝟓𝟓 points more than Chris in the same game. Lebron scored the same number of points as Dwayne’s previous personal best. Let 𝒅𝒅 represent the number of points Dwayne scored during his previous personal best and 𝒄𝒄 represent the number of Chris’s points. a. How many points did Chris score during the game? Equation for Dwayne’s Tape Diagram: 𝒅𝒅+ 𝟏𝟏𝟏𝟏= 𝟓𝟓𝟓𝟓 Equation for Lebron’s Tape Diagram: 𝒄𝒄+ 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟓𝟓𝟓𝟓 𝒄𝒄+ 𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓 𝒄𝒄+ 𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓−𝟐𝟐𝟐𝟐 𝒄𝒄= 𝟑𝟑𝟑𝟑 Therefore, Chris scored 𝟑𝟑𝟑𝟑 points in the game. b. If these are the only three players who scored, what was the team’s total number of points at the end of the game? Dwayne scored 𝟓𝟓𝟓𝟓 points. Chris scored 𝟑𝟑𝟑𝟑 points. Lebron scored 𝟒𝟒𝟒𝟒 points (answer to Dwayne’s equation). Therefore, the total number of points scored is 𝟓𝟓𝟓𝟓+ 𝟑𝟑𝟑𝟑+ 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏𝟏𝟏. Dwayne Lebron A STORY OF RATIOS 313 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 𝒃𝒃 𝒅𝒅 𝟏𝟏𝟏𝟏 𝒃𝒃 𝒃𝒃 𝒃𝒃 𝟏𝟏𝟏𝟏𝟏𝟏 2. The number of customers at Yummy Smoothies varies throughout the day. During the lunch rush on Saturday, there were 𝟏𝟏𝟏𝟏𝟏𝟏 customers at Yummy Smoothies. The number of customers at Yummy Smoothies during dinner time was 𝟏𝟏𝟏𝟏 customers fewer than the number during breakfast. The number of customers at Yummy Smoothies during lunch was 𝟑𝟑 times more than during breakfast. How many people were at Yummy Smoothies during breakfast? How many people were at Yummy Smoothies during dinner? Let 𝒅𝒅 represent the number of customers at Yummy Smoothies during dinner and 𝒃𝒃 represent the number of customers at Yummy Smoothies during breakfast. Tape Diagram for Lunch: Tape Diagram for Dinner: Equation for Lunch’s Tape Diagram: 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑÷ 𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏÷ 𝟑𝟑 𝒃𝒃= 𝟒𝟒𝟒𝟒 Now that we know 𝟒𝟒𝟒𝟒 customers were at Yummy Smoothies for breakfast, we can use this information and the tape diagram for dinner to determine how many customers were at Yummy Smoothies during dinner. 𝒅𝒅+ 𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒 𝒅𝒅+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟒𝟒𝟒𝟒−𝟏𝟏𝟏𝟏 𝒅𝒅= 𝟑𝟑𝟑𝟑 Therefore, 𝟑𝟑𝟑𝟑 customers were at Yummy Smoothies during dinner and 𝟒𝟒𝟒𝟒 customers during breakfast. 3. Karter has 𝟐𝟐𝟐𝟐 T-shirts. Karter has 𝟖𝟖 fewer pairs of shoes than pairs of pants. If the number of T-shirts Karter has is double the number of pants he has, how many pairs of shoes does Karter have? Let 𝒑𝒑 represent the number of pants Karter has and 𝒔𝒔 represent the number of pairs of shoes he has. Tape Diagram for T-Shirts: 𝒑𝒑 𝒑𝒑 Tape Diagram for Shoes: 𝒔𝒔 𝟖𝟖 Equation for T-Shirts Tape Diagram: 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐÷ 𝟐𝟐= 𝟐𝟐𝟐𝟐÷ 𝟐𝟐 𝒑𝒑= 𝟏𝟏𝟏𝟏 Equation for Shoes Tape Diagram: 𝒔𝒔+ 𝟖𝟖= 𝟏𝟏𝟏𝟏 𝒔𝒔+ 𝟖𝟖−𝟖𝟖= 𝟏𝟏𝟏𝟏−𝟖𝟖 𝒔𝒔= 𝟒𝟒 Karter has 𝟒𝟒 pairs of shoes. 𝒑𝒑 𝟐𝟐𝟐𝟐 A STORY OF RATIOS 314 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations 4. Darnell completed 𝟑𝟑𝟑𝟑 push-ups in one minute, which is 𝟖𝟖 more than his previous personal best. Mia completed 𝟔𝟔 more push-ups than Katie. If Mia completed the same amount of push-ups as Darnell completed during his previous personal best, how many push-ups did Katie complete? Let 𝒅𝒅 represent the number of push-ups Darnell completed during his previous personal best and 𝒌𝒌 represent the number of push-ups Katie completed. 𝒅𝒅 𝟖𝟖 𝒌𝒌 𝟔𝟔 𝟖𝟖 𝒌𝒌 𝟏𝟏𝟏𝟏 𝒅𝒅+ 𝟖𝟖= 𝟑𝟑𝟑𝟑 𝒌𝒌+ 𝟔𝟔+ 𝟖𝟖= 𝟑𝟑𝟑𝟑 𝒌𝒌+ 𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑 𝒌𝒌+ 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑−𝟏𝟏𝟏𝟏 𝒌𝒌= 𝟐𝟐𝟐𝟐 Katie completed 𝟐𝟐𝟐𝟐 push-ups. 5. Justine swims freestyle at a pace of 𝟏𝟏𝟏𝟏𝟏𝟏 laps per hour. Justine swims breaststroke 𝟐𝟐𝟐𝟐 laps per hour slower than she swims butterfly. If Justine’s freestyle speed is three times faster than her butterfly speed, how fast does she swim breaststroke? Let 𝒃𝒃 represent Justine’s butterfly speed in laps per hour and 𝒓𝒓 represent Justine’s breaststroke speed in laps per hour. Tape Diagram for Breaststroke: 𝒓𝒓 𝟐𝟐𝟐𝟐 Tape Diagram for Freestyle: 𝒃𝒃 𝒃𝒃 𝒃𝒃 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑÷ 𝟑𝟑= 𝟏𝟏𝟏𝟏𝟏𝟏÷ 𝟑𝟑 𝒃𝒃= 𝟓𝟓𝟓𝟓 Therefore, Justine swims butterfly at a pace of 𝟓𝟓𝟓𝟓 laps per hour. 𝒓𝒓+ 𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓 𝒓𝒓+ 𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓−𝟐𝟐𝟐𝟐 𝒓𝒓= 𝟑𝟑𝟑𝟑 Therefore, Justine swims breaststroke at a pace of 𝟑𝟑𝟑𝟑 laps per hour. 𝟑𝟑𝟑𝟑 𝒃𝒃 𝟏𝟏𝟏𝟏𝟏𝟏 A STORY OF RATIOS 315 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Addition of Decimals II—Round 1 Directions: Evaluate each expression. 1. 2.5 + 4 23. 4.5 + 3.1 2. 2.5 + 0.4 24. 4.5 + 0.31 3. 2.5 + 0.04 25. 4.5 + 0.031 4. 2.5 + 0.004 26. 0.45 + 0.031 5. 2.5 + 0.0004 27. 0.045 + 0.031 6. 6 + 1.3 28. 12 + 0.36 7. 0.6 + 1.3 29. 1.2 + 3.6 8. 0.06 + 1.3 30. 1.2 + 0.36 9. 0.006 + 1.3 31. 1.2 + 0.036 10. 0.0006 + 1.3 32. 0.12 + 0.036 11. 0.6 + 13 33. 0.012 + 0.036 12. 7 + 0.2 34. 0.7 + 3 13. 0.7 + 0.02 35. 0.7 + 0.3 14. 0.07 + 0.2 36. 0.07 + 0.03 15. 0.7 + 2 37. 0.007 + 0.003 16. 7 + 0.02 38. 5 + 0.5 17. 6 + 0.3 39. 0.5 + 0.5 18. 0.6 + 0.03 40. 0.05 + 0.05 19. 0.06 + 0.3 41. 0.005 + 0.005 20. 0.6 + 3 42. 0.11 + 19 21. 6 + 0.03 43. 1.1 + 1.9 22. 0.6 + 0.3 44. 0.11 + 0.19 Number Correct: A STORY OF RATIOS 316 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Addition of Decimals II—Round 1 [KEY] Directions: Evaluate each expression. 1. 2.5 + 4 𝟔𝟔. 𝟓𝟓 23. 4.5 + 3.1 𝟕𝟕. 𝟔𝟔 2. 2.5 + 0.4 𝟐𝟐. 𝟗𝟗 24. 4.5 + 0.31 𝟒𝟒. 𝟖𝟖𝟖𝟖 3. 2.5 + 0.04 𝟐𝟐. 𝟓𝟓𝟓𝟓 25. 4.5 + 0.031 𝟒𝟒. 𝟓𝟓𝟓𝟓𝟓𝟓 4. 2.5 + 0.004 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓 26. 0.45 + 0.031 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 5. 2.5 + 0.0004 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 27. 0.045 + 0.031 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 6. 6 + 1.3 𝟕𝟕. 𝟑𝟑 28. 12 + 0.36 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑 7. 0.6 + 1.3 𝟏𝟏. 𝟗𝟗 29. 1.2 + 3.6 𝟒𝟒. 𝟖𝟖 8. 0.06 + 1.3 𝟏𝟏. 𝟑𝟑𝟑𝟑 30. 1.2 + 0.36 𝟏𝟏. 𝟓𝟓𝟓𝟓 9. 0.006 + 1.3 𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑 31. 1.2 + 0.036 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 10. 0.0006 + 1.3 𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 32. 0.12 + 0.036 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏 11. 0.6 + 13 𝟏𝟏𝟏𝟏. 𝟔𝟔 33. 0.012 + 0.036 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 12. 7 + 0.2 𝟕𝟕. 𝟐𝟐 34. 0.7 + 3 𝟑𝟑. 𝟕𝟕 13. 0.7 + 0.02 𝟎𝟎. 𝟕𝟕𝟕𝟕 35. 0.7 + 0.3 𝟏𝟏 14. 0.07 + 0.2 𝟎𝟎. 𝟐𝟐𝟐𝟐 36. 0.07 + 0.03 𝟎𝟎. 𝟏𝟏 15. 0.7 + 2 𝟐𝟐. 𝟕𝟕 37. 0.007 + 0.003 𝟎𝟎. 𝟎𝟎𝟎𝟎 16. 7 + 0.02 𝟕𝟕. 𝟎𝟎𝟎𝟎 38. 5 + 0.5 𝟓𝟓. 𝟓𝟓 17. 6 + 0.3 𝟔𝟔. 𝟑𝟑 39. 0.5 + 0.5 𝟏𝟏 18. 0.6 + 0.03 𝟎𝟎. 𝟔𝟔𝟔𝟔 40. 0.05 + 0.05 𝟎𝟎. 𝟏𝟏 19. 0.06 + 0.3 𝟎𝟎. 𝟑𝟑𝟑𝟑 41. 0.005 + 0.005 𝟎𝟎. 𝟎𝟎𝟎𝟎 20. 0.6 + 3 𝟑𝟑. 𝟔𝟔 42. 0.11 + 19 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏 21. 6 + 0.03 𝟔𝟔. 𝟎𝟎𝟎𝟎 43. 1.1 + 1.9 𝟑𝟑 22. 0.6 + 0.3 𝟎𝟎. 𝟗𝟗 44. 0.11 + 0.19 𝟎𝟎. 𝟑𝟑 A STORY OF RATIOS 317 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Addition of Decimals II—Round 2 Directions: Evaluate each expression. 1. 7.4 + 3 23. 3.6 + 2.3 2. 7.4 + 0.3 24. 3.6 + 0.23 3. 7.4 + 0.03 25. 3.6 + 0.023 4. 7.4 + 0.003 26. 0.36 + 0.023 5. 7.4 + 0.0003 27. 0.036 + 0.023 6. 6 + 2.2 28. 0.13 + 56 7. 0.6 + 2.2 29. 1.3 + 5.6 8. 0.06 + 2.2 30. 1.3 + 0.56 9. 0.006 + 2.2 31. 1.3 + 0.056 10. 0.0006 + 2.2 32. 0.13 + 0.056 11. 0.6 + 22 33. 0.013 + 0.056 12. 7 + 0.8 34. 2 + 0.8 13. 0.7 + 0.08 35. 0.2 + 0.8 14. 0.07 + 0.8 36. 0.02 + 0.08 15. 0.7 + 8 37. 0.002 + 0.008 16. 7 + 0.08 38. 0.16 + 14 17. 5 + 0.4 39. 1.6 + 1.4 18. 0.5 + 0.04 40. 0.16 + 0.14 19. 0.05 + 0.4 41. 0.016 + 0.014 20. 0.5 + 4 42. 15 + 0.15 21. 5 + 0.04 43. 1.5 + 1.5 22. 5 + 0.4 44. 0.15 + 0.15 Number Correct: Improvement: A STORY OF RATIOS 318 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 28 Lesson 28: Two-Step Problems—All Operations Addition of Decimals II—Round 2 [KEY] Directions: Evaluate each expression. 1. 7.4 + 3 𝟏𝟏𝟏𝟏. 𝟒𝟒 23. 3.6 + 2.3 𝟓𝟓. 𝟗𝟗 2. 7.4 + 0.3 𝟕𝟕. 𝟕𝟕 24. 3.6 + 0.23 𝟑𝟑. 𝟖𝟖𝟖𝟖 3. 7.4 + 0.03 𝟕𝟕. 𝟒𝟒𝟒𝟒 25. 3.6 + 0.023 𝟑𝟑. 𝟔𝟔𝟔𝟔𝟔𝟔 4. 7.4 + 0.003 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒 26. 0.36 + 0.023 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 5. 7.4 + 0.0003 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 27. 0.036 + 0.023 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 6. 6 + 2.2 𝟖𝟖. 𝟐𝟐 28. 0.13 + 56 𝟓𝟓𝟓𝟓. 𝟏𝟏𝟏𝟏 7. 0.6 + 2.2 𝟐𝟐. 𝟖𝟖 29. 1.3 + 5.6 𝟔𝟔. 𝟗𝟗 8. 0.06 + 2.2 𝟐𝟐. 𝟐𝟐𝟐𝟐 30. 1.3 + 0.56 𝟏𝟏. 𝟖𝟖𝟖𝟖 9. 0.006 + 2.2 𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐 31. 1.3 + 0.056 𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑 10. 0.0006 + 2.2 𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 32. 0.13 + 0.056 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏 11. 0.6 + 22 𝟐𝟐𝟐𝟐. 𝟔𝟔 33. 0.013 + 0.056 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 12. 7 + 0.8 𝟕𝟕. 𝟖𝟖 34. 2 + 0.8 𝟐𝟐. 𝟖𝟖 13. 0.7 + 0.08 𝟎𝟎. 𝟕𝟕𝟕𝟕 35. 0.2 + 0.8 𝟏𝟏 14. 0.07 + 0.8 𝟎𝟎. 𝟖𝟖𝟖𝟖 36. 0.02 + 0.08 𝟎𝟎. 𝟏𝟏 15. 0.7 + 8 𝟖𝟖. 𝟕𝟕 37. 0.002 + 0.008 𝟎𝟎. 𝟎𝟎𝟎𝟎 16. 7 + 0.08 𝟕𝟕. 𝟎𝟎𝟎𝟎 38. 0.16 + 14 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏 17. 5 + 0.4 𝟓𝟓. 𝟒𝟒 39. 1.6 + 1.4 𝟑𝟑 18. 0.5 + 0.04 𝟎𝟎. 𝟓𝟓𝟓𝟓 40. 0.16 + 0.14 𝟎𝟎. 𝟑𝟑 19. 0.05 + 0.4 𝟎𝟎. 𝟒𝟒𝟒𝟒 41. 0.016 + 0.014 𝟎𝟎. 𝟎𝟎𝟎𝟎 20. 0.5 + 4 𝟒𝟒. 𝟓𝟓 42. 15 + 0.15 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏 21. 5 + 0.04 𝟓𝟓. 𝟎𝟎𝟎𝟎 43. 1.5 + 1.5 𝟑𝟑 22. 5 + 0.4 𝟓𝟓. 𝟒𝟒 44. 0.15 + 0.15 𝟎𝟎. 𝟑𝟑 A STORY OF RATIOS 319 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 Lesson 29: Multi-Step Problems—All Operations Student Outcomes Students use their knowledge of simplifying expressions, order of operations, and properties of equality to calculate the solution of multi-step equations. Students use tables to determine their answers. Students check to determine if their solutions make the equations true. Classwork Example (20 minutes) Students participate in the discussion by answering the teacher’s questions and following along in their student materials. Example The school librarian, Mr. Marker, knows the library has 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 books but wants to reorganize how the books are displayed on the shelves. Mr. Marker needs to know how many fiction, nonfiction, and resource books are in the library. He knows that the library has four times as many fiction books as resource books and half as many nonfiction books as fiction books. If these are the only types of books in the library, how many of each type of book are in the library? Give students time to work individually or with a partner in order to attempt to make sense of the problem. Students may attempt to solve the problem on their own prior to the following discussion. Draw a tape diagram to represent the total number of books in the library. Draw two more tape diagrams, one to represent the number of fiction books in the library and one to represent the number of resource books in the library. Resource Books: Fiction Books: What variable should we use throughout the problem? We should use 𝒓𝒓 to represent the number of resource books in the library because it represents the fewest amount of books. Choosing the variable to represent a different type of book would create fractions throughout the problem. Write the relationship between resource books and fiction books algebraically. If we let 𝒓𝒓 represent the number of resource books, then 𝟒𝟒𝟒𝟒 represents the number of fiction books. Draw a tape diagram to represent the number of nonfiction books. Nonfiction Books: MP.1 A STORY OF RATIOS 320 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations How did you decide how many sections this tape diagram would have? There are half as many nonfiction books as fiction books. Since the fiction book tape diagram has four sections, the nonfiction book tape diagram should have two sections. Represent the number of nonfiction books in the library algebraically. 𝟐𝟐𝟐𝟐 because that is half as many as fiction books (𝟒𝟒𝟒𝟒). Use the tape diagrams we drew to solve the problem. We know that combining the tape diagrams for each type of book will leave us with 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 total books. Write an equation that represents the tape diagram. 𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐+ 𝒓𝒓= 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 Determine the value of 𝒓𝒓. We can gather like terms and then solve the equation. 𝟕𝟕𝟕𝟕= 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕÷ 𝟕𝟕= 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒÷ 𝟕𝟕 𝒓𝒓= 𝟐𝟐𝟐𝟐𝟐𝟐 What does this 200 mean? There are 200 resource books in the library because 𝑟𝑟 represented the number of resource books. How many fiction books are in the library? There are 𝟖𝟖𝟖𝟖𝟖𝟖 fiction books in the library because 𝟒𝟒(𝟐𝟐𝟐𝟐𝟐𝟐) = 𝟖𝟖𝟖𝟖𝟖𝟖. How many nonfiction books are in the library? There are 𝟒𝟒𝟒𝟒𝟒𝟒 nonfiction books in the library because 𝟐𝟐(𝟐𝟐𝟐𝟐𝟐𝟐) = 𝟒𝟒𝟒𝟒𝟒𝟒. We can use a different math tool to solve the problem as well. If we were to make a table, how many columns would we need? 4 Why do we need four columns? We need to keep track of the number of fiction, nonfiction, and resource books that are in the library, but we also need to keep track of the total number of books. Set up a table with four columns, and label each column. Fiction Nonfiction Resource Total 𝒓𝒓 𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 A STORY OF RATIOS 321 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations Highlight the important information from the word problem that will help us fill out the second row in our table. The school librarian, Mr. Marker, knows the library has 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 books but wants to reorganize how the books are displayed on the shelves. Mr. Marker needs to know how many fiction, nonfiction, and resource books are in the library. He knows that the library has four times as many fiction books as resource books and half as many nonfiction books as fiction books. If these are the only types of books in the library, how many of each type of book are in the library? Fill out the second row of the table using the algebraic representations. Fiction Nonfiction Resource Total 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝒓𝒓 𝟕𝟕𝟕𝟕 If 𝑟𝑟= 1, how many of each type of book would be in the library? Fiction Nonfiction Resource Total 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝒓𝒓 𝟕𝟕𝟕𝟕 𝟒𝟒 𝟐𝟐 𝟏𝟏 𝟕𝟕 How can we fill out another row of the table? Substitute different values in for 𝑟𝑟. Substitute 5 in for 𝑟𝑟. How many of each type of book would be in the library then? Fiction Nonfiction Resource Total 𝟒𝟒 𝟐𝟐 𝟏𝟏 𝟕𝟕 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟑𝟑𝟑𝟑 Does the library have four times as many fiction books as resource books? Yes, because 5 ∙4 = 20. Does the library have half as many nonfiction books as fiction books? Yes, because half of 20 is 10. How do we determine how many of each type of book is in the library when there are 1,400 books in the library? Continue to multiply the rows by the same value, until the total column has 1,400 books. At this point, allow students to work individually to determine how many fiction, nonfiction, and resource books are in the library if there are 1,400 total books. Each table may look different because students may choose different values to multiply by. A sample answer is shown below. Fiction Nonfiction Resource Total 𝟒𝟒 𝟐𝟐 𝟏𝟏 𝟕𝟕 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑𝟑𝟑 𝟖𝟖𝟖𝟖𝟖𝟖 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 How many fiction books are in the library? 𝟖𝟖𝟖𝟖𝟖𝟖 A STORY OF RATIOS 322 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations How many nonfiction books are in the library? 𝟒𝟒𝟒𝟒𝟒𝟒 How many resource books are in the library? 𝟐𝟐𝟐𝟐𝟐𝟐 Let us check and make sure that our answers fit the relationship described in the word problem. Does the library have four times as many fiction books as resource books? Yes, because 𝟐𝟐𝟐𝟐𝟐𝟐∙𝟒𝟒= 𝟖𝟖𝟖𝟖𝟖𝟖. Does the library have half as many nonfiction books as fiction books? Yes, because half of 𝟖𝟖𝟖𝟖𝟖𝟖 is 𝟒𝟒𝟒𝟒𝟒𝟒. Does the library have 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 books? Yes, because 𝟖𝟖𝟖𝟖𝟖𝟖+ 𝟒𝟒𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐𝟐𝟐= 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒. Exercises (15 minutes) Students work in small groups to answer the following problems using tables and algebraic methods. Exercises Solve each problem below using tables and algebraic methods. Then, check your answers with the word problems. 1. Indiana Ridge Middle School wanted to add a new school sport, so they surveyed the students to determine which sport is most popular. Students were able to choose among soccer, football, lacrosse, or swimming. The same number of students chose lacrosse and swimming. The number of students who chose soccer was double the number of students who chose lacrosse. The number of students who chose football was triple the number of students who chose swimming. If 𝟒𝟒𝟒𝟒𝟒𝟒 students completed the survey, how many students chose each sport? Soccer Football Lacrosse Swimming Total 𝟐𝟐 𝟑𝟑 𝟏𝟏 𝟏𝟏 𝟕𝟕 The rest of the table will vary. Soccer Football Lacrosse Swimming Total 𝟐𝟐 𝟑𝟑 𝟏𝟏 𝟏𝟏 𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔 𝟔𝟔𝟔𝟔 𝟒𝟒𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 students chose soccer, 𝟏𝟏𝟏𝟏𝟏𝟏 students chose football, 𝟔𝟔𝟔𝟔 students chose lacrosse, and 𝟔𝟔𝟔𝟔 students chose swimming. We can confirm that these numbers satisfy the conditions of the word problem because lacrosse and swimming were chosen by the same number of students. 𝟏𝟏𝟏𝟏𝟏𝟏 is double 𝟔𝟔𝟔𝟔, so soccer was chosen by double the number of students as lacrosse, and 𝟏𝟏𝟏𝟏𝟏𝟏 is triple 𝟔𝟔𝟔𝟔, so football was chosen by 𝟑𝟑 times as many students as swimming. Also, 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟔𝟔𝟔𝟔+ 𝟔𝟔𝟔𝟔= 𝟒𝟒𝟒𝟒𝟒𝟒. Algebraically: Let 𝒔𝒔 represent the number of students who chose swimming. Then, 𝟐𝟐𝟐𝟐 is the number of students who chose soccer, 𝟑𝟑𝟑𝟑 is the number of students who chose football, and 𝒔𝒔 is the number of students who chose lacrosse. 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑+ 𝒔𝒔+ 𝒔𝒔= 𝟒𝟒𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕= 𝟒𝟒𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕÷ 𝟕𝟕= 𝟒𝟒𝟒𝟒𝟒𝟒÷ 𝟕𝟕 𝒔𝒔= 𝟔𝟔𝟔𝟔 Therefore, 𝟔𝟔𝟔𝟔 students chose swimming, and 𝟔𝟔𝟔𝟔 students chose lacrosse. 𝟏𝟏𝟏𝟏𝟏𝟏 students chose soccer because 𝟐𝟐(𝟔𝟔𝟔𝟔) = 𝟏𝟏𝟏𝟏𝟏𝟏, and 𝟏𝟏𝟏𝟏𝟏𝟏 students chose football because 𝟑𝟑(𝟔𝟔𝟔𝟔) = 𝟏𝟏𝟏𝟏𝟏𝟏. MP.1 A STORY OF RATIOS 323 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations 2. At Prairie Elementary School, students are asked to pick their lunch ahead of time so the kitchen staff will know what to prepare. On Monday, 𝟔𝟔 times as many students chose hamburgers as chose salads. The number of students who chose lasagna was one third the number of students who chose hamburgers. If 𝟐𝟐𝟐𝟐𝟐𝟐 students ordered lunch, how many students chose each option if hamburger, salad, and lasagna were the only three options? Hamburger Salad Lasagna Total 𝟔𝟔 𝟏𝟏 𝟐𝟐 𝟗𝟗 The rest of the table will vary. Hamburger Salad Lasagna Total 𝟔𝟔 𝟏𝟏 𝟐𝟐 𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 students chose a hamburger for lunch, 𝟐𝟐𝟐𝟐 students chose a salad, and 𝟓𝟓𝟓𝟓 students chose lasagna. We can confirm that these numbers satisfy the conditions of the word problem because 𝟐𝟐𝟐𝟐∙𝟔𝟔= 𝟏𝟏𝟏𝟏𝟏𝟏, so hamburgers were chosen by 𝟔𝟔 times more students than salads. Also, 𝟏𝟏 𝟑𝟑∙𝟏𝟏𝟏𝟏𝟏𝟏= 𝟓𝟓𝟓𝟓, which means lasagna was chosen by one third of the number of students who chose hamburgers. Finally, 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟐𝟐𝟐𝟐+ 𝟓𝟓𝟓𝟓= 𝟐𝟐𝟐𝟐𝟐𝟐, which means 𝟐𝟐𝟐𝟐𝟐𝟐 students completed the survey. Algebraically: Let 𝒔𝒔 represent the number of students who chose a salad. Then, 𝟔𝟔𝟔𝟔 represents the number of students who chose hamburgers, and 𝟐𝟐𝟐𝟐 represents the number of students who chose lasagna. 𝟔𝟔𝟔𝟔+ 𝒔𝒔+ 𝟐𝟐𝟐𝟐= 𝟐𝟐𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗= 𝟐𝟐𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗÷ 𝟗𝟗= 𝟐𝟐𝟐𝟐𝟐𝟐÷ 𝟗𝟗𝟗𝟗 𝒔𝒔= 𝟐𝟐𝟐𝟐 This means that 𝟐𝟐𝟐𝟐 students chose salad, 𝟏𝟏𝟏𝟏𝟏𝟏 students chose hamburgers because 𝟔𝟔(𝟐𝟐𝟐𝟐) = 𝟏𝟏𝟏𝟏𝟏𝟏, and 𝟓𝟓𝟓𝟓 students chose lasagna because 𝟐𝟐(𝟐𝟐𝟐𝟐) = 𝟓𝟓𝟓𝟓. 3. The art teacher, Mr. Gonzalez, is preparing for a project. In order for students to have the correct supplies, Mr. Gonzalez needs 𝟏𝟏𝟏𝟏 times more markers than pieces of construction paper. He needs the same number of bottles of glue as pieces of construction paper. The number of scissors required for the project is half the number of pieces of construction paper. If Mr. Gonzalez collected 𝟒𝟒𝟒𝟒𝟒𝟒 items for the project, how many of each supply did he collect? Markers Construction Paper Glue Bottles Scissors Total 𝟐𝟐𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟏𝟏 𝟐𝟐𝟐𝟐 The rest of the table will vary. Markers Construction Paper Glue Bottles Scissors Total 𝟐𝟐𝟐𝟐 𝟐𝟐 𝟐𝟐 𝟏𝟏 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒𝟒𝟒 Mr. Gonzalez collected 𝟑𝟑𝟑𝟑𝟑𝟑 markers, 𝟑𝟑𝟑𝟑 pieces of construction paper, 𝟑𝟑𝟑𝟑 glue bottles, and 𝟏𝟏𝟏𝟏 scissors for the project. We can confirm that these numbers satisfy the conditions of the word problem because Mr. Gonzalez collected the same number of pieces of construction paper and glue bottles. Also, 𝟑𝟑𝟑𝟑∙𝟏𝟏𝟏𝟏= 𝟑𝟑𝟑𝟑𝟑𝟑, so Mr. Gonzalez collected 𝟏𝟏𝟏𝟏 times more markers than pieces of construction paper and glue bottles. Mr. Gonzalez only collected 𝟏𝟏𝟏𝟏 pairs of scissors, which is half of the number of pieces of construction paper. The supplies collected add up to 𝟒𝟒𝟒𝟒𝟒𝟒 supplies, which is the number of supplies indicated in the word problem. Algebraically: Let 𝒔𝒔 represent the number of scissors needed for the project, which means 𝟐𝟐𝟐𝟐𝟐𝟐 represents the number of markers needed, 𝟐𝟐𝟐𝟐 represents the number of pieces of construction paper needed, and 𝟐𝟐𝟐𝟐 represents the number of glue bottles needed. 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐+ 𝒔𝒔= 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐= 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝒔𝒔= 𝟏𝟏𝟏𝟏 This means that 𝟏𝟏𝟏𝟏 pairs of scissors, 𝟑𝟑𝟑𝟑𝟑𝟑 markers, 𝟑𝟑𝟑𝟑 pieces of construction paper, and 𝟑𝟑𝟑𝟑 glue bottles are required for the project. MP.1 A STORY OF RATIOS 324 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations 4. The math teacher, Ms. Zentz, is buying appropriate math tools to use throughout the year. She is planning on buying twice as many rulers as protractors. The number of calculators Ms. Zentz is planning on buying is one quarter of the number of protractors. If Ms. Zentz buys 𝟔𝟔𝟔𝟔 items, how many protractors does Ms. Zentz buy? Rulers Protractors Calculators Total 𝟖𝟖 𝟒𝟒 𝟏𝟏 𝟏𝟏𝟏𝟏 The rest of the table will vary. Rulers Protractors Calculators Total 𝟖𝟖 𝟒𝟒 𝟏𝟏 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝟓𝟓 𝟔𝟔𝟔𝟔 Ms. Zentz will buy 𝟐𝟐𝟐𝟐 protractors. We can confirm that this number satisfies the conditions of the word problem because the number of protractors is half of the number of rulers, and the number of calculators is one fourth of the number of protractors. Also, 𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐+ 𝟓𝟓= 𝟔𝟔𝟔𝟔, so the total matches the total supplies that Ms. Zentz bought. Algebraically: Let 𝒄𝒄 represent the number of calculators Ms. Zentz needs for the year. Then, 𝟖𝟖𝟖𝟖 represents the number of rulers, and 𝟒𝟒𝟒𝟒 represents the number of protractors Ms. Zentz will need throughout the year. 𝟖𝟖𝟖𝟖+ 𝟒𝟒𝟒𝟒+ 𝒄𝒄= 𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏= 𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏= 𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏 𝒄𝒄= 𝟓𝟓 Therefore, Ms. Zentz will need 𝟓𝟓 calculators, 𝟒𝟒𝟒𝟒 rulers, and 𝟐𝟐𝟐𝟐 protractors throughout the year. Allow time to answer student questions and discuss answers. In particular, encourage students to compare solution methods with one another, commenting on the accuracy and efficiency of each. Closing (5 minutes) Pam says she only needed two rows in her table to solve each of the problems. How was she able to do this? Answers will vary. Pam only needed two rows on her table because she found the scale factor from the total in the first row and the total given in the problem. Once this scale factor is determined, it can be used for all the columns in the table because each table is a ratio table. Is there a more efficient way to get to the answer than choosing random values by which to multiply each row? Find out how many groups of one set of materials it will take to obtain the total amount desired. Then, multiply the entire row by this number. Students may need to see a demonstration to fully understand the reasoning. Use the exercises to further explain. Relate this problem-solving strategy to the ratio tables discussed throughout Module 1. Exit Ticket (5 minutes) MP.1 A STORY OF RATIOS 325 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations Name Date Lesson 29: Multi-Step Problems—All Operations Exit Ticket Solve the problem using tables and equations, and then check your answer with the word problem. Try to find the answer only using two rows of numbers on your table. A pet store owner, Byron, needs to determine how much food he needs to feed the animals. Byron knows that he needs to order the same amount of bird food as hamster food. He needs four times as much dog food as bird food and needs half the amount of cat food as dog food. If Byron orders 600 packages of animal food, how much dog food does he buy? Let 𝑏𝑏 represent the number of packages of bird food Byron purchased for the pet store. A STORY OF RATIOS 326 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations Exit Ticket Sample Solutions Solve the problem using tables and equations, and then check your answer with the word problem. Try to find the answer only using two rows of numbers on your table. A pet store owner, Byron, needs to determine how much food he needs to feed the animals. Byron knows that he needs to order the same amount of bird food as hamster food. He needs four times as much dog food as bird food and needs half the amount of cat food as dog food. If Byron orders 𝟔𝟔𝟔𝟔𝟔𝟔 packages of animal food, how much dog food does he buy? Let 𝒃𝒃 represent the number of packages of bird food Byron purchased for the pet store. Bird Food Hamster Food Dog Food Cat Food Total 𝟏𝟏 𝟏𝟏 𝟒𝟒 𝟐𝟐 𝟖𝟖 The rest of the table will vary (unless they follow suggestions from the Closing). Bird Food Hamster Food Dog Food Cat Food Total 𝟏𝟏 𝟏𝟏 𝟒𝟒 𝟐𝟐 𝟖𝟖 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕 𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔𝟔𝟔 Byron would need to order 𝟑𝟑𝟑𝟑𝟑𝟑 packages of dog food. The answer makes sense because Byron ordered the same amount of bird food and hamster food. The table also shows that Byron ordered four times as much dog food as bird food, and the amount of cat food he ordered is half the amount of dog food. The total amount of pet food Byron ordered was 𝟔𝟔𝟔𝟔𝟔𝟔 packages, which matches the word problem. Algebraically: Let 𝒃𝒃 represent the number of packages of bird food Byron purchased for the pet store. Therefore, 𝒃𝒃 also represents the amount of hamster food, 𝟒𝟒𝟒𝟒 represents the amount of dog food, and 𝟐𝟐𝟐𝟐 represents the amount of cat food required by the pet store. 𝒃𝒃+ 𝒃𝒃+ 𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐= 𝟔𝟔𝟔𝟔𝟔𝟔 𝟖𝟖𝟖𝟖= 𝟔𝟔𝟔𝟔𝟔𝟔 𝟖𝟖𝟖𝟖÷ 𝟖𝟖= 𝟔𝟔𝟔𝟔𝟔𝟔÷ 𝟖𝟖 𝒃𝒃= 𝟕𝟕𝟕𝟕 Therefore, Byron will order 𝟕𝟕𝟕𝟕 pounds of bird food, which results in 𝟑𝟑𝟑𝟑𝟑𝟑 pounds of dog food because 𝟒𝟒(𝟕𝟕𝟕𝟕) = 𝟑𝟑𝟎𝟎𝟎𝟎. Problem Set Sample Solutions Create tables to solve the problems, and then check your answers with the word problems. 1. On average, a baby uses three times the number of large diapers as small diapers and double the number of medium diapers as small diapers. a. If the average baby uses 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗 diapers, size large and small, how many of each size would be used? Small Medium Large Total 𝟑𝟑 𝟐𝟐 𝟏𝟏 𝟔𝟔 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 𝟗𝟗𝟗𝟗𝟗𝟗 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗 An average baby would use 𝟒𝟒𝟒𝟒𝟒𝟒 small diapers, 𝟗𝟗𝟗𝟗𝟗𝟗 medium diapers, and 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 large diapers. The answer makes sense because the number of large diapers is 𝟑𝟑 times more than small diapers. The number of medium diapers is double the number of small diapers, and the total number of diapers is 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗. A STORY OF RATIOS 327 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations b. Support your answer with equations. Let 𝒔𝒔 represent the number of small diapers a baby needs. Therefore, 𝟐𝟐𝟐𝟐 represents the number of medium diapers, and 𝟑𝟑𝟑𝟑 represents the amount of large diapers a baby needs. 𝒔𝒔+ 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟑𝟑= 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗 𝟔𝟔𝟔𝟔= 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗 𝟔𝟔𝟔𝟔 𝟔𝟔= 𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗 𝟔𝟔 𝒔𝒔= 𝟒𝟒𝟒𝟒𝟒𝟒 Therefore, a baby requires 𝟒𝟒𝟒𝟒𝟒𝟒 small diapers, 𝟗𝟗𝟗𝟗𝟗𝟗 medium diapers (because 𝟐𝟐(𝟒𝟒𝟒𝟒𝟒𝟒) = 𝟗𝟗𝟗𝟗𝟗𝟗), and 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒 large diapers (because 𝟑𝟑(𝟒𝟒𝟒𝟒𝟒𝟒) = 𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒), which matches the answer in part (a). 2. Tom has three times as many pencils as pens but has a total of 𝟏𝟏𝟏𝟏𝟏𝟏 writing utensils. a. How many pencils does Tom have? Pencils Pens Total 𝟑𝟑 𝟏𝟏 𝟒𝟒 𝟕𝟕𝟕𝟕 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 b. How many more pencils than pens does Tom have? 𝟕𝟕𝟕𝟕−𝟐𝟐𝟐𝟐= 𝟓𝟓𝟓𝟓. Tom has 𝟓𝟓𝟓𝟓 more pencils than pens. 3. Serena’s mom is planning her birthday party. She bought balloons, plates, and cups. Serena’s mom bought twice as many plates as cups. The number of balloons Serena’s mom bought was half the number of cups. a. If Serena’s mom bought 𝟖𝟖𝟖𝟖 items, how many of each item did she buy? Balloons Plates Cups Total 𝟏𝟏 𝟒𝟒 𝟐𝟐 𝟕𝟕 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐 𝟖𝟖𝟖𝟖 Serena’s mom bought 𝟏𝟏𝟏𝟏 balloons, 𝟒𝟒𝟒𝟒 plates, and 𝟐𝟐𝟐𝟐 cups. b. Tammy brought 𝟏𝟏𝟏𝟏 balloons to the party. How many total balloons were at Serena’s birthday party? 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏= 𝟐𝟐𝟐𝟐. There were 𝟐𝟐𝟐𝟐 total balloons at the party. c. If half the plates and all but four cups were used during the party, how many plates and cups were used? 𝟏𝟏 𝟐𝟐∙𝟒𝟒𝟒𝟒= 𝟐𝟐𝟐𝟐. Twenty-four plates were used during the party. 𝟐𝟐𝟐𝟐−𝟒𝟒= 𝟐𝟐𝟐𝟐. Twenty cups were used during the party. 4. Elizabeth has a lot of jewelry. She has four times as many earrings as watches but half the number of necklaces as earrings. Elizabeth has the same number of necklaces as bracelets. a. If Elizabeth has 𝟏𝟏𝟏𝟏𝟏𝟏 pieces of jewelry, how many earrings does she have? Earrings Watches Necklaces Bracelets Total 𝟒𝟒 𝟏𝟏 𝟐𝟐 𝟐𝟐 𝟗𝟗 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 Elizabeth has 𝟓𝟓𝟓𝟓 earrings, 𝟏𝟏𝟏𝟏 watches, 𝟐𝟐𝟐𝟐 necklaces, and 𝟐𝟐𝟐𝟐 bracelets. A STORY OF RATIOS 328 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 29 Lesson 29: Multi-Step Problems—All Operations b. Support your answer with an equation. Let 𝒘𝒘 represent the number of watches Elizabeth has. Therefore, 𝟒𝟒𝟒𝟒 represents the number of earrings Elizabeth has, and 𝟐𝟐𝟐𝟐 represents both the number of necklaces and bracelets she has. 𝟒𝟒𝟒𝟒+ 𝒘𝒘+ 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗𝟗𝟗= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗𝟗𝟗 𝟗𝟗= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗 𝒘𝒘= 𝟏𝟏𝟏𝟏 Therefore, Elizabeth has 𝟏𝟏𝟏𝟏 watches, 𝟓𝟓𝟓𝟓 earrings because 𝟒𝟒(𝟏𝟏𝟏𝟏) = 𝟓𝟓𝟓𝟓, and 𝟐𝟐𝟐𝟐 necklaces and bracelets each because 𝟐𝟐(𝟏𝟏𝟏𝟏) = 𝟐𝟐𝟐𝟐. 5. Claudia was cooking breakfast for her entire family. She made double the amount of chocolate chip pancakes as she did regular pancakes. She only made half as many blueberry pancakes as she did regular pancakes. Claudia also knows her family loves sausage, so she made triple the amount of sausage as blueberry pancakes. a. How many of each breakfast item did Claudia make if she cooked 𝟗𝟗𝟗𝟗 items in total? Chocolate Chip Pancakes Regular Pancakes Blueberry Pancakes Sausage Total 𝟒𝟒 𝟐𝟐 𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟗𝟗 𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗 Claudia cooked 𝟑𝟑𝟑𝟑 chocolate chip pancakes, 𝟏𝟏𝟏𝟏 regular pancakes, 𝟗𝟗 blueberry pancakes, and 𝟐𝟐𝟐𝟐 pieces of sausage. b. After everyone ate breakfast, there were 𝟒𝟒 chocolate chip pancakes, 𝟓𝟓 regular pancakes, 𝟏𝟏 blueberry pancake, and no sausage left. How many of each item did the family eat? The family ate 𝟑𝟑𝟑𝟑 chocolate chip pancakes, 𝟏𝟏𝟏𝟏 regular pancakes, 𝟖𝟖 blueberry pancakes, and 𝟐𝟐𝟐𝟐 pieces of sausage during breakfast. 6. During a basketball game, Jeremy scored triple the number of points as Donovan. Kolby scored double the number of points as Donovan. a. If the three boys scored 𝟑𝟑𝟑𝟑 points, how many points did each boy score? Jeremy Donovan Kolby Total 𝟑𝟑 𝟏𝟏 𝟐𝟐 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 Jeremy scored 𝟏𝟏𝟏𝟏 points, Donovan scored 𝟔𝟔 points, and Kolby scored 𝟏𝟏𝟏𝟏 points. b. Support your answer with an equation. Let 𝒅𝒅 represent the number of points Donovan scored, which means 𝟑𝟑𝟑𝟑 represents the number of points Jeremy scored, and 𝟐𝟐𝟐𝟐 represents the number of points Kolby scored. 𝟑𝟑𝟑𝟑+ 𝒅𝒅+ 𝟐𝟐𝟐𝟐= 𝟑𝟑𝟑𝟑 𝟔𝟔𝟔𝟔= 𝟑𝟑𝟑𝟑 𝟔𝟔𝟔𝟔 𝟔𝟔= 𝟑𝟑𝟑𝟑 𝟔𝟔 𝒅𝒅= 𝟔𝟔 Therefore, Donovan scored 𝟔𝟔 points, Jeremy scored 𝟏𝟏𝟏𝟏 points because 𝟑𝟑(𝟔𝟔) = 𝟏𝟏𝟏𝟏, and Kolby scored 𝟏𝟏𝟏𝟏 points because 𝟐𝟐(𝟔𝟔) = 𝟏𝟏𝟏𝟏. A STORY OF RATIOS 329 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 Topic H: Applications of Equations GRADE 6 • MODULE 4 6 GRADE Mathematics Curriculum Topic H Applications of Equations 6.EE.B.5, 6.EE.B.6, 6.EE.B.7, 6.EE.B.8, 6.EE.C.9 Focus Standards: 6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form 𝑥𝑥+ 𝑝𝑝= 𝑞𝑞 and 𝑝𝑝𝑝𝑝= 𝑞𝑞 for cases in which 𝑝𝑝, 𝑞𝑞, and 𝑥𝑥 are all nonnegative rational numbers. 6.EE.B.8 Write an inequality of the form 𝑥𝑥> 𝑐𝑐 or 𝑥𝑥< 𝑐𝑐 to represent a constraint or condition in a real-world mathematical problem. Recognize that inequalities of the form 𝑥𝑥> 𝑐𝑐 or 𝑥𝑥< 𝑐𝑐 have infinitely many solutions; represent solutions of such inequalities on number line diagrams. 6.EE.C.9 Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation 𝑑𝑑= 65𝑡𝑡 to represent the relationship between distance and time. A STORY OF RATIOS 330 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Topic H Topic H: Applications of Equations Instructional Days: 5 Lesson 30: One-Step Problems in the Real World (P)1 Lesson 31: Problems in Mathematical Terms (P) Lesson 32: Multi-Step Problems in the Real World (P) Lesson 33: From Equations to Inequalities (P) Lesson 34: Writing and Graphing Inequalities in Real-World Problems (P) In Topic H, students apply their knowledge from the entire module to solve equations in real-world, contextual problems. In Lesson 30, students use prior knowledge from Grade 4 to solve missing angle problems. Students write and solve one-step equations in order to determine a missing angle. Lesson 31 involves students using their prior knowledge from Module 1 to construct tables of independent and dependent values in order to analyze equations with two variables from real-life contexts. They represent equations by plotting values from the tables on a coordinate grid in Lesson 32. The module concludes with Lessons 33 and 34, where students refer to true and false number sentences in order to move from solving equations to writing inequalities that represent a constraint or condition in real-life or mathematical problems. Students understand that inequalities have infinitely many solutions and represent those solutions on number line diagrams. 1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson A STORY OF RATIOS 331 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Lesson 30: One-Step Problems in the Real World Student Outcomes Students calculate missing angle measures by writing and solving equations. Lesson Notes This is an application lesson based on understandings developed in Grade 4. The three standards applied in this lesson include the following: 4.MD.C.5 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement: a. An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles. b. An angle that turns through 𝑛𝑛 one-degree angles is said to have an angle measure of 𝑛𝑛 degrees. 4.MD.C.6 Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. 4.MD.C.7 Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real-world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. This lesson focuses, in particular, on 4.MD.C.7. Classwork Fluency Exercise (5 minutes): Subtraction of Decimals Sprint: Refer to Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening Exercise (3 minutes) Students start the lesson with a review of key angle terms from Grade 4. Opening Exercise Draw an example of each term, and write a brief description. Acute Less than 𝟗𝟗𝟗𝟗° A STORY OF RATIOS 332 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Obtuse Between 𝟗𝟗𝟗𝟗° and 𝟏𝟏𝟏𝟏𝟏𝟏° Right Exactly 𝟗𝟗𝟗𝟗° Straight Exactly 𝟏𝟏𝟏𝟏𝟏𝟏° Reflex Between 𝟏𝟏𝟏𝟏𝟏𝟏° and 𝟑𝟑𝟑𝟑𝟑𝟑° Example 1 (3 minutes) Example 1 ∠𝑨𝑨𝑨𝑨𝑨𝑨 measures 𝟗𝟗𝟗𝟗°. The angle has been separated into two angles. If one angle measures 𝟓𝟓𝟓𝟓°, what is the measure of the other angle? In this lesson, we will be using algebra to help us determine unknown measures of angles. How are these two angles related? The two angles have a sum of 𝟗𝟗𝟗𝟗°. What equation could we use to solve for 𝒙𝒙? 𝒙𝒙° + 𝟓𝟓𝟓𝟓° = 𝟗𝟗𝟗𝟗° Now, let’s solve. 𝒙𝒙° + 𝟓𝟓𝟓𝟓° −𝟓𝟓𝟓𝟓° = 𝟗𝟗𝟗𝟗° −𝟓𝟓𝟓𝟓° 𝒙𝒙° = 𝟑𝟑𝟑𝟑° The measure of the unknown angle is 𝟑𝟑𝟑𝟑°. MP.4 A STORY OF RATIOS 333 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Example 2 (3 minutes) Example 2 Michelle is designing a parking lot. She has determined that one of the angles should be 𝟏𝟏𝟏𝟏𝟏𝟏°. What is the measure of angle 𝒙𝒙 and angle 𝒚𝒚? How is angle 𝒙𝒙 related to the 𝟏𝟏𝟏𝟏𝟏𝟏° angle? The two angles form a straight line. Therefore, they should add up to 𝟏𝟏𝟏𝟏𝟏𝟏°. What equation would we use to show this? 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° How would you solve this equation? 𝟏𝟏𝟏𝟏𝟏𝟏° was added to angle 𝒙𝒙, so I will take away 𝟏𝟏𝟏𝟏𝟏𝟏° to get back to angle 𝒙𝒙. 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° = 𝟔𝟔𝟔𝟔° The angle next to 𝟏𝟏𝟏𝟏𝟏𝟏°, labeled with an 𝒙𝒙, is equal to 𝟔𝟔𝟔𝟔°. How is angle 𝒚𝒚 related to the angle that measures 𝟏𝟏𝟏𝟏𝟏𝟏°? These two angles also form a straight line and must add up to 𝟏𝟏𝟏𝟏𝟏𝟏°. Therefore, angles 𝒙𝒙 and 𝒚𝒚 must both be equal to 𝟔𝟔𝟔𝟔°. Example 3 (3 minutes) Example 3 A beam of light is reflected off a mirror. Below is a diagram of the reflected beam. Determine the missing angle measure. How are the angles in this question related? There are three angles that, when all placed together, form a straight line. This means that the three angles have a sum of 𝟏𝟏𝟏𝟏𝟏𝟏°. What equation could we write to represent the situation? 𝟓𝟓𝟓𝟓° + 𝒙𝒙° + 𝟓𝟓𝟓𝟓° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° 𝒚𝒚° MP.4 𝒙𝒙° 𝟓𝟓𝟓𝟓° 𝟓𝟓𝟓𝟓° A STORY OF RATIOS 334 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World How would you solve an equation like this? We can combine the two angles that we do know. 𝟓𝟓𝟓𝟓° + 𝟓𝟓𝟓𝟓° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟏𝟏𝟏𝟏𝟏𝟏° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° = 𝟕𝟕𝟕𝟕° The angle of the bounce is 𝟕𝟕𝟕𝟕°. Example 4 (3 minutes) Example 4 ∠𝑨𝑨𝑨𝑨𝑨𝑨 measures 𝟗𝟗𝟗𝟗°. It has been split into two angles, ∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑫𝑫𝑫𝑫𝑫𝑫. The measure of ∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑫𝑫𝑫𝑫𝑫𝑫 is in a ratio of 𝟒𝟒: 𝟏𝟏. What are the measures of each angle? Use a tape diagram to represent the ratio 𝟒𝟒: 𝟏𝟏. What is the measure of each angle? 𝟓𝟓 units = 𝟗𝟗𝟗𝟗° 𝟏𝟏 unit = 𝟏𝟏𝟏𝟏° 𝟒𝟒 units = 𝟕𝟕𝟕𝟕° ∠𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟕𝟕𝟕𝟕°. ∠𝑫𝑫𝑫𝑫𝑫𝑫 is 𝟏𝟏𝟏𝟏°. How can we represent this situation with an equation? 𝟒𝟒𝟒𝟒° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° Solve the equation to determine the measure of each angle. 𝟒𝟒𝟒𝟒° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° 𝟓𝟓𝟓𝟓° = 𝟗𝟗𝟗𝟗° 𝟓𝟓𝟓𝟓° ÷ 𝟓𝟓= 𝟗𝟗𝟗𝟗° ÷ 𝟓𝟓 𝒙𝒙° = 𝟏𝟏𝟏𝟏° 𝟒𝟒𝟒𝟒° = 𝟒𝟒(𝟏𝟏𝟏𝟏°) = 𝟕𝟕𝟕𝟕° The measure of ∠𝑫𝑫𝑫𝑫𝑫𝑫 is 𝟏𝟏𝟏𝟏° and the measure of ∠𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟕𝟕𝟕𝟕°. ∠𝑨𝑨𝑨𝑨𝑨𝑨 ∠𝑫𝑫𝑫𝑫𝑫𝑫 MP.4 A STORY OF RATIOS 335 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World 𝟔𝟔𝟔𝟔° 𝒙𝒙° 𝟑𝟑𝟑𝟑° Exercises (15 minutes) Students work independently. Exercises Write and solve an equation in each of the problems. 1. ∠𝑨𝑨𝑨𝑨𝑨𝑨 measures 𝟗𝟗𝟗𝟗°. It has been split into two angles, ∠𝑨𝑨𝑨𝑨𝑨𝑨 and ∠𝑫𝑫𝑫𝑫𝑫𝑫. The measure of the two angles is in a ratio of 𝟐𝟐: 𝟏𝟏. What are the measures of each angle? 𝒙𝒙° + 𝟐𝟐𝟐𝟐° = 𝟗𝟗𝟗𝟗⁰ 𝟑𝟑𝟑𝟑° = 𝟗𝟗𝟗𝟗° 𝟑𝟑𝟑𝟑° 𝟑𝟑= 𝟗𝟗𝟗𝟗° 𝟑𝟑 𝒙𝒙° = 𝟑𝟑𝟑𝟑° One of the angles measures 𝟑𝟑𝟑𝟑°, and the other measures 𝟔𝟔𝟔𝟔°. 2. Solve for 𝒙𝒙. 𝒙𝒙° + 𝟔𝟔𝟔𝟔° + 𝟑𝟑𝟑𝟑° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° = 𝟕𝟕𝟕𝟕° 3. Candice is building a rectangular piece of a fence according to the plans her boss gave her. One of the angles is not labeled. Write an equation, and use it to determine the measure of the unknown angle. 𝒙𝒙° 𝟒𝟒𝟒𝟒° 𝒙𝒙° + 𝟒𝟒𝟒𝟒° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟒𝟒𝟒𝟒° −𝟒𝟒𝟒𝟒° = 𝟗𝟗𝟗𝟗° −𝟒𝟒𝟒𝟒° 𝒙𝒙° = 𝟒𝟒𝟒𝟒° A STORY OF RATIOS 336 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World 𝟑𝟑𝟑𝟑° 𝒙𝒙° 𝟑𝟑𝟑𝟑° 𝟑𝟑𝟑𝟑˚ 𝟐𝟐𝟐𝟐˚ 𝒙𝒙˚ 4. Rashid hit a hockey puck against the wall at a 𝟑𝟑𝟑𝟑° angle. The puck hit the wall and traveled in a new direction. Determine the missing angle in the diagram. 𝟑𝟑𝟑𝟑° + 𝒙𝒙° + 𝟑𝟑𝟑𝟑° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟕𝟕𝟕𝟕° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟕𝟕𝟕𝟕° −𝟕𝟕𝟕𝟕° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟕𝟕𝟕𝟕° 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° The measure of the missing angle is 𝟏𝟏𝟏𝟏𝟏𝟏°. 5. Jaxon is creating a mosaic design on a rectangular table. He has added two pieces to one of the corners. The first piece has an angle measuring 𝟑𝟑𝟑𝟑° and is placed in the corner. A second piece has an angle measuring 𝟐𝟐𝟐𝟐° and is also placed in the corner. Draw a diagram to model the situation. Then, write an equation, and use it to determine the measure of the unknown angle in a third piece that could be added to the corner of the table. 𝒙𝒙° + 𝟑𝟑𝟑𝟑° + 𝟐𝟐𝟐𝟐° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟔𝟔𝟔𝟔° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟔𝟔𝟔𝟔° −𝟔𝟔𝟔𝟔° = 𝟗𝟗𝟗𝟗° −𝟔𝟔𝟔𝟔° 𝒙𝒙° = 𝟐𝟐𝟐𝟐° The measure of the unknown angle is 𝟐𝟐𝟐𝟐°. Closing (3 minutes) Explain how you determined the equation you used to solve for the missing angle or variable. I used the descriptions in the word problems. For example, if it said “the sum of the angles,” I knew to add the measures together. I also used my knowledge of angles to know the total angle measure. For example, I know a straight angle has a measure of 180°, and a right angle or a corner has a measure of 90°. Exit Ticket (7 minutes) A STORY OF RATIOS 337 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Name Date Lesson 30: One-Step Problems in the Real World Exit Ticket Write an equation, and solve for the missing angle in each question. 1. Alejandro is repairing a stained glass window. He needs to take it apart to repair it. Before taking it apart, he makes a sketch with angle measures to put it back together. Write an equation, and use it to determine the measure of the unknown angle. 2. Hannah is putting in a tile floor. She needs to determine the angles that should be cut in the tiles to fit in the corner. The angle in the corner measures 90°. One piece of the tile will have a measure of 38°. Write an equation, and use it to determine the measure of the unknown angle. 𝑥𝑥° 38° 𝑥𝑥° 40° 30° A STORY OF RATIOS 338 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World 𝒙𝒙˚ 𝟓𝟓𝟓𝟓˚ Exit Ticket Sample Solutions Write an equation, and solve for the missing angle in each question. 1. Alejandro is repairing a stained glass window. He needs to take it apart to repair it. Before taking it apart, he makes a sketch with angle measures to put it back together. Write an equation, and use it to determine the measure of the unknown angle. 𝟒𝟒𝟒𝟒° + 𝒙𝒙° + 𝟑𝟑𝟑𝟑° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟒𝟒𝟒𝟒° + 𝟑𝟑𝟑𝟑° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟕𝟕𝟕𝟕° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟕𝟕𝟕𝟕° −𝟕𝟕𝟕𝟕° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟕𝟕𝟕𝟕° 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° The missing angle measures 𝟏𝟏𝟏𝟏𝟏𝟏°. 2. Hannah is putting in a tile floor. She needs to determine the angles that should be cut in the tiles to fit in the corner. The angle in the corner measures 𝟗𝟗𝟗𝟗°. One piece of the tile will have a measure of 𝟑𝟑𝟑𝟑°. Write an equation, and use it to determine the measure of the unknown angle. 𝒙𝒙° + 𝟑𝟑𝟑𝟑° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟑𝟑𝟑𝟑° −𝟑𝟑𝟑𝟑° = 𝟗𝟗𝟗𝟗° −𝟑𝟑𝟑𝟑° 𝒙𝒙° = 𝟓𝟓𝟓𝟓° The measure of the unknown angle is 𝟓𝟓𝟓𝟓°. Problem Set Sample Solutions Write and solve an equation for each problem. 1. Solve for 𝒙𝒙. 𝒙𝒙° + 𝟓𝟓𝟓𝟓° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟓𝟓𝟓𝟓° −𝟓𝟓𝟓𝟓° = 𝟗𝟗𝟗𝟗° −𝟓𝟓𝟓𝟓° 𝒙𝒙° = 𝟑𝟑𝟑𝟑° The measure of the missing angle is 𝟑𝟑𝟑𝟑°. 𝒙𝒙° 𝟑𝟑𝟑𝟑° 𝒙𝒙° 𝟒𝟒𝟒𝟒° 𝟑𝟑𝟑𝟑° A STORY OF RATIOS 339 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟔𝟔𝟔𝟔° 𝒙𝒙° 2. ∠𝑩𝑩𝑩𝑩𝑩𝑩 measures 𝟗𝟗𝟗𝟗°. Solve for 𝒙𝒙. 3. Thomas is putting in a tile floor. He needs to determine the angles that should be cut in the tiles to fit in the corner. The angle in the corner measures 𝟗𝟗𝟗𝟗°. One piece of the tile will have a measure of 𝟐𝟐𝟐𝟐°. Write an equation, and use it to determine the measure of the unknown angle. 𝒙𝒙° + 𝟐𝟐𝟐𝟐° = 𝟗𝟗𝟗𝟗° 𝒙𝒙° + 𝟐𝟐𝟐𝟐° −𝟐𝟐𝟐𝟐° = 𝟗𝟗𝟗𝟗° −𝟐𝟐𝟐𝟐° 𝒙𝒙° = 𝟔𝟔𝟔𝟔° The measure of the unknown angle is 𝟔𝟔𝟔𝟔°. 4. Solve for 𝒙𝒙. 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° + 𝟔𝟔𝟔𝟔° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° + 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° = 𝟏𝟏𝟏𝟏° The measure of the missing angle is 𝟏𝟏𝟏𝟏°. 5. Aram has been studying the mathematics behind pinball machines. He made the following diagram of one of his observations. Determine the measure of the missing angle. 𝟓𝟓𝟓𝟓° + 𝒙𝒙° + 𝟔𝟔𝟔𝟔° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟏𝟏𝟏𝟏𝟏𝟏° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟏𝟏𝟏𝟏𝟏𝟏° + 𝒙𝒙° −𝟏𝟏𝟏𝟏𝟏𝟏° = 𝟏𝟏𝟏𝟏𝟏𝟏° −𝟏𝟏𝟏𝟏𝟏𝟏° 𝒙𝒙° = 𝟔𝟔𝟔𝟔° The measure of the missing angle is 𝟔𝟔𝟔𝟔°. 6. The measures of two angles have a sum of 𝟗𝟗𝟗𝟗°. The measures of the angles are in a ratio of 𝟐𝟐: 𝟏𝟏. Determine the measures of both angles. 𝟐𝟐𝟐𝟐° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° 𝟑𝟑𝟑𝟑° = 𝟗𝟗𝟗𝟗° 𝟑𝟑𝟑𝟑° 𝟑𝟑= 𝟗𝟗𝟗𝟗° 𝟑𝟑 𝒙𝒙° = 𝟑𝟑𝟑𝟑° The angles measure 𝟑𝟑𝟑𝟑° and 𝟔𝟔𝟔𝟔°. 𝟏𝟏𝟏𝟏° + 𝒙𝒙° + 𝟐𝟐𝟐𝟐° = 𝟗𝟗𝟗𝟗° 𝟏𝟏𝟏𝟏° + 𝟐𝟐𝟐𝟐° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° 𝟒𝟒𝟒𝟒° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° 𝟒𝟒𝟒𝟒° −𝟒𝟒𝟒𝟒° + 𝒙𝒙° = 𝟗𝟗𝟗𝟗° −𝟒𝟒𝟒𝟒° 𝒙𝒙° = 𝟓𝟓𝟓𝟓° 𝟓𝟓𝟓𝟓° 𝟔𝟔𝟔𝟔° 𝒙𝒙° A STORY OF RATIOS 340 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World 7. The measures of two angles have a sum of 𝟏𝟏𝟏𝟏𝟏𝟏°. The measures of the angles are in a ratio of 𝟓𝟓: 𝟏𝟏. Determine the measures of both angles. 𝟓𝟓𝟓𝟓° + 𝒙𝒙° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟔𝟔𝟔𝟔° = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝟔𝟔𝟔𝟔° 𝟔𝟔= 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔° 𝒙𝒙° = 𝟑𝟑𝟑𝟑˚ The angles measure 𝟑𝟑𝟑𝟑° and 𝟏𝟏𝟏𝟏𝟏𝟏°. A STORY OF RATIOS 341 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Subtraction of Decimals—Round 1 Directions: Evaluate each expression. 1. 55 −50 23. 9.9 −5 2. 55 −5 24. 9.9 −0.5 3. 5.5 −5 25. 0.99 −0.5 4. 5.5 −0.5 26. 0.99 −0.05 5. 88 −80 27. 4.7 −2 6. 88 −8 28. 4.7 −0.2 7. 8.8 −8 29. 0.47 −0.2 8. 8.8 −0.8 30. 0.47 −0.02 9. 33 −30 31. 8.4 −1 10. 33 −3 32. 8.4 −0.1 11. 3.3 −3 33. 0.84 −0.1 12. 1 −0.3 34. 7.2 −5 13. 1 −0.03 35. 7.2 −0.5 14. 1 −0.003 36. 0.72 −0.5 15. 0.1 −0.03 37. 0.72 −0.05 16. 4 −0.8 38. 8.6 −7 17. 4 −0.08 39. 8.6 −0.7 18. 4 −0.008 40. 0.86 −0.7 19. 0.4 −0.08 41. 0.86 −0.07 20. 9 −0.4 42. 5.1 −4 21. 9 −0.04 43. 5.1 −0.4 22. 9 −0.004 44. 0.51 −0.4 Number Correct: A STORY OF RATIOS 342 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Subtraction of Decimals—Round 1 [KEY] Directions: Evaluate each expression. 1. 55 −50 𝟓𝟓 23. 9.9 −5 𝟒𝟒. 𝟗𝟗 2. 55 −5 𝟓𝟓𝟓𝟓 24. 9.9 −0.5 𝟗𝟗. 𝟒𝟒 3. 5.5 −5 𝟎𝟎. 𝟓𝟓 25. 0.99 −0.5 𝟎𝟎. 𝟒𝟒𝟒𝟒 4. 5.5 −0.5 𝟓𝟓 26. 0.99 −0.05 𝟎𝟎. 𝟗𝟗𝟗𝟗 5. 88 −80 𝟖𝟖 27. 4.7 −2 𝟐𝟐. 𝟕𝟕 6. 88 −8 𝟖𝟖𝟖𝟖 28. 4.7 −0.2 𝟒𝟒. 𝟓𝟓 7. 8.8 −8 𝟎𝟎. 𝟖𝟖 29. 0.47 −0.2 𝟎𝟎. 𝟐𝟐𝟐𝟐 8. 8.8 −0.8 𝟖𝟖 30. 0.47 −0.02 𝟎𝟎. 𝟒𝟒𝟒𝟒 9. 33 −30 𝟑𝟑 31. 8.4 −1 𝟕𝟕. 𝟒𝟒 10. 33 −3 𝟑𝟑𝟑𝟑 32. 8.4 −0.1 𝟖𝟖. 𝟑𝟑 11. 3.3 −3 𝟎𝟎. 𝟑𝟑 33. 0.84 −0.1 𝟎𝟎. 𝟕𝟕𝟕𝟕 12. 1 −0.3 𝟎𝟎. 𝟕𝟕 34. 7.2 −5 𝟐𝟐. 𝟐𝟐 13. 1 −0.03 𝟎𝟎. 𝟗𝟗𝟗𝟗 35. 7.2 −0.5 𝟔𝟔. 𝟕𝟕 14. 1 −0.003 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗 36. 0.72 −0.5 𝟎𝟎. 𝟐𝟐𝟐𝟐 15. 0.1 −0.03 𝟎𝟎. 𝟎𝟎𝟎𝟎 37. 0.72 −0.05 𝟎𝟎. 𝟔𝟔𝟔𝟔 16. 4 −0.8 𝟑𝟑. 𝟐𝟐 38. 8.6 −7 𝟏𝟏. 𝟔𝟔 17. 4 −0.08 𝟑𝟑. 𝟗𝟗𝟗𝟗 39. 8.6 −0.7 𝟕𝟕. 𝟗𝟗 18. 4 −0.008 𝟑𝟑. 𝟗𝟗𝟗𝟗𝟗𝟗 40. 0.86 −0.7 𝟎𝟎. 𝟏𝟏𝟏𝟏 19. 0.4 −0.08 𝟎𝟎. 𝟑𝟑𝟑𝟑 41. 0.86 −0.07 𝟎𝟎. 𝟕𝟕𝟕𝟕 20. 9 −0.4 𝟖𝟖. 𝟔𝟔 42. 5.1 −4 𝟏𝟏. 𝟏𝟏 21. 9 −0.04 𝟖𝟖. 𝟗𝟗𝟗𝟗 43. 5.1 −0.4 𝟒𝟒. 𝟕𝟕 22. 9 −0.004 𝟖𝟖. 𝟗𝟗𝟗𝟗𝟗𝟗 44. 0.51 −0.4 𝟎𝟎. 𝟏𝟏𝟏𝟏 A STORY OF RATIOS 343 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Subtraction of Decimals—Round 2 Directions: Evaluate each expression. 1. 66 −60 23. 6.8 −4 2. 66 −6 24. 6.8 −0.4 3. 6.6 −6 25. 0.68 −0.4 4. 6.6 −0.6 26. 0.68 −0.04 5. 99 −90 27. 7.3 −1 6. 99 −9 28. 7.3 −0.1 7. 9.9 −9 29. 0.73 −0.1 8. 9.9 −0.9 30. 0.73 −0.01 9. 22 −20 31. 9.5 −2 10. 22 −2 32. 9.5 −0.2 11. 2.2 −2 33. 0.95 −0.2 12. 3 −0.4 34. 8.3 −5 13. 3 −0.04 35. 8.3 −0.5 14. 3 −0.004 36. 0.83 −0.5 15. 0.3 −0.04 37. 0.83 −0.05 16. 8 −0.2 38. 7.2 −4 17. 8 −0.02 39. 7.2 −0.4 18. 8 −0.002 40. 0.72 −0.4 19. 0.8 −0.02 41. 0.72 −0.04 20. 5 −0.1 42. 9.3 −7 21. 5 −0.01 43. 9.3 −0.7 22. 5 −0.001 44. 0.93 −0.7 Number Correct: Improvement: A STORY OF RATIOS 344 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 30 Lesson 30: One-Step Problems in the Real World Subtraction of Decimals—Round 2 [KEY] Directions: Evaluate each expression. 1. 66 −60 𝟔𝟔 23. 6.8 −4 𝟐𝟐. 𝟖𝟖 2. 66 −6 𝟔𝟔𝟔𝟔 24. 6.8 −0.4 𝟔𝟔. 𝟒𝟒 3. 6.6 −6 𝟎𝟎. 𝟔𝟔 25. 0.68 −0.4 𝟎𝟎. 𝟐𝟐𝟐𝟐 4. 6.6 −0.6 𝟔𝟔 26. 0.68 −0.04 𝟎𝟎. 𝟔𝟔𝟔𝟔 5. 99 −90 𝟗𝟗 27. 7.3 −1 𝟔𝟔. 𝟑𝟑 6. 99 −9 𝟗𝟗𝟗𝟗 28. 7.3 −0.1 𝟕𝟕. 𝟐𝟐 7. 9.9 −9 𝟎𝟎. 𝟗𝟗 29. 0.73 −0.1 𝟎𝟎. 𝟔𝟔𝟔𝟔 8. 9.9 −0.9 𝟗𝟗 30. 0.73 −0.01 𝟎𝟎. 𝟕𝟕𝟕𝟕 9. 22 −20 𝟐𝟐 31. 9.5 −2 𝟕𝟕. 𝟓𝟓 10. 22 −2 𝟐𝟐𝟐𝟐 32. 9.5 −0.2 𝟗𝟗. 𝟑𝟑 11. 2.2 −2 𝟎𝟎. 𝟐𝟐 33. 0.95 −0.2 𝟎𝟎. 𝟕𝟕𝟕𝟕 12. 3 −0.4 𝟐𝟐. 𝟔𝟔 34. 8.3 −5 𝟑𝟑. 𝟑𝟑 13. 3 −0.04 𝟐𝟐. 𝟗𝟗𝟗𝟗 35. 8.3 −0.5 𝟕𝟕. 𝟖𝟖 14. 3 −0.004 𝟐𝟐. 𝟗𝟗𝟗𝟗𝟗𝟗 36. 0.83 −0.5 𝟎𝟎. 𝟑𝟑𝟑𝟑 15. 0.3 −0.04 𝟎𝟎. 𝟐𝟐𝟐𝟐 37. 0.83 −0.05 𝟎𝟎. 𝟕𝟕𝟕𝟕 16. 8 −0.2 𝟕𝟕. 𝟖𝟖 38. 7.2 −4 𝟑𝟑. 𝟐𝟐 17. 8 −0.02 𝟕𝟕. 𝟗𝟗𝟗𝟗 39. 7.2 −0.4 𝟔𝟔. 𝟖𝟖 18. 8 −0.002 𝟕𝟕. 𝟗𝟗𝟗𝟗𝟗𝟗 40. 0.72 −0.4 𝟎𝟎. 𝟑𝟑𝟑𝟑 19. 0.8 −0.02 𝟎𝟎. 𝟕𝟕𝟕𝟕 41. 0.72 −0.04 𝟎𝟎. 𝟔𝟔𝟔𝟔 20. 5 −0.1 𝟒𝟒. 𝟗𝟗 42. 9.3 −7 𝟐𝟐. 𝟑𝟑 21. 5 −0.01 𝟒𝟒. 𝟗𝟗𝟗𝟗 43. 9.3 −0.7 𝟖𝟖. 𝟔𝟔 22. 5 −0.001 𝟒𝟒. 𝟗𝟗𝟗𝟗𝟗𝟗 44. 0.93 −0.7 𝟎𝟎. 𝟐𝟐𝟐𝟐 A STORY OF RATIOS 345 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms Lesson 31: Problems in Mathematical Terms Student Outcomes Students analyze an equation in two variables to choose an independent variable and a dependent variable. Students determine whether or not the equation is solved for the second variable in terms of the first variable or vice versa. They then use this information to determine which variable is the independent variable and which is the dependent variable. Students create a table by placing the independent variable in the first row or column and the dependent variable in the second row or column. They compute entries in the table by choosing arbitrary values for the independent variable (no constraints) and then determine what the dependent variable must be. Classwork Example 1 (10 minutes) Example 1 Marcus reads for 𝟑𝟑𝟑𝟑 minutes each night. He wants to determine the total number of minutes he will read over the course of a month. He wrote the equation 𝒕𝒕= 𝟑𝟑𝟑𝟑𝟑𝟑 to represent the total amount of time that he has spent reading, where 𝒕𝒕 represents the total number of minutes read and 𝒅𝒅 represents the number of days that he read during the month. Determine which variable is independent and which is dependent. Then, create a table to show how many minutes he has read in the first seven days. Number of Days (𝒅𝒅) Total Minutes Read (𝟑𝟑𝟑𝟑𝟑𝟑) 𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐 𝟔𝟔𝟔𝟔 𝟑𝟑 𝟗𝟗𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕 𝟐𝟐𝟐𝟐𝟐𝟐 When setting up a table, we want the independent variable in the first column and the dependent variable in the second column. What do independent and dependent mean? The independent variable changes, and when it does, it affects the dependent variable. So, the dependent variable depends on the independent variable. In this example, which would be the independent variable, and which would be the dependent variable? The dependent variable is the total number of minutes read because it depends on how many days Marcus reads. The independent variable is the number of days that Marcus reads. How could you use the table of values to determine the equation if it had not been given? The number of minutes read shown in the table is always 30 times the number of days. So, the equation would need to show that the total number of minutes read is equal to the number of days times 30. Independent variable __ Dependent variable __ Number of days Total minutes read MP.1 A STORY OF RATIOS 346 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms Example 2 (5 minutes) Example 2 Kira designs websites. She can create three different websites each week. Kira wants to create an equation that will give her the total number of websites she can design given the number of weeks she works. Determine the independent and dependent variables. Create a table to show the number of websites she can design over the first 𝟓𝟓 weeks. Finally, write an equation to represent the number of websites she can design when given any number of weeks. How did you determine which is the dependent variable and which is the independent variable? Because the number of websites she can make depends on how many weeks she works, I determined that the number of weeks worked was the independent variable, and the number of websites designed was the dependent variable. Does knowing which one is independent and which one is dependent help you write the equation? I can write the equation and solve for the dependent variable by knowing how the independent variable will affect the dependent variable. In this case, I knew that every week 3 more websites could be completed, so then I multiplied the number of weeks by 3. Example 3 (5 minutes) Example 3 Priya streams movies through a company that charges her a $𝟓𝟓 monthly fee plus $𝟏𝟏. 𝟓𝟓𝟓𝟓 per movie. Determine the independent and dependent variables, write an equation to model the situation, and create a table to show the total cost per month given that she might stream between 𝟒𝟒 and 𝟏𝟏𝟏𝟏 movies in a month. Is the flat fee an independent variable, a dependent variable, or neither? The $5 flat fee is neither. It is not causing the change in the dependent value, and it is not changing. Instead, the $5 flat fee is a constant that is added on each month. Independent variable___ Dependent variable___ Equation________ # of Weeks Worked (𝒘𝒘) # of Websites Designed (𝒅𝒅) 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟔𝟔 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 # of weeks worked # of websites designed 𝒅𝒅= 𝟑𝟑𝟑𝟑, where 𝒘𝒘 is the number of weeks worked and 𝒅𝒅 is the number of websites designed. # of Movies (𝒎𝒎) Total Cost Per Month, in dollars (𝒄𝒄) 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟕𝟕 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟗𝟗 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 MP.1 Independent variable___ Dependent variable________ Equation______ # of movies watched per month Total cost per month, in dollars 𝒄𝒄= 𝟏𝟏. 𝟓𝟓𝟓𝟓+ 𝟓𝟓 or 𝒄𝒄= 𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓+ 𝟓𝟓 A STORY OF RATIOS 347 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms Why isn’t the equation 𝑐𝑐= 5𝑚𝑚+ 1.50? The $5 fee is only paid once a month. 𝑚𝑚 is the number of movies watched per month, so it needs to be multiplied by the price per movie, which is $1.50. Exercises (15 minutes) Students work in pairs or independently. Exercises 1. Sarah is purchasing pencils to share. Each package has 𝟏𝟏𝟏𝟏 pencils. The equation 𝒏𝒏= 𝟏𝟏𝟏𝟏𝟏𝟏, where 𝒏𝒏 is the total number of pencils and 𝒑𝒑 is the number of packages, can be used to determine the total number of pencils Sarah purchased. Determine which variable is dependent and which is independent. Then, make a table showing the number of pencils purchased for 𝟑𝟑–𝟕𝟕 packages. The number of packages, 𝒑𝒑, is the independent variable. The total number of pencils, 𝒏𝒏, is the dependent variable. # of Packages (𝒑𝒑) Total # of Pencils (𝒏𝒏= 𝟏𝟏𝟏𝟏𝟏𝟏) 𝟑𝟑 𝟑𝟑𝟑𝟑 𝟒𝟒 𝟒𝟒𝟒𝟒 𝟓𝟓 𝟔𝟔𝟔𝟔 𝟔𝟔 𝟕𝟕𝟕𝟕 𝟕𝟕 𝟖𝟖𝟖𝟖 2. Charlotte reads 𝟒𝟒 books each week. Let 𝒃𝒃 be the number of books she reads each week, and let 𝒘𝒘 be the number of weeks that she reads. Determine which variable is dependent and which is independent. Then, write an equation to model the situation, and make a table that shows the number of books read in under 𝟔𝟔 weeks. The number of weeks, 𝒘𝒘, is the independent variable. The number of books, 𝒃𝒃, is the dependent variable. 𝒃𝒃= 𝟒𝟒𝟒𝟒 # of Weeks (𝒘𝒘) # of Books (𝒃𝒃= 𝟒𝟒𝟒𝟒) 𝟏𝟏 𝟒𝟒 𝟐𝟐 𝟖𝟖 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟐𝟐𝟐𝟐 MP.1 A STORY OF RATIOS 348 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms 3. A miniature golf course has a special group rate. You can pay $𝟐𝟐𝟐𝟐 plus $𝟑𝟑 per person when you have a group of 𝟓𝟓 or more friends. Let 𝒇𝒇 be the number of friends and 𝒄𝒄 be the total cost. Determine which variable is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 𝟓𝟓 to 𝟏𝟏𝟏𝟏 friends. The number of friends, 𝒇𝒇, is the independent variable. The total cost in dollars, 𝒄𝒄, is the dependent variable. 𝒄𝒄= 𝟑𝟑𝟑𝟑+ 𝟐𝟐𝟐𝟐 # of Friends (𝒇𝒇) Total Cost, in dollars (𝒄𝒄= 𝟑𝟑𝟑𝟑+ 𝟐𝟐𝟐𝟐) 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟔𝟔 𝟑𝟑𝟑𝟑 𝟕𝟕 𝟒𝟒𝟒𝟒 𝟖𝟖 𝟒𝟒𝟒𝟒 𝟗𝟗 𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 4. Carlos is shopping for school supplies. He bought a pencil box for $𝟑𝟑, and he also needs to buy notebooks. Each notebook is $𝟐𝟐. Let 𝒕𝒕 represent the total cost of the supplies and 𝒏𝒏 be the number of notebooks Carlos buys. Determine which variable is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 𝟏𝟏 to 𝟓𝟓 notebooks. The total number of notebooks, 𝒏𝒏, is the independent variable. The total cost in dollars, 𝒕𝒕, is the dependent variable. 𝒕𝒕= 𝟐𝟐𝟐𝟐+ 𝟑𝟑 # of Notebooks (𝒏𝒏) Total Cost, in dollars (𝒕𝒕= 𝟐𝟐𝟐𝟐+ 𝟑𝟑) 𝟏𝟏 𝟓𝟓 𝟐𝟐 𝟕𝟕 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 Closing (5 minutes) Use this time for partners to share their answers from the exercises with another set of partners. How can you determine which variable is independent and which variable is dependent? The dependent variable is affected by changes in the independent variable. I can write a sentence stating that one variable depends on another. For example, the amount of money earned depends on the number of hours worked. So, the money earned is the dependent variable, and the number of hours worked is the independent variable. Exit Ticket (5 minutes) A STORY OF RATIOS 349 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms Name Date Lesson 31: Problems in Mathematical Terms Exit Ticket For each problem, determine the independent and dependent variables, write an equation to represent the situation, and then make a table with at least 5 values that models the situation. 1. Kyla spends 60 minutes of each day exercising. Let 𝑑𝑑 be the number of days that Kyla exercises, and let 𝑚𝑚 represent the total minutes of exercise in a given time frame. Show the relationship between the number of days that Kyla exercises and the total minutes that she exercises. 2. A taxicab service charges a flat fee of $8 plus an additional $1.50 per mile. Show the relationship between the total cost and the number of miles driven. Independent variable Dependent variable Equation Independent variable Dependent variable Equation A STORY OF RATIOS 350 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms Exit Ticket Sample Solutions For each problem, determine the independent and dependent variables, write an equation to represent the situation, and then make a table with at least 𝟓𝟓 values that models the situation. 1. Kyla spends 𝟔𝟔𝟔𝟔 minutes of each day exercising. Let 𝒅𝒅 be the number of days that Kyla exercises, and let 𝒎𝒎 represent the total minutes of exercise in a given time frame. Show the relationship between the number of days that Kyla exercises and the total minutes that she exercises. Tables may vary. # of Days # of Minutes 𝟎𝟎 𝟎𝟎 𝟏𝟏 𝟔𝟔𝟔𝟔 𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒 𝟐𝟐𝟐𝟐𝟐𝟐 2. A taxicab service charges a flat fee of $𝟖𝟖 plus an additional $𝟏𝟏. 𝟓𝟓𝟓𝟓 per mile. Show the relationship between the total cost and the number of miles driven. Tables may vary. # of Miles Total Cost, in dollars 𝟎𝟎 𝟖𝟖. 𝟎𝟎𝟎𝟎 𝟏𝟏 𝟗𝟗. 𝟓𝟓𝟓𝟓 𝟐𝟐 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟑𝟑 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟒𝟒 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 Problem Set Sample Solutions 1. Jaziyah sells 𝟑𝟑 houses each month. To determine the number of houses she can sell in any given number of months, she uses the equation 𝒕𝒕= 𝟑𝟑𝟑𝟑, where 𝒕𝒕 is the total number of houses sold and 𝒎𝒎 is the number of months. Name the independent and dependent variables. Then, create a table to show how many houses she sells in fewer than 𝟔𝟔 months. The independent variable is the number of months. The dependent variable is the total number of houses sold. # of Months Total Number of Houses 𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟔𝟔 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 Independent variable Dependent variable Equation 𝒄𝒄= 𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓+ 𝟖𝟖 Number of miles Total cost, in dollars 𝒎𝒎= 𝟔𝟔𝟔𝟔𝟔𝟔 Number of Days Total Number of Minutes Independent variable Dependent variable Equation A STORY OF RATIOS 351 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms 2. Joshua spends 𝟐𝟐𝟐𝟐 minutes of each day reading. Let 𝒅𝒅 be the number of days that he reads, and let 𝒎𝒎 represent the total minutes of reading. Determine which variable is independent and which is dependent. Then, write an equation that models the situation. Make a table showing the number of minutes spent reading over 𝟕𝟕 days. The number of days, 𝒅𝒅, is the independent variable. The total number of minutes of reading, 𝒎𝒎, is the dependent variable. 𝒎𝒎= 𝟐𝟐𝟐𝟐𝟐𝟐 # of Days # of Minutes 𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐 𝟓𝟓𝟓𝟓 𝟑𝟑 𝟕𝟕𝟕𝟕 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕 𝟏𝟏𝟏𝟏𝟏𝟏 3. Each package of hot dog buns contains 𝟖𝟖 buns. Let 𝒑𝒑 be the number of packages of hot dog buns and 𝒃𝒃 be the total number of buns. Determine which variable is independent and which is dependent. Then, write an equation that models the situation, and make a table showing the number of hot dog buns in 𝟑𝟑 to 𝟖𝟖 packages. The number of packages, 𝒑𝒑, is the independent variable. The total number of hot dog buns, 𝒃𝒃, is the dependent variable. 𝒃𝒃= 𝟖𝟖𝟖𝟖 # of Packages Total # of Hot Dog Buns 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟒𝟒 𝟑𝟑𝟑𝟑 𝟓𝟓 𝟒𝟒𝟒𝟒 𝟔𝟔 𝟒𝟒𝟒𝟒 𝟕𝟕 𝟓𝟓𝟓𝟓 𝟖𝟖 𝟔𝟔𝟔𝟔 4. Emma was given 𝟓𝟓 seashells. Each week she collected 𝟑𝟑 more. Let 𝒘𝒘 be the number of weeks Emma collects seashells and 𝒔𝒔 be the number of seashells she has total. Which variable is independent, and which is dependent? Write an equation to model the relationship, and make a table to show how many seashells she has from week 𝟒𝟒 to week 𝟏𝟏𝟏𝟏. The number of weeks, 𝒘𝒘, is the independent variable. The total number of seashells, 𝒔𝒔, is the dependent variable. 𝒔𝒔= 𝟑𝟑𝟑𝟑+ 𝟓𝟓 # of Weeks Total # of Seashells 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟐𝟐𝟐𝟐 𝟔𝟔 𝟐𝟐𝟐𝟐 𝟕𝟕 𝟐𝟐𝟐𝟐 𝟖𝟖 𝟐𝟐𝟐𝟐 𝟗𝟗 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 A STORY OF RATIOS 352 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 31 Lesson 31: Problems in Mathematical Terms 5. Emilia is shopping for fresh produce at a farmers’ market. She bought a watermelon for $𝟓𝟓, and she also wants to buy peppers. Each pepper is $𝟎𝟎. 𝟕𝟕𝟕𝟕. Let 𝒕𝒕 represent the total cost of the produce and 𝒏𝒏 be the number of peppers bought. Determine which variable is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 𝟏𝟏 to 𝟓𝟓 peppers. The number of peppers, 𝒏𝒏, is the independent variable. The total cost in dollars, 𝒕𝒕, is the dependent variable. 𝒕𝒕= 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕+ 𝟓𝟓 # of Peppers Total Cost, in dollars 𝟏𝟏 𝟓𝟓. 𝟕𝟕𝟕𝟕 𝟐𝟐 𝟔𝟔. 𝟓𝟓𝟓𝟓 𝟑𝟑 𝟕𝟕. 𝟐𝟐𝟐𝟐 𝟒𝟒 𝟖𝟖. 𝟎𝟎𝟎𝟎 𝟓𝟓 𝟖𝟖. 𝟕𝟕𝟕𝟕 6. A taxicab service charges a flat fee of $𝟕𝟕 plus an additional $𝟏𝟏. 𝟐𝟐𝟐𝟐 per mile driven. Show the relationship between the total cost and the number of miles driven. Which variable is independent, and which is dependent? Write an equation to model the relationship, and make a table to show the cost of 𝟒𝟒 to 𝟏𝟏𝟏𝟏 miles. The number of miles driven, 𝒎𝒎, is the independent variable. The total cost in dollars, 𝒄𝒄, is the dependent variable. 𝒄𝒄= 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐+ 𝟕𝟕 # of Miles Total Cost, in dollars 𝟒𝟒 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟓𝟓 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝟔𝟔 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟕𝟕 𝟏𝟏𝟏𝟏. 𝟕𝟕𝟕𝟕 𝟖𝟖 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎 𝟗𝟗 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 A STORY OF RATIOS 353 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟖𝟖 Number of Packages Beverages for the Party 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟔𝟔 Total Number of Beverages Lesson 32: Multi-Step Problems in the Real World Student Outcomes Students analyze an equation in two variables, choose an independent variable and a dependent variable, make a table, and make a graph for the equation by plotting the points in the table. For the graph, the independent variable is usually represented by the horizontal axis, and the dependent variable is usually represented by the vertical axis. Classwork Opening Exercise (5 minutes) Opening Exercise Xin is buying beverages for a party that come in packs of 𝟖𝟖. Let 𝒑𝒑 be the number of packages Xin buys and 𝒕𝒕 be the total number of beverages. The equation 𝒕𝒕= 𝟖𝟖𝟖𝟖 can be used to calculate the total number of beverages when the number of packages is known. Determine the independent and dependent variables in this scenario. Then, make a table using whole number values of 𝒑𝒑 less than 𝟔𝟔. The total number of beverages is the dependent variable because the total number of beverages depends on the number of packages purchased. Therefore, the independent variable is the number of packages purchased. Number of Packages (𝒑𝒑) Total Number of Beverages (𝒕𝒕= 𝟖𝟖𝒑𝒑) 𝟎𝟎 𝟎𝟎 𝟏𝟏 𝟖𝟖 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟒𝟒 𝟑𝟑𝟑𝟑 𝟓𝟓 𝟒𝟒𝟒𝟒 Example 1 (7 minutes) Example 1 Make a graph for the table in the Opening Exercise. A STORY OF RATIOS 354 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 Number of Hours 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 Number of Miles Road Trip To make a graph, we must determine which variable is measured along the horizontal axis and which variable is measured along the vertical axis. Generally, the independent variable is measured along the 𝑥𝑥-axis. Which axis is the 𝑥𝑥-axis? The 𝑥𝑥-axis is the horizontal axis. Where would you put the dependent variable? On the 𝑦𝑦-axis. It travels vertically, or up and down. We want to show how the number of beverages changes when the number of packages changes. To check that you have set up your graph correctly, try making a sentence out of the labels on the axes. Write your sentence using the label from the 𝑦𝑦-axis first followed by the label from the 𝑥𝑥-axis. The total number of beverages depends on the number of packages purchased. Example 2 (3 minutes) Example 2 Use the graph to determine which variable is the independent variable and which is the dependent variable. Then, state the relationship between the quantities represented by the variables. The number of miles driven depends on how many hours they drive. Therefore, the number of miles driven is the dependent variable, and the number of hours is the independent variable. This graph shows that they can travel 𝟓𝟓𝟓𝟓 miles every hour. So, the total number of miles driven increases by 𝟓𝟓𝟓𝟓 every time the number of hours increases by 𝟏𝟏. MP.2 A STORY OF RATIOS 355 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟔𝟔 𝟕𝟕 𝟖𝟖 𝟗𝟗 Number of Weeks 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗𝟗𝟗 𝟔𝟔𝟔𝟔 𝟑𝟑𝟑𝟑 Money Saved Number of Weeks Number of Books Exercise (20 minutes) Students work individually. Exercises 1. Each week Quentin earns $𝟑𝟑𝟑𝟑. If he saves this money, create a graph that shows the total amount of money Quentin has saved from week 𝟏𝟏 through week 𝟖𝟖. Write an equation that represents the relationship between the number of weeks that Quentin has saved his money, 𝒘𝒘, and the total amount of money in dollars he has saved, 𝒔𝒔. Then, name the independent and dependent variables. Write a sentence that shows this relationship. 𝒔𝒔= 𝟑𝟑𝟑𝟑𝟑𝟑 The amount of money saved in dollars, 𝒔𝒔, is the dependent variable, and the number of weeks, 𝒘𝒘, is the independent variable. Number of Weeks Total Saved ($) 𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐 𝟔𝟔𝟔𝟔 𝟑𝟑 𝟗𝟗𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕 𝟐𝟐𝟐𝟐𝟐𝟐 𝟖𝟖 𝟐𝟐𝟐𝟐𝟐𝟐 Therefore, the amount of money Quentin has saved increases by $𝟑𝟑𝟑𝟑 for every week he saves money. 2. Zoe is collecting books to donate. She started with 𝟑𝟑 books and collects two more each week. She is using the equation 𝒃𝒃= 𝟐𝟐𝟐𝟐+ 𝟑𝟑, where 𝒃𝒃 is the total number of books collected and 𝒘𝒘 is the number of weeks she has been collecting books. Name the independent and dependent variables. Then, create a graph to represent how many books Zoe has collected when 𝒘𝒘 is 𝟓𝟓 or less. The number of weeks is the independent variable. The number of books collected is the dependent variable. Number of Weeks Number of Books Collected 𝟎𝟎 𝟑𝟑 𝟏𝟏 𝟓𝟓 𝟐𝟐 𝟕𝟕 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 Amount of Money ($) A STORY OF RATIOS 356 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World Total Cost of Ride ($) Number of Miles 3. Eliana plans to visit the fair. She must pay $𝟓𝟓 to enter the fairgrounds and an additional $𝟑𝟑 per ride. Write an equation to show the relationship between 𝒓𝒓, the number of rides, and 𝒕𝒕, the total cost in dollars. State which variable is dependent and which is independent. Then, create a graph that models the equation. 𝒕𝒕= 𝟑𝟑𝟑𝟑+ 𝟓𝟓 The number of rides is the independent variable, and the total cost in dollars, is the dependent variable. # of Rides Total Cost (in dollars) 𝟎𝟎 𝟓𝟓 𝟏𝟏 𝟖𝟖 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏 Closing (5 minutes) Imagine that you are helping a friend with his math work. Here is the problem he was solving: Henry is taking a taxicab home. The cab company charges an initial fee of $5 plus $2 for each additional mile. Henry uses the equation 𝑡𝑡= 2𝑚𝑚+ 5 to calculate the cost of the ride, where 𝑡𝑡 is the total cost and 𝑚𝑚 is the number of miles. Your friend states that 𝑡𝑡 is the dependent variable and 𝑚𝑚 is the independent variable. Then, the friend starts to sketch a graph. What would you tell your friend when looking over her work? I would tell my friend that the dependent variable should go on the vertical axis or the 𝑦𝑦-axis. Then your graph will show that the total cost of the ride depends on how many miles you travel in the taxicab. Exit Ticket (5 minutes) A STORY OF RATIOS 357 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World Name Date Lesson 32: Multi-Step Problems in the Real World Exit Ticket Determine which variable is the independent variable and which variable is the dependent variable. Write an equation, make a table, and plot the points from the table on the graph. Enoch can type 40 words per minute. Let 𝑤𝑤 be the number of words typed and 𝑚𝑚 be the number of minutes spent typing. Independent variable Dependent variable Equation A STORY OF RATIOS 358 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World 𝟒𝟒𝟒𝟒 𝟖𝟖𝟖𝟖 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟔𝟔 # of Minutes Spent Typing 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐 Words Typed Per Minute # of Words Typed Exit Ticket Sample Solutions Determine which variable is the independent variable and which variable is the dependent variable. Write an equation, make a table, and plot the points from the table on the graph. Enoch can type 𝟒𝟒𝟒𝟒 words per minutes. Let 𝒘𝒘 be the number of words typed and 𝒎𝒎 be the number of minutes spent typing. The independent variable is the number of minutes spent typing. The dependent variable is the number of words typed. The equation is 𝒘𝒘= 𝟒𝟒𝟒𝟒𝟒𝟒. # of Minutes # of Words 𝟎𝟎 𝟎𝟎 𝟏𝟏 𝟒𝟒𝟒𝟒 𝟐𝟐 𝟖𝟖𝟖𝟖 𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓 𝟐𝟐𝟐𝟐𝟐𝟐 Problem Set Sample Solutions 1. Caleb started saving money in a cookie jar. He started with $𝟐𝟐𝟐𝟐. He adds $𝟏𝟏𝟏𝟏 to the cookie jar each week. Write an equation where 𝒘𝒘 is the number of weeks Caleb saves his money and 𝒕𝒕 is the total amount in dollars in the cookie jar. Determine which variable is the independent variable and which is the dependent variable. Then, graph the total amount in the cookie jar for 𝒘𝒘 being less than 𝟔𝟔 weeks. 𝒕𝒕= 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟐𝟐𝟐𝟐 The total amount, 𝒕𝒕, is the dependent variable. The number of weeks, 𝒘𝒘, is the independent variable. # of Weeks Total Amount in Cookie Jar ($) 𝟎𝟎 𝟐𝟐𝟐𝟐 𝟏𝟏 𝟑𝟑𝟑𝟑 𝟐𝟐 𝟒𝟒𝟒𝟒 𝟑𝟑 𝟓𝟓𝟓𝟓 𝟒𝟒 𝟔𝟔𝟔𝟔 𝟓𝟓 𝟕𝟕𝟕𝟕 Total Amount in Cookie Jar ($) # of Weeks Money in Cookie Jar A STORY OF RATIOS 359 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World # of Weeks Total Amount ($) Total Amount Saved 2. Kevin is taking a taxi from the airport to his home. There is a $𝟔𝟔 flat fee for riding in the taxi. In addition, Kevin must also pay $𝟏𝟏 per mile. Write an equation where 𝒎𝒎 is the number of miles and 𝒕𝒕 is the total cost in dollars of the taxi ride. Determine which variable is independent and which is dependent. Then, graph the total cost for 𝒎𝒎 being less than 𝟔𝟔 miles. 𝒕𝒕= 𝟏𝟏𝟏𝟏+ 𝟔𝟔 The total cost in dollars, 𝒕𝒕, is the dependent variable. The number of miles, 𝒎𝒎, is the independent variable. # of Miles Total Cost ($) 𝟎𝟎 𝟔𝟔 𝟏𝟏 𝟕𝟕 𝟐𝟐 𝟖𝟖 𝟑𝟑 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 3. Anna started with $𝟏𝟏𝟏𝟏. She saved an additional $𝟓𝟓 each week. Write an equation that can be used to determine the total amount saved in dollars, 𝒕𝒕, after a given number of weeks, 𝒘𝒘. Determine which variable is independent and which is dependent. Then, graph the total amount saved for the first 𝟖𝟖 weeks. 𝒕𝒕= 𝟓𝟓𝟓𝟓+ 𝟏𝟏𝟏𝟏 The total amount saved in dollars, 𝒕𝒕, is the dependent variable. The number of weeks, 𝒘𝒘, is the independent variable. # of Weeks Total Amount ($) 𝟎𝟎 𝟏𝟏𝟏𝟏 𝟏𝟏 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟒𝟒 𝟑𝟑𝟑𝟑 𝟓𝟓 𝟑𝟑𝟑𝟑 𝟔𝟔 𝟒𝟒𝟒𝟒 𝟕𝟕 𝟒𝟒𝟒𝟒 𝟖𝟖 𝟓𝟓𝟓𝟓 Total Cost ($) Total Cost of a Taxi Ride Number of Miles A STORY OF RATIOS 360 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 32 Lesson 32: Multi-Step Problems in the Real World 4. Aliyah is purchasing produce at the farmers’ market. She plans to buy $𝟏𝟏𝟏𝟏 worth of potatoes and some apples. The apples cost $𝟏𝟏. 𝟓𝟓𝟓𝟓 per pound. Write an equation to show the total cost of the produce, where 𝑻𝑻 is the total cost, in dollars, and 𝒂𝒂 is the number of pounds of apples. Determine which variable is dependent and which is independent. Then, graph the equation on the coordinate plane. 𝑻𝑻= 𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓+ 𝟏𝟏𝟏𝟏 The total cost in dollars is the dependent variable. The number of pounds of apples is the independent variable. # of Pounds of Apples Total Cost ($) 𝟎𝟎 𝟏𝟏𝟏𝟏 𝟏𝟏 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 𝟒𝟒 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓 # of lb. of Apples Total Cost at the Farmers’ Market Total Cost ($) A STORY OF RATIOS 361 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Lesson 33: From Equations to Inequalities Student Outcomes Students understand that an inequality with numerical expressions is either true or false. It is true if the numbers calculated on each side of the inequality sign result in a correct statement and is false otherwise. Students understand solving an inequality is answering the question of which values from a specified set, if any, make the inequality true. Classwork Example 1 (8 minutes) Students review their work from Lesson 23 and use this throughout the lesson. Example 1 What value(s) does the variable have to represent for the equation or inequality to result in a true number sentence? What value(s) does the variable have to represent for the equation or inequality to result in a false number sentence? a. 𝒚𝒚+ 𝟔𝟔= 𝟏𝟏𝟏𝟏 The number sentence is true when 𝒚𝒚 is 𝟏𝟏𝟏𝟏. The sentence is false when 𝒚𝒚 is any number other than 𝟏𝟏𝟏𝟏. b. 𝒚𝒚+ 𝟔𝟔> 𝟏𝟏𝟏𝟏 The number sentence is true when 𝒚𝒚 is any number greater than 𝟏𝟏𝟏𝟏. The sentence is false when 𝒚𝒚 is 𝟏𝟏𝟏𝟏 or any number less than 𝟏𝟏𝟏𝟏. c. 𝒚𝒚+ 𝟔𝟔≥𝟏𝟏𝟏𝟏 The number sentence is true when 𝒚𝒚 is 𝟏𝟏𝟏𝟏 or any number greater than 𝟏𝟏𝟏𝟏. The sentence is false when 𝒚𝒚 is a number less than 𝟏𝟏𝟏𝟏. d. 𝟑𝟑𝟑𝟑= 𝟏𝟏𝟏𝟏 The number sentence is true when 𝒈𝒈 is 𝟓𝟓. The number sentence is false when 𝒈𝒈 is any number other than 𝟓𝟓. e. 𝟑𝟑𝟑𝟑< 𝟏𝟏𝟏𝟏 The number sentence is true when 𝒈𝒈 is any number less than 𝟓𝟓. The number sentence is false when 𝒈𝒈 is 𝟓𝟓 or any number greater than 𝟓𝟓. f. 𝟑𝟑𝟑𝟑≤𝟏𝟏𝟏𝟏 The number sentence is true when 𝒈𝒈 is 𝟓𝟓 or any number less than 𝟓𝟓. The number sentence is false when 𝒈𝒈 is any number greater than 𝟓𝟓. MP.6 A STORY OF RATIOS 362 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Example 2 (12 minutes) Students move from naming the values that make the sentence true or false to using a set of numbers and determining whether or not the numbers in the set make the equation or inequality true or false. Example 2 Which of the following number(s), if any, make the equation or inequality true: {𝟎𝟎, 𝟑𝟑, 𝟓𝟓, 𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏}? a. 𝒎𝒎+ 𝟒𝟒= 𝟏𝟏𝟏𝟏 𝒎𝒎= 𝟖𝟖 or {𝟖𝟖} b. 𝒎𝒎+ 𝟒𝟒< 𝟏𝟏𝟏𝟏 {𝟎𝟎, 𝟑𝟑, 𝟓𝟓} How does the answer to part (a) compare to the answer to part (b)? In part (a), 8 is the only number that will result in a true number sentence. But in part (b), any number in the set that is less than 8 will make the number sentence true. c. 𝒇𝒇−𝟒𝟒= 𝟐𝟐 None of the numbers in the set will result in a true number sentence. d. 𝒇𝒇−𝟒𝟒> 𝟐𝟐 {𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏} Is there a number that we could include in the set so that part (c) will have a solution? Yes. The number 6 will make the equation in part (c) true. Would 6 be part of the solution set in part (d)? No. The 6 would not make part (d) a true number sentence because 6 −4 is not greater than 2. How could we change part (d) so that 6 would be part of the solution? Answers will vary; If the > was changed to a ≥, we could include 6 in the solution set. e. 𝟏𝟏 𝟐𝟐𝒉𝒉= 𝟖𝟖 None of the numbers in the set will result in a true number sentence. f. 𝟏𝟏 𝟐𝟐𝒉𝒉≥𝟖𝟖 None of the numbers in the set will result in a true number sentence. Which whole numbers, if any, make the inequality in part (f) true? Answers will vary; 16 and any number greater than 16 will make the number sentence true. MP.6 A STORY OF RATIOS 363 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Exercises (16 minutes) Students practice either individually or in pairs. Exercises Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: {𝟎𝟎, 𝟏𝟏, 𝟓𝟓, 𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏}. 1. 𝒎𝒎+ 𝟓𝟓= 𝟔𝟔 𝒎𝒎= 𝟏𝟏 or {𝟏𝟏} 2. 𝒎𝒎+ 𝟓𝟓≤𝟔𝟔 {𝟎𝟎, 𝟏𝟏} 3. 𝟓𝟓𝟓𝟓= 𝟒𝟒𝟒𝟒 𝒉𝒉= 𝟖𝟖 or {𝟖𝟖} 4. 𝟓𝟓𝟓𝟓> 𝟒𝟒𝟒𝟒 {𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏} 5. 𝟏𝟏 𝟐𝟐𝒚𝒚= 𝟓𝟓 There is no solution in the set. 6. 𝟏𝟏 𝟐𝟐𝒚𝒚≤𝟓𝟓 {𝟎𝟎, 𝟏𝟏, 𝟓𝟓, 𝟖𝟖} 7. 𝒌𝒌−𝟑𝟑= 𝟐𝟐𝟐𝟐 There is no solution in the set. 8. 𝒌𝒌−𝟑𝟑> 𝟐𝟐𝟐𝟐 There is no solution in the set. A STORY OF RATIOS 364 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Closing (3 minutes) In some of the equations and inequalities we worked within this lesson, none of the numbers in the given set were solutions. What does this mean? Are there numbers that will make the number sentences true that are not in the set? None of the numbers in the set resulted in a true number sentence. However, there are numbers that could make the number sentence true. For example, in Exercise 5, 𝑦𝑦= 10 would make a true number sentence but was not included in the given set of numbers. Is it possible for every number in a set to result in a true number sentence? Yes, it is possible. For example, if the inequality says 𝑥𝑥> 5 and all the numbers in the set are greater than 5, then all the numbers in the set will result in a true number sentence. Consider the equation 𝑦𝑦+ 3 = 11 and the inequality 𝑦𝑦+ 3 < 11. How does the solution to the equation help you determine the solution set to the inequality? In the equation 𝑦𝑦+ 3 = 11, 𝑦𝑦= 8 will result in a true number sentence. In the inequality, we want 𝑦𝑦+ 3 to be a value less than 11. So, the numbers that will make it true must be less than 8. Exit Ticket (6 minutes) A STORY OF RATIOS 365 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Name Date Lesson 33: From Equations to Inequalities Exit Ticket Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: {3, 4, 7, 9, 12, 18, 32}. 1. 1 3 𝑓𝑓= 4 2. 1 3 𝑓𝑓< 4 3. 𝑚𝑚+ 7 = 20 4. 𝑚𝑚+ 7 ≥20 A STORY OF RATIOS 366 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities Exit Ticket Sample Solutions Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: {𝟑𝟑, 𝟒𝟒, 𝟕𝟕, 𝟗𝟗, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑}. 1. 𝟏𝟏 𝟑𝟑𝒇𝒇= 𝟒𝟒 𝒇𝒇= 𝟏𝟏𝟏𝟏 or {𝟏𝟏𝟏𝟏} 2. 𝟏𝟏 𝟑𝟑𝒇𝒇< 𝟒𝟒 {𝟑𝟑, 𝟒𝟒, 𝟕𝟕, 𝟗𝟗} 3. 𝒎𝒎+ 𝟕𝟕= 𝟐𝟐𝟐𝟐 There is no number in the set that will make this equation true. 4. 𝒎𝒎+ 𝟕𝟕≥𝟐𝟐𝟐𝟐 {𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑} Problem Set Sample Solutions Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: {𝟎𝟎, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟗𝟗, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐}. 1. 𝒉𝒉−𝟖𝟖= 𝟓𝟓 𝒉𝒉= 𝟏𝟏𝟏𝟏 or {𝟏𝟏𝟏𝟏} 2. 𝒉𝒉−𝟖𝟖< 𝟓𝟓 {𝟎𝟎, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟗𝟗} 3. 𝟒𝟒𝟒𝟒= 𝟑𝟑𝟑𝟑 𝒈𝒈= 𝟗𝟗 or {𝟗𝟗} 4. 𝟒𝟒𝟒𝟒≥𝟑𝟑𝟑𝟑 {𝟗𝟗, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐} 5. 𝟏𝟏 𝟒𝟒𝒚𝒚= 𝟕𝟕 There is no number in the set that will make this equation true. 6. 𝟏𝟏 𝟒𝟒𝒚𝒚> 𝟕𝟕 There is no number in the set that will make this inequality true. A STORY OF RATIOS 367 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 33 Lesson 33: From Equations to Inequalities 7. 𝒎𝒎−𝟑𝟑= 𝟏𝟏𝟏𝟏 𝒎𝒎= 𝟏𝟏𝟏𝟏 or {𝟏𝟏𝟏𝟏} 8. 𝒎𝒎−𝟑𝟑≤𝟏𝟏𝟏𝟏 {𝟎𝟎, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟗𝟗, 𝟏𝟏𝟏𝟏} A STORY OF RATIOS 368 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems Lesson 34: Writing and Graphing Inequalities in Real-World Problems Student Outcomes Students recognize that inequalities of the form 𝑥𝑥< 𝑐𝑐 and 𝑥𝑥> 𝑐𝑐, where 𝑥𝑥 is a variable and 𝑐𝑐 is a fixed number, have infinitely many solutions when the values of 𝑥𝑥 come from a set of rational numbers. Classwork Example 1 (10 minutes) Begin with a discussion of what each of these statements means. Have students share possible amounts of money that could fit the given statement to build toward a graph and an inequality. Example 1 Statement Inequality Graph a. Caleb has at least $𝟓𝟓. b. Tarek has more than $𝟓𝟓. c. Vanessa has at most $𝟓𝟓. d. Li Chen has less than $𝟓𝟓. How much money could Caleb have? He could have $5, $5.01, $5.90, $6, $7, $8, $9, …. More simply, he could have $5 or any amount greater than $5. How would we show this as an inequality? 𝑐𝑐≥5, where 𝑐𝑐 is the amount of money that Caleb has in dollars What numbers on the graph do we need to show as a solution? 5 is a solution and everything to the right. Because we want to include 5 in the solution, we will draw a solid circle over the 5 and then an arrow to the right to show that all the numbers 5 and greater are part of the solution. 𝒕𝒕> 𝟓𝟓 𝒗𝒗≤ 𝟓𝟓 𝒄𝒄≥𝟓𝟓 𝑳𝑳< 𝟓𝟓 MP.4 A STORY OF RATIOS 369 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems How does the statement about Tarek differ from the statement about Caleb? Tarek has more than $5, but he cannot have exactly $5, where Caleb might have had exactly $5. So, how would we show this as an inequality? 𝑡𝑡> 5, where 𝑡𝑡 is the amount of money Tarek has in dollars When we graph the inequality for Tarek, we still want a circle on the 5, but this time it will not be solid to show that 5 is not included in the solution. What does “at most” mean in Vanessa’s example? Vanessa could have $5 but no more than 5. So, she could have less than $5, including $4, $3, $2, $1, $0, or even a negative amount if she owes someone money. How would we write this as an inequality? 𝑣𝑣≤5, where 𝑣𝑣 is the amount of money Vanessa has in dollars How would you show this on the graph? We would put a circle on the 5 and then an arrow toward the smaller numbers. Would we have a solid or an open circle? It would be solid to show that 5 is part of the solution. Would the inequality and graph for Li Chen be the same as Vanessa’s solution? Why or why not? No. They would be similar but not exactly the same. Li Chen cannot have $5 exactly. So, the circle in the graph would be open, and the inequality would be 𝐿𝐿< 5, where 𝐿𝐿 represents the amount of money Li Chen has in dollars. Example 2 (5 minutes) Example 2 Kelly works for Quick Oil Change. If customers have to wait longer than 𝟐𝟐𝟐𝟐 minutes for the oil change, the company does not charge for the service. The fastest oil change that Kelly has ever done took 𝟔𝟔 minutes. Show the possible customer wait times in which the company charges the customer. 𝟔𝟔≤𝒙𝒙≤𝟐𝟐𝟐𝟐 How is this example different from the problems in Example 1? This one is giving a range of possible values. The number of minutes he takes to change the oil should be somewhere between two values instead of greater than just one or less than just one. Let’s start with the first bit of information. What does the second sentence of the problem tell us about the wait times for paying customers? The oil change must take 20 minutes or less. How would we show this on a number line? Because 20 minutes is part of the acceptable time limit, we will use a solid circle and shade to the left. MP.4 A STORY OF RATIOS 370 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems Now, let’s look at the other piece of information. The fastest Kelly has ever completed the oil change is 6 minutes. What does this mean about the amount of time it will take? This means that it will take 6 minutes or more to complete an oil change. How would we show this on a number line? Because 6 minutes is a possible amount of time, we will use a solid circle. Then, we will shade to the right. Now, we need to put both of these pieces of information together to make one model of the inequality. How could we show both of these on one number line? Instead of an arrow, we would have two circles, and we would shade in between. Should the circles be open or solid? Because he has to change the oil in 20 minutes or less, the 20 is part of the solution, and the circle will be closed. The 6 minutes is also part of the solution because it is an actual time that Kelly has completed the work. The circle at 6 should also be closed. Example 3 (5 minutes) Example 3 Gurnaz has been mowing lawns to save money for a concert. Gurnaz will need to work for at least six hours to save enough money, but he must work fewer than 𝟏𝟏𝟏𝟏 hours this week. Write an inequality to represent this situation, and then graph the solution. 𝟔𝟔 ≤𝒙𝒙< 𝟏𝟏𝟏𝟏 How would we represent Gurnaz working at least six hours? “At least” tells us that Gurnaz must work 6 hours or more. 𝑥𝑥≥6 What inequality would we use to show that he must work fewer than 16 hours? “Fewer than” means that Gurnaz cannot actually work 16 hours. So, we will use 𝑥𝑥< 16. MP.4 A STORY OF RATIOS 371 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems Exercises 1–5 (15 minutes) Students work individually. Exercises 1–5 Write an inequality to represent each situation. Then, graph the solution. 1. Blayton is at most 𝟐𝟐 meters above sea level. 𝒃𝒃≤𝟐𝟐, where 𝒃𝒃 is Blayton’s position in relationship to sea level in meters 2. Edith must read for a minimum of 𝟐𝟐𝟐𝟐 minutes. 𝑬𝑬≥𝟐𝟐𝟐𝟐, where 𝑬𝑬 is the number of minutes Edith reads 3. Travis milks his cows each morning. He has never gotten fewer than 𝟑𝟑 gallons of milk; however, he always gets fewer than 𝟗𝟗 gallons of milk. 𝟑𝟑≤𝒙𝒙< 𝟗𝟗, where 𝒙𝒙 represents the gallons of milk 4. Rita can make 𝟖𝟖 cakes for a bakery each day. So far, she has orders for more than 𝟑𝟑𝟑𝟑 cakes. Right now, Rita needs more than four days to make all 𝟑𝟑𝟑𝟑 cakes. 𝒙𝒙> 𝟒𝟒, where 𝒙𝒙 is the number of days Rita has to bake the cakes 5. Rita must have all the orders placed right now done in 𝟕𝟕 days or fewer. How will this change your inequality and your graph? 𝟒𝟒< 𝒙𝒙≤𝟕𝟕 Our inequality will change because there is a range for the number of days Rita has to bake the cakes. The graph has changed because Rita is more limited in the amount of time she has to bake the cakes. Instead of the graph showing any number larger than 𝟒𝟒, the graph now has a solid circle at 𝟕𝟕 because Rita must be done baking the cakes in a maximum of 𝟕𝟕days. Possible Extension Exercises 6–10 The following problems combine the skills used to solve equations in previous lessons within this module and inequalities. Possible Extension Exercises 6–10 6. Kasey has been mowing lawns to save up money for a concert. He earns $𝟏𝟏𝟏𝟏 per hour and needs at least $𝟗𝟗𝟗𝟗 to go to the concert. How many hours should he mow? 𝟏𝟏𝟏𝟏𝟏𝟏≥𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏≥𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏 𝒙𝒙≥𝟔𝟔 Kasey will need to mow for 𝟔𝟔 or more hours. A STORY OF RATIOS 372 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems 7. Rachel can make 𝟖𝟖 cakes for a bakery each day. So far, she has orders for more than 𝟑𝟑𝟑𝟑 cakes. How many days will it take her to complete the orders? 𝟖𝟖𝟖𝟖> 𝟑𝟑𝟑𝟑 𝟖𝟖𝟖𝟖 𝟖𝟖> 𝟑𝟑𝟑𝟑 𝟖𝟖 𝒙𝒙> 𝟒𝟒 Rachel will need to work more than 𝟒𝟒 days. 8. Ranger saves $𝟕𝟕𝟕𝟕 each week. He needs to save at least $𝟐𝟐, 𝟖𝟖𝟖𝟖𝟖𝟖 to go on a trip to Europe. How many weeks will he need to save? 𝟕𝟕𝟕𝟕𝟕𝟕≥𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟕𝟕𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕≥𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟕𝟕𝟕𝟕 𝒙𝒙≥𝟒𝟒𝟒𝟒 Ranger needs to save for at least 𝟒𝟒𝟒𝟒 weeks. 9. Clara has less than $𝟕𝟕𝟕𝟕. She wants to buy 𝟑𝟑 pairs of shoes. What price shoes can Clara afford if all the shoes are the same price? 𝟑𝟑𝟑𝟑< 𝟕𝟕𝟕𝟕 𝟑𝟑𝟑𝟑 𝟑𝟑< 𝟕𝟕𝟕𝟕 𝟑𝟑 𝒙𝒙< 𝟐𝟐𝟐𝟐 Clara can afford shoes that are greater than $𝟎𝟎 and less than $𝟐𝟐𝟐𝟐. 10. A gym charges $𝟐𝟐𝟐𝟐 per month plus $𝟒𝟒 extra to swim in the pool for an hour. If a member only has $𝟒𝟒𝟒𝟒 to spend each month, at most how many hours can the member swim? 𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐≤𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒+ 𝟐𝟐𝟐𝟐−𝟐𝟐𝟐𝟐≤𝟒𝟒𝟒𝟒−𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒≤𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒 𝟒𝟒≤𝟐𝟐𝟐𝟐 𝟒𝟒 𝒙𝒙≤𝟓𝟓 The member can swim in the pool for 𝟓𝟓 hours. However, we also know that the total amount of time the member spends in the pool must be greater than or equal to 𝟎𝟎 hours because the member may choose not to swim. 𝟎𝟎≤𝒙𝒙≤𝟓𝟓 Closing (5 minutes) How are inequalities different from equations? Inequalities can have a range of possible values that make the statement true, where equations do not. Does the phrase “at most” refer to being less than or greater than something? Give an example to support your answer. “At most” means that you can have that amount or less than that amount. You cannot go over. My mom says that I can watch at most 3 TV shows after I do my homework. This means that I can watch 3 or fewer than 3 TV shows. Exit Ticket (5 minutes) A STORY OF RATIOS 373 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems Name Date Lesson 34: Writing and Graphing Inequalities in Real-World Problems Exit Ticket For each question, write an inequality. Then, graph your solution. 1. Keisha needs to make at least 28 costumes for the school play. Since she can make 4 costumes each week, Keisha plans to work on the costumes for at least 7 weeks. 2. If Keisha has to have the costumes complete in 10 weeks or fewer, how will our solution change? A STORY OF RATIOS 374 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems Exit Ticket Sample Solutions For each question, write an inequality. Then, graph your solution. 1. Keisha needs to make at least 𝟐𝟐𝟐𝟐 costumes for the school play. Since she can make 𝟒𝟒 costumes each week, Keisha plans to work on the costumes for at least 𝟕𝟕 weeks. 𝒙𝒙≥𝟕𝟕 Keisha should plan to work on the costumes for 𝟕𝟕 or more weeks. 2. If Keisha has to have the costumes complete in 𝟏𝟏𝟏𝟏 weeks or fewer, how will our solution change? Keisha had 𝟕𝟕 or more weeks in Problem 𝟏𝟏. It will still take her at least 𝟕𝟕 weeks, but she cannot have more than 𝟏𝟏𝟏𝟏 weeks. 𝟕𝟕≤𝒙𝒙≤𝟏𝟏𝟏𝟏 Problem Set Sample Solutions Write and graph an inequality for each problem. 1. At least 𝟏𝟏𝟏𝟏 𝒙𝒙≥𝟏𝟏𝟏𝟏 2. Less than 𝟕𝟕 𝒙𝒙< 𝟕𝟕 3. Chad will need at least 𝟐𝟐𝟐𝟐 minutes to complete the 𝟓𝟓K race. However, he wants to finish in under 𝟑𝟑𝟑𝟑 minutes. 𝟐𝟐𝟐𝟐≤𝒙𝒙< 𝟑𝟑𝟑𝟑 4. Eva saves $𝟔𝟔𝟔𝟔 each week. Since she needs to save at least $𝟐𝟐, 𝟒𝟒𝟒𝟒𝟒𝟒 to go on a trip to Europe, she will need to save for at least 𝟒𝟒𝟒𝟒 weeks. 𝒙𝒙≥𝟒𝟒𝟒𝟒 A STORY OF RATIOS 375 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 Lesson 34 Lesson 34: Writing and Graphing Inequalities in Real-World Problems 5. Clara has $𝟏𝟏𝟏𝟏𝟏𝟏. She wants to buy 𝟒𝟒 pairs of the same pants. Due to tax, Clara can afford pants that are less than $𝟐𝟐𝟐𝟐. Clara must spend less than $𝟐𝟐𝟐𝟐, but we also know that Clara will spend more than $𝟎𝟎 when she buys pants at the store. 𝟎𝟎< 𝒙𝒙< 𝟐𝟐𝟐𝟐 6. A gym charges $𝟑𝟑𝟑𝟑 per month plus $𝟒𝟒 extra to swim in the pool for an hour. Because a member has just $𝟓𝟓𝟓𝟓 to spend at the gym each month, the member can swim at most 𝟓𝟓 hours. The member can swim in the pool for 𝟓𝟓 hours. However, we also know that the total amount of time the member spends in the pool must be greater than or equal to 𝟎𝟎 hours because the member may choose not to swim. 𝟎𝟎≤𝒙𝒙≤𝟓𝟓 A STORY OF RATIOS 376 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations Name Date 1. Gertrude is deciding which cell phone plan is the best deal for her to buy. Super Cell charges a monthly fee of $10 and also charges $0.15 per call. She makes a note that the equation is 𝑀𝑀= 0.15𝐶𝐶+ 10, where 𝑀𝑀 is the monthly charge, in dollars, and 𝐶𝐶 is the number of calls placed. Global Cellular has a plan with no monthly fee but charges $0.25 per call. She makes a note that the equation is 𝑀𝑀= 0.25𝐶𝐶, where 𝑀𝑀 is the monthly charge, in dollars, and 𝐶𝐶 is the number of calls placed. Both companies offer unlimited text messages. a. Make a table for both companies showing the cost of service, 𝑀𝑀, for making from 0 to 200 calls per month. Use multiples of 20. Cost of Services, 𝑴𝑴, in Dollars Number of Calls, 𝑪𝑪 Super Cell 𝑴𝑴= 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 Global Cellular 𝑴𝑴= 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐 A STORY OF RATIOS 377 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations b. Construct a graph for the two equations on the same graph. Use the number of calls, 𝐶𝐶, as the independent variable and the monthly charge, in dollars, 𝑀𝑀, as the dependent variable. c. Which cell phone plan is the best deal for Gertrude? Defend your answer with specific examples. A STORY OF RATIOS 378 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 2. Sadie is saving her money to buy a new pony, which costs $600. She has already saved $75. She earns $50 per week working at the stables and wonders how many weeks it will take to earn enough for a pony of her own. a. Make a table showing the week number, 𝑊𝑊, and total savings, in dollars, 𝑆𝑆, in Sadie’s savings account. Number of Weeks Total Savings (in dollars) b. Show the relationship between the number of weeks and Sadie’s savings using an expression. c. How many weeks will Sadie have to work to earn enough to buy the pony? A STORY OF RATIOS 379 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 3. The elevator at the local mall has a weight limit of 1,800 pounds and requires that the maximum person allowance be no more than nine people. a. Let 𝑥𝑥 represent the number of people. Write an inequality to describe the maximum allowance of people allowed in the elevator at one time. b. Draw a number line diagram to represent all possible solutions to part (a). c. Let 𝑤𝑤 represent the amount of weight, in pounds. Write an inequality to describe the maximum weight allowance in the elevator at one time. d. Draw a number line diagram to represent all possible solutions to part (c). A STORY OF RATIOS 380 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 4. Devin’s football team carpools for practice every week. This week is his parents’ turn to pick up team members and take them to the football field. While still staying on the roads, Devin’s parents always take the shortest route in order to save gasoline. Below is a map of their travels. Each gridline represents a street and the same distance. Devin’s father checks his mileage and notices that he drove 18 miles between his house and Stop 3. a. Create an equation, and determine the amount of miles each gridline represents. b. Using this information, determine how many total miles Devin’s father will travel from home to the football field, assuming he made every stop. Explain how you determined the answer. c. At the end of practice, Devin’s father dropped off team members at each stop and went back home. How many miles did Devin’s father travel all together? Stop 4 Stop 3 Stop 2 Stop 1 Devin’s House Football Field A STORY OF RATIOS 381 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 5. For a science experiment, Kenneth reflects a beam off a mirror. He is measuring the missing angle created when the light reflects off the mirror. (Note: The figure is not drawn to scale.) Use an equation to determine the missing angle, labeled 𝑥𝑥 in the diagram. 51° 𝑥𝑥° 51° A STORY OF RATIOS 382 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations A Progression Toward Mastery Assessment Task Item STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem. STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem. STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, OR an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem. STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem. 1 a 6.EE.C.9 Student is not able to make a table with multiples of 20 calls or cannot calculate the monthly charge based on the number of calls. Student correctly completes the Number of Calls column but is not able to use the equations to accurately complete the other two columns. Student accurately calculates one column, but the other two columns have errors. Student accurately calculates all of the columns. b 6.EE.C.9 6.EE.B.6 Student is not able to graph the data from the table. Student attempts to graph the data from the table. Several mistakes or omissions are present. Student graphs the data from the table but has minor mistakes or omissions. Student includes at least three of the four criteria of a perfect graph. Student graphs the data from the table without error (i.e., graph is titled, axes are labeled, units are included on axes, and points are plotted accurately). c 6.EE.C.9 Student cannot conclude which plan is better or chooses a plan without evidentiary support. Student chooses one plan or the other, showing support for that plan. Student does not recognize that the best deal depends on the number of calls placed in a month. Student concludes that the best deal depends on the number of calls placed in a month. Student does not describe a complete analysis. Student concludes that the best deal depends on the number of calls placed in a month. The break-even point, 100 calls costing $25, is identified. Answer specifically states that Super Cell is the better deal if the number of calls per month is < 100, and Global Cellular is the better deal if the number of calls per month is > 100. A STORY OF RATIOS 383 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 2 a 6.EE.C.9 Student is not able to make the table or attempts to make the table but has many errors. Student is able to make the table and calculate most rows accurately. Compounding errors may be present. Titles or variables may be missing. Student accurately makes the table and accurately computes the total savings for each week. Titles or variables may be missing. OR Student begins with 50 and 75 every week. Student accurately makes the table and accurately computes the total savings for each week. Titles and variables are present. b 6.EE.C.9 Student is not able to write the expression. Student attempts to write the expression but is inaccurate (perhaps writing 75𝑊𝑊+ 50). Student writes the expression 50𝑊𝑊+ 75 but does not include a description of what the variable represents. Student accurately writes the expression 50𝑊𝑊+ 75 and includes a description of what the variable represents. c 6.EE.C.9 Student cannot make a conclusion about how many weeks are needed for the purchase, or an answer is provided that is not supported by the student’s table. Student concludes that some number of weeks other than 11 weeks will be needed for the purchase. Student concludes that 11 weeks of work will be needed for the purchase. Student concludes that 11 weeks of work will be needed for the purchase and that Sadie will have $625, which is $25 more than the cost of the pony. 3 a 6.EE.B.8 Student does not write an inequality or tries writing the inequality with incorrect information (e.g., 1,800). Student writes the inequality 𝑥𝑥≥9 or 𝑥𝑥> 9. Student writes the inequality 𝑥𝑥< 9 because student does not realize there can be 9 people on the elevator. Student writes the inequality 0 ≤𝑥𝑥≤9. b 6.EE.C.9 Student does not draw a number line or draws a line but does not indicate 0 ≤𝑥𝑥≤9. Student draws an accurate number line and uses either a line segment or discrete symbols but does not include 0 and/or 9 in the solution set. Student draws a number line but uses a line segment to indicate continuous points 0 ≤ 𝑥𝑥≤9. Student draws an accurate number line, using discrete symbols indicating whole numbers from 0 to 9. c 6.EE.B.8 Student does not write an inequality or tries writing the inequality with incorrect information (e.g., 9). Student writes the inequality 𝑊𝑊≥1,800 or 𝑊𝑊> 1,800. Student writes the inequality 𝑊𝑊< 1,800 because student does not realize there can be 1,800 pounds on the elevator. Student writes the inequality 0 ≤𝑊𝑊≤1,800. d 6.EE.C.9 Student does not draw a number line or draws a line but does not indicate 0 ≤𝑊𝑊≤1,800. Student draws an accurate number line and uses either a line segment or discrete symbols but does not include 0 and/or 1,800 in the solution set. Student draws a number line but uses discrete symbols indicating whole numbers from 0 to 1,800 are in the solution set 0 ≤𝑊𝑊≤1,800. Student draws an accurate number line, using a line segment indicating all numbers from 0 to 1,800 are in the solution set 0 ≤ 𝑊𝑊≤1,800. A STORY OF RATIOS 384 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 4 a 6.EE.B.7 6.EE.C.9 6.EE.B.5 6.EE.B.6 Student is unable to create an equation and is unable to determine the amount of miles Devin’s father traveled between their house and Stop 3. Student is able to determine the amount of miles Devin’s father traveled between their house and Stop 3 but does not write an equation. Student creates an equation, 9𝐺𝐺= 18 mi, but does not use the equation to determine the amount of miles Devin’s father traveled between their house and Stop 3. Student creates an equation and uses it to determine the number of miles Devin’s father traveled between their house and Stop 3. Let 𝐺𝐺 represent the number of gridlines passed on the map. 9𝐺𝐺= 18 mi 𝐺𝐺 9 = 18 mi 9 𝐺𝐺= 2 mi b 6.EE.A.2 6.EE.B.6 6.EE.B.6 Student does not describe how the answer was derived or leaves the answer blank. Student determines the correct distance (30 miles) but does not explain how the answer was determined or offers an incomplete explanation. Student inaccurately counts the intersections passed (15) but accurately applies the equation with the incorrect count. Explanation is correct and clear. Student accurately counts the intersections passed (15) and applies that to the correct equation: 15 ∙2 mi = 30 mi. c 6.EE.C.9 6.EE.B.5 6.EE.B.6 Student answer does not indicate a concept of round-trip distance being double that of a one-way trip. Student does not double the correct one-way trip distance (30 miles) from part (b) or doubles the number incorrectly. Student does not use the one-way trip distance (30 miles) from part (b) but counts the blocks for the round trip. Student doubles the correct one-way trip distance (30 miles) from part (b) to arrive at the correct round-trip distance (60 miles). 5 6.EE.B.5 6.EE.B.6 6.EE.B.7 Student does not show any of the steps necessary to solve the problem or simply answers 51°. Student adds 51° + 51° to arrive at 102° but does not subtract this from 180° to find the missing angle. Student correctly finds the missing angle, 78°, showing clearly the steps involved but does not use an equation. OR Student makes an arithmetic error, but clear evidence of conceptual understanding is evident. Student correctly finds the missing angle, 78°, by using an equation and clearly showing the steps involved. Student might reference the terms supplementary angles or straight angles or start with 51° + 51° + 𝑥𝑥° = 180° before solving it correctly. A STORY OF RATIOS 385 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations Name Date 1. Gertrude is deciding which cell phone plan is the best deal for her to buy. Super Cell charges a monthly fee of $10 and also charges $0.15 per call. She makes a note that the equation is 𝑀𝑀= 0.15𝐶𝐶+ 10, where 𝑀𝑀 is the monthly charge, in dollars, and 𝐶𝐶 is the number of calls placed. Global Cellular has a plan with no monthly fee but charges $0.25 per call. She makes a note that the equation is 𝑀𝑀= 0.25𝐶𝐶, where 𝑀𝑀 is the monthly charge, in dollars, and 𝐶𝐶 is the number of calls placed. Both companies offer unlimited text messages. a. Make a table for both companies showing the cost of service, 𝑀𝑀, for making from 0 to 200 calls per month. Use multiples of 20. A STORY OF RATIOS 386 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations b. Construct a graph for the two equations on the same graph. Use the number of calls, 𝐶𝐶, as the independent variable and the monthly charge, in dollars, 𝑀𝑀, as the dependent variable. c. Which cell phone plan is the best deal for Gertrude? Defend your answer with specific examples. A STORY OF RATIOS 387 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 2. Sadie is saving her money to buy a new pony, which costs $600. She has already saved $75. She earns $50 per week working at the stables and wonders how many weeks it will take to earn enough for a pony of her own. a. Make a table showing the week number, 𝑊𝑊, and total savings, in dollars, 𝑆𝑆, in Sadie’s savings account. b. Show the relationship between the number of weeks and Sadie’s savings using an expression. c. How many weeks will Sadie have to work to earn enough to buy the pony? A STORY OF RATIOS 388 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 3. The elevator at the local mall has a weight limit of 1,800 pounds and requires that the maximum person allowance be no more than nine people. a. Let 𝑥𝑥 represent the number of people. Write an inequality to describe the maximum allowance of people allowed in the elevator at one time. b. Draw a number line diagram to represent all possible solutions to part (a). c. Let 𝑤𝑤 represent the amount of weight, in pounds. Write an inequality to describe the maximum weight allowance in the elevator at one time. d. Draw a number line diagram to represent all possible solutions to part (c). A STORY OF RATIOS 389 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 4. Devin’s football team carpools for practice every week. This week is his parents’ turn to pick up team members and take them to the football field. While still staying on the roads, Devin’s parents always take the shortest route in order to save gasoline. Below is a map of their travels. Each gridline represents a street and the same distance. Devin’s father checks his mileage and notices that he drove 18 miles between his house and Stop 3. a. Create an equation, and determine the amount of miles each gridline represents. b. Using this information, determine how many total miles Devin’s father will travel from home to the football field, assuming he made every stop. Explain how you determined the answer. c. At the end of practice, Devin’s father dropped off team members at each stop and went back home. How many miles did Devin’s father travel all together? Stop 4 Stop 3 Stop 2 Stop 1 Devin’s House Footba l Field A STORY OF RATIOS 390 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 6•4 End-of-Module Assessment Task Module 4: Expressions and Equations 5. For a science experiment, Kenneth reflects a beam off a mirror. He is measuring the missing angle created when the light reflects off the mirror. (Note: The figure is not drawn to scale.) Use an equation to determine the missing angle, labeled 𝑥𝑥 in the diagram. 51° 𝑥𝑥° 51° A STORY OF RATIOS 391 ©20 15 G re at Min ds. eureka-math.org G6-M4-TE-1.3.0-08.2015 |
4399 | https://openstax.org/books/intermediate-algebra-2e/pages/5-review-exercises | Ch. 5 Review Exercises - Intermediate Algebra 2e | OpenStax
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Intermediate Algebra 2e
Review Exercises
Intermediate Algebra 2eReview Exercises
Contents Contents
Highlights
Table of contents
Preface
1 Foundations
2 Solving Linear Equations
3 Graphs and Functions
4 Systems of Linear Equations
5 Polynomials and Polynomial Functions
Introduction
5.1 Add and Subtract Polynomials
5.2 Properties of Exponents and Scientific Notation
5.3 Multiply Polynomials
5.4 Dividing Polynomials
Chapter Review
Exercises
Review Exercises
Practice Test
6 Factoring
7 Rational Expressions and Functions
8 Roots and Radicals
9 Quadratic Equations and Functions
10 Exponential and Logarithmic Functions
11 Conics
12 Sequences, Series and Binomial Theorem
Answer Key
Index
Search for key terms or text.
Close
Review Exercises
Add and Subtract Polynomials
Types of Polynomials
In the following exercises, determine the type of polynomial.
16 x 2−40 x−25 16 x 2−40 x−25 16 x 2−40 x−25
343.
5 m+9 5 m+9 5 m+9
−15−15−15
345.
y 2+6 y 3+9 y 4 y 2+6 y 3+9 y 4 y 2+6 y 3+9 y 4
Add and Subtract Polynomials
In the following exercises, add or subtract the polynomials.
4 p+11 p 4 p+11 p 4 p+11 p
347.
−8 y 3−5 y 3−8 y 3−5 y 3−8 y 3−5 y 3
(4 a 2+9 a−11)+(6 a 2−5 a+10)(4 a 2+9 a−11)+(6 a 2−5 a+10)(4 a 2+9 a−11)+(6 a 2−5 a+10)
349.
(8 m 2+12 m−5)−(2 m 2−7 m−1)(8 m 2+12 m−5)−(2 m 2−7 m−1)(8 m 2+12 m−5)−(2 m 2−7 m−1)
(y 2−3 y+12)+(5 y 2−9)(y 2−3 y+12)+(5 y 2−9)(y 2−3 y+12)+(5 y 2−9)
351.
(5 u 2+8 u)−(4 u−7)(5 u 2+8 u)−(4 u−7)(5 u 2+8 u)−(4 u−7)
Find the sum of 8 q 3−27 8 q 3−27 8 q 3−27 and q 2+6 q−2.q 2+6 q−2.q 2+6 q−2.
353.
Find the difference of x 2+6 x+8 x 2+6 x+8 x 2+6 x+8 and x 2−8 x+15.x 2−8 x+15.x 2−8 x+15.
In the following exercises, simplify.
17 m n 2−(−9 m n 2)+3 m n 2 17 m n 2−(−9 m n 2)+3 m n 2 17 m n 2−(−9 m n 2)+3 m n 2
355.
18 a−7 b−21 a 18 a−7 b−21 a 18 a−7 b−21 a
2 p q 2−5 p−3 q 2 2 p q 2−5 p−3 q 2 2 p q 2−5 p−3 q 2
357.
(6 a 2+7)+(2 a 2−5 a−9)(6 a 2+7)+(2 a 2−5 a−9)(6 a 2+7)+(2 a 2−5 a−9)
(3 p 2−4 p−9)+(5 p 2+14)(3 p 2−4 p−9)+(5 p 2+14)(3 p 2−4 p−9)+(5 p 2+14)
359.
(7 m 2−2 m−5)−(4 m 2+m−8)(7 m 2−2 m−5)−(4 m 2+m−8)(7 m 2−2 m−5)−(4 m 2+m−8)
(7 b 2−4 b+3)−(8 b 2−5 b−7)(7 b 2−4 b+3)−(8 b 2−5 b−7)(7 b 2−4 b+3)−(8 b 2−5 b−7)
361.
Subtract (8 y 2−y+9)(8 y 2−y+9)(8 y 2−y+9) from (11 y 2−9 y−5)(11 y 2−9 y−5)(11 y 2−9 y−5)
Find the difference of (z 2−4 z−12)(z 2−4 z−12)(z 2−4 z−12) and (3 z 2+2 z−11)(3 z 2+2 z−11)(3 z 2+2 z−11)
363.
(x 3−x 2 y)−(4 x y 2−y 3)+(3 x 2 y−x y 2)(x 3−x 2 y)−(4 x y 2−y 3)+(3 x 2 y−x y 2)(x 3−x 2 y)−(4 x y 2−y 3)+(3 x 2 y−x y 2)
(x 3−2 x 2 y)−(x y 2−3 y 3)−(x 2 y−4 x y 2)(x 3−2 x 2 y)−(x y 2−3 y 3)−(x 2 y−4 x y 2)(x 3−2 x 2 y)−(x y 2−3 y 3)−(x 2 y−4 x y 2)
Evaluate a Polynomial Function for a Given Value of the Variable
In the following exercises, find the function values for each polynomial function.
365.
For the function f(x)=7 x 2−3 x+5 f(x)=7 x 2−3 x+5 f(x)=7 x 2−3 x+5 find:
ⓐf(5)f(5)f(5)ⓑf(−2)f(−2)f(−2)ⓒf(0)f(0)f(0)
For the function g(x)=15−16 x 2,g(x)=15−16 x 2,g(x)=15−16 x 2, find:
ⓐg(−1)g(−1)g(−1)ⓑg(0)g(0)g(0)ⓒg(2)g(2)g(2)
367.
A pair of glasses is dropped off a bridge 640 feet above a river. The polynomial function h(t)=−16 t 2+640 h(t)=−16 t 2+640 h(t)=−16 t 2+640 gives the height of the glasses t seconds after they were dropped. Find the height of the glasses when t=6.t=6.t=6.
A manufacturer of the latest soccer shoes has found that the revenue received from selling the shoes at a cost of p p p dollars each is given by the polynomial R(p)=−5 p 2+360 p.R(p)=−5 p 2+360 p.R(p)=−5 p 2+360 p. Find the revenue received when p=10 p=10 p=10 dollars.
Add and Subtract Polynomial Functions
In the following exercises, find ⓐ (f + g)(x) ⓑ (f + g)(3) ⓒ (f − g)(x) ⓓ (f − g)(−2)
369.
f(x)=2 x 2−4 x−7 f(x)=2 x 2−4 x−7 f(x)=2 x 2−4 x−7 and g(x)=2 x 2−x+5 g(x)=2 x 2−x+5 g(x)=2 x 2−x+5
f(x)=4 x 3−3 x 2+x−1 f(x)=4 x 3−3 x 2+x−1 f(x)=4 x 3−3 x 2+x−1 and g(x)=8 x 3−1 g(x)=8 x 3−1 g(x)=8 x 3−1
Properties of Exponents and Scientific Notation
Simplify Expressions Using the Properties for Exponents
In the following exercises, simplify each expression using the properties for exponents.
371.
p 3⋅p 10 p 3·p 10 p 3·p 10
2⋅2 6 2·2 6 2·2 6
373.
a⋅a 2⋅a 3 a·a 2·a 3 a·a 2·a 3
x⋅x 8 x·x 8 x·x 8
375.
y a⋅y b y a·y b y a·y b
2 8 2 2 2 8 2 2 2 8 2 2
377.
a 6 a a 6 a a 6 a
n 3 n 12 n 3 n 12 n 3 n 12
379.
1 x 5 1 x 5 1 x 5
3 0 3 0 3 0
381.
y 0 y 0 y 0
(14 t)0(14 t)0(14 t)0
383.
12 a 0−15 b 0 12 a 0−15 b 0 12 a 0−15 b 0
Use the Definition of a Negative Exponent
In the following exercises, simplify each expression.
6−2 6−2 6−2
385.
(−10)−3(−10)−3(−10)−3
5⋅2−4 5·2−4 5·2−4
387.
(8 n)−1(8 n)−1(8 n)−1
y−5 y−5 y−5
389.
10−3 10−3 10−3
1 a−4 1 a−4 1 a−4
391.
1 6−2 1 6−2 1 6−2
−5−3−5−3−5−3
393.
(−1 5)−3(−1 5)−3(−1 5)−3
−(1 2)−3−(1 2)−3−(1 2)−3
395.
(−5)−3(−5)−3(−5)−3
(5 9)−2(5 9)−2(5 9)−2
397.
(−3 x)−3(−3 x)−3(−3 x)−3
In the following exercises, simplify each expression using the Product Property.
(y 4)3(y 4)3(y 4)3
399.
(3 2)5(3 2)5(3 2)5
(a 10)y(a 10)y(a 10)y
401.
x−3⋅x 9 x−3·x 9 x−3·x 9
r−5⋅r−4 r−5·r−4 r−5·r−4
403.
(u v−3)(u−4 v−2)(u v−3)(u−4 v−2)(u v−3)(u−4 v−2)
(m 5)−1(m 5)−1(m 5)−1
405.
p 5⋅p−2⋅p−4 p 5·p−2·p−4 p 5·p−2·p−4
In the following exercises, simplify each expression using the Power Property.
(k−2)−3(k−2)−3(k−2)−3
407.
q 4 q 20 q 4 q 20 q 4 q 20
b 8 b−2 b 8 b−2 b 8 b−2
409.
n−3 n−5 n−3 n−5 n−3 n−5
In the following exercises, simplify each expression using the Product to a Power Property.
(−5 a b)3(−5 a b)3(−5 a b)3
411.
(−4 p q)0(−4 p q)0(−4 p q)0
(−6 x 3)−2(−6 x 3)−2(−6 x 3)−2
413.
(3 y−4)2(3 y−4)2(3 y−4)2
In the following exercises, simplify each expression using the Quotient to a Power Property.
(3 5 x)−2(3 5 x)−2(3 5 x)−2
415.
(3 x y 2 z)4(3 x y 2 z)4(3 x y 2 z)4
(4 p−3 q 2)2(4 p−3 q 2)2(4 p−3 q 2)2
In the following exercises, simplify each expression by applying several properties.
417.
(x 2 y)2(3 x y 5)3(x 2 y)2(3 x y 5)3(x 2 y)2(3 x y 5)3
(−3 a−2)4(2 a 4)2(−6 a 2)3(−3 a−2)4(2 a 4)2(−6 a 2)3(−3 a−2)4(2 a 4)2(−6 a 2)3
419.
(3 x y 3 4 x 4 y−2)2(6 x y 4 8 x 3 y−2)−1(3 x y 3 4 x 4 y−2)2(6 x y 4 8 x 3 y−2)−1(3 x y 3 4 x 4 y−2)2(6 x y 4 8 x 3 y−2)−1
In the following exercises, write each number in scientific notation.
2.568 2.568 2.568
421.
5,300,000
0.00814 0.00814 0.00814
In the following exercises, convert each number to decimal form.
423.
2.9×10 4 2.9×10 4 2.9×10 4
3.75×10−1 3.75×10−1 3.75×10−1
425.
9.413×10−5 9.413×10−5 9.413×10−5
In the following exercises, multiply or divide as indicated. Write your answer in decimal form.
(3×10 7)(2×10−4)(3×10 7)(2×10−4)(3×10 7)(2×10−4)
427.
(1.5×10−3)(4.8×10−1)(1.5×10−3)(4.8×10−1)(1.5×10−3)(4.8×10−1)
6×10 9 2×10−1 6×10 9 2×10−1 6×10 9 2×10−1
429.
9×10−3 1×10−6 9×10−3 1×10−6 9×10−3 1×10−6
Multiply Polynomials
Multiply Monomials
In the following exercises, multiply the monomials.
(−6 p 4)(9 p)(−6 p 4)(9 p)(−6 p 4)(9 p)
431.
(1 3 c 2)(30 c 8)(1 3 c 2)(30 c 8)(1 3 c 2)(30 c 8)
(8 x 2 y 5)(7 x y 6)(8 x 2 y 5)(7 x y 6)(8 x 2 y 5)(7 x y 6)
433.
(2 3 m 3 n 6)(1 6 m 4 n 4)(2 3 m 3 n 6)(1 6 m 4 n 4)(2 3 m 3 n 6)(1 6 m 4 n 4)
Multiply a Polynomial by a Monomial
In the following exercises, multiply.
7(10−x)7(10−x)7(10−x)
435.
a 2(a 2−9 a−36)a 2(a 2−9 a−36)a 2(a 2−9 a−36)
−5 y(125 y 3−1)−5 y(125 y 3−1)−5 y(125 y 3−1)
437.
(4 n−5)(2 n 3)(4 n−5)(2 n 3)(4 n−5)(2 n 3)
Multiply a Binomial by a Binomial
In the following exercises, multiply the binomials using:
ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.
(a+5)(a+2)(a+5)(a+2)(a+5)(a+2)
439.
(y−4)(y+12)(y−4)(y+12)(y−4)(y+12)
(3 x+1)(2 x−7)(3 x+1)(2 x−7)(3 x+1)(2 x−7)
441.
(6 p−11)(3 p−10)(6 p−11)(3 p−10)(6 p−11)(3 p−10)
In the following exercises, multiply the binomials. Use any method.
(n+8)(n+1)(n+8)(n+1)(n+8)(n+1)
443.
(k+6)(k−9)(k+6)(k−9)(k+6)(k−9)
(5 u−3)(u+8)(5 u−3)(u+8)(5 u−3)(u+8)
445.
(2 y−9)(5 y−7)(2 y−9)(5 y−7)(2 y−9)(5 y−7)
(p+4)(p+7)(p+4)(p+7)(p+4)(p+7)
447.
(x−8)(x+9)(x−8)(x+9)(x−8)(x+9)
(3 c+1)(9 c−4)(3 c+1)(9 c−4)(3 c+1)(9 c−4)
449.
(10 a−1)(3 a−3)(10 a−1)(3 a−3)(10 a−1)(3 a−3)
Multiply a Polynomial by a Polynomial
In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.
(x+1)(x 2−3 x−21)(x+1)(x 2−3 x−21)(x+1)(x 2−3 x−21)
451.
(5 b−2)(3 b 2+b−9)(5 b−2)(3 b 2+b−9)(5 b−2)(3 b 2+b−9)
In the following exercises, multiply. Use either method.
(m+6)(m 2−7 m−30)(m+6)(m 2−7 m−30)(m+6)(m 2−7 m−30)
453.
(4 y−1)(6 y 2−12 y+5)(4 y−1)(6 y 2−12 y+5)(4 y−1)(6 y 2−12 y+5)
Multiply Special Products
In the following exercises, square each binomial using the Binomial Squares Pattern.
(2 x−y)2(2 x−y)2(2 x−y)2
455.
(x+3 4)2(x+3 4)2(x+3 4)2
(8 p 3−3)2(8 p 3−3)2(8 p 3−3)2
457.
(5 p+7 q)2(5 p+7 q)2(5 p+7 q)2
In the following exercises, multiply each pair of conjugates using the Product of Conjugates.
(3 y+5)(3 y−5)(3 y+5)(3 y−5)(3 y+5)(3 y−5)
459.
(6 x+y)(6 x−y)(6 x+y)(6 x−y)(6 x+y)(6 x−y)
(a+2 3 b)(a−2 3 b)(a+2 3 b)(a−2 3 b)(a+2 3 b)(a−2 3 b)
461.
(12 x 3−7 y 2)(12 x 3+7 y 2)(12 x 3−7 y 2)(12 x 3+7 y 2)(12 x 3−7 y 2)(12 x 3+7 y 2)
(13 a 2−8 b 4)(13 a 2+8 b 4)(13 a 2−8 b 4)(13 a 2+8 b 4)(13 a 2−8 b 4)(13 a 2+8 b 4)
Divide Monomials
Divide Monomials
In the following exercises, divide the monomials.
463.
72 p 12÷8 p 3 72 p 12÷8 p 3 72 p 12÷8 p 3
−26 a 8÷(2 a 2)−26 a 8÷(2 a 2)−26 a 8÷(2 a 2)
465.
45 y 6−15 y 10 45 y 6−15 y 10 45 y 6−15 y 10
−30 x 8−36 x 9−30 x 8−36 x 9−30 x 8−36 x 9
467.
28 a 9 b 7 a 4 b 3 28 a 9 b 7 a 4 b 3 28 a 9 b 7 a 4 b 3
11 u 6 v 3 55 u 2 v 8 11 u 6 v 3 55 u 2 v 8 11 u 6 v 3 55 u 2 v 8
469.
(5 m 9 n 3)(8 m 3 n 2)(10 m n 4)(m 2 n 5)(5 m 9 n 3)(8 m 3 n 2)(10 m n 4)(m 2 n 5)(5 m 9 n 3)(8 m 3 n 2)(10 m n 4)(m 2 n 5)
(42 r 2 s 4)(54 r s 2)(6 r s 3)(9 s)(42 r 2 s 4)(54 r s 2)(6 r s 3)(9 s)(42 r 2 s 4)(54 r s 2)(6 r s 3)(9 s)
Divide a Polynomial by a Monomial
In the following exercises, divide each polynomial by the monomial
471.
(54 y 4−24 y 3)÷(−6 y 2)(54 y 4−24 y 3)÷(−6 y 2)(54 y 4−24 y 3)÷(−6 y 2)
63 x 3 y 2−99 x 2 y 3−45 x 4 y 3 9 x 2 y 2 63 x 3 y 2−99 x 2 y 3−45 x 4 y 3 9 x 2 y 2 63 x 3 y 2−99 x 2 y 3−45 x 4 y 3 9 x 2 y 2
473.
12 x 2+4 x−3−4 x 12 x 2+4 x−3−4 x 12 x 2+4 x−3−4 x
Divide Polynomials using Long Division
In the following exercises, divide each polynomial by the binomial.
(4 x 2−21 x−18)÷(x−6)(4 x 2−21 x−18)÷(x−6)(4 x 2−21 x−18)÷(x−6)
475.
(y 2+2 y+18)÷(y+5)(y 2+2 y+18)÷(y+5)(y 2+2 y+18)÷(y+5)
(n 3−2 n 2−6 n+27)÷(n+3)(n 3−2 n 2−6 n+27)÷(n+3)(n 3−2 n 2−6 n+27)÷(n+3)
477.
(a 3−1)÷(a+1)(a 3−1)÷(a+1)(a 3−1)÷(a+1)
Divide Polynomials using Synthetic Division
In the following exercises, use synthetic Division to find the quotient and remainder.
x 3−3 x 2−4 x+12 x 3−3 x 2−4 x+12 x 3−3 x 2−4 x+12 is divided by x+2 x+2 x+2
479.
2 x 3−11 x 2+11 x+12 2 x 3−11 x 2+11 x+12 2 x 3−11 x 2+11 x+12 is divided by x−3 x−3 x−3
x 4+x 2+6 x−10 x 4+x 2+6 x−10 x 4+x 2+6 x−10 is divided by x+2 x+2 x+2
Divide Polynomial Functions
In the following exercises, divide.
481.
For functions f(x)=x 2−15 x+54 f(x)=x 2−15 x+54 f(x)=x 2−15 x+54 and g(x)=x−9,g(x)=x−9,g(x)=x−9, find ⓐ(f g)(x)(f g)(x)(f g)(x)
ⓑ(f g)(−2)(f g)(−2)(f g)(−2)
For functions f(x)=x 3+x 2−7 x+2 f(x)=x 3+x 2−7 x+2 f(x)=x 3+x 2−7 x+2 and g(x)=x−2,g(x)=x−2,g(x)=x−2, find ⓐ(f g)(x)(f g)(x)(f g)(x)
ⓑ(f g)(3)(f g)(3)(f g)(3)
Use the Remainder and Factor Theorem
In the following exercises, use the Remainder Theorem to find the remainder.
483.
f(x)=x 3−4 x−9 f(x)=x 3−4 x−9 f(x)=x 3−4 x−9 is divided by x+2 x+2 x+2
f(x)=2 x 3−6 x−24 f(x)=2 x 3−6 x−24 f(x)=2 x 3−6 x−24 divided by x−3 x−3 x−3
In the following exercises, use the Factor Theorem to determine if x−c x−c x−c is a factor of the polynomial function.
485.
Determine whether x−2 x−2 x−2 is a factor of x 3−7 x 2+7 x−6 x 3−7 x 2+7 x−6 x 3−7 x 2+7 x−6.
Determine whether x−3 x−3 x−3 is a factor of x 3−7 x 2+11 x+3 x 3−7 x 2+11 x+3 x 3−7 x 2+11 x+3.
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Authors: Lynn Marecek, Andrea Honeycutt Mathis
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Book title: Intermediate Algebra 2e
Publication date: May 6, 2020
Location: Houston, Texas
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