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![\"\"](\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\")
\n\n# 6.1: Fraction Rounding to the Nearest Half\n\nDifficulty Level: At Grade Created by: CK-12\nEstimated2 minsto complete\n%\nProgress\nPractice Fraction Rounding to the Nearest Half\n\nMEMORY METER\nThis indicates how strong in your memory this concept is\nProgress\nEstimated2 minsto complete\n%\nEstimated2 minsto complete\n%\nMEMORY METER\nThis indicates how strong in your memory this concept is\n\nHave you ever measured something? Was the measurement perfect or was it too small?\n\nWell, if you were working on a construction site, you would be doing a lot of measuring so rounding up or down if things weren't perfect would be very important. These part measurements are also fractions, just like you have learned about in other Concepts.\n\nThink about how you would round: three - fourths, one - sixth or five - tenths as you work through this Concept.\n\nIn this Concept, you will learn how to round a fraction to the nearest half.\n\n### Guidance\n\nWe use fractions in everyday life all the time. Remember that when we talk about a fraction, we are talking about a part of a whole. Often times, we need to use an exact fraction, but sometimes, we can use an estimate. If you think back to our earlier work on estimation, you will remember that an estimate is an approximate value that makes sense or is reasonable given the problem.\n\nWhat fraction does this picture represent?\n\nIf we wanted to be exact about this fraction, we could say that there are 1220\\begin{align*}\\frac{12}{20}\\end{align*} shaded boxes. However, it might be simpler to say that about half of the boxes are shaded.\n\nWe call this rounding to the nearest half.\n\nHow do we round to the nearest half?\n\nTo round a fraction to the nearest half, we need to think in terms of halves. We often think in terms of wholes, so this is definitely a change in our thinking. There are three main values to round to when we round a fraction to the nearest half.\n\nThe first is zero. We can think of 0 as 02\\begin{align*}\\frac{0}{2}\\end{align*}, or zero halves. The second value is 12\\begin{align*}\\frac{1}{2}\\end{align*}, or one half. The third value is 1, which can be thought of as 22\\begin{align*}\\frac{2}{2}\\end{align*}, or two halves. When rounding to the nearest half, we round the fraction to whichever half the fraction is closest to on the number line 0, 12\\begin{align*}\\frac{1}{2}\\end{align*}, or 1. If a fraction is equally close to two different halves, we round the fraction up.\n\n56\\begin{align*}\\frac{5}{6}\\end{align*}\n\nTo figure out which value five-sixths is closest to, we must first think in terms of sixths. Since the denominator is six, that means that the whole is divided into six parts. The fraction 06\\begin{align*}\\frac{0}{6}\\end{align*} would be the value of zero, 36\\begin{align*}\\frac{3}{6}\\end{align*} would be the value of 12\\begin{align*}\\frac{1}{2}\\end{align*}, and 66\\begin{align*}\\frac{6}{6}\\end{align*} is the same as 1. The fraction 56\\begin{align*}\\frac{5}{6}\\end{align*} is closest to 66\\begin{align*}\\frac{6}{6}\\end{align*}, so rounding to the nearest half would be rounding to 1.\n\nTry a few of these on your own. Round each fraction to the nearest half.\n\n#### Example A\n\n15\\begin{align*}\\frac{1}{5}\\end{align*}\n\nSolution: 0\n\n#### Example B\n\n38\\begin{align*}\\frac{3}{8}\\end{align*}\n\nSolution:12\\begin{align*}\\frac{1}{2}\\end{align*}\n\n#### Example C\n\n79\\begin{align*}\\frac{7}{9}\\end{align*}\n\nSolution:1\\begin{align*}1\\end{align*}\n\nNow let's go back and think about those measurements from the very beginning of the Concept.\n\nThink about how you would round: three - fourths, one - sixth or five - tenths as you work through this Concept. We can round three - fourths to 1. We can round one - sixth down to 0. We can round five - tenths to one - half.\n\n### Vocabulary\n\nHere are the vocabulary words in this Concept.\n\nFraction\na part of a whole written with a fraction bar, a numerator and a denominator.\nEstimate\nto find an approximate answer that is reasonable and makes sense given the problem.\n\n### Guided Practice\n\nHere is one for you to try on your own.\n\nJessica discovered that 49\\begin{align*}\\frac{4}{9}\\end{align*} of a pan of brownies had been eaten. Is the amount of brownies left closer to one - half or one whole?\n\nIf 49\\begin{align*}\\frac{4}{9}\\end{align*} of the pan had been eaten, then that means that 59\\begin{align*}\\frac{5}{9}\\end{align*} of the pan had not been eaten. This is closer to one - half of the pan of brownies.\n\n### Video Review\n\nHere are videos for review.\n\nEstimating with fractions - This video is a secondary skill to rounding fractions. It involves estimating with fractions.\n\n### Practice\n\nDirections: Round each fraction to the nearest half.\n\n1. 215\\begin{align*}\\frac{2}{15}\\end{align*}\n\n2. 17\\begin{align*}\\frac{1}{7}\\end{align*}\n\n3. 89\\begin{align*}\\frac{8}{9}\\end{align*}\n\n4. 715\\begin{align*}\\frac{7}{15}\\end{align*}\n\n5. 613\\begin{align*}\\frac{6}{13}\\end{align*}\n\n6. 1011\\begin{align*}\\frac{10}{11}\\end{align*}\n\n7. 78\\begin{align*}\\frac{7}{8}\\end{align*}\n\n8. 47\\begin{align*}\\frac{4}{7}\\end{align*}\n\n9. 37\\begin{align*}\\frac{3}{7}\\end{align*}\n\n10. 119\\begin{align*}\\frac{1}{19}\\end{align*}\n\n11. 210\\begin{align*}\\frac{2}{10}\\end{align*}\n\n12. 45\\begin{align*}\\frac{4}{5}\\end{align*}\n\n13. 23\\begin{align*}\\frac{2}{3}\\end{align*}\n\n14. 211\\begin{align*}\\frac{2}{11}\\end{align*}\n\n15. 19\\begin{align*}\\frac{1}{9}\\end{align*}\n\n### Notes/Highlights Having trouble? Report an issue.\n\nColor Highlighted Text Notes\n\n### Vocabulary Language: English\n\nEstimate\n\nTo estimate is to find an approximate answer that is reasonable or makes sense given the problem.\n\nfraction\n\nA fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.\n\nShow Hide Details\nDescription\nDifficulty Level:\nAuthors:\nTags:\nSubjects:<|endoftext|>"},"score":{"kind":"number","value":4.75,"string":"4.75"}}},{"rowIdx":1062,"cells":{"token_count":{"kind":"number","value":2072,"string":"2,072"},"text":{"kind":"string","value":"TANGENT, COTANGENT, COSECANT, AND SECANT FUNCTIONS\n\nExample 1:\n\nGiven , do the following:\n\na. State the EXACT period.\n\nSince b = 2, the period is /2. This means that the representative picture is /2 units in length.\n\nb. Sketch the function on the interval by hand.\n\n(1) Keep in mind the graph and characteristics of the basic cotangent function!\n\n(2) Mark off a distance along the x-axis starting at the origin to represent the period /2\n\n(3) Divide this period into four equal intervals. Each will be of length /2 4 = /2 1/4 = /8\n\n(4) Draw dashed vertical lines at the beginning and end of the period to represent the vertical asymptotes (do not draw the y-axis as a dashed line).\n\n(5) Create the appropriate representative picture by using three points marking the intervals.\n\nNote that the x-values are integer multiples of /8 and the y-values must be calculated using the x-values. Use a calculator!\n\nSpecifically, the points will be (/8, 2), (2/8 = /4, 0), and (3/8, 2).\n\n(6) Connect the points to form the representative picture of the cotangent function.\n\n(7) Copy several more cycles in the same manner to the right and left of the representative picture on the interval\n\n1. Please note the concavities of the graph of the cotangent function.\n\n2. Also, each branch continues to head toward the vertical asymptotes without ever \"getting there.\" Do not draw the branches parallel to the asymptotes.\n\nNote that the units along the x-axis are DIFFERENT from the units along the y-axis! As long as you place numbers along your axes it does not matter \"how long\" your units are!\n\nc. Find the EXACT equations of the two vertical asymptotes that the representative picture approaches.\n\nThe vertical asymptotes occur at the beginning and end of the representative picture. Knowing the period and the location of the representative picture, the EXACT equations of the two vertical asymptotes are\n\nx = 0 and x = /2\n\nExample 2:\n\nGiven is the function , do the following:\n\na. State the period.\n\nSince b = /4, the period is as follows:\n\nThis means that the representative picture is 4 units in length.\n\nb. Sketch the function on the interval [8, 8] by hand.\n\n(1) Keep in mind the graph and characteristics of the basic tangent function!\n\n(2) Mark off a distance along the x-axis starting at the origin to represent the period 4. Note that the representative picture of the tangent function is bisected by the y-axis!\n\n(3) Divide this period into four equal intervals. Each will be of length 1.\n\n(4) Draw dashed vertical lines at the beginning and end of the period to represent the vertical asymptotes (do not draw the y-axis as a dashed line).\n\n(5) Create the appropriate representative picture by using three points marking the intervals.\n\nNote that the x-values are integer multiples of 1 and the y-values must be calculated using the x-values. Use a calculator!\n\nSpecifically, the points will be (1, 2), (0, 0), and (1,2).\n\n(6) Connect the points to form the representative picture of the tangent function.\n\n(7) Copy several more cycles in the same manner to the right and left of the representative picture on the interval [8, 8] .\n\n1. Please note the concavities of the graph of the tangent function.\n\n2. Also, each branch continues to head toward the vertical asymptotes without ever \"getting there.\" Do not draw the branches parallel to the asymptotes.\n\nNote that the units along the x-axis are DIFFERENT from the units along the y-axis! As long as you place numbers along your axes it does not matter \"how long\" your units are!\n\nc. Find the EXACT equations of the two vertical asymptotes that the representative picture approaches.\n\nThe vertical asymptotes occur at the beginning and end of the representative picture. Knowing the period and the location of the representative picture, the EXACT equations of the two vertical asymptotes are\n\nx = 2 and x = 2\n\nExample 3:\n\nGiven , do the following:\n\na. State the period.\n\nSince b = /4, the period of our cotangent function is calculated as follows:\n\nThis means that the representative picture 4 units in length.\n\nb. Sketch the function on the interval by hand.\n\n(1) Keep in mind the graph and characteristics of the basic cotangent function!\n\n(2) Mark off a distance along the x-axis starting at the origin to represent the period 4.\n\n(3) Divide this period into four equal intervals. Each will be of length 1.\n\n(4) Draw dashed vertical lines at the beginning and end of the period to represent the vertical asymptotes (do not draw the y-axis as a dashed line).\n\n(5) Create the appropriate representative picture by using three points marking the intervals.\n\nNote that the x-values are integer multiples of 1 and the y-values must be calculated using the x-values. Use a calculator!\n\nSpecifically, the points will be (1, 1), (2, 0), and (3,1).\n\n(6) Connect the points to form the representative picture of the tangent function.\n\n(7) Copy several more cycles in the same manner to the right and left of the representative picture on the interval\n\n1. Please note the concavities of the graph of the cotangent function.\n\n2. Also, each branch continues to head toward the vertical asymptotes without ever \"getting there.\" Do not draw the branches parallel to the asymptotes.\n\nNote that the units along the x-axis are DIFFERENT from the units along the y-axis! As long as you place numbers along your axes it does not matter \"how long\" your units are!\n\nc. Find the EXACT equations of the two vertical asymptotes that the representative picture approaches.\n\nThe vertical asymptotes occur at the beginning and end of the representative picture. Knowing the period and the location of the representative picture, the EXACT equations of the two vertical asymptotes are\n\nx = 0 and x = 4\n\nExample 4:\n\nGiven is the function , do the following:\n\na. State the period.\n\nSince b = 1/2, the period of our tangent function is calculated as follows:\n\nThis means that the representative picture is 2 units in length.\n\nb. Sketch the function on the interval by hand.\n\n(1) Keep in mind the graph and characteristics of the basic tangent function!\n\n(2) Mark off a distance along the x-axis starting at the origin to represent the period 2. Note that the representative picture of the tangent function is bisected by the y-axis!\n\n(3) Divide this period into four equal intervals. Each will be of length 24 = /2.\n\n(4) Draw dashed vertical lines at the beginning and end of the period to represent the vertical asymptotes (do not draw the y-axis as a dashed line).\n\n(5) Create the appropriate representative picture by using three points marking the intervals.\n\nNote that the x-values are integer multiples of /2 and the y-values must be calculated using the x-values. Use a calculator!\n\nSpecifically, the points will be (/2, 1), (0, 0), and (/2, 1).\n\n(6) Connect the points to form the representative picture of the tangent function.\n\n(7) Copy several more cycles in the same manner to the right and left of the representative picture on the interval\n\n1. Please note the concavities of the graph of the tangent function.\n\n2. Also, each branch continues to head toward the vertical asymptotes without ever \"getting there.\" Do not draw the branches parallel to the asymptotes.\n\nNote that the units along the x-axis are DIFFERENT from the units along the y-axis! As long as you place numbers along your axes it does not matter \"how long\" your units are!\n\nc. Find the EXACT equations of the two vertical asymptotes that the representative picture approaches.\n\nThe vertical asymptotes occur at the beginning and end of the representative picture. Knowing the period and the location of the representative picture, the EXACT equations of the two vertical asymptotes are\n\nx = and x =\n\nExample 5:\n\nUse Desmos at https://www.desmos.com/calculator to draw the graph of .\n\nRequired Graph Characteristics:\n\n1. Show the representative picture and copies of one to the left of it and two to the right of it.\n2. Place numbers on both both axes.\n3. The x-axis numbers must contain . The beginning/ending point of each interval in each copy of the representative picture must show numbers.\n4. The y-axis numbers must show the appropriate values for the x-axis numbers used in (3).\n5. Do not make the y-axis too large. We need to see the concavities clearly!\n\nYou can find instructions for Desmos at http://profstewartmath.com/Math127/A_CONTENTS/desmos.htm\n\nSince b = 2, the period is /2. This means that the representative picture is /2 units in length. This knowledge will allow us to find the required x-values.\n\nConsidering the value a = 2, the y-axis can be between 6 and 6 in the Desmos \"Graph Settings.\"\n\nNOTE:\n\nSometimes, we have to try different settings in the Desmos \"Graph Settings\" window before the required x-axis numbers show up. Additionally, when we export the image, we need to determine which size best shows all required characteristics.\n\nIn the graph above, the Desmos \"Graph Settings\" for the x-axis are between and and the Step is /8.\n\nThe \"Large Square\" size was used for Desmos image export. Incidentally, the other settings refused to show all of the x-axis numbers.<|endoftext|>"},"score":{"kind":"number","value":4.53125,"string":"4.53125"}}},{"rowIdx":1063,"cells":{"token_count":{"kind":"number","value":867,"string":"867"},"text":{"kind":"string","value":"Mission: (Project Description)\nWithin this project, we will bring into question the “skin” of an object, and how alterations to it, alters our own perceptions of the object. Many artists, such as, Meret Oppenheim, Lucas Samaras, and Joseph Beuys, have pursued this concept creating works that question or negate an objects original function. The simple manipulation of a surface can allow strong conceptual narratives to develop, speaking to a greater importance beyond the original object.\nTransform: v. 1 make a thorough or dramatic change in the form, appearance, character, etc.\n- Bring to class three (3 dimensional) objects.\n- Object should not be smaller than 12 inches or larger than 24 inches.\n- A class discussion of objects will occur.\n- Consider the objects conceptual importance relative to its potential transformation.\n- Select one object from the initial three. Select the object with a strong visual presence.\n- Students should select one material in which to cover the chosen object. This material should be an antithesis to the original object, or draw into question the function or accepted meaning of the object.\n- Repetition and a obsessive development of the surface is a critical component of developing this project. Choose a material that will draw emphasis to this concept.\n- Objects must be completely covered by the chosen material. Students should focus on the obsessive nature involved in covering the objects.\nUse the vocabulary on the previous list along with other elements and Principles of Design to describe and interpret your work. The writing assignment should include your critique on the content and context and how these factors contribute to the concept of the work you have created. A general description of formal qualities should be addressed as they pertain to the concept, yet a stronger focus on content, context, and of course the idea (Concept) will foster a stronger discussion.\n- Look at the work analytically.\n- Remove yourself from the work to create new perspectives.\n- Be attentive to your gut reactions to the work.\n- Make notes of your observations” as reference. (In your sketch book)\n- Insert an image or sketch of your work within the word or pages document.\n- Content: Three to four paragraphs of text should be written before adding images.\n- Note: A paragraph is made up of three to five sentences.\nTools and Materials:\n- Eye Protection/Safety Glasses\n- Sketch Book\n- Utility Knife\n- Hobby knife\n- Dust Mask/Particulate Respirator\n- Safety Glasses\n- Focus Object\n- Materials for skin\n- Mixed media\n- Adhesives or other fastening materials\n- Rubber or Latex gloves (optional)\nCreate a label for your project with these specifics:\n- Dimensions: HWD\n- Year Complete:\nTape your label to your project. Use masking tape so that it will not harm your project.\n- Method: Planning, Cutting Safely, Building techniques and strategies\n- Material Characteristics:\n- Three-Dimensional surfaces and skin transformations\n- Aesthetic qualities\n- Sensibility to form: Good design, clean aesthetics\n- Integrity of the artist to their craft\n- Concept from objects\n- Work ethic: Participation in and out of class\n- Scheduling: Students stay on task, project finished by deadline, critique.\n- Craftsmanship and aesthetics.\n- Attention to detail and craft involved in the development of the object’s “skin” will be taken into consideration during grading and critique. Therefore, ample time should be devoted to the completion of the project.\nThis process may use chemicals and materials that might be dangerous for some people. Read the MSDS (Material Safety Data Sheet) for questionable materials. A respirator should be used when materials off gas or fume. A dust mask may be used rather than a respirator. The respirator will provide better protection.\n- Meret Oppenheim\n- Tom Friedman\n- Anne Hamilton\n- Liza Lou\n- Do Ho Suh\n- Gabrielle Kanter\n- Nick Cave\n- Damien Hirst\n- Nicola Bolla\n- Joseph Beys\n- Lucas Samaras\n- Janine Antoni<|endoftext|>"},"score":{"kind":"number","value":3.796875,"string":"3.796875"}}},{"rowIdx":1064,"cells":{"token_count":{"kind":"number","value":634,"string":"634"},"text":{"kind":"string","value":"# Videos\n\nChoose from over 80 mathematical videos.\n\nOur team of top teacher mathematicians have put together all the important topics and explain them with the types of examples that you would find in a normal textbook or exam.\n• #### Solving Inequalities\n\nThis video explains linear and quadratic inequalities and how they can be solved both algebraically and graphically. It includes information on inequalities in which the modulus symbol is used.\nLength: 26 minutes\n\nThis video is about the solution of quadratic equations. These take the form ax2+bx+c = 0. We will look at four methods: factorisation, completing the square, using a formula, and solution using graphs.\nLength: 50 minutes\n• #### Solving Trigonometric Equations\n\nTrigonometric equations. The strategy we adopt in solving trigonometric equations is to find one solution using knowledge of commonly occurring angles and then use the symmetries in the graphs of the trigonometric functions to deduce additional solutions.\nLength: 44 minutes\n• #### Substitution & Formulae\n\nIn mathematics, engineering and science, formulae are used to relate physical quantities to each other. They provide rules so that if we know the values of certain quantities, the values of others can be calculated. This video looks at several formulae and how they are used.\nLength: 20 minutes\n• #### Surds\n\nRoots and powers are closely related, but some roots can be written as whole numbers. These are known as Surds. This video looks at how you can manipulate surds, simplify expressions involving surds and rationalising expressions involving surds.\nLength: 33 minutes\n• #### Tangents and Normals\n\nTangents and normals. This unit explains how differentiation can be used to calculate the equations of the tangent and normal to a curve. The tangent is a straight line which just touches the curve at a given point. The normal is a straight line through the point perpendicular to the tangent.\nLength: 31 minutes\n\nThe addition formulae. There are six so-called addition formulae often needed in the solution of trigonometric problems. In this unit we start with one and derive a second. Then we take another one as given and derive a second one from that. Finally we use these four to help us derive the final two.\nLength: 27 minutes\n• #### The Chain Rule\n\nThe chain rule. This is a special rule that may be used to differentiate a composite function (that is, a function of another function).\nLength: 24 minutes\n• #### The Double Angle Formulae\n\nThe double angle formulae. Double angle formulae are so called because they involve trigonometric functions of double angles e.g. sin 2A, cos 2A and tan 2A.\nLength: 26 minutes\n• #### The Geometry of a Circle\n\nThe geometry of a circle. In this unit the equation of a circle is found from the co-ordinates of its centre and its radius. There are two different forms of the equation. The unit also covers some examples involving tangents to circles.\nLength: 40 minutes<|endoftext|>"},"score":{"kind":"number","value":4.5,"string":"4.5"}}},{"rowIdx":1065,"cells":{"token_count":{"kind":"number","value":226,"string":"226"},"text":{"kind":"string","value":"Write a report about solar power and how it produces electricity to make an efficient house. The report should have around 500 words with pictures and references included.\nThis material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.\nThe Sun radiates enormous amount of energy in the form of heat and light energy which is known as solar energy. The solar energy from the sun passes through the atmosphere and some of this energy is reflected back into the space. Only a small part of this solar energy reaches the earth’s atmosphere and nearly half of it is absorbed by the gases present in the atmosphere while passing through the atmosphere and the rest reaches the earth’s surface. The earth’s surface receives approximately 1000 watts of energy per square meter. We can make use of this energy to produce electricity....\nThis is only a preview of the solution. Please use the purchase button to see the entire solution<|endoftext|>"},"score":{"kind":"number","value":3.75,"string":"3.75"}}},{"rowIdx":1066,"cells":{"token_count":{"kind":"number","value":1291,"string":"1,291"},"text":{"kind":"string","value":"How Do You Solve for the Number of Years in the Compound Interest Formula?\n\nIn another MathFAQ, I looked at how you can find the rate in the compound interest formula. Now let’s look at an example where we solve for the number of years n. This problem is different because what we are looking for appears in a power.\n\nProblem Suppose \\$5000 is deposited in an account that earns 2% compound interest that is done annually. In how many years will there be \\$6000 in the account.\n\nSolution This problem requires the use of the compound interest formula,\n\nThis formula applies when interest is earned on an annual basis and the interest is earned once a year.\n\nLet’s look at the quantities in the problem statement:\n\n• \\$5000 is deposited in an account > P = 5000\n• that earns 2% compound interest that is done annually > r = 0.02\n• Will there be \\$6000 in the account > A = 6000\n\nPutting these values into the formula above gives us\n\nUnlike other problems where we solve for P or r, here we need to solve for the power in the right hand side, n. Solving for a value in the power requires the property of logarithms, log(yx) = x logy. It allows us to move the n in the power and change it to a multiplier. But before we can apply this property, we isolate the factor containing the n:\n\nNow take the logarithm of both sides of the equation:\n\nThis gives us\n\nor n ≈ 9.21 years.\n\nIn WolframAlpha, we could evaluate the logs as follows.\n\nHow Do You Solve For The Rate In The Compound Interest Formula?\n\nProblems that ask you to solve for the rate r in the compound interest formula require the use of roots or creative use of exponents. Let’s look at an example.\n\nProblem Suppose 5000 dollars is deposited in an account that earns compound interest that is done annually. If there is 7000 dollars in the account after 2 years, what is the annual interest rate?\n\nSolution The easiest way to approach this problem is to use the compound interest formula,\n\nThis formula applies when interest is earned on an annual basis and the interest is earned once a year.\n\nLet’s look at the quantities in the problem statement:\n\n• 5000 dollars is deposited in an account > P = 5000\n• If there is 7000 dollars in the account after 2 years > A = 7000 and n = 2\n\nPutting these values into the formula above gives us\n\nWe need to find the annual interest rate r. Since the r is hidden in the parentheses, we start by isolating the parentheses.\n\nTo get at the r, we need to remove the square on the parentheses.\n\nUsing a calculator to do the square root, we get r ≈ 0.183 or 18.3%.\n\nAlthough most calculators have a square root key, when removing powers it is often useful to raise both sides to a power. For instance, we could remove the square by raising both sides to the ½ power.\n\nWhen you raise a power to another power, you multiply the exponents 2 ∙ ½ = 1. The right side simply becomes 1 + r. Now we can solve for r:\n\nUsing the power key on your calculator gives the same answer as before. Make sure the 1/2 power is entirely in the power. You can make sure this happens using parentheses: (7000/5000)^(1/2)-1.\n\nNow what if the interest is earned over six years instead of two years? Instead of a square on the parentheses we now have a sixth power.\n\nTo solve for r in this equation, we follow similar steps.\n\nThe root can be computed on a graphing calculator using the MATH button or put into WolframAlpha:\n\nEither method gives r ≈ 0.577 or 5.77%. Notice that the annual interest is lower when it is earned over a longer period of time.\n\nIf we use a 1/6 power to solve for r, we would carry out the steps below:\n\nUsing a 1/6 power on your calculator gives the same answer as above.\n\nHow Do You Find Compound Interest Future Value In Google Sheets?\n\nSpreadsheets have several built in functions for working with compound interest and annuities. To use these functions, we’ll start with a standard sheet.\n\nThis worksheet contains the variables used throughout Chapter 8. These variables correspond to these letter used in the text.\n\n• Number of periods is n\n• Annual interest rate is r\n• Payment is R\n• Present value is P\n• Future value is A\n• Periods per year is m\n\nValues given in a problem will be entered in column B. Values calculated by the spreadsheet will be entered in column C. We will also assume that amounts paid out are negative and amounts received are positive.\n\nWhat Is The Difference Between Unpaid Balance And Average Balance?\n\nTwo methods of calculating interest on a credit card statement are generally in use: the unpaid balance method and the average daily balance method. To look at the difference between the unpaid balance method and the average balance method, let’s look at the credit card statement below.\n\nIn this statement, the average daily balance method is used to calculate the interest.\n\nIf you calculate with the average balance method, it looks like this.\n\nYou can click on this to make it larger.\n\nHere is how interest is calculated via the unpaid balance method. I noticed that I got a different number for the new balance in the two different classes. I think one of you gave me a bogus number somewhere. This picture is from the MW class and Jim in the TTh class came up with 2149.86. The key is to make sure you add and subtract the numbers appropriately to get the new balance.\n\nYou can click on this to make it larger.\n\nNotice that these look very similar…only how the balance is calculated is different. Once you get that balance, the interest is calculated with I = P r t.\n\nHow Does Add On Interest Work?\n\nWhen you have an add-on loan, the interest on the loan is a percentage of the original amount. Payments are calculated on the loan amount plus the interest (add-on!)\n\nLet’s look at an example.<|endoftext|>"},"score":{"kind":"number","value":4.71875,"string":"4.71875"}}},{"rowIdx":1067,"cells":{"token_count":{"kind":"number","value":780,"string":"780"},"text":{"kind":"string","value":"The Hubble Space Telescope shows the luminous explosion of supernova 1987a within the Large Magellanic Cloud, the neighboring galaxy to the Milky Way.\nCredit: R. Kirshner (Harvard-Smithsonian Center for Astrophysics and Gordon and Betty Moore Foundation), and M. Mutchler and R. Avila (STScI)/NASA/ESA\nOn Feb. 23, 1987, the light from a giant, exploding star reached Earth. The event, which took place in the Large Magellanic Cloud, a small galaxy 168,000 light-years away that circles our Milky Way, was the closest supernova to occur in nearly 400 years, and the first since the invention of modern telescopes.\nMore than 30 years later, a team has used X-ray observations and physical simulations to accurately measure the temperature of elements in the gas around the dead star for the first time. As the hyperfast shockwaves from the heart of the supernova slam into atoms in the surrounding gas, they heat those atoms to hundreds of millions of degrees Fahrenheit.\nGoing out with a bang\nWhen giant stars reach old age, their outer layers slough off and cool into enormous, remnant structures around the star. The star’s core creates a spectacular supernova blast, leaving behind either an ultradense neutron star or a black hole. Shock waves from the explosion travel out at one-tenth the speed of light and hit the surrounding gas, heating it up and making it shine in bright X-rays.\nNASA’s space-based Chandra X-ray telescope has been monitoring the emissions from supernova 1987A, as the dead star is known, since the telescope was launched 20 years ago. In that time, supernova 1987A has surprised researchers time and again, David Burrows, a physicist at The Pennsylvania State University and co-author of the new paper, told Live Science. “One big surprise was the discovery of a series of three rings around it,” he said.\nSince around 1997, the shock wave from supernova 1987A has been interacting with the innermost ring, called the equatorial ring, Burrows said. Using Chandra, he and his group have been monitoring the light created by the shock waves as they interact with the equatorial ring in order to learn how the gas and dust in the ring heats up. They wanted to figure out the temperatures of different elements in the material as the shock front engulfs it, a long-standing issue that has been difficult to determine accurately.\nTo help in the measurements, the team created detailed 3D computer simulations of the supernova that disentangled the many processes at play — the speed of the shock wave, the temperature of the gas and the resolution limits of Chandra’s instruments. From there, they were able to pin down the temperature of a wide range of elements, from light atoms like nitrogen and oxygen, all the way up to heavy ones like silicon and iron, said Burrows. The temperatures ranged from millions to hundreds of millions of degrees.\nThe findings provide important insights into the dynamics of supernova 1987A and help test models of a specific type of shock front, Jacco Vink, a high-energy astrophysicist at the University of Amsterdam in the Netherlands, who was not involved in the work, told Live Science.\nBecause the charged particles from the blast are not hitting atoms in the surrounding gas, but rather scatter the gas atoms using electric and magnetic fields, this shock is known as a collisionless shock, he added. The process is common throughout the universe, and so understanding it better would help researchers with other phenomena, such as the solar wind’s interaction with interstellar material and cosmological simulations about the formation of large-scale structure in the universe.\nOriginally published on Live Science.<|endoftext|>"},"score":{"kind":"number","value":4.09375,"string":"4.09375"}}},{"rowIdx":1068,"cells":{"token_count":{"kind":"number","value":704,"string":"704"},"text":{"kind":"string","value":"Unlike domestic dogs, female coyotes only give birth once a year, in early summer. They can hybridize, though, creating offspring called ‘coydogs’.\nCoyotes live throughout North America, from the humid forests of Panama to the treeless tundra regions of Canada and Alaska. They are most common in the unpopulated desert areas of the south-western United States and northern Mexico. As wolves have declined in numbers across North America, so the smaller and more adaptable coyote has become more numerous. Its range expanded markedly through the twentieth century and now coyotes can be encountered in suburban parts of New York and elsewhere, reaching the east coast only 30 years ago.\nThese dogs look a little like small grey wolves. They are typically found alone. Coyotes may dig their own den or enlarge the burrow of another animal. They are primarily nocturnal, being most active around dawn and dusk, but they do sometimes hunt during the day. They can reach speeds of up to 64 kmh (40 mph) when chasing swift jackrabbits and other prey.\nThese dogs are adaptable opportunistic feeders, and they are able to survive in farmland and suburban regions. They are increasingly coming into conflict with human communities expanding into the desert, which see them as pests. These wild dogs are unpopular with farmers as they attack sheep, although they also prey more commonly on jackrabbits, which can harm the grazing.\nCoyote fur varies from grey to yellow. The head and legs may have reddish hair on them. A black line runs along the back. The bushy tail is about half as long as the rest of the body. Coyotes are much smaller than wolves, but significantly larger than foxes.\nCoyote pairs mate in late winter, and a litter of 5 to 10 young is born after a gestation of 63 to 65 days. The male brings all the food for the female and young at first, but later both parents hunt for food.\nDistribution: Occurs throughout North America, from Alaska southwards to Mexico. Present in all US mainland states. Only absent from north-central and western Canada.\nHabitat: Prairies, desert, open woodland, forest and tundra.\nWeight: 7 - 21 kg (15 - 46 lb); males are heavier, as are those of northern races.\nLength: 105 - 140 cm (41 - 55 in); up to 87 cm (34 in) tall.\nMaturity: 12 months.\nGestation Period: 60 - 63 days.\nBreeding: 1-19, average 6; weaned by 35 days.\nFood: Omnivorous, opportunistic hunters, mainly of small mammals, larger insects and birds, as well as scavengers and consumers of vegetable matter.\nLifespan: 11 - 12 years.\nThe ears are broad and, when held erect, help pinpoint sounds, allowing hunting under cover of darkness.\nThe snout is quite narrow, with black nostrils.\nWatchful and alert, the eyes are green. Coyotes have excellent night vision.\nCoyotes from desert areas have reddish coats, while those inhabiting woodland areas are more grey.\nCoyotes live on their own or in pairs.\nHOWLING AT THE MOON\nCoyotes regularly howl and bark at night. This is also the time of day when they are most active.<|endoftext|>"},"score":{"kind":"number","value":4,"string":"4"}}},{"rowIdx":1069,"cells":{"token_count":{"kind":"number","value":8373,"string":"8,373"},"text":{"kind":"string","value":"Courses\n\n# NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev\n\n## Class 8 : NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev\n\nThe document NCERT Solutions(Part- 1)- Squares and Square Roots Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.\nAll you need of Class 8 at this link: Class 8\n\nQ.1. Find the perfect square numbers between\n(i) 30 and 40\n(ii) 50 and 60.\nSolution.\n(i) Since,\n⇒ 1 *1 = 1\n⇒ 2 * 2 = 4\n⇒ 3 * 3 = 9\n⇒ 4 * 4 = 16\n⇒ 5 * 5 = 25\n⇒ 6 * 6 = 36\n⇒ 7 * 7 = 49\nThus, 36 is a perfect square number between 30 and 40.\n\n(ii) Since, 7 * 7 = 49 and 8 * 8 = 64.\nIt means there is no perfect number between 49 and 64, and thus there is no perfect number between 50 and 60.\n\nQ.2. Can we say whether the following numbers are perfect squares? How do we know?\n\n(i) 1057\n(ii) 23453\n(iii) 7928\n(iv) 222222\n(v) 1069\n(vi) 2061\n\nWrite five numbers which you can decide by looking at their one’s digit that they are not square numbers.\nSolution.\n(i) 1057\n\n• The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9)\n• 1057 cannot be a square number.\n\n(ii) 23453\n\n• The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, and 9).\n• 23453 cannot be a square number.\n\n(iii) 7928\n\n• The ending digit is 8 (which is not one of 0, 1, 4, 5, 6 and 9).\n• 7928 cannot be a square number.\n\n(iv) 222222\n\n• The ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).\n• 222222 cannot be a square number.\n\n(v) 1069\n\n• The ending digit is 9.\n• It may or may not be a square number.\n\nAlso,\n\n• 30 * 30 = 900\n• 31 * 31 = 961\n• 32 * 32 = 1024\n• 33 * 33 = 1089\n\ni.e. No natural number between 1024 and 1089 is a square number.\n\n∴ 1069 cannot be a square number.\n\n(vi) 2061\n\n• The ending digit is 1\n• It may or may not be a square number.\n\n45 * 45 = 2025\n\nand 46 * 46 = 2116\n\ni.e. No natural number between 2025 and 2116 is a square number.\n\n∴ 2061 is not a square number.\n\nWe can write many numbers which do not end with 0, 1, 4, 5, 6 or 9. (i.e. which are not square number).\nFive such numbers can be:\n\n1234, 4312, 5678, 87543, 1002007.\n\nQ.3. Write five numbers that you cannot decide just by looking at their unit’s digit (or one’s place) whether they are square numbers or not.\nSolution. Any natural number ending in 0, 1, 4, 5, 6 or 9 can be or cannot be a square-number.\nFive such numbers are:\n\n56790, 3671, 2454, 76555, 69209\n\nProperty 1. If a number has 1 or 9 in the unit’s place, then its square ends in 1.\nExample: (1)= 1, (9)2 = 81, (11)2 = 121, (19)2 = 361, (21)2 = 441.\n\nQ.4. Which of 1232, 772, 822, 1612, 1092 would end with digit 1?\nSolution. The squares of those numbers end in 1 which ends in either 1 or 9.\n\n∴ The squares of 161 and 109 would end in 1.\n\nProperty 2. When a square number ends in 6, then the number whose square it will have 4 or 6 in its unit place.\n\nQ.5. Which of the following numbers would have digit 6 at unit place.\n(i) 19\n(ii) 242\n(iii) 262\n(iv) 362\n(v) 342\nSolution.\n(i) 192: Unit’s place digit = 9\n\n∴ 19would not have unit’s digit as 6.\n\n(ii) 242: Unit’s place digit = 4\n\n∴ 242 would have unit’s digit as 6.\n\n(iii) 262: Unit’s place digit = 6\n\n∴ 262 would have 6 as unit’s place.\n\n(iv) 362: Unit place digit = 6\n\n∴ 362 would end in 6.\n\n(v) 342: Since the unit place digit is 4\n\n∴ 342 would have unit place digit as 6.\n\nQ.6. What will be the “one’s digit” in the square of the following numbers?\n\n(i) 1234\n(ii) 26387\n(iii) 52698\n(iv) 99880\n(v) 21222\n(vi) 9106\nSolution.\n(i) Since Ending digit = 4 and 42 = 16\n\n∴ (1234)2 will have 6 as the one’s digit.\n\n(ii) Since Ending digit is 7 and 7= 49\n\n∴ (26387)2 will have 9 as the one’s digit.\n\n(iii) Since Ending digit is 8, and 82 = 64\n\n∴ (52692)2 will end in 4.\n\n(iv) Since Ending digit is 0.\n\n∴ (99880)2 will end in 0.\n\n(v) Since Ending digit = 2 and 22 = 4\n\n∴ Ending digit of (21222)2 is 4.\n\n(vi) ∵ 62 = 36\n\n∴ Ending digit of (9106)is 6.\n\nProperty 3. A square number can only have even number of zeros at the end.\n\nProperty 4. The squares of odd numbers are odd and the squares of even numbers are even.\n\nQ.7. The square of which of the following numbers would be an odd number/an even number? Why?\n\n(i) 727\n(ii) 158\n(iii) 269\n(iv) 1980\n\nSolution.\n\n(i) 727\n\nSince 727 is an odd number.\n\n∴ Its square is also an odd number.\n\n(ii) 158\n\nSince 158 is an even number.\n\n∴ Its square is also an even number.\n\n(iii) 269\n\nSince 269 is an odd number.\n\n∴ Its square is also an odd number.\n\n(iv) 1980\n\nSince 1980 is an even number.\n\n∴ Its square is also an even number.\n\nQ.8. What will be the number of zeros in the square of the following numbers?\n\n(i) 60\n(ii) 400\n\nSolution.\n\n(i) In 60, the number of zero is 1.\n\n∴ Its square will have 2 zeros.\n\n(ii) Since there are 2 zeroes in 400.\n\n∴ Its square will have 4 zeros.\n\nProperty 5. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.\n\nProperty 6. There are 2n non-perfect square numbers between the squares of the numbers n and n + 1.\n\nQ.9. How many natural numbers lie between 92 and 102? Between 112 and 122?\n\nSolution.\n\n(a) Between 92 and 102\n\nHere, n = 9 and n + 1 = 10\n\n∴ Natural number between 92 and 102 are (2 * n) or 2 * 9, i.e. 18.\n\n(b) Between 112 and 122\n\nHere, n = 11 and n + 1 = 12\n\n∴ Natural numbers between 11and 122 are (2 * n) or 2 *11, i.e. 22.\n\nQ.10. How many non-square numbers lie between the following pairs of numbers:\n\n(i) 100and 1012\n(ii) 902 and 912\n(iii) 10002 and 10012\n\nSolution.\n(i) Between 1002 and 1012\n\nHere, n = 100\n∴ n * 2 = 100 * 2 = 200\n\n∴ 200 non-square numbers lie between 1002 and 1012.\n\n(ii) Between 902 and 912\n\nHere, n = 90\n∴ 2 * n = 2 * 90 or 180\n\n∴ 180 non-square numbers lie between 90 and 91.\n\n(iii) Between 10002 and 10012\n\nHere, n = 1000\n∴ 2 * n = 2 * 1000 or 2000\n\n∴ 2000 non-square numbers lie between 10002 and 10012.\n\nProperty 7. The sum of first n odd natural numbers is n2.\nOR\nIf there is a square number, it has to be the sum of the successive odd numbers starting from 1.\n\nQ.11. Find whether each of the following numbers is a perfect square or not?\n\n(i) 121\n(ii) 55\n(iii) 81\n(iv) 49\n(v) 69\n\nSolution. Remember: If a natural number cannot be expressed as a sum of successive odd natural numbers starting from 1, then it is not a perfect square.\n\n(i) 121\n\nSince,\n\n121 – 1 = 120\n120 – 3 = 117\n117 – 5 = 112\n112 – 7 = 105\n105 – 9 = 96\n96 – 11 = 85\n\n85 – 13 = 72\n72 – 15 = 57\n57 – 17 = 40\n40 – 19 = 21\n21 – 21 = 0\n\ni.e. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. Thus, 121 is a perfect square.\n\n(ii) 55\n\nSince,\n\n55 – 1 = 54\n\n54 – 3 = 51\n\n51 – 5 = 46\n\n46 – 7 = 39\n\n39 – 9 = 30\n\n30 – 11 = 19\n\n19 – 13 = 6\n6 – 15 = –9\n\nSince, 55 cannot be expressed as the sum of successive odd numbers starting from 1. Thus, 55 is not a perfect square.\n\n(iii) 81\n\nSince,\n81 – 1 = 80\n80 – 3 = 77\n77 – 5 = 72\n72 – 7 = 65\n65 – 9 = 56\n56 – 11 = 45\n45 – 13 = 32\n32 – 15 = 17\n17 – 17 = 0\n\n∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17. Thus, 81 is a perfect square.\n\n(iv) 49\n\nSince,\n\n49 – 1 = 48\n\n48 – 3 = 45\n\n45 – 5 = 40\n\n40 – 7 = 33\n\n33 – 9 = 24\n\n24 – 11 = 13\n\n13 – 13 = 0\n\n∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13. Thus, 49 is a perfect square.\n\n(v) 69\n\nSince,\n69 – 1 = 68\n68 – 3 = 65\n65 – 5 = 60\n60 – 7 = 53\n53 – 9 = 44\n44 – 11 = 33\n33 – 13 = 20\n20 – 15 = 5\n5 – 17 = –12\n\n∴ 69 cannot be expressed as the sum of consecutive odd numbers starting from 1. Thus, 69 is not a perfect square.\n\nProperty 8. The square of an odd number can be expressed as the sum of two consecutive natural numbers.\n\nTry These\nQ1. Express the following as the sum of two consecutive integers.\n(i) 212\n(ii) 132\n(iii) 112\n(iv) 192\nSolution:\n(i) n = 21\n\nor 212 = 220 + 221 = 441\n\n(ii) n = 13\n\n∴ 132 = 85 + 84 = 169\n\nSimilarly,\n\n(iii) 112 = 60 + 61 = 121\n\n(iv) 19= 180 + 181 = 361\n\nQ2. Do you think the reverse is also true, i.e. is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer.\nSolution: No, it is not always true.\nExample:\n(i) 5 + 6 = 11, 11 is not a perfect square.\n(ii) 21 + 22 = 43, 43 is not a perfect square.\n\nProperty 10. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.\n\nExamples:\n\n92 – 82 = 81 – 64 = 17 = 9 + 8\n\n102 – 92 = 100 – 81 = 19 = 10 + 9\n\n15– 142 = 225 – 196 = 29 = 15 + 14\n\n1012 – 1002 = 10201 – 10000 = 201 = 101 + 100\n\nProperty 11. If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) X (n – 1) = n2 – 1.\n\nExample:\n\n10 * 12 = (11 – 1) * (11 + 1) = 112 – 1\n\n11 * 13 = (12 – 1) * (12 + 1) = 122 – 1\n\n25 * 27 = (26 – 1) * (26 + 1) = 262 – 1\n\nQ3. Write the square, making use of the above pattern.\n\n(i) 1111112\n(ii) 11111112\n\nSolution: Using above pattern we can write:\n\n(i) (111111)2 = 12345654321\n\n(ii) (1111111)2 = 1234567654321\n\nQ4. Can you find the square of the following numbers using the above pattern?\n(i) 66666672\n(ii) 666666672\nSolution: Using the above pattern, we can have:\n(i) 6666667= 44444448888889\n(ii) 666666672 = 4444444488888889\n\nExercise 6.1\nQ1. What will be the unit digit of the squares of the following numbers?\n(i) 81\n\n(ii) 272\n(iii) 799\n(iv) 3853\n(v) 1234\n(vi) 26387\n(vii) 52698\n(viii) 99880\n(ix) 12796\n(x) 55555\n\nSolution:\n\n(i) Since, 1 * 1 = 1\n\n∴ The unit’s digit of (81)2 will be 1.\n\n(ii) Since, 2 * 2 = 4\n\n∴ The unit’s digits of (272)will be 4.\n\n(iii) Since, 9 * 9 = 81\n\n∴ The unit’s digit of (799)2 will be 1.\n\n(iv) Since, 3 * 3 = 9\n\n∴ The unit’s digit of (3853)will be 9.\n\n(v) Since, 4 * 4 = 16\n\n∴ The unit’s digit of (1234)2 will be 6.\n\n(vi) Since 7 * 7 = 49\n\n∴ The unit’s digit of (26387)2 will be 9.\n\n(vii) Since, 8 * 8 = 64\n\n∴ The unit’s digit of (52698)2 will be 4.\n\n(viii) Since 0 * 0 = 0\n\n∴ The unit’s digit of (99880)2 will be 0.\n\n(ix) Since 6 * 6 = 36\n\n∴ The unit’s digit of (12796)2 will be 6.\n\n(x) Since, 5 * 5 = 25\n\n∴ The unit’s digit of (55555)will be 5.\n\nQ2. The following numbers are obviously not perfect squares. Give reason.\n(i) 1057\n\n(ii) 23453\n\n(iii) 7928\n\n(iv) 222222\n(v) 64000\n\n(vi) 89722\n(vii) 222000\n(viii) 505050\nSolution:\n(i) 1057\n\nSince, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)\n\n∴ 1057 is not a perfect square.\n\n(ii) 23453\n\nSince, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9).\n\n∴ 23453 is not a perfect square.\n\n(iii) 7928\n\nSince, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).\n\n∴ 7928 is not a perfect square.\n\n(iv) 222222\n\nSince, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).\n\n∴ 222222 is not a perfect square.\n\n(v) 64000\n\nSince the number of zeros is odd.\n\n∴ 64000 is not a perfect square.\n\n(vi) 89722\n\nSince, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).\n\n∴ 89722 is not a perfect square.\n\n(viii) 222000\n\nSince the number of zeros is odd.\n\n∴ 222000 is not a perfect square.\n\n(viii) 505050\n\nThe unit’s digit is odd zero.\n\n∴ 505050 can not be a perfect square.\n\nQ3. The squares of which of the following would be odd numbers?\n(i) 431\n(ii) 2826\n(iii) 7779\n(iv) 82004\n\nSolution: Since the square of an odd natural number is odd and that of an even number is an even number.\n\n(i) The square of 431 is an odd number.\n[∵ 431 is an odd number.]\n(ii) The square of 2826 is an even number.\n[∵ 2826 is an even number.]\n\n(iii) The square of 7779 is an odd number.\n[∵ 7779 is an odd number.]\n\n(iv) The square of 82004 is an even number.\n[∵ 82004 is an even number.]\n\nQ4. Observe the following pattern and find the missing digits.\n\n112 = 121\n\n1012 = 10201\n\n10012 = 1002001\n\n1000012 = 1 … 2 …1\n\n100000012 = …\nSolution: Observing the above pattern, we have:\n(i) (100001)2 = 10000200001\n\n(ii) (10000001)2 = 100000020000001\n\nQ5. Observe the following pattern and supply the missing number.\n\n112 = 121\n\n1012 = 10201\n\n101012 = 102030201\n\n10101012 = ………….\n\n………….= 10203040504030201\nSolution: Observing the above, we have:\n(i) (1010101)= 1020304030201\n\n(ii) 10203040504030201 = (101010101)2\n\nQ6. Using the given pattern, find the missing numbers.\n\n12 + 22 + 22 = 32\n\n22 + 32 + 62 = 72\n\n32 + 42 + 122 = 132\n\n42 + 52 + __2 = 212\n\n52 + __2 + 302 = 312\n\n62 + 72 + __2 = __2\n\nNote: To find pattern:\n\nThird number is related to first and second number. How?\n\nFourth number is related to third number. How?\nSolution: The missing numbers are:\n\n(i) 42 + 52 + 202 = 212\n\n(ii) 52 + 62 + 302 = 312\n\n(iii) 62 + 72 + 422 = 432\n\nQ7. Without adding, find the sum.\n(i) 1 + 3 + 5 + 7 + 9\n\n(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19\n\n(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23\nSolution:\n(i) The sum of first 5 odd numbers = 52 = 25\n\n(ii) The sum of first 10 odd numbers = 102 = 100\n\n(iii) The sum of first 12 odd numbers = 12= 144\n\nQ8. (i) Express 49 as the sum of 7 odd numbers.\n\n(ii) Express 121 as the sum of 11 odd numbers.\nSolution:\n(i) 49 =7= Sum of first 7 odd numbers\n\n= 1 + 3 + 5 + 7 + 9 + 11 + 13\n\n(ii) 121 = 112 = Sum of first 11 odd numbers\n\n= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\n\nQ9. How many numbers lie between squares of the following numbers?\n\n(i) 12 and 13\n(ii) 25 and 26\n(iii) 99 and 100\nSolution: Since between n2 and (n + 1)2, there are 2n non-square numbers.\n(i) Between 12and 132, there are 2 *12, i.e. 24 numbers.\n\n(ii) Between 252 and 262, there are 2 * 25, i.e. 50 numbers.\n\n(iii) Between 992 and 1002, there are 2 * 99, i.e. 198 numbers.\n\nFinding the Square of a number\n\nExample. Find the square of 27.\n\nSolution: (27)= (20 + 7)2\n\nUsing the formula (a + b)2 = a+ 2ab + b2, we have\n\n(20 + 7)2 = (20)2 + 2 * (20) * (7) + (7)2\n\n= 400 + 280 + 49\n\n= 729\n\nThus, (27)2 = 729\n\nNote: For any number ending with 5, the square is a(a + 1) hundred + 25.\n\nFor example,\n\n(25)2 = 2(2 + 1) * 100 + 25 = 625\n\n(35)2 = 3(3 + 1) * 100 + 25 = 1225\n\n(65)2 = 6(6 + 1) *100 + 25 = 4225\n\n(125)2 = 12(12 + 1) * 100 + 25 = 15625\n\nPythagorean Triplets\n\nIf three numbers a, b and c are such that a2 + b2 = c2, then they are called Pythagorean Triplets and they represent the sides of a right triangle.\n\nExample:\n\n(i) 3, 4, 5 form a Pythagorean triplet.\n\n[∵ 32 + 42 = 52]\n\n(ii) 8, 15, 17 form a Pythagorean triplet.\n\n[∵ 82 + 152 = 172]\n\nNote: For any natural number n, (n > 1), we have\n\n(2n)2 + (n2 – 1)2 = (n2 + 1)2\n\nsuch that 2n, n2 – 1 and n2 + 1 are a Pythagorean triplet.\n\nExample. Write a Pythagorean triplet whose one member is 15.\n\nSolution: Since, a Pythagorean triplet is given by 2n, n2 – 1 and n2 + 1.\n\n∴ 2n = 15 or n = 15/2 is not an integer.\n\nSo, let us assume that\n\nn2 – 1 = 15\n\nor n2 = 15 + 1 = 16\n\nor n2 = 42, i.e. n = 4\n\nNow, the required Pythagorean triplet is\n\n2n, n2 – 1 and n2 + 1\n\nor 2(4), 4– 1 and 42 + 1\n\nor 8, 15 and 17\n\nRemember\n\nAll Pythagorean triplets may not be obtained using the above form.\n\nQuestion: Find the square of the following numbers containing 5 in unit’s place.\n\n(i) 15 (ii) 95 (iii) 105 (iv) 205\n\nSolution:\n\n(i) (15)2 = 1 * (1 + 1) * 100 + 25\n\n= 1 * 2 * 100 + 25\n\n= 200 + 25 = 225\n\n(ii) (95)2 = 9(9 + 1) * 100 + 25\n\n= 9 * 10 * 100 + 25\n\n= 9000 + 25 = 9025\n\n(iii) (105)2 = 10 * (10 + 1) * 100 + 25\n\n= 10 *11 * 100 + 25\n\n= 11000 + 25 = 11025\n\n(iv) (205)2 = 20 * (20 + 1) * 100 = 25\n\n= 20 * 21 * 100 + 25\n\n= 42000 + 25 = 42025\n\nExercise 6.2\n\nQue 1. Find the square of the following numbers.\n\n(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46\n\nSolution:\n\n(i) (32)2 = (30 + 2)2\n\n= 302 + 2(30)(2) + (2)2\n\n= 900 + 120 + 4 = 1024\n\n(ii) (35)= (30 + 5)2\n\n= (30)2 + 2(30)(5) + (5)2\n\n= 900 + 300 + 25\n\n= 1200 + 25 = 1225\n\nSecond method\n\n352 = 3 * (3 + 1) * 100 + 25\n\n= 3 * 4 * 100 + 25\n\n= 1200 + 25 = 1225\n\n(iii) (86)2 = (80 + 6)2\n\n= (80)2 + 2(80)(6) + (6)2\n\n= 6400 + 960 + 36 = 7396\n\n(iv) (93)2 = (90 + 3)2\n\n= (90)2 + 2(90)(3) + (3)2\n\n= 8100 + 540 + 9 = 8649\n\n(v) (71)2 = (70 + 1)2\n\n= (70)+ 2(70)(1) + (1)2\n\n= 4900 + 140 + 1 = 5041\n\n(vi) (46)2 = (40 + 6)2\n\n= (40)2 + 2(40)(6) + (6)2\n\n= 1600 + 480 + 36 = 2116\n\nQue 2. Write a Pythagorean triplet whose one member is\n\n(i) 6 (ii) 14 (iii) 16 (iv) 18\n\nSolution:\n\n(i) Let 2n = 6 ∴n = 3\n\nNow, n2 – 1 = 32 – 1 = 8\n\nand n2 + 1 = 32 + 1 = 10\n\nThus, the required Pythagorean triplet is 6, 8, 10.\n\n(ii) Let 2n = 14 ∴ n = 7\n\nNow, n2 – 1 = 72 – 1 = 48\n\nand n2 + 1 = 72 + 1 = 50\n\nThus, the required Pythagorean triplet is 14, 48, 50.\n\n(iii) Let 2n = 16 ∴n = 8\n\nNow, n2 – 1 = 82 – 1\n\n= 64 – 1 = 63\n\nand n2 + 1 = 82 + 1\n\n= 64 + 1 = 65\n\nThus, the required Pythagorean triplet is 16, 63, 65.\n\n(iv) Let 2n = 18 ∴n = 9\n\nNow, n2 – 1 = 92 – 1\n\n= 81 – 1 = 80\n\nand n+ 1 = 92 + 1\n\n= 81 + 1 = 82\n\nThus, the required Pythagorean triple is 18, 80, 82.\n\n Q. Write all the square numbers between 100 and 300.Ans: 121, 144, 169, 196, 225, 256 and 289 Q. Which of the following would end with digit 9: 1232, 772, 842, 1612, and 102. Ans: (123)2 and (77)2Q. Write all the non-square numbers between 42 and 52.Ans: 17, 18, 19, 20, 21, 22, 23 and 24 Q. Fill in the blanks:(i) 1 + 3 + 5 + 7 + 9 + … = 52(ii) 1 + 3 + 5 + 7 + 9 + … = 62Ans: (i) 11 (ii) 11 + 13 Q. Fill in the blanks: (i) 112 = …… + …… (ii) (…..)2 = 112 + 113 Ans: (i) 60 + 61 (ii) (15)2Q. Find the square of the following numbers actual multiplication: (i) 39 (ii) 42Ans: (i) 1521 (ii) 1764Q. Find the square of: (i) 35 (ii) 105 Ans: (i) 1225 (ii) 11025Q. Write the Pythagorean triplet whose smallest number is 8.Ans: 8, 15, 17Q. Find a Pythagorean triplet in which one member is 12.Ans: 12, 35, 37\n\nSquare Roots\n\nFinding the square root of a number is just the opposite operation of squaring it. For example, the square of 5 is 25.\n\n∴ Square root of 25 is 5.\n\nNote: We kwon that (–2) * (–2) = 4, then we say that (–2) is also the square root of 4. Similarly, square roots of 100 are 10 and (–10).\n\nBut in this chapter, we shall be studying about positive square roots only\n\nFinding Square Root Through Prime Factorisation\n\nWe know that a factor that occurs once in the prime factorisation of a number, occurs twice in the prime factorisation of its square. Thus, we can use this fact of prime factorisation of a number to find the square root of a perfect square.\n\nNote: A perfect square has complete pairs of its prime factors.\n\nExample. Is 1008 a perfect square? If not, find the smallest multiply of 1008 which is a perfect square end then find the square root of the new number.\n\nSolution: We have\n\n1008 = 2 * 2 * 2 * 2 * 3 * 3 * 7\n\nAs the prime factor 7 has no pair.\n\n∴1008 is not a perfect square. Obviously, if 7 gets a pair, then the number will become a perfect square.\n\n∴1008 * 7 = [2 * 2 * 2 * 2 * 3 * 3 * 7] * 7\n\nor 7056 = 2 * 2 * 2 * 2 * 3 * 3 * 7\n\nThus, 7056 is the required multiple of 1008 which is a perfect square.\n\nNow, = 2 * 2 * 3 * 7 = 84\n\nQuestion: (i) 112 = 121. What is the square root of 121?\n(ii) 142 = 196. What is the square root of 196?\n\nSolution: (i) The square root of 121 is 11.\n(ii) The square root of 196 is 14.\n\nThink, Discuss and Write\n\nQuestion:\n\n(–1)2 = 1. Is –1, a square root of 1?\n\n(–2)2 = 4. Is –2, a square root of 4?\n\n(–9)2 = 81. Is –9, a square root of 81?\n\nSolution:\n\n(i) Since (–1) * (–1) = 1\n\ni.e. (–1)2 = 1\n\n∴ Square root of 1 can also be –1.\n\nSimilarly,\n\n(ii) Yes (–2) is a square root of 4.\n\n(iii) Yes (–9) is a square root of 81.\n\nSince, we have to consider the positive square roots only and the symbol for a positive square\nroot is √\n\n= 4 [and not (–4)]\n\nSimilarly, means, the positive square root of 25, i.e. 5.\n\nNote: We can also find the square root by subtracting successive odd numbers starting from 1.\n\nQuestion: By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root.\n\n(i) 121 (ii) 55 (iii) 36 (iv) 49 (v) 90\n\nSolution:\n\n(i) Subtracting the successive odd numbers from 121, we have\n\n121 – 1 = 120 120 – 3 = 117\n\n117 – 5 = 112 112 – 7 = 105\n\n105 – 9 = 96 96 – 11 = 85\n\n85 – 13 = 72 72 – 15 = 57\n\n57 – 17 = 40 40 – 19 = 21\n\n21 – 21 = 0\n\n∴ = 11. [∵ We had to subtract the first 11 odd numbers.]\n\n(ii)\n\n∵ 55 – 1 = 54 54 – 3 = 51\n51 – 5 = 46 46 – 7 = 39\n39 – 9 = 30 30 – 11 = 19\n19 – 13 = 6 6 – 15 = –9\n\nand we do not reach to 0. ∴ 55 is not a perfect square.\n\n(iii)\n\n∵ 36 – 1 = 35 35 – 3 = 32\n32 – 5 = 27 27 – 7 = 20\n20 – 9 = 11 11 – 11 = 0\n\nand we have obtained 0 after subtracting 6 successive odd numbers.\n\n∴ 36 is a perfect square.\n\nThus, = 6.\n\n(iv) We have\n\n49 – 1 = 48 48 – 3 = 45\n45 – 5 = 40 40 – 7 = 33\n33 – 9 = 24 24 – 11 = 13\n13 – 13 = 0\n\n∵ We have obtained 0 after successive subtraction of 7 odd numbers.\n\n∴ 49 is a perfect square,\n\nThus, = 7.\n\n(v) We have:\n\n90 – 1 = 89 89 – 3 = 86\n86 – 5 = 81 81 – 7 = 74\n74 – 9 = 65 65 – 11 = 54\n54 – 13 = 41 41 – 15 = 26\n26 – 17 = 9 9 – 19 = –10\n\nSince, we can not reach to 0 after subtracting successive odd numbers.\n\n∴ 90 is not a perfect square.\n\nOffer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!\n\n## Mathematics (Maths) Class 8\n\n135 videos|321 docs|48 tests\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n;<|endoftext|>"},"score":{"kind":"number","value":4.8125,"string":"4.8125"}}},{"rowIdx":1070,"cells":{"token_count":{"kind":"number","value":1630,"string":"1,630"},"text":{"kind":"string","value":"# Fibonacci sequence\n\nThis is the first activity of Computerization algorithm team. We introduced the method to find the $n$th term of the Fibonacci sequence, which mainly uses matrix exponentiation.\n\n## Problem\n\nThe Fibonacci sequence:\n$F_{n}= \\begin{cases} 0,&n=0\\\\ 1,&n=1\\\\ F_{n-2}+F_{n-1},&n>1 \\end{cases}$\n\nGiven $n$, find $F_{n}\\text{ mod }10^9+7$\n\nInput constraintsMemory limitExecution time\n$0\\le n\\le 10^{19}$64MB1.0s\n\n## Solution\n\nThe input size of $10^{19}$ obviously prohibits any attempt to solve it with loops. Is there a better way than a simple $\\mathcal{O}(n)$? It turns out that with matrix exponentiation, we can achieve $\\mathcal{O}(\\log n)$. We observe that:\n\n$\\begin{pmatrix}F_{n+1}\\\\F_{n+2}\\end{pmatrix}=\\begin{pmatrix}F_{n+1}\\\\F_{n}+F_{n+1}\\end{pmatrix}=\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}\\begin{pmatrix}F_{n}\\\\F_{n+1}\\end{pmatrix}$\n\nThis step is applicable to all recursive sequences, so it should be easily reached for an experienced candidate. Generally, for $F_{n+2}=aF_{n}+bF_{n+1}$, we have\n\n$\\begin{pmatrix}F_{n+1}\\\\F_{n+2}\\end{pmatrix}=\\begin{pmatrix}F_{n+1}\\\\aF_{n}+bF_{n+1}\\end{pmatrix}=\\begin{pmatrix}0&1\\\\a&b\\end{pmatrix}\\begin{pmatrix}F_{n}\\\\F_{n+1}\\end{pmatrix}$\n\nFrom the recursive definition,\n\n$\\begin{pmatrix}F_{n+m}\\\\F_{n+m+1}\\end{pmatrix}=\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^m\\begin{pmatrix}F_{n}\\\\F_{n+1}\\end{pmatrix}$\n\nSubstituting $n=0$,\n\n$\\begin{pmatrix}F_{m}\\\\F_{m+1}\\end{pmatrix}=\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^m\\begin{pmatrix}F_0\\\\F_1\\end{pmatrix}$\n\nNow the problem is transformed into finding the matrix raised to the $m$th power. If $m=2^0a_0+2^1a_1+2^2a_2+\\dots$ (representation in binary), then\n\n$\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^m=\\left(\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{1}\\right)^{a_0}\\times \\left(\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{2}\\right)^{a_1}\\times \\left(\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{4}\\right)^{a_2}\\dots$\n\nThe $2^k$th powers of the original matrix can, in fact, be preprocessed. When $m<10^{19}$, $k<\\log_2 10^{19}<64$, so we only need to store at most 63 matrices. In addition,\n\n$\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{2^k}=\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{2^{k-1}}\\times \\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}^{2^{k-1}}$\n\nwhich implies that the powers can be attained within $\\mathcal{O}(\\log m)$ time. This is the idea of fast matrix exponentiation: compute all $2^k$th powers, and put those needed together.\n\n## Program\n\nBelow is the C++ code, where the most intractable part is probably implementation of matrix multiplication:\n\n#include