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# Difference between revisions of "1960 IMO Problems/Problem 1" ## Problem Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$. ## Solutions ### Solution 1 Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$. Then $$100a + 10b+c = 11m$$ for some $m$. We also have $m=a^2+b^2+c^2$. Substituting this into the first equation and simplification, we get $$100a+10b+c = 11a^2 +11b^2 +11c^2$$ For an integer divisible by $11$, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$. Thus we get: $b = a + c$ or $b = a + c - 11$. Case $1$: Let $b=a+c$. We get $$100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2$$ $$10a+c = 2a^2+2ac+2c^2$$ Since the right side is even, the left side must also be even. Let $c=2q$ for some $q = 0,1,2,3,4$. Then $$10a+2q=2a^2+4aq+8q^2$$$$5a+q=a^2+2aq+4q^2$$ Substitute $q=0,1,2,3,4$ into the last equation and then solve for $a$. When $q=0$, we get $a=5$. Thus $c=0$ and $b=5$. We get that $N=550$ which works. When $q=1$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=2$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=3$, we get that $a$ is not an integer. There is no $N$ for this case. When $q=4$, we get that $a$ is not an integer. There is no $N$ for this case. Case $2$: Let $b = a + c - 11$. We get $$100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)$$ $$10a+c=2a^2+2c^2+2ac-22a-22c+131$$ $$2(a-8)^2+2(c-\frac{23}{4})^2+2ac-\frac{505}{8}=0$$ Now we test all $c=0\rightarrow10$. When $c=0,1,2,4,5,6,7,8,9$, we get no integer solution to $a$. Thus, for these values of $c$, there is no valid $N$. However, when $c=3$, we get $$2(a-8)^2+2(3-\frac{23}{4})^2+6a-\frac{505}{8}=0$$ $$2(a-8)^2+6a-48 = 0$$ We get that $a=8$ is a valid solution. For this case, we get $a=8,b=0,c=3$, so $N=803$, and this is a valid value. Thus, the answers are $\boxed{N=550,803}$. ### Solution 2 Define a ten to be all ten positive integers which begin with a fixed tens digit. We can make a systematic approach to this: By inspection, $\dfrac{N}{11}$ must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well. For a given ten, the sum of the squares of the digits of $N$ increases faster than $\dfrac{N}{11}$, so we can have at most one number in every ten that works. We check the first ten: $1111=121$ $1^2+2^2+1^2=4$ $1211=132$ $1^2+3^2+2^2=14$ 11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work. We try the second ten: $2111=231$ $2^2+3^2+1^2=14$ $2211=242$ $2^2+4^2+2^2=24$ Therefore, no numbers in the second ten work. We continue, to find out that 50 and 73 are the only ones that works. $N=5011=550$, $N=7311=803$ so there are two $N$ that works. ### Solution 3 Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting $\boxed{N=550,803}$.
Suggested languages for you: Americas Europe Q30. Expert-verified Found in: Page 427 ### Geometry Book edition Student Edition Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen Pages 227 pages ISBN 9780395977279 # A wooden fence high and 220ft long is to be painted on both sides.What is the total area to be painted?A gallon of a certain type paint will cover only $200f{t}^{2}$ of area for the first coat, but on the second coat a gallon of the same paint will cover $300f{t}^{2}.$ If the fence is to be given two coats of paint, how many gallons of paint should be bought? 1. The area to be painted is $2640f{t}^{2}$ 2. Total paint required is22 gallon. See the step by step solution ## Step 1. Given information. Length of the wooden fence$=6ft$ Height of the wooden ffence $=220ft$ ## Step 2. Concept Used. If bis the length of the wooden fence and h is the height of the fence. Then, the area of the fence is $A=b×h$ ## Step 3. Let’s calculate the area of the fence. Put the value of length and height into the formula and find the area. Area of the fence :$\begin{array}{c}A=220×6\\ =1320f{t}^{2}\end{array}$ Since, it is to be painted on both sides the painted area gets doubled i.e. $\begin{array}{l}A=2×1320\\ A=2640f{t}^{2}\end{array}$ So, area to be painted is $2640f{t}^{2}$ In the first coat $200f{t}^{2}=1$ gallon Then,$2640f{t}^{2}=\left(\frac{2640}{200}\right)=13.2$ gallon In the second coat$300f{t}^{2}=1$ gallon Then,$2640f{t}^{2}=\left(\frac{2640}{300}\right)=8.8$ gallon Thus$=\left(13.2+8.8\right)=22$ Total paint gallon Therefore, the area to be painted is$2640f{t}^{2}$ and total paint required is 22gallon.
# Sampling Distributions ## Presentation on theme: "Sampling Distributions"— Presentation transcript: Sampling Distributions Chapter 8 Sampling Distributions Parameter A number that describes the population Symbols we will use for parameters include m - mean s – standard deviation p – proportion (p) b0 – y-intercept of LSRL b1 – slope of LSRL Statistic A number that that can be computed from sample data without making use of any unknown parameter Symbols we will use for statistics include x – mean s – standard deviation p – proportion b0 – y-intercept of LSRL b1 – slope of LSRL Identify the boldface values as parameter or statistic. A carload lot of ball bearings has mean diameter cm. This is within the specifications for acceptance of the lot by the purchaser. By chance, an inspector chooses 100 bearings from the lot that have mean diameter cm. Because this is outside the specified limits, the lot is mistakenly rejected. Why do we take samples instead of taking a census? A census is not always accurate. Census are difficult or impossible to do. Census are very expensive to do. A distribution is all the values that a variable can be. The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population. What is the mean and standard deviation of this population? Consider the population – the length of fish (in inches) in my pond - consisting of the values 2, 7, 10, 11, 14 What is the mean and standard deviation of this population? mx = 8.8 sx = Let’s take samples of size 2 (n = 2) from this population: How many samples of size 2 are possible? 5C2 = 10 mx = 8.8 Find all 10 of these samples and record the sample means. What is the mean and standard deviation of the sample means? sx = Repeat this procedure with sample size n = 3 How many samples of size 3 are possible? 5C3 = 10 mx = 8.8 What is the mean and standard deviation of the sample means? Find all of these samples and record the sample means. sx = What do you notice? mx = m as n sx The mean of the sampling distribution EQUALS the mean of the population. As the sample size increases, the standard deviation of the sampling distribution decreases. mx = m as n sx A statistic used to estimate a parameter is unbiased if the mean of its sampling distribution is equal to the true value of the parameter being estimated. General Properties mx = m s sx = n Rule 1: Rule 2: This rule is approximately correct as long as no more than 5% (10%) of the population is included in the sample mx = m s n sx = General Properties Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n. Activity – drawing samples General Properties Rule 4: Central Limit Theorem When n is sufficiently large, the sampling distribution of x is well approximated by a normal curve, even when the population distribution is not itself normal. CLT can safely be applied if n exceeds 30. Remember the army helmets . . . EX) The army reports that the distribution of head circumference among soldiers is approximately normal with mean 22.8 inches and standard deviation of 1.1 inches. a) What is the probability that a randomly selected soldier’s head will have a circumference that is greater than 23.5 inches? P(X > 23.5) = .2623 What normal curve are you now working with? b) What is the probability that a random sample of five soldiers will have an average head circumference that is greater than 23.5 inches? Do you expect the probability to be more or less than the answer to part (a)? Explain What normal curve are you now working with? P(X > 23.5) = .0774 Suppose a team of biologists has been studying the Pinedale children’s fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches. What is the probability that a single trout taken at random from the pond is between 8 and 12 inches long? P(8 < X < 12) = .8427 P(8< x <12) = .9978 x = 11.23 inches What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long? What sample mean would be at the 95th percentile? (Assume n = 5) Do you expect the probability to be more or less than the answer to part (a)? Explain P(8< x <12) = .9978 x = inches A soft-drink bottler claims that, on average, cans contain 12 oz of soda. Let x denote the actual volume of soda in a randomly selected can. Suppose that x is normally distributed with s = .16 oz. Sixteen cans are randomly selected and a mean of 12.1 oz is calculated. What is the probability that the mean of 16 cans will exceed 12.1 oz? P(x >12.1) = .0062 A hot dog manufacturer asserts that one of its brands of hot dogs has a average fat content of 18 grams per hot dog with standard deviation of 1 gram. Consumers of this brand would probably not be disturbed if the mean was less than 18 grams, but would be unhappy if it exceeded 18 grams. An independent testing organization is asked to analyze a random sample of 36 hot dogs. Suppose the resulting sample mean is 18.4 grams. What is the probability that the sample mean is greater than 18.4 grams? P(x >12.1) = .0082 Does this result indicate that the manufacturer’s claim is incorrect? Yes, not likely to happen by chance alone.
1 / 112 # Chapter 6 Chapter 6. Describing Polygons. Q. A polygon is a figure that is: -formed by 3 or more segments called sides , such that no 2 sides with a common endpoint are collinear - each side intersects exactly 2 other sides, one at each endpoint. Each endpoint of the side is called a vertex. ## Chapter 6 E N D ### Presentation Transcript 1. Chapter 6 2. Describing Polygons Q A polygon is a figure that is: -formed by 3 or more segments called sides, such that no 2 sides with a common endpoint are collinear - each side intersects exactly 2 other sides, one at each endpoint. Each endpoint of the side is called a vertex. Vertex P R Side T S Vertex 3. Naming Polygons 4. Identifying Convex and Concave A polygon is convex if no line that contains a side of the polygon contains a point in the interior of the polygon. A polygon is that is not convex is called nonconvex or concave. 5. Definitions: A polygon is equilateral if all of its sides are congruent. A polygon is equiangular if all of its interior angles are congruent. A polygon is regular if it is both equilateral and equiangular. 6. Theorem 6.1: Polygon Angle-Sum Theorem The sum of the measures of the interior angles of an n-gon is (n-2)180. If you draw a diagonal in a polygon, you create triangles. Using the Triangle Sum Theorem you can conclude that the sum of the measures of the interior angles of a quadrilateral is 2(180)=360°. 7. Corollary to Polygon Angle-Sum Theorem The measure of each of the interior angles of a regular polygon is (Where n is the number of sides.) Theorem 6.2 Polygon Exterior Angle-Sum Theorem The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360 degrees. 8. PARALLELOGRAM: • A quadrilateral with both pairs of opposite sides parallel. 9. Theorem 6.3: • If a quadrilateral is a parallelogram, then its opposite sides are congruent. 10. Theorem 6.4: • If a quadrilateral is a parallelogram, then its consecutive angles are supplementary. 11. Theorem 6.5: • If a quadrilateral is a parallelogram, then its opposite angles are congruent. 12. Theorem 6.6: • If a quadrilateral is a parallelogram, then its diagonals bisect each other. 13. Special Parallelograms A square is a parallelogram with 4 right angles and 4 congruent sides A rhombus is a parallelogram with 4 congruent sides. A rectangle is a parallelogram with 4 right angles Rectangle Square Rhombus 14. Diagonals of Special Parallelograms Thm. 6.13: If a parallelogram is a rhombus, then its diagonals are perpendicular. Thm. 6.14: If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles. • Thm. 6.15: If a parallelogram is a rectangle, then its diagonals are congruent. 16. Properties of Trapezoids • A trapezoid is a quadrilateral with exactly one pair of parallel sides. Trapezoid Terminology: The parallel sides are called BASES. The nonparallel sides are called LEGS. There are two pairs of base angles, the two touching the top base, and the two touching the bottom base. 17. ISOSCELES TRAPEZOID If the legs of a trapezoid are congruent, then the trapezoid is an isosceles trapezoid. Thm. 6.19: If a quadrilateral is an isosceles trapezoid, then each pair of base angles is congruent. Thm. 6.20: If a quadrilateral is an isosceles trapezoid, then its diagonals are congruent. 18. Midsegment of a Trapezoid The midsegment of a trapezoid connects the midpoints of its legs. Thm. 6.21: If a quadrilateral is a trapezoid, then… • The midsegment is parallel to both bases and • The length of the midsegment is half the sum of the lengths of the bases. 19. Kites A kite is a quadrilateral that has 2 pairs of consecutive congruent sides, but opposite sides are not congruent. Thm. 6.22: If a quadrilateral is a kite, then its diagonals are perpendicular. Interesting fact: If a quadrilateral is a kite, then exactly one pair of opposite angles are congruent. 20. Relationships Among Quadrilaterals • Fill in Chart 21. Formulas and the Coordinate Plane Formula Distance Formula Midpoint Formula Slope Formula • When to Use it…. • To determine whether… • Sides are congruent • Diagonals are congruent • To determine … • The coordinates of the midpoint of a side • Whether diagonals bisect each other • To determine whether… • Opposite sides are parallel • Diagonals are perpendicular • Sides are perpendicular 22. Chapter 7 23. Reminders on Ratios: • It is a comparison of two quantities by division • Notation: or , read a to b. • Measured in same units. • Denominator can not be zero. • Usually expressed in simplified form: 6:8 simplified to 3:4 24. An equation that equals two ratios. Proportion Extremes Means 25. Similar Polygons Two polygons are SIMILAR if and only if: 1-their corresponding angles are congruent, 2- the measures of their corresponding sides are proportional. 26. Similar Polygons • Symbol to indicate similarity: ~ • ABCD ~ GHIJ (ratio of the lengths of two corresponding sides) 27. Angle-Angle Similarity Postulate Postulate 7.1 Angle-Angle Similarity (AA~) Postulate: If two angles of one triangle are congruent to 2 angles of another triangle, then the two triangles are similar. Hint: From earlier information we know that if we have two congruent angles then we also know the third angles are congruent. Thus AA is the same as AAA. This is the most common proof of two triangles to be similar. 28. THEOREMS X M P N Z Y XY MN ZX PM If XM and= THEOREM 7.1 Side-Angle-Side (SAS~) Similarity Theorem If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar. then XYZ ~ MNP. 29. THEOREMS P A AB PQ BC QR CA RP Q R If = = B C THEOREM 7.2 Side-Side-Side (SSS~) Similarity Theorem If the corresponding sides of two triangles are proportional, then the triangles are similar. then ABC ~ PQR. 30. Theorem 7.3 • The altitude to the hypotenuse of a right triangle divides the triangle into 2 triangles that are similar to the original triangle and to each other. ΔCBD ~ ΔABC ΔACD ~ ΔABC ΔCBD ~ ΔACD 31. The GEOMETRIC MEAN between two positive numbers a and b is the positive number x where • What is the geometric mean of 5 and 12? • What is the geometric mean of 6 and 16? 32. Theorem 7.4: Side-Splitter Theorem If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. C E A B D 33. Theorem 7.5 Triangle-Angle Bisector Theorem If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that are proportional to the other two sides of the triangle. 34. Chapter 8 35. The Pythagorean Theorem Remember the Pythagorean Theorem is only true for RIGHT triangles!! The hypotenuse (c) is always the longest side and opposite the right angle! The legs (a and b) are the two sides that form the right angle. 36. Pythagorean Triple 37. Theorem 8.3: If , then the triangle is obtuse. Theorem 8.4: If , then the triangle is acute. Classifying Triangles: 38. Certain triangles possess "special" properties that allow us to use "short cut formulas" in arriving at information about their measures. These formulas let us arrive at the answer very quickly. Theorem 8-5 - 45º-45º-90º 39. Theorem 8.6 Hypotenuse = 2 x shorter leg Longer Leg = x shorter leg 40. Trigonometric Ratios A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The three basic trigonometric ratios are sine, cosine, and tangent, which are abbreviated as sin, cos, and tan, respectively. 41. Writing Trigonometric Ratios What are the ratios for angle G? 42. Angle of Elevation/Depression: The angle formed by a horizontal line and the line of sight to an object either above or below the horizontal line. 43. Vectors: • Vector: any quantity with both magnitude (size) and direction • The magnitude corresponds to the distance from the initial point to the terminal point of the vector • The direction corresponds to which way the arrow is pointed 44. Describing a Vector • You can indicate a vector by using an ordered pair. For example, <-2, 4> is a vector with its initial point at the origin and its terminal point at (-2, 4). • We use brackets to represent a vector (called Component Form) 45. Direction of a Vector: • You can use a compass arrangement on the coordinate grid to describe a vector’s direction. This vector is 30 south of east. This vector is 40 east of north. 46. Magnitude of a Vector • The magnitude of a vector is its length. You can use the distance formula to determine the length, or magnitude, of a vector. 47. Adding Two Vectors • The sum of two vectors is called the Resultant More Related
# What is 135/91 as a decimal? ## Solution and how to convert 135 / 91 into a decimal 135 / 91 = 1.484 Convert 135/91 to 1.484 decimal form by understanding when to use each form of the number. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. After deciding on which representation is best, let's dive into how we can convert fractions to decimals. ## 135/91 is 135 divided by 91 Converting fractions to decimals is as simple as long division. 135 is being divided by 91. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 135 divided by 91. To solve the equation, we must divide the numerator (135) by the denominator (91). Here's 135/91 as our equation: ### Numerator: 135 • Numerators sit at the top of the fraction, representing the parts of the whole. Overall, 135 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Now let's explore the denominator of the fraction. ### Denominator: 91 • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 91 is a large number which means you should probably use a calculator. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Let's start converting! ## Converting 135/91 to 1.484 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 91 \enclose{longdiv}{ 135 }$$ To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well. ### Step 2: Solve for how many whole groups you can divide 91 into 135 $$\require{enclose} 00.1 \\ 91 \enclose{longdiv}{ 135.0 }$$ Since we've extended our equation we can now divide our numbers, 91 into 1350 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply this number by 91, the denominator to get the first part of your answer! ### Step 3: Subtract the remainder $$\require{enclose} 00.1 \\ 91 \enclose{longdiv}{ 135.0 } \\ \underline{ 91 \phantom{00} } \\ 1259 \phantom{0}$$ If you don't have a remainder, congrats! You've solved the problem and converted 135/91 into 1.484 If you have a remainder over 91, go back. Your solution will need a bit of adjustment. If you have a number less than 91, continue! ### Step 4: Repeat step 3 until you have no remainder Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 135/91 and 1.484 bring clarity and value to numbers in every day life. Here are just a few ways we use 135/91, 1.484 or 148% in our daily world: ### When you should convert 135/91 into a decimal Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.148 hours, not 1 and 135/91 hours. Percentage format is also used in contracts as well. ### When to convert 1.484 to 135/91 as a fraction Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches. ### Practice Decimal Conversion with your Classroom • If 135/91 = 1.484 what would it be as a percentage? • What is 1 + 135/91 in decimal form? • What is 1 - 135/91 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 1.484 + 1/2?
Question # In a seminar, the number of participants in Hindi, English and Mathematics are $60,84$ and $108$. Find the number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject. Hint: To find the number of rooms required to seat all the participants, firstly find the number of participants to be seated in each room. As an equal number of participants are to be seated in each room, find the HCF of the number of participants, i.e., HCF of $60,84,108$. Divide the HCF of three numbers by each of the numbers to get the number of rooms needed for participants of each subject. Add the number of rooms required for each subject to find the total number of rooms required to seat all the participants. We have $60,84,108$ participants of Hindi, English and Mathematics. We have to find the number of rooms required to seat all the participants such that the number of participants is the same in all the rooms and all the participants of the same subject are seated in one room. Firstly, we will calculate the number of rooms needed to seat all the participants of a particular subject. As the same number of participants are to be seated in each room, we will evaluate the HCF of $60,84,108$. To find the HCF of $60,84,108$, we will write the prime factorization of each of the numbers. We will choose the common prime factors of all the three numbers. The number obtained by choosing the common prime factors of all the numbers will be the HCF of three numbers. We will now write prime factorization of all the three numbers. The prime factorization of $60$ is $60={{2}^{2}}\times 3\times 5$. The prime factorization of $84$ is $84={{2}^{2}}\times 3\times 7$. The prime factorization of $108$ is $108={{2}^{2}}\times {{3}^{3}}$. The terms common to prime factorization of all the three numbers is ${{2}^{2}}\times 3=12$. Thus, the HCF of $60,84,108$ is $12$. We will now calculate the number of rooms required to seat all the participants of each subject. To do so, we will divide each of the numbers by their HCF. Number of rooms required to seat Hindi participants $=\dfrac{60}{12}=5$. Number of rooms required to seat English participants $=\dfrac{84}{12}=7$. Number of rooms required to seat Mathematics participants $=\dfrac{108}{12}=9$. So, the total number of rooms required to seat all the participants $=5+7+9=21$. Hence, we need $21$ rooms to seat all the participants such that $12$ participants will be seated in all the rooms. Note: To write the prime factorization of any number, start by dividing the number by the first prime number, which is $2$ and then continue to divide by $2$ until you get a number which is not divisible by $2$ (which means that you get a decimal or remainder on dividing the number by $2$). Then start dividing the number by the next prime number which is $3$. Continue dividing the number by $3$ until you get a number which is not divisible by $3$. Thus, continuing the process, keep dividing the numbers by series of prime numbers $5,7,...$ until the only numbers left are prime numbers. Write the given number as a product of all the prime numbers (considering the fact to count each prime number as many times as it divides the given number) to get the prime factorization of the given number.
# Fraction Operations ## Objective Decompose fractions as a sum of unit fractions and as a sum of smaller fractions. ## Common Core Standards ### Core Standards ? • 4.NF.B.3.B — Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8. ? • 3.NF.A.1 • 3.NF.A.2 ## Criteria for Success ? 1. Decompose a fraction less than or equal to 1 into a sum of unit fractions, recording the decomposition with an equation. 2. Decompose a fraction less than or equal to 1 into a sum of fractions with the same denominator in more than one way, recording each decomposition with an equation. 3. Justify decompositions with a visual model, such as a tape diagram or number line. ## Tips for Teachers ? • The following material is needed for today's lesson: • Throughout the unit, students will rely on tape diagrams and number lines to represent fraction operations. They won’t be formally introduced to the use of an area model to perform these operations until Grade 5, where they are particularly useful for helping to find a common unit when the units are unrelated to one another (e.g., 5 and 7, where they share no common factors). However, if students naturally choose to represent these computations with an area model or gravitate toward them, their use should not be discouraged. • Before the Problem Set, you could have students play around with manipulatives to decompose fractions into a sum of unit fractions or non-unit fractions, similar to Joe Schwartz’s blog post “Building Towers”. Just make sure students are building fractions less than or equal to 1. If you don’t have fraction towers or even fraction tiles, you can create them from paper. #### Remote Learning Guidance If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Task 3 (benefits from worked example). You could consolidate this task with Lesson 2's Anchor Task 3. Find more guidance on adapting our math curriculum for remote learning here. #### Fishtank Plus • Problem Set • Student Handout Editor • Vocabulary Package ? ### Problem 1 1. Use the strips of paper given to you to fill in the following outline. What fraction of the whole outline is each piece that you used to fill it? 1. Come up with as many possible ways to fill the outline. Write all of the combinations of fractions you used to fill it. ### Problem 2 Decompose the fraction represented below into a sum of unit fractions and in at least one other way. Record the decompositions with equations. ### Problem 3 1. Decompose the following fractions into a sum of unit fractions. Write an equation to record your decomposition. i. ${{3\over4}}$ ii. ${{10\over10}}$ 1. Find another way to decompose the fractions in part (a). Justify your decomposition with a tape diagram or number line. ## Problem Set & Homework #### Discussion of Problem Set • How else could you have decomposed any of the tape diagrams or number lines in #2? (You could choose to discuss which based on what’s best for your students.) • Why does #3(a) start with “${1 =}$” as opposed to a fraction? What equivalent fraction could we use instead of $1$? • What other decompositions could have been possible answers in #4? • In #5, how would you record the fraction’s decomposition into unit fractions? How can you draw parentheses around a certain number of unit fractions to model another decomposition you wrote (e.g., $\left({1\over8}+{1\over8}\right)+\left({1\over8}+{1\over8}\right) + {1\over8}$ to model ${2\over8}+{2\over8}+{1\over8}$)? • What relationship does the unit fraction have with the number of units in a whole? • How did you decompose numbers in earlier grades? How is that similar to and different from how you decomposed fractions today? ? Decompose the following fraction in three different ways, including as a sum of a unit of fractions. Record the decompositions with equations. ${{9\over10}}$ #### References EngageNY Mathematics Grade 4 Mathematics > Module 5 > Topic A > Lesson 2Exit Ticket Grade 4 Mathematics > Module 5 > Topic A > Lesson 2 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m.. Modified by The Match Foundation, Inc. ?
# Golden ratio The golden section is a line segment sectioned into two according to the golden ratio. The total length a + b is to the longer segment a as a is to the shorter segment b. In mathematics and the arts, two quantities are in the golden ratio if the ratio between the sum of those quantities and the larger one is the same as the ratio between the larger one and the smaller. The golden ratio is an irrational mathematical constant, approximately 1.6180339887.[1] At least since the Renaissance, many artists and architects have proportioned their works to approximate the golden ratio—especially in the form of the golden rectangle, in which the ratio of the longer side to the shorter is the golden ratio—believing this proportion to be aesthetically pleasing. Mathematicians have studied the golden ratio because of its unique and interesting properties. The golden ratio is often denoted by the Greek letter phi (Φ or φ). The figure of a golden section illustrates the geometric relationship that defines this constant. Expressed algebraically: $\frac{a+b}{a} = \frac{a}{b} = \varphi\,.$ This equation has as its unique positive solution the algebraic irrational number $\varphi = \frac{1+\sqrt{5}}{2}\approx 1.61803\,39887\ldots\,$ [1] Other names frequently used for or closely related to the golden ratio are golden section (Latin: sectio aurea), golden mean, golden number, and the Greek letter phi (Φ).[2][3][4] Other terms encountered include extreme and mean ratio,[5] medial section, divine proportion, divine section (Latin: sectio divina), golden proportion, golden cut,[6] and mean of Phidias.[7][8][9] Construction of a golden rectangle: 1. Construct a unit square (red). 2. Draw a line from the midpoint of one side to an opposite corner. 3. Use that line as the radius to draw an arc that defines the long dimension of the rectangle. ## Calculation List of numbers γ - ζ(3) - √2 - √3 - √5 - φ - α - e - π - δ Binary 1.1001111000110111011… Decimal 1.6180339887498948482… Hexadecimal 1.9E3779B97F4A7C15F39… Continued fraction $1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}}}$ Algebraic form $\frac{1 + \sqrt{5}}{2}$ Two quantities a and b are said to be in the golden ratio φ if: $\frac{a+b}{a} = \frac{a}{b} = \varphi\,.$ This equation unambiguously defines φ. The right equation shows that a = bφ, which can be substituted in the left part, giving $\frac{b\varphi+b}{b\varphi}=\frac{b\varphi}{b}\,.$ Dividing out b yields $\frac{\varphi+1}{\varphi}=\varphi.$ Multiplying both sides by φ and rearranging terms leads to: ${\varphi}^2 - \varphi - 1 = 0.$ The only positive solution to this quadratic equation is $\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,$ ## History Mathematician Mark Barr proposed using the first letter in the name of Greek sculptor Phidias, phi, to symbolize the golden ratio. Usually, the lowercase form (φ) is used. Sometimes, the uppercase form (Φ) is used for the reciprocal of the golden ratio, 1/φ. Michael Maestlin, first to publish a decimal approximation of the golden ratio, in 1597. The golden ratio has fascinated Western intellectuals of diverse interests for at least 2,400 years: Some of the greatest mathematical minds of all ages, from Pythagoras and Euclid in ancient Greece, through the medieval Italian mathematician Leonardo of Pisa (Fibonacci) and the Renaissance astronomer Johannes Kepler, to present-day scientific figures such as Oxford physicist Roger Penrose, have spent endless hours over this simple ratio and its properties. But the fascination with the Golden Ratio is not confined just to mathematicians. Biologists, artists, musicians, historians, architects, psychologists, and even mystics have pondered and debated the basis of its ubiquity and appeal. In fact, it is probably fair to say that the Golden Ratio has inspired thinkers of all disciplines like no other number in the history of mathematics. Mario LivioThe Golden Ratio: The Story of Phi, The World's Most Astonishing Number Ancient Greek mathematicians first studied what we now call the golden ratio because of its frequent appearance in geometry. The division of a line into "extreme and mean ratio" (the golden section) is important in the geometry of regular pentagrams and pentagons. The Greeks usually attributed discovery of this concept to Pythagoras or his followers. The regular pentagram, which has a regular pentagon inscribed within it, was the Pythagoreans' symbol. Euclid's Elements (Greek: Στοιχεῖα) provides the first known written definition of what is now called the golden ratio: "A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less."[5] Euclid explains a construction for cutting (sectioning) a line "in extreme and mean ratio", i.e. the golden ratio.[10] Throughout the Elements, several propositions (theorems in modern terminology) and their proofs employ the golden ratio.[11] Some of these propositions show that the golden ratio is an irrational number. The name "extreme and mean ratio" was the principal term used from the 3rd century BC[5] until about the 18th century. The modern history of the golden ratio starts with Luca Pacioli's Divina Proportione of 1509, which captured the imagination of artists, architects, scientists, and mystics with the properties, mathematical and otherwise, of the golden ratio. The first known approximation of the (inverse) golden ratio by a decimal fraction, stated as "about 0.6180340," was written in 1597 by Prof. Michael Maestlin of the University of Tübingen in a letter to his former student Johannes Kepler.[12] Since the twentieth century, the golden ratio has been represented by the Greek letter Φ or φ (phi, after Phidias, a sculptor who is said to have employed it) or less commonly by τ (tau, the first letter of the ancient Greek root τομή—meaning cut). ### Timeline Timeline according to Priya Hemenway[13]. • Phidias (490–430 BC) made the Parthenon statues that seem to embody the golden ratio. • Plato (427–347 BC), in his Timaeus, describes five possible regular solids (the Platonic solids, the tetrahedron, cube, octahedron, dodecahedron and icosahedron), some of which are related to the golden ratio.[14] • Euclid (c. 325–c. 265 BC), in his Elements, gave the first recorded definition of the golden ratio, which he called, as translated into English, "extreme and mean ratio" (Greek: ακρος και μεσος λογος).[5] • Fibonacci (1170–1250) mentioned the numerical series now named after him in his Liber Abaci; the Fibonacci sequence is closely related to the golden ratio. • Luca Pacioli (1445–1517) defines the golden ratio as the "divine proportion" in his Divina Proportione. • Johannes Kepler (1571–1630) describes the golden ratio as a "precious jewel": "Geometry has two great treasures: one is the Theorem of Pythagoras, and the other the division of a line into extreme and mean ratio; the first we may compare to a measure of gold, the second we may name a precious jewel." These two treasures are combined in the Kepler triangle. • Charles Bonnet (1720–1793) points out that in the spiral phyllotaxis of plants going clockwise and counter-clockwise were frequently two successive Fibonacci series. • Martin Ohm (1792–1872) is believed to be the first to use the term goldener Schnitt (golden section) to describe this ratio, in 1835.[15] • Edouard Lucas (1842–1891) gives the numerical sequence now known as the Fibonacci sequence its present name. • Mark Barr (20th century) suggests the Greek letter phi (φ), the initial letter of Greek sculptor Phidias's name, as a symbol for the golden ratio.[16] • Roger Penrose (b.1931) discovered a symmetrical pattern that uses the golden ratio in the field of aperiodic tilings, which led to new discoveries about quasicrystals. ## Aesthetics Beginning in the Renaissance, a body of literature on the aesthetics of the golden ratio has developed. As a result, architects, artists, book designers, and others have been encouraged to use the golden ratio in the dimensional relationships of their works. The first and most influential of these was De Divina Proportione by Luca Pacioli, a three-volume work published in 1509. Pacioli, a Franciscan friar, was known mostly as a mathematician, but he was also trained and keenly interested in art. De Divina Proportione explored the mathematics of the golden ratio. Though it is often said that Pacioli advocated the golden ratio's application to yield pleasing, harmonious proportions, Livio points out that that interpretation has been traced to an error in 1799, and that Pacioli actually advocated the Vitruvian system of rational proportions.[2] Pacioli also saw Catholic religious significance in the ratio, which led to his work's title. Containing illustrations of regular solids by Leonardo Da Vinci, Pacioli's longtime friend and collaborator, De Divina Proportione was a major influence on generations of artists and architects alike. ### Architecture Some studies of the Acropolis, including the Parthenon, conclude that many of its proportions approximate the golden ratio. The Parthenon's facade as well as elements of its facade and elsewhere can be circumscribed by golden rectangles.[17] To the extent that classical buildings or their elements are proportioned according to the golden ratio, this might indicate that their architects were aware of the golden ratio and consciously employed it in their designs. Alternatively, it is possible that the architects used their own sense of good proportion, and that this led to some proportions that closely approximate the golden ratio. On the other hand, such retrospective analyses can always be questioned on the ground that the investigator chooses the points from which measurements are made or where to superimpose golden rectangles, and that these choices affect the proportions observed. Some scholars deny that the Greeks had any aesthetic association with golden ratio. For example, Midhat J. Gazalé says, "It was not until Euclid, however, that the golden ratio's mathematical properties were studied. In the Elements (308 B.C.) the Greek mathematician merely regarded that number as an interesting irrational number, in connection with the middle and extreme ratios. Its occurrence in regular pentagons and decagons was duly observed, as well as in the dodecahedron (a regular polyhedron whose twelve faces are regular pentagons). It is indeed exemplary that the great Euclid, contrary to generations of mystics who followed, would soberly treat that number for what it is, without attaching to it other than its factual properties."[18] And Keith Devlin says, "Certainly, the oft repeated assertion that the Parthenon in Athens is based on the golden ratio is not supported by actual measurements. In fact, the entire story about the Greeks and golden ratio seems to be without foundation. The one thing we know for sure is that Euclid, in his famous textbook Elements, written around 300 B.C., showed how to calculate its value."[19] Near-contemporary sources like Vitruvius exclusively discuss proportions that can be expressed in whole numbers, i.e. commensurate as opposed to irrational proportions. A geometrical analysis of the Great Mosque of Kairouan reveals a consistent application of the golden ratio throughout the design, according to Boussora and Mazouz.[20] It is found in the overall proportion of the plan and in the dimensioning of the prayer space, the court, and the minaret. Boussora and Mazouz also examined earlier archaeological theories about the mosque, and demonstrate the geometric constructions based on the golden ratio by applying these constructions to the plan of the mosque to test their hypothesis. The Swiss architect Le Corbusier, famous for his contributions to the modern international style, centered his design philosophy on systems of harmony and proportion. Le Corbusier's faith in the mathematical order of the universe was closely bound to the golden ratio and the Fibonacci series, which he described as "rhythms apparent to the eye and clear in their relations with one another. And these rhythms are at the very root of human activities. They resound in man by an organic inevitability, the same fine inevitability which causes the tracing out of the Golden Section by children, old men, savages and the learned."[21] Le Corbusier explicitly used the golden ratio in his Modulor system for the scale of architectural proportion. He saw this system as a continuation of the long tradition of Vitruvius, Leonardo da Vinci's "Vitruvian Man", the work of Leon Battista Alberti, and others who used the proportions of the human body to improve the appearance and function of architecture. In addition to the golden ratio, Le Corbusier based the system on human measurements, Fibonacci numbers, and the double unit. He took Leonardo's suggestion of the golden ratio in human proportions to an extreme: he sectioned his model human body's height at the navel with the two sections in golden ratio, then subdivided those sections in golden ratio at the knees and throat; he used these golden ratio proportions in the Modulor system. Le Corbusier's 1927 Villa Stein in Garches exemplified the Modulor system's application. The villa's rectangular ground plan, elevation, and inner structure closely approximate golden rectangles.[22] Another Swiss architect, Mario Botta, bases many of his designs on geometric figures. Several private houses he designed in Switzerland are composed of squares and circles, cubes and cylinders. In a house he designed in Origlio, the golden ratio is the proportion between the central section and the side sections of the house.[23] ### Painting Leonardo Da Vinci's illustration from De Divina Proportione applies geometric proportions to the human face. Leonardo da Vinci's illustrations in De Divina Proportione (On the Divine Proportion) and his views that some bodily proportions exhibit the golden ratio have led some scholars to speculate that he incorporated the golden ratio in his own paintings. Some suggest that his Mona Lisa, for example, employs the golden ratio in its geometric equivalents.[24] Whether Leonardo proportioned his paintings according to the golden ratio has been the subject of intense debate. The secretive Leonardo seldom disclosed the bases of his art, and retrospective analysis of the proportions in his paintings can never be conclusive. Salvador Dalí explicitly used the golden ratio in his masterpiece, The Sacrament of the Last Supper. The dimensions of the canvas are a golden rectangle. A huge dodecahedron, with edges in golden ratio to one another, is suspended above and behind Jesus and dominates the composition.[25][2] Mondrian used the golden section extensively in his geometrical paintings.[26] A statistical study on 565 works of art of different great painters, performed in 1999, found that these artists had not used the golden ratio in the size of their canvases. The study concluded that the average ratio of the two sides of the paintings studied is 1.34, with averages for individual artists ranging from 1.04 (Goya) to 1.46 (Bellini).[27] On the other hand, Pablo Tosto listed over 350 works by well-known artists, including more than 100 which have canvasses with golden rectangle and root-5 proportions, and others with proportions like root-2, 3, 4, and 6.[28] ### Book design Depiction of the proportions in a medieval manuscript. According to Jan Tschichold: "Page proportion 2:3. Margin proportions 1:1:2:3. Text area proportioned in the Golden Section."[29] According to Jan Tschichold,[30] "There was a time when deviations from the truly beautiful page proportions 2:3, 1:√3, and the Golden Section were rare. Many books produced between 1550 and 1770 show these proportions exactly, to within half a millimetre." ### Perceptual studies Studies by psychologists, starting with Fechner, have been devised to test the idea that the golden ratio plays a role in human perception of beauty. While Fechner found a preference for rectangle ratios centered on the golden ratio, later attempts to carefully test such a hypothesis have been, at best, inconclusive.[2][31] ### Music James Tenney reconceived his piece For Ann (rising), which consists of up to twelve computer-generated upwardly glissandoing tones (see Shepard tone), as having each tone start so it is the golden ratio (in between an equal tempered minor and major sixth) below the previous tone, so that the combination tones produced by all consecutive tones are a lower or higher pitch already, or soon to be, produced. Ernő Lendvai analyzes Béla Bartók's works as being based on two opposing systems, that of the golden ratio and the acoustic scale,[32] though other music scholars reject that analysis.[2] In Bartok's Music for Strings, Percussion and Celesta the xylophone progression occurs at the intervals 1:2:3:5:8:5:3:2:1.[33] French composer Erik Satie used the golden ratio in several of his pieces, including Sonneries de la Rose+Croix. The golden ratio is also apparent in the organisation of the sections in the music of Debussy's Image, Reflections in Water, in which "the sequence of keys is marked out by the intervals 34, 21, 13 and 8, and the main climax sits at the phi position."[33] The musicologist Roy Howat has observed that the formal boundaries of La Mer correspond exactly to the golden section.[34] Trezise finds the intrinsic evidence "remarkable," but cautions that no written or reported evidence suggests that Debussy consciously sought such proportions.[35] Also, many works of Chopin, mainly Etudes (studies) and Nocturnes, are formally based on the golden ratio. This results in the biggest climax of both musical expression and technical difficulty after about 2/3 of the piece. Pearl Drums positions the air vents on its Masters Premium models based on the golden ratio. The company claims that this arrangement improves bass response and has applied for a patent on this innovation.[36] In the opinion of author Leon Harkleroad, "Some of the most misguided attempts to link music and mathematics have involved Fibonacci numbers and the related golden ratio."[37] ## Nature Adolf Zeising, whose main interests were mathematics and philosophy, found the golden ratio expressed in the arrangement of branches along the stems of plants and of veins in leaves. He extended his research to the skeletons of animals and the branchings of their veins and nerves, to the proportions of chemical compounds and the geometry of crystals, even to the use of proportion in artistic endeavors. In these phenomena he saw the golden ratio operating as a universal law.[38] Zeising wrote in 1854: [The Golden Ratio is a universal law] in which is contained the ground-principle of all formative striving for beauty and completeness in the realms of both nature and art, and which permeates, as a paramount spiritual ideal, all structures, forms and proportions, whether cosmic or individual, organic or inorganic, acoustic or optical; which finds its fullest realization, however, in the human form. [39] ## Mathematics ### Golden ratio conjugate The negative root of the quadratic equation for φ (the "conjugate root") is 1 − ϕ ≈ −0.618. The absolute value of this quantity (≈ 0.618) corresponds to the length ratio taken in reverse order (shorter segment length over longer segment length, b / a), and is sometimes referred to as the golden ratio conjugate.[40] It is denoted here by the capital Phi (Φ): $\Phi = {1 \over \varphi} \approx 0.61803\,39887\,.$ Alternatively, Φ can be expressed as $\Phi = \varphi -1\,.$ This illustrates the unique property of the golden ratio among positive numbers, that ${1 \over \varphi} = \varphi - 1\,$ or its inverse: ${1 \over \Phi} = \Phi + 1\,.$ ### Short proofs of irrationality #### Contradiction from an expression in lowest terms Recall that: the whole is the longer part plus the shorter part; the whole is to the longer part as the longer part is to the shorter part. If we call the whole n and the longer part m, then the second statement above becomes n is to m as m is to n − m, or, algebraically $\frac nm = \frac{m}{n-m}.\qquad ()$ To say that φ is rational means that φ is a fraction n/m where n and m are integers. We may take n/m to be in lowest terms and n and m to be positive. But if n/m is in lowest terms, then the identity labeled () above says m/(n − m) is in still lower terms. That is a contradiction that follows from the assumption that φ is rational. #### Derivation from irrationality of √5 Another short proof—perhaps more commonly known—of the irrationality of the golden ratio makes use of the closure of rational numbers under addition and multiplication. If $\textstyle\frac{1 + \sqrt{5}}{2}$ is rational, then $\textstyle2\left(\frac{1 + \sqrt{5}}{2} - \frac{1}{2}\right) = \sqrt{5}$ is also rational, which is a contradiction if it is already known that the square root of a non-square natural number is irrational. ### Alternate forms The formula φ = 1 + 1/φ can be expanded recursively to obtain a continued fraction for the golden ratio:[41] $\varphi = [1; 1, 1, 1, \dots] = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}$ and its reciprocal: $\varphi^{-1} = [0; 1, 1, 1, \dots] = 0 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}\,.$ The convergents of these continued fractions (1, 2, 3/2, 5/3, 8/5, 13/8, … , or 1, 1/2, 2/3, 3/5, 5/8, 8/13, …) are ratios of successive Fibonacci numbers. The equation φ2 = 1 + φ likewise produces the continued square root form: $\varphi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}\,.$ Also: $\varphi = 1+2\sin(\pi/10) = 1 + 2\sin 18^\circ$ $\varphi = {1 \over 2}\csc(\pi/10) = {1 \over 2}\csc 18^\circ$ $\varphi = 2\cos(\pi/5)=2\cos 36^\circ.\,$ These correspond to the fact that the length of the diagonal of a regular pentagon is φ times the length of its side, and similar relations in a pentagram. If x agrees with φ to n decimal places, then $\frac{x^2+2x}{x^2+1}$ agrees with it to 2n decimal places. This relationship depends upon the choice of the degree as the measure of angle, and will not hold when using other units of angular measure. ### Geometry The number φ turns up frequently in geometry, particularly in figures with pentagonal symmetry. The length of a regular pentagon's diagonal is φ times its side. The vertices of a regular icosahedron are those of three mutually orthogonal golden rectangles. There is no known general algorithm to arrange a given number of nodes evenly on a sphere, for any of several definitions of even distribution (see, for example, Thomson problem). However, a useful approximation results from dividing the sphere into parallel bands of equal area and placing one node in each band at longitudes spaced by a golden section of the circle, i.e. 360°/φ 222.5°. This method was used to arrange the 1500 mirrors of the student-participatory satellite Starshine-3.[42]. #### Golden triangle, pentagon and pentagram ##### Golden triangle Golden triangle The golden triangle can be characterised as an isosceles triangle ABC with the property that bisecting the angle C produces a new triangle CXB which is a similar triangle to the original. If angle BCX = α, then XCA = α because of the bisection, and CAB = α because of the similar triangles; ABC = 2α from the original isosceles symmetry, and BXC = 2α by similarity. The angles in a triangle add up to 180°, so 5α = 180, giving α = 36°. So the angles of the golden triangle are thus 36°-72°-72°. The angles of the remaining obtuse isosceles triangle AXC (sometimes called the golden gnomon) are 36°-36°-108°. Suppose XB has length 1, and we call BC length φ. Because of the isosceles triangles BC=XC and XC=XA, so these are also length φ. Length AC = AB, therefore equals φ+1. But triangle ABC is similar to triangle CXB, so AC/BC = BC/BX, and so AC also equals φ2. Thus φ2 = φ+1, confirming that φ is indeed the golden ratio. ##### Pentagram A pentagram colored to distinguish its line segments of different lengths. The four lengths are in golden ratio to one another. The golden ratio plays an important role in regular pentagons and pentagrams. Each intersection of edges sections other edges in the golden ratio. Also, the ratio of the length of the shorter segment to the segment bounded by the 2 intersecting edges (a side of the pentagon in the pentagram's center) is φ, as the four-color illustration shows. The pentagram includes ten isosceles triangles: five acute and five obtuse isosceles triangles. In all of them, the ratio of the longer side to the shorter side is φ. The acute triangles are golden triangles. The obtuse isosceles triangles are golden gnomon. ##### Ptolemy's theorem The golden ratio in a regular pentagon can be computed using Ptolemy's theorem. The golden ratio can also be confirmed by applying Ptolemy's theorem to the quadrilateral formed by removing one vertex from a regular pentagon. If the quadrilateral's long edge and diagonals are b, and short edges are a, then Ptolemy's theorem gives b2 = a2 + ab which yields ${b \over a}={{(1+\sqrt{5})}\over 2}\,.$ #### Scalenity of triangles Consider a triangle with sides of lengths a, b, and c in decreasing order. Define the "scalenity" of the triangle to be the smaller of the two ratios a/b and b/c. The scalenity is always less than φ and can be made as close as desired to φ.[43] ### Relationship to Fibonacci sequence Approximate and true golden spirals. The green spiral is made from quarter-circles tangent to the interior of each square, while the red spiral is a Golden Spiral, a special type of logarithmic spiral. Overlapping portions appear yellow. The length of the side of a larger square to the next smaller square is in the golden ratio. A Fibonacci spiral that approximates the golden spiral, using Fibonacci sequence square sizes up to 34. The mathematics of the golden ratio and of the Fibonacci sequence are intimately interconnected. The Fibonacci sequence is: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, … The closed-form expression (known as Binet's formula, even though it was already known by Abraham de Moivre) for the Fibonacci sequence involves the golden ratio: $F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}} = {{\varphi^n-(-\varphi)^{-n}} \over {\sqrt 5}}\,.$ The golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence (or any Fibonacci-like sequence): $\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\varphi.$ Therefore, if a Fibonacci number is divided by its immediate predecessor in the sequence, the quotient approximates φ; e.g., 987/610 ≈ 1.6180327868852. These approximations are alternately lower and higher than φ, and converge on φ as the Fibonacci numbers increase, and: $\sum_{n=1}^{\infty}\|F(n)\varphi-F(n+1)\| = \varphi\,.$ Furthermore, the successive powers of φ obey the Fibonacci recurrence: $\varphi^{n+1} = \varphi^n + \varphi^{n-1}\,.$ This identity allows any polynomial in φ to be reduced to a linear expression. For example: \begin{align} 3\varphi^3 - 5\varphi^2 + 4 & = 3(\varphi^2 + \varphi) - 5\varphi^2 + 4 \\ & = 3[(\varphi + 1) + \varphi] - 5(\varphi + 1) + 4 \\ & = \varphi + 2 \approx 3.618. \end{align} ### Other properties The golden ratio has the simplest expression (and slowest convergence) as a continued fraction expansion of any irrational number (see Alternate forms above). It is, for that reason, one of the worst cases of the Lagrange's approximation theorem. This may be the reason angles close to the golden ratio often show up in phyllotaxis (the growth of plants). The defining quadratic polynomial and the conjugate relationship lead to decimal values that have their fractional part in common with φ: $\varphi^2 = \varphi + 1 = 2.618\dots\,$ ${1 \over \varphi} = \varphi - 1 = 0.618\dots\,.$ The sequence of powers of φ contains these values 0.618…, 1.0, 1.618…, 2.618…; more generally, any power of φ is equal to the sum of the two immediately preceding powers: $\varphi^n = \varphi^{n-1} + \varphi^{n-2} = \varphi \cdot \operatorname{F}_n + \operatorname{F}_{n-1}\,.$ As a result, one can easily decompose any power of φ into a multiple of φ and a constant. The multiple and the constant are always adjacent Fibonacci numbers. This leads to another property of the positive powers of φ: If $\lfloor n/2 - 1 \rfloor = m$, then: $\!\ \varphi^n = \varphi^{n-1} + \varphi^{n-3} + \cdots + \varphi^{n-1-2m} + \varphi^{n-2-2m}.$ When the golden ratio is used as the base of a numeral system (see Golden ratio base, sometimes dubbed phinary or φ-nary), every integer has a terminating representation, despite φ being irrational, but every fraction has a non-terminating representation. The golden ratio is the fundamental unit of the algebraic number field $\mathbb{Q}(\sqrt{5})$ and is a Pisot-Vijayaraghavan number. Also, ${\varphi + 1 \over \varphi - 1} = \varphi ^3.$ ### Decimal expansion The golden ratio's decimal expansion can be calculated directly from the expression $\varphi = {1+\sqrt{5} \over 2},$ with √5 ≈ 2.2360679774997896964. The square root of 5 can be calculated with the Babylonian method, starting with an initial estimate such as xφ = 2 and iterating $x_{n+1} = \frac{(x_n + 5/x_n)}{2}$ for n = 1, 2, 3, …, until the difference between xn and xn−1 becomes zero, to the desired number of digits. The Babylonian algorithm for √5 is equivalent to Newton's method for solving the equation x2 − 5 = 0. In its more general form, Newton's method can be applied directly to any algebraic equation, including the equation x2 − x − 1 = 0 that defines the golden ratio. This gives an iteration that converges to the golden ratio itself, $x_{n+1} = \frac{x_n^2 + 1}{2x_n - 1},$ for an appropriate initial estimate xφ such as xφ = 1. A slightly faster method is to rewrite the equation as x − 1 − 1/x = 0, in which case the Newton iteration becomes $x_{n+1} = \frac{x_n^2 + 2x_n}{x_n^2 + 1}.$ These iterations all converge quadratically; that is, each step roughly doubles the number of correct digits. The golden ratio is therefore relatively easy to compute with arbitrary precision. The time needed to compute n digits of the golden ratio is proportional to the time needed to divide two n-digit numbers. This is considerably faster than known algorithms for the transcendental numbers π and e. An easily programmed alternative using only integer arithmetic is to calculate two large consecutive Fibonacci numbers and divide them. The ratio of Fibonacci numbers F25001 and F25000, each over 5000 digits, yields over 10,000 significant digits of the golden ratio. Millions of digits of φ are available (sequence A001622 in OEIS). See the web page of Alexis Irlande for the 17,000,000,000 first digits[44]. ## Pyramids A regular square pyramid is determined by its medial right triangle, whose edges are the pyramid's apothem (a), semi-base (b), and height (h); the face inclination angle is also marked. Mathematical proportions b:h:a of $1:\sqrt{\varphi}:\varphi$ and $3:4:5\$ and $1:4/\pi:1.61899\$ are of particular interest in relation to Egyptian pyramids. Both Egyptian pyramids and those mathematical regular square pyramids that resemble them can be analyzed with respect to the golden ratio and other ratios. ### Mathematical pyramids and triangles A pyramid in which the apothem (slant height along the bisector of a face) is equal to φ times the semi-base (half the base width) is sometimes called a golden pyramid. The isosceles triangle that is the face of such a pyramid can be constructed from the two halves of a diagonally split golden rectangle (of size semi-base by apothem), joining the medium-length edges to make the apothem. The height of this pyramid is $\sqrt{\varphi}$ times the semi-base (that is, the slope of the face is $\sqrt{\varphi}$); the square of the height is equal to the area of a face, φ times the square of the semi-base. The medial right triangle of this "golden" pyramid (see diagram), with sides $1:\sqrt{\varphi}:\varphi$ is interesting in its own right, demonstrating via the Pythagorean theorem the relationship $\sqrt{\varphi} = \sqrt{\varphi^2 - 1}$ or $\varphi = \sqrt{1 + \varphi}$. This "Kepler triangle"[45] is the only right triangle proportion with edge lengths in geometric progression,[46] just as the 3–4–5 triangle is the only right triangle proportion with edge lengths in arithmetic progression. The angle with tangent $\sqrt{\varphi}$ corresponds to the angle that the side of the pyramid makes with respect to the ground, 51.827… degrees (51° 49' 38").[47] A nearly similar pyramid shape, but with rational proportions, is described in the Rhind Mathematical Papyrus (the source of a large part of modern knowledge of ancient Egyptian mathematics), based on the 3:4:5 triangle;[48] the face slope corresponding to the angle with tangent 4/3 is 53.13 degrees (53 degrees and 8 minutes).[49] The slant height or apothem is 5/3 or 1.666… times the semi-base. The Rhind papyrus has another pyramid problem as well, again with rational slope (expressed as run over rise). Egyptian mathematics did not include the notion of irrational numbers,[50] and the rational inverse slope (run/rise, multiplied by a factor of 7 to convert to their conventional units of palms per cubit) was used in the building of pyramids.[48] Another mathematical pyramid with proportions almost identical to the "golden" one is the one with perimeter equal to 2π times the height, or h:b = 4:π. This triangle has a face angle of 51.854° (51°51'), very close to the 51.827° of the Kepler triangle. This pyramid relationship corresponds to the coincidental relationship $\sqrt{\varphi} \approx 4/\pi$. Egyptian pyramids very close in proportion to these mathematical pyramids are known.[49] ### Egyptian pyramids In the mid nineteenth century, Röber studied various Egyptian pyramids including Khafre, Menkaure and some of the Gizeh, Sakkara and Abusir groups, and was interpreted as saying that half the base of the side of the pyramid is the middle mean of the side, forming what other authors identified as the Kepler triangle; many other mathematical theories of the shape of the pyramids have also been explored.[46] One Egyptian pyramid is remarkably close to a "golden pyramid" – the Great Pyramid of Giza (also known as the Pyramid of Cheops or Khufu). Its slope of 51° 52' is extremely close to the "golden" pyramid inclination of 51° 50' and the π-based pyramid inclination of 51° 51'; other pyramids at Giza (Chephren, 52° 20', and Mycerinus, 50° 47')[48] are also quite close. Whether the relationship to the golden ratio in these pyramids is by design or by accident remains controversial. Several other Egyptian pyramids are very close to the rational 3:4:5 shape.[49] Adding fuel to controversy over the architectural authorship of the Great Pyramid, Eric Temple Bell, mathematician and historian, claimed in 1950 that Egyptian mathematics would not have supported the ability to calculate the slant height of the pyramids, or the ratio to the height, except in the case of the 3:4:5 pyramid, since the 3:4:5 triangle was the only right triangle known to the Egyptians and they did not know the Pythagorean theorem nor any way to reason about irrationals such as π or φ.[51] Michael Rice[52] asserts that principal authorities on the history of Egyptian architecture have argued that the Egyptians were well acquainted with the golden ratio and that it is part of mathematics of the Pyramids, citing Giedon (1957).[53] Historians of science have always debated whether the Egyptians had any such knowledge or not, contending rather that its appearance in an Egyptian building is the result of chance.[54] In 1859, the pyramidologist John Taylor claimed that, in the Great Pyramid of Giza, the golden ratio is represented by the ratio of the length of the face (the slope height), inclined at an angle θ to the ground, to half the length of the side of the square base, equivalent to the secant of the angle θ.[55] The above two lengths were about 186.4 and 115.2 meters respectively. The ratio of these lengths is the golden ratio, accurate to more digits than either of the original measurements. Similarly, Howard Vyse, according to Matila Ghyka,[56] reported the great pyramid height 148.2 m, and half-base 116.4 m, yielding 1.6189 for the ratio of slant height to half-base, again more accurate than the data variability. ## Disputed sightings Examples of disputed observations of the golden ratio include the following: • Historian John Man states that the pages of the Gutenberg Bible were "based on the golden section shape". However, according to Man's own measurements, the ratio of height to width was 1.45.[57] • In 1991, Jean-Claude Perez proposed a connection between DNA base sequences and gene sequences and the golden ratio.[58][59] Another such connection, between the Fibonacci numbers and golden ratio and Chargaff's second rule concerning the proportions of nucleobases in the human genome, was proposed in 2007.[60] The sculpture Ratio by Andrew Rogers in Jerusalem is proportioned according to Fibonacci numbers; some call it Golden Ratio. • Australian sculptor Andrew Rogers's 50-ton stone and gold sculpture entitled Ratio, installed outdoors in Jerusalem.[61] Despite the sculpture's sometimes being referred to as "Golden Ratio,"[62] it is not proportioned according to the golden ratio, and the sculptor does not call it that: the height of each stack of stones, beginning from either end and moving toward the center, is the beginning of the Fibonacci sequence: 1, 1, 2, 3, 5, 8. His sculpture Ascend in Sri Lanka, also in his Rhythms of Life series, is similarly constructed, with heights 1, 1, 2, 3, 5, 8, 13, but no descending side.[61] • It is sometimes claimed that the number of bees in a beehive divided by the number of drones yields the golden ratio.[63] In reality, the proportion of drones in a beehive varies greatly by beehive, by bee race, by season, and by beehive health status; normal hive populations range from 5,000 to 20,000 bees, while drone numbers range "from none in the winter to as many as 1,500 in the spring and summer" (Graham, 1992, pp 350),[64] thus the ratio is normally much greater than the golden ratio. * This misunderstanding may arise because in theory bees have approximately this ratio of male to female ancestors (See The Bee Ancestry Code) - the caveat being that ancestry can trace back to the same drone by more than one route, so the actual numbers of bees do not need to match the formula. • Some specific proportions in the bodies of many animals (including humans[65][66]) and parts of the shells of mollusks[4] and cephalopods are often claimed to be in the golden ratio. There is actually a large variation in the real measures of these elements in specific individuals, and the proportion in question is often significantly different from the golden ratio.[65] The ratio of successive phalangeal bones of the digits and the metacarpal bone has been said to approximate the golden ratio.[66] The Nautilus shell, the construction of which proceeds in a logarithmic spiral, is often cited, usually with the idea that any logarithmic spiral is related to the golden ratio, but sometimes with the claim that each new chamber is proportioned by the golden ratio relative to the previous one;[63] however, measurements of Nautilus shells do not support this claim.[67] • The proportions of different plant components (numbers of leaves to branches, diameters of geometrical figures inside flowers) are often claimed to show the golden ratio proportion in several species.[68] In practice, there are significant variations between individuals, seasonal variations, and age variations in these species. While the golden ratio may be found in some proportions in some individuals at particular times in their life cycles, there is no consistent ratio in their proportions. • In investing, some practitioners of technical analysis use the golden ratio to indicate support of a price level, or resistance to price increases, of a stock or commodity; after significant price changes up or down, new support and resistance levels are supposedly found at or near prices related to the starting price via the golden ratio.[69] The use of the golden ratio in investing is also related to more complicated patterns described by Fibonacci numbers; see, e.g. Elliott wave principle. However, other market analysts have published analyses suggesting that these percentages and patterns are not supported by the data.[70] • An equation derived in 1994 connects the golden ratio to the Number of the Beast (666):[2] $-\frac{\varphi}{2}=\sin666^\circ=\cos(6\cdot 6 \cdot 6^\circ).{}$ Which can be combined into the expression: $-\varphi=\sin666^\circ+\cos(6\cdot 6 \cdot 6^\circ).{}$ ## References and footnotes 1. ^ a b The golden ratio can be derived by the quadratic formula, by starting with the first number as 1, then solving for 2nd number x, where the ratios (x + 1)/x = x/1 or (multiplying by x) yields: x + 1 = x2, or thus a quadratic equation: x2 − x − 1 = 0. Then, by the quadratic formula, for positive x = (−b + √(b2 − 4ac))/(2a) with a = 1, b = −1, c = −1, the solution for x is: (−(−1) + √((−1)2 − 4·1·(−1)))/(2·1) or (1 + √(5))/2. 2. ^ a b c d e f Livio, Mario (2002). The Golden Ratio: The Story of Phi, The World's Most Astonishing Number. New York: Broadway Books. ISBN 0-7679-0815-5. 3. ^ Piotr Sadowski, The Knight on His Quest: Symbolic Patterns of Transition in Sir Gawain and the Green Knight, Cranbury NJ: Associated University Presses, 1996 4. ^ a b Richard A Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific Publishing, 1997 5. ^ a b c d Euclid, Elements, Book 6, Definition 3. 6. ^ Summerson John, Heavenly Mansions: And Other Essays on Architecture (New York: W.W. Norton, 1963) pp.37 . "And the same applies in architecture, to the rectangles representing these and other ratios (e.g. the 'golden cut'). The sole value of these ratios is that they are intellectually fruitful and suggest the rhythms of modular design." 7. ^ Jay Hambidge, Dynamic Symmetry: The Greek Vase, New Haven CT: Yale University Press, 1920 8. ^ William Lidwell, Kritina Holden, Jill Butler, Universal Principles of Design: A Cross-Disciplinary Reference, Gloucester MA: Rockport Publishers, 2003 9. ^ Pacioli, Luca. De divina proportione, Luca Paganinem de Paganinus de Brescia (Antonio Capella) 1509, Venice. 10. ^ Euclid, Elements, Book 6, Proposition 30. 11. ^ Euclid, Elements, Book 2, Proposition 11; Book 4, Propositions 10–11; Book 13, Propositions 1–6, 8–11, 16–18. 12. ^ "The Golden Ratio". The MacTutor History of Mathematics archive. Retrieved on 2007-09-18. 13. ^ Hemenway, Priya (2005). Divine Proportion: Phi In Art, Nature, and Science. New York: Sterling. pp. 20–21. ISBN 1-4027-3522-7. 14. ^ Plato (360 BC) (Benjamin Jowett trans.). "Timaeus". The Internet Classics Archive. Retrieved on May 30 2006. 15. ^ Underwood Dudley (1999). Die Macht der Zahl: Was die Numerologie uns weismachen will. Springer. pp. p.245. ISBN 3764359781. 16. ^ Cook, Theodore Andrea (1979) [1914]. The Curves of Life. New York: Dover Publications. ISBN 0-48623-701-X. 17. ^ Van Mersbergen, Audrey M., "Rhetorical Prototypes in Architecture: Measuring the Acropolis", Philosophical Polemic Communication Quarterly, Vol. 46, 1998. 18. ^ Midhat J. Gazalé , Gnomon, Princeton University Press, 1999. ISBN 0-691-00514-1 19. ^ Keith J. Devlin The Math Instinct: Why You're A Mathematical Genius (Along With Lobsters, Birds, Cats, And Dogs) New York: Thunder's Mouth Press, 2005, ISBN 1-56025-672-9 20. ^ Boussora, Kenza and Mazouz, Said, The Use of the Golden Section in the Great Mosque of Kairouan, Nexus Network Journal, vol. 6 no. 1 (Spring 2004), Available online 21. ^ Le Corbusier, The Modulor p. 25, as cited in Padovan, Richard, Proportion: Science, Philosophy, Architecture (1999), p. 316, Taylor and Francis, ISBN 0-419-22780-6 22. ^ Le Corbusier, The Modulor, p. 35, as cited in Padovan, Richard, Proportion: Science, Philosophy, Architecture (1999), p. 320. Taylor & Francis. ISBN 0-419-22780-6: "Both the paintings and the architectural designs make use of the golden section". 23. ^ Urwin, Simon. Analysing Architecture (2003) pp. 154-5, ISBN 0-415-30685-X 24. ^ Livio, Mario. "The golden ratio and aesthetics". Retrieved on 2008-03-21. 25. ^ Hunt, Carla Herndon and Gilkey, Susan Nicodemus. Teaching Mathematics in the Block pp. 44, 47, ISBN 1-883001-51-X 26. ^ Bouleau, Charles, The Painter's Secret Geometry: A Study of Composition in Art (1963) pp.247-8, Harcourt, Brace & World, ISBN 0-87817-259-9 27. ^ Olariu, Agata, Golden Section and the Art of Painting Available online 28. ^ Tosto, Pablo, La composición áurea en las artes plásticas - El número de oro, Librería Hachette, 1969, p. 134 -144 29. ^ Ibid. Tschichold, pp.43 Fig 4. "Framework of ideal proportions in a medieval manuscript without multiple columns. Determined by Jan Tschichold 1953. Page proportion 2:3. margin proportions 1:1:2:3, Text area proportioned in the Golden Section. The lower outer corner of the text area is fixed by a diagonal as well." 30. ^ Jan Tschichold, The Form of the Book, Hartley & Marks (1991), ISBN 0-88179-116-4. 31. ^ The golden ratio and aesthetics, by Mario Livio 32. ^ Lendvai, Ernő (1971). Béla Bartók: An Analysis of His Music. London: Kahn and Averill. 33. ^ a b Smith, Peter F. The Dynamics of Delight: Architecture and Aesthetics (New York: Routledge, 2003) pp 83, ISBN 0-415-30010-X 34. ^ Roy Howat (1983). Debussy in Proportion: A Musical Analysis. Cambridge University Press. ISBN 0521311454. 35. ^ Simon Trezise (1994). Debussy: La Mer. Cambridge University Press. pp. p.53. ISBN 0521446562. 36. ^ "Pearl Masters Premium". Pearl Corporation. Retrieved on December 2 2007. 37. ^ Leon Harkleroad (2006). The Math Behind the Music. Cambridge University Press. ISBN 0521810957. 38. ^ Ibid. Padovan, R. Proportion: Science, Philosophy, Architecture , pp. 305-06 39. ^ Zeising, Adolf, Neue Lehre van den Proportionen des meschlischen Körpers, Leipzig, 1854, preface. 40. ^ Eric W. Weisstein, Golden Ratio Conjugate at MathWorld. 41. ^ Max. Hailperin, Barbara K. Kaiser, and Karl W. Knight (1998). Concrete Abstractions: An Introduction to Computer Science Using Scheme. Brooks/Cole Pub. Co. ISBN 0534952119. 42. ^ "A Disco Ball in Space". NASA. 2001-10-09. Retrieved on 2007-04-16. 43. ^ American Mathematical Monthly, pp. 49-50, 1954. 44. ^ The golden number to 17 000 000 000 digits. Universidad Nacional de Colombia. 2008. 45. ^ The Best of Astraea: 17 Articles on Science, History and Philosophy. Astrea Web Radio. 2006. ISBN 1425970400. 46. ^ a b Roger Herz-Fischler (2000). The Shape of the Great Pyramid. Wilfrid Laurier University Press. ISBN 0889203245. 47. ^ Midhat Gazale, Gnomon: From Pharaohs to Fractals, Princeton Univ. Press, 1999 48. ^ a b c Eli Maor, Trigonometric Delights, Princeton Univ. Press, 2000 49. ^ a b c "The Great Pyramid, The Great Discovery, and The Great Coincidence". Retrieved on 2007-11-25. 50. ^ Lancelot Hogben, Mathematics for the Million, London: Allen & Unwin, 1942, p. 63., as cited by Dick Teresi, Lost Discoveries: The Ancient Roots of Modern Science—from the Babylonians to the Maya, New York: Simon & Schuster, 2003, p.56 51. ^ Eric Temple Bell, The Development of Mathematics, New York: Dover, 1940, p.40 52. ^ Rice, Michael, Egypt's Legacy: The Archetypes of Western Civilisation, 3000 to 30 B.C pp. 24 Routledge, 2003, ISBN 0-415-26876-1 53. ^ S. Giedon, 1957, The Beginnings of Architecture, The A.W. Mellon Lectures in the Fine Arts, 457, as cited in Rice, Michael, Egypt's Legacy: The Archetypes of Western Civilisation, 3000 to 30 B.C pp.24 Routledge, 2003 54. ^ Markowsky, George (January 1992). "Misconceptions about the Golden Ratio" (PDF). College Mathematics Journal 23 (1): 1. 55. ^ Taylor, The Great Pyramid: Why Was It Built and Who Built It?, 1859 56. ^ Matila Ghyka The Geometry of Art and Life, New York: Dover, 1977 57. ^ Man, John, Gutenberg: How One Man Remade the World with Word (2002) pp. 166-67, Wiley, ISBN 0-471-21823-5. "The half-folio page (30.7 x 44.5 cm) was made up of two rectangles—the whole page and its text area—based on the so called 'golden section', which specifies a crucial relationship between short and long sides, and produces an irrational number, as pi is, but is a ratio of about 5:8." 58. ^ J.C. Perez (1991), "Chaos DNA and Neuro-computers: A Golden Link", in Speculations in Science and Technology vol. 14 no. 4, ISSN 0155-7785 59. ^ Perez, Jean-claude (1997). L'ADN décrypté. Embourg (Belgium): Marco Pietteur. ISBN 2-87211-017-8. 60. ^ Yamagishi, Michel E.B., and Shimabukuro, Alex I. (2007), "Nucleotide Frequencies in Human Genome and Fibonacci Numbers", in Bulletin of Mathematical Biology, ISSN 0092-8240 (print), ISSN 1522-9602 (online). 61. ^ a b "Sculptures by Andrew Rogers". 62. ^ ""Golden Ratio" in Jerusalem". 63. ^ a b Ivan Moscovich, Ivan Moscovich Mastermind Collection: The Hinged Square & Other Puzzles, New York: Sterling, 2004 64. ^ Joe M. Graham, "The Hive and the Honey Bee," Hamilton Illinois: Dadant & Sons, 1992 65. ^ a b Stephen Pheasant, Bodyspace, London: Taylor & Francis, 1998 66. ^ a b Walter van Laack, A Better History Of Our World: Volume 1 The Universe, Aachen: van Laach GmbH, 2001. 67. ^ Peterson, Ivars, "Sea shell spirals", Science News . 68. ^ Derek Thomas, Architecture and the Urban Environment: A Vision for the New Age, Oxford: Elsevier, 2002 69. ^ For instance, Osler writes that "38.2 percent and 61.8 percent retracements of recent rises or declines are common," in Osler, Carol (2000). "Support for Resistance: Technical Analysis and Intraday Exchange Rates" (PDF). Federal Reserve Bank of New York Economic Policy Review 6 (2): 53–68. 70. ^ Roy Batchelor and Richard Ramyar, "Magic numbers in the Dow," 25th International Symposium on Forecasting, 2005, p. 13, 31. "Not since the 'big is beautiful' days have giants looked better", Tom Stevenson, The Daily Telegraph, Apr. 10, 2006, and "Technical failure", The Economist, Sep. 23, 2006, are both popular-press accounts of Batchelor and Ramyar's research. • Doczi, György (2005) [1981]. The Power of Limits: Proportional Harmonies in Nature, Art, and Architecture. Boston: Shambhala Publications. ISBN 1-590-30259-1. • Huntley, H. E. (1970). The Divine Proportion: A Study in Mathematical Proportion. New York: Dover Publications. ISBN 0-486-22254-3. • Joseph, George G. (2000) [1991]. The Crest of the Peacock: The Non-European Roots of Mathematics (New Ed. ed.). Princeton, NJ: Princeton University Press. ISBN 0-691-00659-8. • Sahlqvist, Leif (2008). Cardinal Alignments and the Golden Section: Principles of Ancient Cosmography and Design (3rd Rev. Ed. ed.). Charleston, SC: BookSurge. ISBN 1-4196-2157-2. • Schneider, Michael S. (1994). A Beginner's Guide to Constructing the Universe: The Mathematical Archetypes of Nature, Art, and Science. New York: HarperCollins. ISBN 0-060-16939-7. • Walser, Hans (2001) [Der Goldene Schnitt 1993]. The Golden Section. Peter Hilton trans.. Washington, DC: The Mathematical Association of America. ISBN 0-88385-534-8.
Can someone help me understand percentages? You get a different result because they are different operations: dividing by .85 is like multiplying by 1/.85, which is 1.1765. What makes you think dividing by .85 should be equal to multiplying by 1.15? If you're trying to calculate 15% more or less of something: \- multplying by 1 + percent will give you the increment (1.15 is 15% more) \- multiplying by 1 - percent will give you the decrement (0.85 is 15% less) by 115 is 15% more than 100, but 100 is not 15% less than 115. 15% less than 115 would be 115 * 0.85 = 97.75. 100 is ~13.04% less than 115. Let's say something costs $1.00. The price rises by 15%. That's 15% of$1.00 Let's say something costs $1.15. The price drops by 15%. That's 15% of$1.15. 15% of $1.15 is more than 15% of$1.00. You shouldn't expect "15%" to always represent the same number, it depends on what you're taking 15% of. >Ex: 100/.85 equals 117.6470588 100\1.15 equals 115 Here you're asking the question, "what is 100 85% of? What would you take 15% off from to end up at $1.00?" As I noted, 15% of$1.15 is more than 0.15. If you discount 15% from $1.15 you end up with less than$1.00. You have to start with a larger number than $1.15 in order to take off 15% and end up with$1.00. Because 15% of $1.15 is more than 0.15. So that's why it's something like$1.18. Whenever you use portions, be clear about what is the whole from which these portions come. They’re not just portions, per se. They’re portions of something. For example, let’s say we have 16 pencils and we wish to get rid of a fourth (or a quarter) of them. One fourth of 16 is 4. So we get rid of 4 pencils. Now we only have 12 pencils. Here comes Eliana. Hi, Eliana! We ask Eliana to add a fourth (or a quarter). She sees 12 pencils. She correctly calculates that a fourth of 12 is 3. So she adds 3 pencils. Now we have 15 pencils! Maybe you can now see what’s happening here. When we subtracted a fourth, we could have asked: “A fourth of what?” And the answer would have been: “A fourth of 16.” This means that, for us, 16 is the whole from which we were taking a fourth. When Eliana added a fourth, we could have asked: “A fourth of what?” And Eliana would have answered: “A fourth of 12.” This means that, for Eliana, 12 is the whole to which we are adding a fourth. A fourth of 16 is not the same as a fourth of 12. Percentages are portions. And the same rule applies. We can present the same example as percentages. From our 100% of 16 pencils, we got rid of 25% of them, so we were left with 75% of what we had, that is: 12 pencils. Then Eliana added 25% to that. 25% of 12 is 3. So now we have 15 pencils. We didn’t end up with 16 pencils because 25% of 16 is not the same as 25% of 12. Well the percentage is always relative to something. In absokute units, you can tell for example, that 7 is greater than 4 by 3 units - vice versa 4 is less than 7 by the same 3 units. But if you use percentages, you take a percentage of one of the two numbers here - either 7 or 4. So if we take 4 as 100%, than 7 should be 7/4 = 1.75 = 175%, so 75% greater. On the other side, if we take 7 as 100%, then 4 is 4/7 ≈ 0.5714 = 57.14%, so 42.86% less. The percentage is never the same, it's a relative unit. By the way, here it's either 75% of 4 (which equals 3 by the way) or 42.86% of 7 (which is also 3, exactly 7 - 4). In my opinion, percentages are extremely confusing, but it can help to take it to an extreme: think about 100% more of something -- thats 2x the amount you started with. but 100% less of something is nothing -- so 100% more and 100% less are not opposites. Similarly 15% more and 15% less are not opposites, but the effect is smaller 117.64... \ 0.85 = 100. Ie take 15% of 117.64... and subtract it from 117.64... and you get 100. In other words: 117.64... \* 0.15 - 17.64... Let's rearrange your equation to see how it is similar to my operation above! You said 100 / 0.85 = 117.64... Multiply both sides by 0.85. 100 = 117.64... \* 0.85. That's just what I did above! I.e. 100/0.85 finds the number for which subtracting 15% gets you 100. This is something I do all the time because I work in food production. If my knowns are that spice blend X is 90% salt, and I have 50 kgs of salt, then I can figure out the number for which subtracting 10% gets me 50 kg, or more to the point the number for which 90% is 50 kg, or even more to the point how much spice blend I can make! 50 kgs / 0.90 = 55.55... kg. 100 \* 1.15 to be pedantic is equivalent to (100 \* 1) + (100 \* 0.15). Ie you're adding 15% of what you have to what you have. We call that 15% more. Whenever you're working with percentages, changes in terms of percentages, and changes of percentages there can be difficulty, misinterpretation, and miscommunication. You just have to make yourself clear on the math and then be wary of interpreting the language! If for some reason you just really wanted to divide by a decimal you would have to do something equivalent to 100 \* 1.15. That would be 100 / (1/1.15) = 100 / .8696... That's probably not something you would want to do, but as you wrestle with converting language to math it's good to carry out these algebraic manipulations just so you can get comfortable with what's going on where and when and why to do it one way and not another! 100÷(20/23)=115 (1/1.15=20/23) Dividing by 0.85 is equivalent to multiplying by 1 3/17
# Using Scientific Notation Updated September 7, 2021 \| Infoplease Staff Sometimes, especially when you are using a calculator, you may come up with a very long number. It might be a big number, like 2,890,000,000. Or it might be a small number, like 0.0000073. Scientific notation is a way to make these numbers easier to work with. In scientific notation, you move the decimal place until you have a number between 1 and 10. Then you add a power of ten that tells how many places you moved the decimal. In scientific notation, 2,890,000,000 becomes 2.89 x 109. How? • Remember that any whole number can be written with a decimal point. For example: 2,890,000,000 = 2,890,000,000.0 • Now, move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the left in 2,890,000,000 you will get 2.89. • Next, count how many places you moved the decimal point. You had to move it 9 places to the left to change 2,890,000,000 to 2.89. You can show that you moved it 9 places to the left by noting that the number should be multiplied by 109. 2.89 x 109 = 2.89 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 2.89 x 109 = 2,890,000,000 Scientific notation can be used to turn 0.0000073 into 7.3 x 10-6. • First, move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the right in 0.0000073 you will get 7.3. • Next, count how many places you moved the decimal point. You had to move it 6 places to the right to change 0.0000073 to 7.3. You can show that you moved it 6 places to the right by noting that the number should be multiplied by 10-6. 7.3 x 10-6 = 0.0000073 Remember: in a power of ten, the exponent—the small number above and to the right of the 10—tells which way you moved the decimal point. • A power of ten with a positive exponent, such as 105, means the decimal was moved to the left. • A power of ten with a negative exponent, such as 10-5, means the decimal was moved to the right. Powers of Ten billions 109 = 1,000,000,000 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000,000 millions 106 = 1,000,000 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000 hundred thousands 105 = 100,000 10 x 10 x 10 x 10 x 10 = 100,000 ten thousands 104 = 10,000 10 x 10 x 10 x 10 = 10,000 thousands 103 = 1,000 10 x 10 x 10 = 1,000 hundreds 102 = 100 10 x 10 = 100 tens 101 = 10 ones 100 = 1 tenths 10–1 = 1/10 1/10 = 0.1 hundredths 10–2 = 1/102 1/102 = 0.01 thousandths 10–3 = 1/103 1/103 = 0.001 ten thousandths 10–4 = 1/104 1/104 = 0.0001 hundred thousandths 10–5 = 1/105 1/105 = 0.00001 millionths 10–6 = 1/106 1/106 = 0.000001 billionths 10–9 = 1/109 1/109 = 0.000000001 Finding Square Roots Numbers and Formulas Factorials Numbers and Formulas Sources +
# HP 35s Instruction Manual page 7 Also See for HP 35s: hp calculators HP 35s Using Algebraic Mode The permutation of 69 items taken 2 at a time is 4,692. Note that it was not necessary to move beyond the closing parenthesis before pressing Ï. Percentage functions are used in algebraic mode and are explained in the business training guide on Percentages. Working with complex numbers, with pairs of numbers representing rectangular or polar coordinates and with pairs of numbers in statistical operations are described in training aids on these subjects. Algebraic operator precedence In algebraic mode the HP 35s calculates using "operator precedence". This means that a combination of several Ù and Ã operations (or several ¸ and ¯ operations) is calculated from left to right, but ¸ and ¯ have a higher "precedence" and are carried out before Ù and Ã. Example 5: What is the result of calculating 1 ÷ 2 ÷ 3 in algebraic mode? Solution: Type the keys 1¯2¯3Ï. The result is equal to 1 ÷ 6, in other words the calculation goes from left to right, 1 ÷ 2, and the result divided by 3. If the calculation went the other way, it would be 1 ÷ (2 ÷ 3), or 3 ÷ 2, giving 1.5 as the result. The rules of precedence are that algebraic calculations are carried out in the following order. 1. Expressions in parentheses. 2. All single number functions, also complex number functions, percentages, and co-ordinate transformations. 3. The two-number functions ' and ). 4. The other two-number functions, {, x and p. 5. Multiplication, division and integer quotient and remainder, ¸, ¯, b and `. Operations with the same level of precedence are carried out from left to right, as this example showed. Example 6: In what order is the calculation 1 + (0.5 + 1.5) × 3 Solution: Type the keys 1Ù40Ë5Ù1Ë5Õ¸3)4º*Ï. hp calculators carried out? 4! - 7 - Figure 14 Figure 15 HP 35s Using Algebraic Mode - Version 1.0
# The value of Question: $f(x)=\left\{\begin{array}{ll}\|x\| \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 0^{-}} f(x)=\|x\| \cos \frac{1}{x}$ $=\lim _{h \rightarrow 0}\|0-h\| \cos \frac{1}{(0-h)}=\lim _{h \rightarrow 0} h \cos \frac{1}{h}$ $=0 \quad\left[\because \cos \frac{1}{x}\right.$ oscillate between $-1$ and 1$]$ $\lim _{x \rightarrow 0^{+}} f(x)=\|x\| \cos \frac{1}{x}$ $=\lim _{h \rightarrow 0}\|0+h\| \cos \frac{1}{(0+h)}=\lim _{h \rightarrow 0} h \cdot \cos \frac{1}{h}=0$ $\lim _{x \rightarrow 0} f(x)=0$ $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} f(x)=0$ $\lim _{x \rightarrow 0^{-}} f(x)=\|x\| \cos \frac{1}{x}$ $=\lim _{h \rightarrow 0}\|0-h\| \cos \frac{1}{(0-h)}=\lim _{h \rightarrow 0} h \cos \frac{1}{h}$ $=0 \quad\left[\because \cos \frac{1}{x}\right.$ oscillate between $-1$ and 1$]$ $\lim _{x \rightarrow 0^{+}} f(x)=\|x\| \cos \frac{1}{x}$ $=\lim _{h \rightarrow 0}\|0+h\| \cos \frac{1}{(0+h)}=\lim _{h \rightarrow 0} h \cdot \cos \frac{1}{h}=0$ $\lim _{x \rightarrow 0} f(x)=0$ $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} f(x)=0$ Thus, the given function f(x) is continuous at x = 0.
# The equation of the directrix of the parabola y2 + 4x + 4y + 2 is This question was previously asked in UP TGT Mathematics 2021 Official Paper View all UP TGT Papers > 1. x = -1 2. x = 1 3. $$x=-\dfrac{3}{2}$$ 4. $$x=\dfrac{3}{2}$$ Option 4 : $$x=\dfrac{3}{2}$$ ## Detailed Solution Concept: parabola refers to an equation of a curve, such that a point on the curve is equidistant from a fixed point, and a fixed-line. The fixed point is called the focus of the parabola, and the fixed line is called the directrix of the parabola. Also, an important point to note is that the fixed point does not lie on the fixed-line. A locus of any point which is equidistant from a given point (focus) and a given line (directrix) is called a parabola. The standard equation of a regular parabola is y2 = 4ax. Here, • Coordinates of vertex: (0, 0) • Coordinates of focus: (a, 0) • Equation of the directrix: x = -a • Equation of axis: y = 0 • Length of the latus rectum: 4a • Focal distance of a point P(x, y): a + x Formula used: (a + b)2 = a2 + b2 + 2ab Calculation: Given equation of the parabola is- y2 + 4x + 4y + 2 = 0 ⇒ (y2 + 4y + 2) + 4x - 2 = 0 ⇒ (y + 2)2 = -4x + 2 (∵ (a + b)2 = a2 + b2 + 2ab) $$⇒ (y + 2)^2 = -4\left (x - \frac{1}{2} \right )$$ Let y + 2 = Y and (x - 1/2) = X ⇒ Y2 = -4X On comparing this equation with y2 = -4ax, we get a = 1 ⇒ Equation of directrix = (X = a) $$⇒ \left (x - \frac{1}{2} \right ) = 1$$ $$\Rightarrow x=\dfrac{3}{2}$$ Hence, The equation of the directrix of the parabola y2 + 4x + 4y + 2 = 0 is $$x=\dfrac{3}{2}$$.
# If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)] Question: If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)] is also equal to y. $\left[\right.$ Hint : Express $\left.\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\right]$ Solution: According to the question, y =2sinα /(1+cosα+sinα) Multiplying numerator and denominator by (1 – cos α + sin α), We get, $\Rightarrow y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha} \times \frac{1-\cos \alpha+\sin \alpha}{1-\cos \alpha+\sin \alpha}$ $=\frac{2 \sin \alpha}{(1+\sin \alpha)+\cos \alpha} \times \frac{(1+\sin \alpha)-\cos \alpha}{(1+\sin \alpha)-\cos \alpha}$ Using $(a+b)(a-b)=a^{2}-b^{2}$, we get: $=\frac{2 \sin \alpha\{(1+\sin \alpha)-\cos \alpha\}}{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}$ $=\frac{2 \sin \alpha(1+\sin \alpha)-2 \sin \alpha \cos \alpha}{1+\sin ^{2} \alpha+2 \sin \alpha-\cos ^{2} \alpha}$ Since, $1-\cos ^{2} \alpha=\sin ^{2} \alpha$ $\therefore y=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{\sin ^{2} \alpha+2 \sin \alpha+\sin ^{2} \alpha}$ $=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{2 \sin \alpha(1+\sin \alpha)}$ $\Rightarrow y=\frac{(1+\sin \alpha-\cos \alpha)}{(1+\sin \alpha)}$ $\Rightarrow y=\frac{(1-\cos \alpha+\sin \alpha)}{(1+\sin \alpha)}$ Hence Proved
Newton's Laws: Chapter Outline \|\| About the Tutorial \|\| Tutorial Topics \|\| Usage Policy \|\| Feedback ## Lesson 3: Newton's Second Law of Motion ### Finding Acceleration As learned earlier in Lesson 3 (as well as in Lesson 2), the net force is the vector sum of all the individual forces. In Lesson 2, we learned how to determine the net force if the magnitudes of all the individual forces are known. In this lesson, we will learn how to determine the acceleration of an object if the magnitudes of all the individual forces are known. The three major equations which will be useful are the equation for net force (Fnet = ma), the equation for gravitational force (Fgrav = mg), and the equation for frictional force (Ffrict = "mu"*Fnorm). The process of determining the acceleration of an object demands that the mass and the net force are known. If mass (m) and net force (Fnet) are known, then the acceleration is determined by use of the equation. Thus, the task involves using the above equations, the given information, and your understanding of Newton's laws to determine the acceleration. To gain a feel for how this method is applied, try the following practice problems. Once you have solved the problems, click the button to check your answers. ### Practice #1 An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (Neglect air resistance.) ### Practice #2 An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the coefficient of friction ("mu") between the object and the surface, the mass, and the acceleration of the object. (Neglect air resistance.) ### Practice #3 A 5-kg object is sliding to the right and encountering a friction force which slows it down. The coefficient of friction ("mu") between the object and the surface is 0.1. Determine the force of gravity, the normal force, the force of friction, the net force, and the acceleration. (Neglect air resistance.) A couple more practice problems are provided below. You should make an effort to solve as many problems as you can without the assistance of notes, solutions, teachers, and other students. Commit yourself to individually solving the problems. In the meantime, an important caution is worth mentioning: Avoid forcing a problem into the form of a previously solved problem. Problems in physics will seldom look the same. Instead of solving problems by rote or by mimicry of a previously solved problem, utilize your conceptual understanding of Newton's laws to work towards solutions to problems. Use your understanding of weight and mass to find the m or the Fgrav in a problem. Use your conceptual understanding of net force (vector sum of all the forces) to find the value of Fnet or the value of an individual force. Do not divorce the solving of physics problems from your understanding of physics concepts. If you are unable to solve physics problems like those above, it is does not necessarily mean that you are having math difficulties. It is likely that you are having a physics concepts difficulty. 1. Edwardo applies a 4.25-N rightward force to a 0.765-kg book to accelerate it across a table top. The coefficient of friction between the book and the tabletop is 0.410. Determine the acceleration of the book. 2. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N rightward force on a 0.500-kg cart to accelerate it across a low-friction track. If the total resistance force to the motion of the cart is 0.72 N, then what is the cart's acceleration? ### Lesson 3: Newton's Second Law of Motion Newton's Laws: Chapter Outline \|\| About the Tutorial \|\| Tutorial Topics \|\| Usage Policy \|\| Feedback
Technology # How to Find the Unit Rate ## Understanding Unit Rates and Why They Are Important Unit rates are ratios that are simplified so that they describe the relationship between a quantity and one unit of that quantity. For example, if you drive 120 miles in 3 hours, the unit rate of your speed is 40 miles per hour. This means that you traveled 40 miles for every hour you drove. Unit rates are important because they allow us to compare two quantities that are measured in different units. They also help us to make predictions and solve problems in real-life situations. For example, unit rates can be used to calculate how much gas is needed to drive a certain distance, or how much time it takes to complete a task at a certain rate. By understanding how to find and use unit rates, you can improve your mathematical skills and make more informed decisions in your daily life. ## Method 1: Using Proportions to Find Unit Rates One method to find unit rates is by using proportions. To do this, set up a fraction with the given quantity as the numerator and the unit of measurement as the denominator. Then, set this fraction equal to a second fraction with the unknown quantity as the numerator and the desired unit of measurement as the denominator. For example, suppose you need to find the unit rate of a car that traveled 240 miles in 6 hours. Start by setting up the fraction 240 miles/6 hours. To find the unit rate in miles per hour, set this fraction equal to x miles/1 hour. 240 miles / 6 hours = x miles / 1 hour Solve for x by cross-multiplying: 240 miles * 1 hour = 6 hours * x miles 240 miles = 6x miles x = 40 miles per hour Therefore, the unit rate of the car’s speed is 40 miles per hour. Using proportions can be a helpful method for finding unit rates, especially when dealing with more complex situations where other methods may be less straightforward. ## Method 2: Using Division to Find Unit Rates Another method for finding unit rates is by using division. To do this, simply divide the given quantity by the unit of measurement. This will give you the unit rate. For example, suppose you need to find the unit rate of a bike that traveled 12 miles in 1 hour. Divide 12 miles by 1 hour to find the unit rate: 12 miles Ã· 1 hour = 12 miles per hour Therefore, the unit rate of the bike’s speed is 12 miles per hour. This method is straightforward and quick, making it a good choice for simple calculations. However, it may not be as useful when dealing with more complex situations that require proportional reasoning. ## Applying Unit Rates to Real-World Situations Unit rates are commonly used in real-world situations to make predictions, solve problems, and analyze data. Here are a few examples: • Gas Mileage: When you fill up your car’s gas tank, you can use the unit rate of miles per gallon (mpg) to calculate how far you can travel on a full tank. For example, if your car gets 30 mpg and your gas tank holds 12 gallons, you can travel up to 360 miles on a full tank (30 mpg x 12 gallons = 360 miles). • Cooking: When following a recipe, you can use the unit rate of teaspoons per tablespoon to convert measurements. For example, if a recipe calls for 3 tablespoons of sugar and you only have teaspoons, you can use the unit rate of 3 teaspoons per 1 tablespoon to measure out 9 teaspoons of sugar (3 tablespoons x 3 teaspoons per tablespoon = 9 teaspoons). • Speed and Distance: When planning a trip, you can use the unit rate of miles per hour to calculate how long it will take to travel a certain distance. For example, if you need to travel 120 miles and your car’s speed is 60 miles per hour, it will take you 2 hours to reach your destination (120 miles Ã· 60 miles per hour = 2 hours). By understanding how to apply unit rates in real-world situations, you can make more informed decisions and solve problems more efficiently. ## Common Mistakes to Avoid When Finding Unit Rates While finding unit rates is a straightforward process, there are some common mistakes that can lead to incorrect answers. Here are a few mistakes to avoid: • Forgetting to Simplify: When setting up a proportion, be sure to simplify the fractions before solving for the unknown quantity. Failing to do so can lead to incorrect answers. • Using the Wrong Units: When dividing, be sure to use the correct units of measurement for both the numerator and denominator. Using the wrong units can result in incorrect unit rates. • Mixing Up Numerators and Denominators: When setting up a proportion, be sure to place the given quantity in the numerator and the unit of measurement in the denominator. Mixing up these values can lead to incorrect answers. • Rounding Too Soon: When working with decimals, be sure to carry out all calculations before rounding. Rounding too soon can lead to imprecise answers. By being aware of these common mistakes, you can avoid errors and find accurate unit rates.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use. # Scale Factor to Find Actual Dimensions ## Use a ratio to find actual dimensions. Estimated12 minsto complete % Progress Practice Scale Factor to Find Actual Dimensions MEMORY METER This indicates how strong in your memory this concept is Progress Estimated12 minsto complete % Use Scale Factor When Problem Solving License: CC BY-NC 3.0 Lifan’s driveway has a length of 24 feet. If the scale is \begin{align}2 \ \text{inches} : 4 \ \text{feet}\end{align}, what is the scale factor? In a diagram, how many inches would Lifan draw to represent his driveway? In this concept, you will learn to use scale factors when problem solving. ### Scale Factor The scale can be used to help you with scale dimensions or actual dimensions. This scale is key in problem solving. If you look at the scale \begin{align}2:1\end{align}, you can use this information to determine the scale factor. The scale factor is the relationship between the scale dimension and the measurement comparison between the scale measurement of the model and the actual length. In this case the scale factor is \begin{align}\frac{1}{2}\end{align}. Let’s look at an example. What is the scale factor if 3 inches is equal to 12 feet? First, write the ratio. \begin{align}\frac{3}{12}\end{align} Next, simplify the fraction. \begin{align}\frac{3}{12} = \frac{1}{4}\end{align} The answer is \begin{align}\frac{1}{4}\end{align}. The scale factor is \begin{align}1:4\end{align}. Now let’s look at a problem where you are applying this information. If the scale dimension is 4, then you can figure out the actual dimension. Look at this proportion: \begin{align}1:2 = 4: x \end{align} First, put the proportion in fraction form. \begin{align}\frac{1}{2} = \frac{4}{x}\end{align} Next, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{1}{2} &=& \frac{4}{x} \\ 1x &=& 2 \times 4 \\ x &=& 8 \end{array}\end{align} The answer is 8. This is the actual dimension. Let’s look at a real world problem. The plans for a flower garden show that it is 6 inches wide on the plan. If the scale for the flower garden is \begin{align}1:12\end{align}, what is the actual width of the flower garden? First, write the proportion. \begin{align}1:12 = 6:x\end{align} Next, put the proportion in fraction form. \begin{align}\frac{1}{12} = \frac{6}{x}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{1}{12} &=& \frac{6}{x} \\ 1x &=& 12 \times 6 \\ x &=& 72 \end{array}\end{align} The answer is 72. The actual width of the flower garden is 72 inches. ### Examples #### Example 1 Earlier, you were given a problem about Lifan and his long driveway. The scale is \begin{align}2 \text{ inches} : 4 \text{ feet}\end{align} and the driveway is 24 feet long. First, write the proportion. Note that \begin{align}2:4\end{align} is the scale factor. \begin{align}2:4 = x: 24\end{align} Next, put the proportion in fraction form. \begin{align}\frac{2}{4} = \frac{x}{24}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{2}{4} &=& \frac{x}{24} \\ 4x &=& 2 \times 24 \\ 4x &=& 48 \end{array}\end{align} Then, divide both sides by 4 to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} 4x &=& 48 \\ \frac{4x}{4} &=& \frac{48}{4} \\ x &=& 12 \end{array}\end{align} The answer is 12. The scale dimension of Lifan’s driveway is 12 inches. #### Example 2 Find the missing actual dimension if the scale factor is \begin{align}2^{{\prime}{\prime}}: 3^{\prime}\end{align} and the scale measurement is \begin{align}6^{{\prime}{\prime}}\end{align}. First, write the proportion. \begin{align}2:3 = 6:x\end{align} Next, put the proportion in fraction form. \begin{align}\frac{2}{3} = \frac{6}{x}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{2}{3} &=& \frac{6}{x} \\ 2x &=& 3 \times 6 \\ 2x &=& 18 \end{array}\end{align} Then, divide both sides by 2 to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} 2x &=& 18 \\ \frac{2x}{2} &=& \frac{18}{2} \\ x &=& 9 \end{array}\end{align} The answer is 9. The actual dimension is 9 feet. #### Example 3 Find the missing actual dimension if the scale factor is \begin{align}{\frac{1}{4}}^{{\prime}{\prime}}: 4^{\prime}\end{align} and the scale measurement is \begin{align}8^{{\prime}{\prime}}\end{align}. First, write the proportion. \begin{align}\frac{1}{4}:4 = 8:x\end{align} Next, put the proportion in fraction form. \begin{align}\begin{array}{rcl} \frac{\frac{1}{4}}{4} &=& \frac{8}{x} \\ \frac{1}{16} &=& \frac{8}{x} \end{array}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{1}{16} &=& \frac{8}{x} \\ 1x &=& 8 \times 16 \\ x &=& 128 \end{array}\end{align} The answer is 128. The actual dimension is 128 feet. #### Example 4 Find the missing actual dimension if the scale factor is \begin{align}{\frac{1}{4}}^{{\prime}{\prime}}:4^{\prime}\end{align} and the scale measurement is \begin{align}12^{\prime}\end{align}. First, write the proportion. \begin{align}\frac{1}{4}:4 = x:12\end{align} Next, put the proportion in fraction form. \begin{align}\begin{array}{rcl} \frac{\frac{1}{4}}{4} &=& \frac{x}{12} \\ \frac{1}{16} &=& \frac{x}{12} \end{array}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{1}{16} &=& \frac{x}{12} \\ 16x &=& 1 \times 12 \\ 16x &=& 12 \end{array}\end{align} Then, divide both sides by 16 to solve for \begin{align}x\end{align} . \begin{align}\begin{array}{rcl} 16x &=& 12 \\ \frac{16x}{16} &=& \frac{12}{16} \\ x &=& \frac{12}{16} \\ x &=& \frac{3}{4} \end{array}\end{align} The answer is \begin{align}\frac{3}{4}\end{align}. The actual dimension is \begin{align}\frac{3}{4}\end{align} inches. #### Example 5 Find the missing actual dimension if the scale factor is \begin{align}{\frac{1}{4}}^{{\prime}{\prime}}:4^{\prime}\end{align} and the scale measurement is \begin{align}16^{{\prime}{\prime}}\end{align}. First, write the proportion. \begin{align}\frac{1}{4}:4 = 16:x\end{align} Next, put the proportion in fraction form. \begin{align}\begin{array}{rcl} \frac{\frac{1}{4}}{4} &=& \frac{16}{x} \\ \frac{1}{16} &=& \frac{16}{x} \end{array}\end{align} Then, cross multiply to solve for \begin{align}x\end{align}. \begin{align}\begin{array}{rcl} \frac{1}{16} &=& \frac{16}{x} \\ 1x &=& 16 \times 16 \\ x &=& 256 \end{array}\end{align} The answer is 256. The actual dimension is 256 feet. ### Review Figure out each scale factor. 1. \begin{align}\frac{2 \text{ inches}}{8 \text{ feet}}\end{align} 2. \begin{align}\frac{13 \text{ inches}}{12 \text{ feet}}\end{align} 3. \begin{align}\frac{6 \text{ inches}}{24 \text{ feet}}\end{align} 4. \begin{align}\frac{11 \text{ inches}}{33 \text{ feet}}\end{align} 5. \begin{align}\frac{16 \text{ inches}}{32 \text{ feet}}\end{align} 6. \begin{align}\frac{18 \text{ inches}}{36 \text{ feet}}\end{align} 7. \begin{align}\frac{6 \text{ inches}}{48 \text{ feet}}\end{align} 8. \begin{align}\frac{6 \text{ inches}}{12 \text{ feet}}\end{align} Solve each problem. 9. A rectangle has a width of 2 inches. A similar rectangle has a width of 9 inches. What scale factor could be used to convert the larger rectangle to the smaller rectangle? 10. A drawing of a man is 4 inches high. The actual man is 64 inches tall. What is the scale factor for the drawing? 11. A map has a scale of \begin{align}1 \text{ inch}= 4 \text{ feet}\end{align}. What is the scale factor of the map? 12. A drawing of a box has dimensions that are 2 inches, 3 inches, and 5 inches. The dimensions of the actual box will be \begin{align}3\frac{1}{4}\end{align} times the dimensions in the drawing. What are the dimensions of the actual box? 13. A room has a length of 10 feet. Hadley is drawing a scale drawing of the room, using the scale factor \begin{align}\frac{1}{50}\end{align}. How long will the room be in Hadley’s drawing? 14. The distance from Anna’s room to the kitchen is 15 meters. Anna is making a diagram of her house using the scale factor of \begin{align}\frac{1}{150}\end{align}. What will be the distance on the diagram from Anna’s room to the kitchen? 15. On a map of Cameron’s town, his house is 9 inches from his school. If the scale of the map is \begin{align}\frac{1}{400}\end{align}, what is the actual distance, in feet, from Cameron’s house to his school? ### Review (Answers) To see the Review answers, open this PDF file and look for section 4.7. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show More ### Vocabulary Language: English Actual Dimension The actual dimensions are the real–life measures of the object or building. Proportion A proportion is an equation that shows two equivalent ratios. Scale Dimension A scale dimension is the measurement used to represent actual dimensions in a drawing or on a map. ### Image Attributions 1. [1]^ License: CC BY-NC 3.0 ### Explore More Sign in to explore more, including practice questions and solutions for Scale Factor to Find Actual Dimensions. Please wait... Please wait...
# 9. Sequences And Series “Respect Is The Key Determinant Of High-Performance Leadership. How Much People Respect You Determines How Well They Perform.” By a sequence, we mean an arrangement of number in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, ….k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence. Let a1, a2, a3, … be the sequence, then the sum expressed as a1 + a2 + a3 + … is called series. A series is called finite series if it has got finite number of terms. An arithmetic progression (A.P.) is a sequence in which terms increase or decrease regularly by the same constant. This constant is called common difference of the A.P. Usually, we denote the first term of A.P. by a, the common difference by d and the last term by l. The general term or the nth term of the A.P. is given by an= a + (n – 1) d. The sum Sn of the first n terms of an A.P. is given by S_n= \frac{n}{2}(2a + (n – 1) d) The arithmetic mean A of any two numbers a and b is given by \frac{a + b}{2} i.e., the sequence a, A, b is in A.P. A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by an= arn – 1 The sum Sn of the first n terms of G.P. is given by S_n= \frac{a(r^n – 1)}{(r – 1)} = \frac{a(1 – r^n)}{(1 – r)}, {\rm if } r \ne 1 . The geometric mean (G.M.) of any two positive numbers a and b is given by \sqrt{ab} i.e., the sequence a, G, b is G.P.
# #55. How to Solve a Nonlinear System an Example with Four Solutions \| Summary and Q&A 913 views October 14, 2020 by The Math Sorcerer #55. How to Solve a Nonlinear System an Example with Four Solutions ## TL;DR Learn how to solve a non-linear system of equations using the addition or elimination method. ## Key Insights • 🚱 The addition or elimination method is a useful technique for solving non-linear systems of equations. • 🍉 Multiplying one equation by a number and adding it to another cancels out specific terms, simplifying the system. • 🧑‍🏭 Factoring out variables helps simplify equations and determine possible solutions. • 🚱 Checking both equations in a non-linear system is essential to ensure consistency and accuracy in finding the solutions. ## Transcript in this problem we have a non-linear system of equations and we have to find the solutions so there's a couple ways to do this i think maybe we should try to do it with what's called the addition or elimination method so the way that works is basically you multiply one equation by a number and then add it to another equation let's maybe multiply th... Read More ## Questions & Answers ### Q: What is the addition or elimination method used for in solving non-linear equations? The addition or elimination method is a technique to simplify and solve non-linear equations by multiplying one equation by a number and adding it to another equation to cancel out terms. ### Q: How do you solve for x in a non-linear system of equations using the addition or elimination method? To solve for x, you multiply one equation by a number and add it to the other equation. This eliminates certain terms, allowing you to solve for x and obtain possible x-values as solutions. ### Q: What is the purpose of factoring out an x in the equation x^2 + 5x = 0? Factoring out an x helps simplify the equation by expressing it in the form x(x + 5) = 0, which allows you to set each factor equal to zero and find the possible x-values that satisfy the equation. ### Q: Why is it necessary to check both equations when finding the solutions to the non-linear system? Both equations need to be checked because they both contain y^2 terms. This means that when finding the square root of y^2, there are two possible answers, and checking both equations ensures consistency and accuracy in determining the solutions. ## Summary & Key Takeaways • The content provides a step-by-step explanation of solving a non-linear system of equations using the addition or elimination method. • The method involves multiplying one equation by a number and adding it to another equation to cancel out certain terms. • The solutions to the system of equations are ordered pairs (x, y), which are obtained by finding the values of x and substituting them back into the equations to solve for y.
Flat 50% Off on Assignment Bookings • Internal Code : • Subject Code : • University : • Subject Name : Science and Maths ## Understanding of Concepts- Task 1 A prime number is a number that is more noteworthy than 1 that cannot be made by duplicating other whole numbers. These numbers can only be multiplied by themselves and 1. For example, the below given are a few numbers that are prime numbers: • 2 • 3 • 5 • 7 • 11 • 13 • 17 • 19 • 23…. The numbers that are made out of these prime numbers are called composite numbers. For example- if we multiply a prime number to a prime number we get a composite number. Like: 22=4 In this case, 2 is a prime number while 4 is a composite number. Prime Factorisation is the method used to find a solution to multiplying two prime numbers and getting a single original number. By Multiplying: Prime factorisation of 90? Solution: 90= 910 Prime factors: 332*5 ## Understanding of Concepts- Task 2 For multiplication of two fractional numbers, we must multiply the numerator of the first number with the numerator of the second number and the same shall be done with the denominator. We are taught that when we are confronted with a troublesome inquiry, for example, 3/5 ÷ 4/7, we should reverse the second part and then multiply (reciprocal). It would positively appear to be more instinctual just to partition the numerator of the first fractional number by the numerator of the second and afterwards correspondingly to isolate one denominator by the other. Would this instinctive strategy for tackling the difficult work? Not exclusively is the appropriate response "yes," utilizing this technique for isolating the fractional numbers really reveals some insight into why we are taught to do reciprocal of the other number and then multiply. Let’s try 35÷47 as an example. While we are aware of the fact that the number 1 is an identity number so we are not supposed the multiply the first fractional number’s division with the element 1, therefore, we multiply the number by 4/4. To be serious, the solution remains the same when we reciprocate the fractional number and multiply subsequently. However, the intuitive method requires a lot more work to achieve the same results. Let us study about the next method. Dividend Divisor Method After dividing the numerator of the first number with the numerator of the second number, i.e dividing 3 with 4 we need to multiply 3 with the denominator of the second number i.e. 3 multiplied with 7. Now, this would be applicable in the case of the denominator as well. We multiply the first fractional number with the reciprocal of the second number is just a shortcut! ## Understanding of Concepts- Task 3 In mathematics, a fraction is a value, which defines the part of a whole. In other words, the fraction is a ratio of two numbers. Whereas, the decimal is a number, whose whole number part and the fractional part are separated by a decimal point. The decimal number can be classified into different types, such as terminating and non-terminating decimals, repeating and non-repeating decimals. While solving many mathematical problems, the conversion of decimal to the fractional value is preferred, as we can easily simplify the fractional values. In this article, we are going to discuss how to convert repeating decimals to fractions in an easy way. Terminating and Non-Terminating Decimals · A terminating decimal is a decimal that has an end digit. It is a decimal, which has a finite number of digits (or terms). Example: 0.15, 0.86, etc. · Non-terminating decimals are the one that does not have an end term. It has an infinite number of terms. Example: 0.5444444….., 0.1111111….., etc. Repeating and Non-Repeating Decimals · Repeating decimals are the one, which has a set of terms in decimal to be repeated uniformly. Example: 0.666666…., 0.123123…., etc. · Non-repeating decimals are the one that does have repeated terms. Repeating Decimals to Fraction Conversion Let us now learn the conversion of repeating decimals into the fractional form. Now, we are going to discuss the two different cases of the repeating fraction. Case 1: Fraction of type 0.abcd¯¯¯¯¯¯¯¯¯¯ The formula to convert this type of repeating decimal to a fraction is given by: abcd¯¯¯¯¯¯¯¯¯¯ = Repeated term / Number of 9’s for the repeated term Example 1: Convert 0.7¯¯¯ to the fractional form. Solution: Here, the number of repeated term is 7 only. Thus the number of times 9 to be repeated in the denominator is only once. 0.7¯¯¯=79 Case 2: Fraction of type 0.ab..cd¯¯¯¯¯ The formula to convert this type of repeating decimal to the fraction is given by: 0.ab..cd¯¯¯¯¯=(ab….cd…..)–ab……Number of time 9′s the repeating term followed by the number of times 0′s for the non−repeated terms. Remember, at the center of any academic work, lies clarity and evidence. Should you need further assistance, do look up to our Science Assignment Help ## Get It Done! Today • 1,212,718Orders • 4.9/5Rating • 5,063Experts ### Highlights • 21 Step Quality Check • 2000+ Ph.D Experts • Live Expert Sessions • Dedicated App • Earn while you Learn with us • Confidentiality Agreement • Money Back Guarantee • Customer Feedback ### Just Pay for your Assignment • Turnitin Report \$10.00 \$9.00Per Page • Consultation with Expert \$35.00Per Hour • Live Session 1-on-1 \$40.00Per 30 min. • Quality Check \$25.00 • Total Free • Let's Start Get 500 Words Free
# LarPCalcLim2_12_05 Views: Category: Education ## Presentation Description No description available. ## Presentation Transcript ### What You Should Learn: Find limits of summations. Use rectangles to approximate areas of plane regions. Use limits of summations to find areas of plane regions. What You Should Learn ### Slide 4: Limits of Summations ### Limits of Summations: Limits of Summations We have used the concept of a limit to obtain a formula for the sum S of an infinite geometric series Using limit notation, this sum can be written as ### Limits of Summations: Limits of Summations The following summation formulas and properties are used to evaluate finite and infinite summations. ### Example 1 – Evaluating a Summation: Example 1 – Evaluating a Summation Evaluate the summation. Solution: Using the second summation formula with n = 200, you can write ### Example 1 – Solution: Example 1 – Solution . cont’d The Area Problem ### The Area Problem: The Area Problem You now have the tools needed to solve the second basic problem of calculus: the area problem. The problem is to find the area of the region R bounded by the graph of a nonnegative, continuous function f , the x -axis, and the vertical lines x = a and x = b , as shown in Figure 12.33. Figure 12.33 ### The Area Problem: The Area Problem If the region R is a square, a triangle, a trapezoid, or a semicircle, you can find its area by using a geometric formula. For more general regions, however, you must use a different approach—one that involves the limit of a summation. The basic strategy is to use a collection of rectangles of equal width that approximates the region R , as illustrated in Example 4. ### Example 4 – Approximating the Area of a Region: Example 4 – Approximating the Area of a Region Use the five rectangles in Figure 12.34 to approximate the area of the region bounded by the graph of f ( x ) = 6 – x 2 , the x -axis, and the lines x = 0 and x = 2. Figure 12.34 ### Example 4 – Solution: Example 4 – Solution Because the length of the interval along the x -axis is 2 and there are five rectangles, the width of each rectangle is . The height of each rectangle can be obtained by evaluating f at the right endpoint of each interval. The five intervals are as follows. Notice that the right endpoint of each interval is for i = 1, 2, 3, 4, 5. ### Example 4 – Solution: Example 4 – Solution The sum of the areas of the five rectangles is cont’d Height Width ### Example 4 – Solution: Example 4 – Solution So, you can approximate the area of R as 8.48 square units. cont’d ### The Area Problem: The Area Problem By increasing the number of rectangles used in Example 4, you can obtain closer and closer approximations of the area of the region. For instance, using 25 rectangles of width each, you can approximate the area to be A  9.17 square units. The following table shows even better approximations. ### The Area Problem: The Area Problem Based on the procedure illustrated in Example 4, the exact area of a plane region R is given by the limit of the sum of n rectangles as n approaches . ### Example 5 – Finding the Area of a Region: Example 5 – Finding the Area of a Region Find the area of the region bounded by the graph of f ( x ) = x 2 and the x -axis between x = 0 and x = 1, as shown in Figure 12.35. Figure 12.35 ### Example 5 – Solution: Example 5 – Solution Begin by finding the dimensions of the rectangles. Width: Height: ### Example 5 – Solution: Example 5 – Solution Next, approximate the area as the sum of the areas of n rectangles. cont’d Summation form ### Example 5 – Solution: Example 5 – Solution Finally, find the exact area by taking the limit as n approaches . cont’d Rational form
# Dividing Fractions Dividing fractions is an important part of fractions. However, before moving onto this important subcategory of fractions, it is important that the kid has a clear understanding of fractions. BYJU’S offers various dividing fractions worksheets that your child can use for practice which will make sure that he has a clear understanding of dividing fractions. This practice will help them to solve dividing fractions word problems. How to Do Division of Fractions Division of fractions might sound difficult, but it is quite an easy thing to do. In the case of the division of fractions, the sign of division is converted into multiplication by reversing the second fraction. For example, divide 3/5 ÷ 4/5; you write it as 3/5 * 5/4, which means the result in simplified terms will be 3/4. This happens if both the numbers are in a fraction. Now to make sure that the children have understood the concept of division of fractions, teachers often ask them to divide whole numbers with fractions. Even when it’s asked to divide the whole number with fractions, it is quite easy. All one needs to do is write the whole number in a fraction form and then reverse the second fraction and find the result. For example, if the question reads divide 5 ÷ 3/4, then you can write it as 5/1 ÷ 3/4 = 5/1 * 4/3, which will give 20/3 as a result. If one has a better understanding of fractions, one can easily write the result in mixed fraction form too. Another question that can be framed from the division of fractions is dividing fractions with decimals. If a question is asked, divide 1.5 ÷ 2/5, then you can write it as 15/10 ÷ 2/5, then reverse the sign and the last fraction, i.e. 15/10 * 5/2, and the answer will be 15/4. You can download the dividing fractions worksheet pdf from our BYJU’S website and give it to your child for practising. These worksheets will help them to clear their understanding and develop a better grip over the concept.
How to Do Mixed Integer Computations The term integers represent the meaning “intact” or “whole”. So, you can generally refer an integer as a whole number, except integers can be negative also! What is an Integer? An integer is a whole number (not any decimal or fraction numbers) which can be zero, positive or negative numbers. Some examples of integers can be $$7 \ , \ 3 \ , \ 0 \ , \ -5 \ , \ -15,$$ etc. Moreover, we can represent integers by the denotation $$Z$$ which comprises of: Positive Integers: As the name suggests, any integer that is greater than zero is termed as a positive integer. Negative Integer: As from the name, any integer that is less than zero is termed as a negative integer. Zero: Zero is neither a positive integer or a negative integer. It is just a whole number. So, we can write $$Z = \{….,-5 \ , \ -4 \ , \ -3 \ , \ -2 \ , \ -1 \ , \ 0 \ , \ 1 \ , \ 2 \ , \ 3 \ , \ 4 \ , \ 5,……\}$$ Also, we can place all integers on a number line where the negative ones are placed on the left of $$“0”$$ and the positive ones on the right. Moreover, we can perform the 4 basic mathematic properties with integers. They are: • Subtraction • Multiplication • Division We often see that negative integers are always written as $$-5 \ , \ -9$$ and so on. But it is not generally considered necessary to write positive integers like $$+5 \ , \ +9$$ and so on. So, when we write just $$5,$$ we mean $$+5$$. Another thing to note is, that an absolute value of any integer is always positive. So, $$\lvert -6 \rvert=6$$ and $$\lvert 6 \rvert$$ is also $$6$$. How to Add and Subtract Integers • In the first case, if two integers have the same sign (either both are positive or both are negative), add up those integers and put the common sign. • In the second case, if two integers have an opposite sign (one is positive and the other one negative), then subtract them and put the sign of the bigger number. Example: • $$-12 +13 = +1,$$ • $$-12 -13 = -25,$$ • $$12 – 13 = -1$$ and so on. How to Multiply and Divide Integers To Multiply or Divide two integers, follow these steps: • Firstly, perform general multiplication or division between the two integers and ignore their sign. • Next, we have to decide the sign. So, if both signs are opposite, then we must always put a negative sign. Also, if both signs of the integers are same, we must use a positive sign. Example: • $$-12 \times +7 = -84,$$ • $$-12 \div +6 = -2,$$ • $$12 \div 3 = 4$$ and so on. Exercises for Mixed Integer Computations 1) $$(-4) \ \times \ (-3) \ =$$ 2) $$36 \ \div \ (-4) \ =$$ 3) $$-14 \ \div \ 2 \ =$$ 4) $$36 \ \div \ 6 \ =$$ 5) $$-24 \ \div \ (-3) \ =$$ 6) $$2 \ \div \ 2 \ =$$ 7) $$-56 \ \div \ (-8) \ =$$ 8) $$24 \ \div \ (-4) \ =$$ 9) $$8 \ \div \ (-1) \ =$$ 10) $$4 \ \times \ (-7) \ =$$ 1) $$(-4) \ \times \ (-3) \ = \color{red}{12}$$ 2) $$36 \ \div \ (-4) \ = \color{red}{-9}$$ 3) $$-14 \ \div \ 2 \ = \color{red}{-7}$$ 4) $$36 \ \div \ 6 \ = \color{red}{6}$$ 5) $$-24 \ \div \ (-3) \ = \color{red}{8}$$ 6) $$2 \ \div \ 2 \ = \color{red}{1}$$ 7) $$-56 \ \div \ (-8) \ = \color{red}{7}$$ 8) $$24 \ \div \ (-4) \ = \color{red}{-6}$$ 9) $$8 \ \div \ (-1) \ = \color{red}{-8}$$ 10) $$4 \ \times \ (-7) \ = \color{red}{-28}$$ Mixed Integer Computations Practice Quiz STAAR Grade 8 Math Study Guide $19.99$14.99 TSIA2 Math Practice Workbook $25.99$14.99 5 Full Length ATI TEAS 6 Math Practice Tests $18.99$12.99 CLEP College Algebra Preparation 2020 – 2021 $21.99$16.99
Top # Subtracting Scientific Notation Subtracting numbers in scientific notation can be done in few ways. One way is to change the numbers out of scientific notation and work the problem normally. Example: change 9.3 x 103 - 8.63 x 102 into 9300 - 863. Then we subtract normally to get 8.437 x 103. Its an simplest way of working when the numbers are not very large or not very small. Scientific notation is a form, where the scientific terms that expressing very large or very small numbers in an easiest form. Numbers written in scientific notation can be used in computations (x, ÷, +, –) with ease. Writing the numbers in the form of K x m n is called as scientific notation, which is also known as exponential notation. Related Calculators Subtracting Scientific Notation Calculator Adding Scientific Notation Calculator Calculate Scientific Notation Dividing Scientific Notation Calculator ## How to Subtract Scientific Notation The following are the steps involved in Subtracting Numbers with Scientific Notation Step 1: Adjust the coefficient of both the numbers to make exponent equal for both the numbers. (It is better to easily adjust the smaller index to equal the larger index). Step 2: Now subtract the smaller coefficient from the larger coefficient of the numbers. Step 3: Write the difference in scientific notation. ## Subtracting Scientific Notation Example Problems Below are the examples on subtracting numbers using scientific notations: ### Solved Examples Question 1: Subtract scientific notation 2.3 x 10 4 from 4.5 x 10 5 Solution: Step 1: Adjust the co-efficient of both the numbers to make exponent equal for both the numbers. Here 2.3 x 10 4 having the smaller power. We can write 2.3 x 10 4 as 0.23 x 10 5 Now 4.5 x 10 5 and 0.23 x 10 5 have the same powers. Step 2: Subtract the coefficient of 0.23 x 10 5 from the coefficient of 4.5 x 10 5 4.5 x 10 5 - 0.23 x 10 5 = 4.5 - 0.23 = 4.27 Step 3: Writing the result in scientific form Hence, Subtracting scientific notation 4.5 x 10 5- 2.3 x 10 4 = 4.27 x 105 Question 2: Subtract scientific notation 0.625 x 102 from 1.15 x 103 Solution: Step 1: Adjust the coefficient of both the numbers to make exponent equal for both the numbers. Here 0.625 x 102 having the smaller power. We can 0.625 x 102 as 0.0625 x 10 3 Now, 1.15 x 103 and 0.0625 x 10 3 have same powers. Step 2: Greater coefficient is 1.15 and lower coefficient is 0.0625, Subtract 0.00625 from 1.15 1.15 - 0.0625 = 1.0875 Step 3: Writing the result in scientific form 1.15 x 103 - 0.625 x 102 = 1.0875 x 103 Question 3: Subtract scientific notation 15.2 x 105 from 34.3 x 105 Solution: Step 1: Adjust the coefficient of both the numbers to make exponent equal for both the numbers. Here, 15.2 x 105 and 34.3 x 105 both has the same power Step 2: Subtract the coefficient of 15.2 x 105 from the coefficient of 34.3 x 105 34.3 - 15.2 = 19.1 Step 3: Writing the result in scientific form 34.3 x 105 - 15.2 x 105 = 19.1 x 105 ### Practice Problem Question: Subtract scientific notation 15.2 x 105 from 34.3 x 105 Related Topics Math Help Online Online Math Tutor *AP and SAT are registered trademarks of the College Board.
1.17: Comparison of Problem-Solving Models Difficulty Level: Basic Created by: CK-12 Estimated15 minsto complete % Progress Practice Problem-Solving Models MEMORY METER This indicates how strong in your memory this concept is Progress Estimated15 minsto complete % Estimated15 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Lisa's grandmother gave her some money for her birthday. Lisa decided to save it to help pay for a trip with her friends this summer. Since her birthday, Lisa has saved an additional $5 every week! It's been 8 weeks and Lisa has$95. How much money did Lisa's grandmother give her for her birthday? In this concept, you will learn how to choose an appropriate strategy for solving a real world problem. Comparing Problem-Solving Models To start solving a real world problem it can help to ask yourself the following key questions. Key Questions: 1. What am I trying to find out? 2. What do I know? 3. How can I solve the problem? By reading the problem a few times you should be able to answer the first two key questions. However, sometimes answering the third key question is not that easy. It is often not obvious how to solve a problem! Here are some commonly used strategies for solving problems. 1. Find a Pattern - Best used when there is a series of numbers and/or when you are being asked for a later quantity. For example, find the number in the tenth step. 2. Guess and Check - Best used when you are looking for one or two numbers and you think one of them might work. You can take a guess, try out a number and then adjust your answer from there. 3. Work Backwards - Best used when you are given a total or final amount and you are looking for some partial amount or original amount. 4. Draw a Picture - Best used when the problem describes some sort of visual. 5. Write an Equation - Best used when there is a missing quantity that needs to be figured out. Then you can write an equation and solve for the answer. 6. Use a Formula - At this point, best used for area and perimeter problems. The more you practice solving problems, the quicker you will become at identifying the most appropriate strategy to use. Here is an example. Ron arranged his herb garden rows in the following order: 2 plants, 5 plants, 11 plants, 23 plants. How many plants will be in the fifth row? First, ask yourself “what am I trying to find out?” You are trying to find out how many plants will be in the fifth row of Ron's herb garden. Next, ask yourself “what do I know?” You know the following pieces of information: • The first row has 2 plants. • The second row has 5 plants. • The third row has 11 plants. • The fourth row has 23 plants. Then, ask yourself “how can I solve the problem?” Notice that in this problem you were given a list of numbers and asked about a later quantity. This is a perfect time to use the find a pattern strategy. First, find the pattern and describe the pattern rule. Then, extend the pattern to figure out how many plants are in the fifth row. Now, implement your plan. So far the pattern is 2, 5, 11, 23, . . . The pattern rule is multiply by 2 and add 1. Next, extend the pattern. The answer is that there will be 47 plants in the fifth row. Remember that once you have an answer, you need to make sure you have actually answered the original question that was asked and that your answer seems realistic. You were trying to find out how many plants are in the fifth row and that's what you did. 47 plants is a lot, but it fits the pattern so your answer makes sense. Examples Example 1 Earlier, you were given a problem about Lisa and her saved money. After her grandmother gave her some money for her birthday, Lisa has been saving $5 a week for 8 weeks. She now has$95! You wonder how much money Lisa got from her grandmother for her birthday. First, ask yourself “what am I trying to find out?” You are trying to find out how much money Lisa got from her grandmother for her birthday. Next, ask yourself “what do I know?” You know the following pieces of information: • Currently Lisa has $95 dollars. • Lisa is saving$5 each week. • It has been 8 weeks. • Lisa started out with just the money from her grandmother. Then, ask yourself “how can I solve the problem?” Notice that in this problem you were given a final amount of $95 and asked about an initial amount. This is a good time to use the work backwards strategy. First, figure out how much money Lisa has saved over the 8 weeks since she got the money from her grandmother. Then, work backwards from the$95 to figure out how much money she must have started with from her grandmother. Now, implement your plan. Lisa has saved $5 a week for 8 weeks. So over the 8 weeks Lisa has saved an additional$40. Next, work backwards. Lisa has $95 now. She added$40 to the money from her grandmother to get to the $95. That means you can subtract$40 form $95 to figure out how much money Lisa started with from her grandmother. The answer is Lisa got$55 from her grandmother. Make sure you have actually answered the original question that was asked and that your answer seems realistic. You were trying to find out how much money Lisa's grandmother gave her for her birthday and you found out that Lisa got $55 from her grandmother.$55 is a realistic amount of money for a birthday present so your answer makes sense. Example 2 Ms. Powell wants to hang a large rectangular tapestry lengthwise on her living room wall. The tapestry has a perimeter of 42 feet and a width of 9 feet. Ms. Powell’s wall is 10 feet high. Will the length of the tapestry fit against the height of Ms. Powell’s ceiling? First, ask yourself “what am I trying to find out?” You are trying to find out if the length of the tapestry is less than the height of Ms. Powell's ceiling which is 10 feet. Next, ask yourself “what do I know?” You know the following pieces of information: • The tapestry is a rectangle. • The tapestry has a perimeter of 42 feet. • The tapestry has a width of 9 feet. • Ms. Powell's ceilings are 10 feet high. • The tapestry will be hung lengthwise. Then, ask yourself “how can I solve the problem?” Notice that this problem referenced perimeter and a rectangle, so this is a great time to use the use a formula strategy. First, use the rectangle perimeter formula to find the length of the tapestry. Then, see if the length of the tapestry is less than 10 feet. Now, implement your plan. The rectangle perimeter formula is . You know and . You want to figure out the length, . Substitute the values for and into the rectangle perimeter formula. Now, solve for . “What number plus 18 is equal to 42?” You know that 24 plus 18 is equal to 42, so must be equal to 24. Next you can solve for the length . “2 times what number is equal to 24?” You know that 2 times 12 is equal to 24 so must be equal to 12. Ms. Powell's tapestry is 12 feet long. Since her ceilings are 10 feet high, the tapestry is too long to fit on the wall. The answer is that Ms. Powell's tapestry is too long to fit on the wall. Now make sure you answered the question. The question asked if the length of the tapestry was less than the height of Ms. Powell's ceiling. You determined that no, the length of the tapestry is not less than the height of the ceiling. 12 feet for the length of a tapestry that you would put on the wall is realistic, so your answer makes sense. Example 3 Melissa has 144 cookies she wants to put evenly into 8 gift bags. How many cookies will go into each bag? First, ask yourself “what am I trying to find out?”. You are trying to find out how many cookies will go into each gift bag. Next, ask yourself “what do I know?”. You know the following pieces of information: • There are 144 cookies total. • There are 8 gift bags. • Each gift bag needs to have the same number of cookies. Then, ask yourself “how can I solve the problem?” Notice that we have a missing quantity to figure out, the number of cookies in each bag. This is a good time to use the write an equation strategy. First, write an equation to show the relationship between the different numbers in the problem. Then, solve the equation. Now, implement your plan. You know the number of cookies in each bag times the number of bags will equal the total number of cookies. The unknown quantity is the number of cookies in each bag so that will be your variable. Let the number of cookies in each bag. Now, solve the equation. “What number times 8 is equal to 144?” If you need to, you can use your calculator to divide 144 by 8 to get the answer. 18 times 8 is equal to 144 so is equal to 18. Next, make sure you have actually answered the original question that was asked and that your answer seems realistic. You were trying to find out how many cookies were in each bag and you did. 18 cookies in a bag is realistic so your answer makes sense. Example 4 A list of numbers is found in the problem. Which strategy might you use? When there is a list of numbers, the find a pattern strategy is a useful one to try. The answer is the find a pattern strategy. Example 5 A final quantity is given, but the initial amount is missing. Which strategy might you use? When you know the final amount but not the starting amount, the work backwards strategy is a good one to try. The answer is the work backwards strategy. Review Solve the following problems. 1. Mary went to the music store with her babysitting money. She bought two CDs for $12.50 each and two magazines for$4.25 each. She left the store with $10.25. How much money did she start with? 2. Since he began his fitness routine, Mr. Trigg has measured his weight every week. His weights for the first six weeks are as follows: 236, 230, 232, 226, 228, 222. If the pattern continues, how much will he weigh in the tenth week? 3. The area of City Park is . The length of the park is 15 feet. What is the perimeter of the park? 4. A farmer planted corn, wheat, and cotton in a total of 88 fields. He planted 10 rows of corn. If he planted even numbers of rows for the other crops, how many rows of wheat did he plant? 5. Mrs. Whitaker is mailing a pair of shoes to her daughter. She wants to fit the rectangular shoebox inside a larger square box. The area of the shoebox is ; the length of one side is 12 inches. One side of the larger square box measures 14 inches. Will the shoebox fit in the larger box? How do you know? 6. After a pin-ball game, the score board showed that the combined points of Peter, Ella, and Ned is 728. Ella scored half the points of Ned and Peter scored one-fourth the points of Ned. How many points did each player score? 7. Tami made a total of$47 babysitting on New Year’s Eve. She made her hourly rate plus a $7 tip. If she worked 5 hours, what is her hourly rate? 8. A weightlifter lifts weights in the following order: 0.5 lb, 1.5 lb, 4.5 lb, 13.5 lb. How many pounds will he lift next? 9. Figure A is a square, with a side that measures 9 cm. Figure B is a square with a side that measures 6 cm. Which figure has the greater area, Figure A or Figure B? 10. Mr. Rowe and Mrs. Rowe are driving 959 miles to a beach vacation. They want to split the distance over 4 days, driving the exact same amount on the first three days and the remainder on the fourth day. If they drive 119 miles on the fourth day, how many miles will they drive on the first day? 11. Cedric spent$27.75 on pizza for his friends. Each cheese pizza cost $8 and each extra topping cost$0.75. If Cedric bought 3 cheese pizzas, how many extra toppings did he get? 12. What was the total cost without the toppings? 13. If 29 people went to the zoo the first day and double went the second, how many went in both days combined? 14. If on the third day, double the people from the second day went, how many people went to the zoo? 15. If ten less went to the zoo on the fourth day verses the third day, how many people went to the zoo on the fourth day? Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English TermDefinition Difference The result of a subtraction operation is called a difference. Problem Solving Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language. Product The product is the result after two amounts have been multiplied. Quotient The quotient is the result after two amounts have been divided. Sum The sum is the result after two or more amounts have been added together. Show Hide Details Description Difficulty Level: Basic Authors: Tags: Subjects:
# Mean Deviation #### Topics • Mean deviation for grouped data • Mean deviation for ungrouped data ## Notes An absolute measure of dispersion is the mean of these deviations. A measure of central tendency lies between the maximum and the minimum values of the set of observations. Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations from mean ( x ) is zero. M.D(a) = ("Sum of absolute values of deviations from a")/ ("Number of observations") Remark : Mean deviation may be obtained from any measure of central tendency. However, mean deviation from mean and median are commonly used in statistical studies. 1) Mean deviation for ungrouped data : Let n observations be x_1, x_2, x_3, ...., x_n. The following steps are involved in the calculation of mean deviation about mean or median: Step 1 - Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘a’. Step 2- Find the deviation of each xi from a, i.e., x_1 – a, x_2 – a, x_3 – a,. . . , x_n– a Step 3- Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is there, i.e., \|x_1 - a\| , \|x_2 - a\| , \| x_3 - a \|, ..., \|x_n-a\| Step 4- Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e., M.D.(a) = $\displaystyle\sum_{i=1}^{n} \|x_i - a\|$ Thus M.D.(bar x) = 1/n $\displaystyle\sum_{i=1}^{n} \|x_i - \bar{x} \|$ , where bar x = Mean and M.D.(M) = 1/n $\displaystyle\sum_{i=1}^{n} \|x_i - M\|$, where M = Median For example : The mean deviation about the mean for the following data : 12 , 3 , 18 , 17, 4,9,17 ,19,20,15,8,17,2,3,16,11,3,1,0,5 find the mean ( x ) of the given data bar x = 1/20 sum _(i=1)^20 x_i = 200 / 20 = 10 The absolute values of the deviations from mean, i.e., \|x_i - bar x\| are 2,7,8,7,6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5 Therefore sum_(i=1)^20 \|x_i - bar x\| = 124 and M.D.(bar x) = 124/20 = 6.2 2) Mean deviation for grouped data : Data can be grouped into two ways : (a) Discrete frequency distribution, (b) Continuous frequency distribution. Let us discuss the method of finding mean deviation for both types of the data. Discrete frequency distribution: Let the given data consist of n distinct values x_1, x_2, ..., x_n occurring with frequencies f_1, f_2 , ..., f_n respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution: x : x_1 x_2 x_3 ... x_n f : f_1 f_2 f_3 ... f_n First of all we find the mean bar x of the given data by using the formula$\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} x_i f_i }{ \displaystyle\sum_{i=1}^{n} f_i\ }= \frac {1}{N} \displaystyle\sum_{i=1}^{n} x_i f_i$ , where summation x_i f_i denotes the sum of the products of observations xi with their respective frequenciesf_i and N =$\displaystyle\sum_{i=1}^{n} f_i$ is the sum of the frequencies.The mean of the absolute values of the deviations, which is the required mean deviation about the mean. M.D.$\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_i \|x_i - \bar{x}\| }{ \displaystyle\sum_{i=1}^{n} f_i\ }= \frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i \|x_i - \bar{x}\|$ , To find mean deviation about median, we find the median of the given discrete frequency distribution. For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify the observation whose cumulative frequency is equal to or just greater than N/2, where N is the sum of frequencies. This value of the observation lies in the middle of the data, therefore, it is the required median. After finding median, we obtain the mean of the absolute values of the deviations from median.Thus M.D.(M) =$\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i \|x_i -M\|$ (b) Continuous frequency distribution: A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies. The assumption that the frequency in each class is centred at its mid-point. The mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation. The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations. The data is first arranged in ascending order. Then, the median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula Median = l + (N/2 - C)/f xx h where median class is the class interval whose cumulative frequency is just greater than or equal to N/2 , N is the sum of frequencies , l , f , h and C are , respectively the lower limit , the frequency, the width of the median class and C the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values of the deviations of mid-point xi of each class from the median i.e. \|x_i - M\|are obtained. M.D.(M) = $\frac {1}{N} \displaystyle\sum_{i=1}^{n} f_i \|x_i -M\|$ 3) Limitations of mean deviation: In a series, where the degree of variability is very high, the median is not a representative central tendency. The sum of the deviations from the mean (minus signs ignored) is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion. 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Home Math Fixing Two-Step Equations ### Fixing Two-Step Equations When fixing two-step equations, these equations can have the next linear type: ax + b = c or ax − b = c, the place a known as coefficient and b and c are known as constants. After all, solely two steps are wanted as proven under: Step 1 Isolate ax Step 2 Isolate x Bear in mind additionally the Golden Rule of Algebra. No matter you do to at least one aspect of an equation, you have to additionally do the identical to the opposite aspect of the equation to be able to maintain the equation balanced. ## Fixing two-step equations of the shape ax – b = c Instance #1 Clear up 2x – 2 = 6 Step 1: Isolate 2x by including 2 to each side of 2x – 2 = 6 2x – 2 + 2 = 6 + 2 2x = 8 Step 2: Divide each side of 2x = 8 by 2 to isolate x 2x / 2 = 8 / 2 x = 4 You may examine to see if x = 4 is certainly an answer by changing 4 within the unique equation. 2(4) – 2 = 8 – 2 = 6 Usually talking, when fixing equations of the shape ax − b = c, you possibly can add b to each side of the equation to isolate ax (step 1) after which divide each side by a to isolate x (step 2). ax – b = c Step 1: ax – b + b = c + b ax + 0 = c + b ax = c + b Step 2: ax/a = (c + b)/a x = (c + b)/a ## Fixing two-step equations of the shape ax + b = c Instance #2 Clear up 6x + 2 = 20 Step 1: Isolate 6x by subtracting 2 from each side of 6x + 2 = 20 6x + 2 – 2 = 20 – 2 6x = 18 Step 2: Divide each side of 6x = 18 by 6 to isolate x 6x / 6 = 18 / 6 x = 3 Typically, when fixing equations of the shape ax + b = c, you possibly can subtract b from each side of the equation to isolate ax (step 1) after which divide each side by a to isolate x (step 2). ax + b = c Step 1: ax + b – b = c – b ax + 0 = c – b ax = c – b Step 2: ax/a = (c – b)/a x = (c – b)/a There’s nothing particular about subtracting b or including b to each side after which divide by a! In observe, mathematicians want so as to add or subtract one thing from each side after which divide by a as a result of calculations are often simpler. Discover although that there’s one other solution to resolve instance #1 above with simply two steps as effectively. 1. Step one is to divide each side of the equation by a 2. The second step is to both add one thing to each side of the brand new equation or subtract one thing from each side of the brand new equation. Clear up 2x – 2 = 6 by following the 2 steps outlined instantly above Step 1: Divide each side of 2x – 2 = 6 by 2 (2x – 2) / 2 = 6 / 2 x – 1 = 3 Step 2: Add 1 to each side of x – 1 = 3 x – 1 + 1 = 3 + 1 x = 4 ## Fixing two-step equations involving fractions Instance #3 Clear up (2/5)x + 4 = 14 Step 1: Isolate (2/5)x by subtracting 4 from each side of (2/5)x + 4 = 14 (2/5)x + 4 – 4 = 14 – 4 (2/5)x = 10 Step 2: Multiply each side of (2/5)x = 10 by the reciprocal of two/5 or 5/2. (5/2)(2/5)x = 10(5/2) (10/10)x = 50/2 x = 25 Once more, we will additionally resolve instance #3 by dividing each side by 2/5 first. (2/5)x + 4 = 14 Divide each side by 2/5 [(2/5) ÷ (2/5)]x + 4 ÷ 2/5 = 14 ÷ 2/5 [1]x + 4/1 ÷ 2/5 = 14/1 ÷ 2/5 x + 4/1 × 5/2 = 14/1 × 5/2 x + 20/2 = 70/2 x + 10 = 35 Subtract 10 from each side x + 10 – 10 = 35 – 10 x + 0 = 25 x = 25 ## Fixing two-step equations involving decimals The steps to comply with are the identical when fixing two-step equations involving decimals. Instance #4 Clear up 3.1x + 1.2 = 7.4 Subtract 1.2 from each side of 3.1x + 1.2 = 7.4 to isolate 3.1x 3.1x + 1.2 – 1.2 = 7.4 – 1.2 3.1x = 6.2 Divide each side of three.1x = 6.2 by 3.1 to isolate x (3.1x)/3.1 = 6.2/3.1 x = 2 ## The tremendous quick solution to resolve two-step equations. When taking a take a look at, there is no such thing as a must comply with all of the steps above if all it’s essential to do is to decide on the suitable reply from an inventory of a number of decisions. To resolve ax + b = c, simply use x = (c – b)/a You may resolve instance #2 or 6x + 2 = 20 very quick. x = (20 – 2)/6 = 18/6 = 3 Equally, you possibly can resolve instance #1 or 2x – 2 = 6 very quick. x = (c + b)/a = (6 + 2)/2 = 8/2 = 4 Linear equations calculator Simply resolve equations of the shape ax + b = c with the clicking of a button!
My Math Forum Problem Algebra Pre-Algebra and Basic Algebra Math Forum June 2nd, 2015, 10:27 PM #41 Banned Camp Joined: May 2015 From: Casablanca Posts: 36 Thanks: 7 $2x-7=5x+6$. We'll solve this equation in steps: 1- Get rid of $2x$ by subtracting $2x$ in both sides: $(2x-7)-2x=(5x+6)-2x$ 2-Get rid of brackets and put numbers having x's one next the other: $2x-2x-7=5x-2x+6$ 3-You see that $2x-2x=0$ and $-7$ is left alone. In the other side $5x-2x=3x$. 4-From step $3$ the equation rearranges to: $-7=3x+6$. 5-We must cancel out $6$ in this step. To do so, we add the opposite of $6$ in both sides, which is $-6$: $-7-6=3x+6-6$. Note that $6-6=0$. 6-The equation becomes: $-13=3x$ or $3x=-13$ 7-Get rid of $3$. To do so, divide both sides by $3$: $\dfrac{3x}{3}=-\dfrac{13}{3}$. Note that $\dfrac{3x}{3}=\dfrac{3}{3}x=1x=x$. $x=-\dfrac{13}{3}$ Hope that helped. Last edited by Assayid; June 2nd, 2015 at 10:35 PM. June 3rd, 2015, 08:15 AM #42 Member Joined: May 2015 From: O Posts: 34 Thanks: 0 Quote: Originally Posted by Assayid $2x-7=5x+6$. We'll solve this equation in steps: 1- Get rid of $2x$ by subtracting $2x$ in both sides: $(2x-7)-2x=(5x+6)-2x$ 2-Get rid of brackets and put numbers having x's one next the other: $2x-2x-7=5x-2x+6$ 3-You see that $2x-2x=0$ and $-7$ is left alone. In the other side $5x-2x=3x$. 4-From step $3$ the equation rearranges to: $-7=3x+6$. 5-We must cancel out $6$ in this step. To do so, we add the opposite of $6$ in both sides, which is $-6$: $-7-6=3x+6-6$. Note that $6-6=0$. 6-The equation becomes: $-13=3x$ or $3x=-13$ 7-Get rid of $3$. To do so, divide both sides by $3$: $\dfrac{3x}{3}=-\dfrac{13}{3}$. Note that $\dfrac{3x}{3}=\dfrac{3}{3}x=1x=x$. $x=-\dfrac{13}{3}$ Hope that helped. Why you got banned? June 3rd, 2015, 08:36 AM #43 Member Joined: Jun 2015 From: Casablanca Posts: 47 Thanks: 3 Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam. June 3rd, 2015, 08:47 AM #44 Member Joined: May 2015 From: O Posts: 34 Thanks: 0 Quote: Originally Posted by Mohajir Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam. Ah I see. WB! Thanks for the help! June 3rd, 2015, 10:17 AM #45 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Well Mathlover, I see you "live nowhere"; perhaps you can move to Banned Camp" too?! June 3rd, 2015, 12:39 PM #46 Member Joined: May 2015 From: O Posts: 34 Thanks: 0 Quote: Originally Posted by Mohajir Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam. I'm confuse why and how would you know if the answer is a negative? June 3rd, 2015, 12:41 PM #47 Member Joined: May 2015 From: O Posts: 34 Thanks: 0 Quote: Originally Posted by Denis Well Mathlover, I see you "live nowhere"; perhaps you can move to Banned Camp" too?! Why is that? June 3rd, 2015, 01:05 PM #48 Member Joined: Jun 2015 From: Casablanca Posts: 47 Thanks: 3 Quote: Originally Posted by Mathlover44 I'm confuse why and how would you know if the answer is a negative? I got a message from mods. Hey, Mathlover44, don't bother yourself with it anymore. And thanks for your concern. June 3rd, 2015, 01:22 PM #49 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Please...can someone close this thread... Tags problem Thread Tools Display Modes Linear Mode Contact - Home - Forums - Cryptocurrency Forum - Top
# Calculate geometric probabilities. - PowerPoint PPT Presentation Calculate geometric probabilities. 1 / 16 Calculate geometric probabilities. ## Calculate geometric probabilities. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Objectives Calculate geometric probabilities. Use geometric probability to predict results in real-world situations. 2. If every outcome in the sample space is equally likely, the theoretical probability of an event is Remember that in probability, the set of all possible outcomes of an experiment is called the sample space. Any set of outcomes is called an event. 3. Remember! If an event has a probability p of occurring, the probability of the event not occurring is 1 – p. 4. The point is on RS. Example 1A: Using Length to Find Geometric Probability A point is chosen randomly on PS. Find the probability of each event. 5. The point is not on QR. Subtract from 1 to find the probability that the point is not on QR. Example 1B: Using Length to Find Geometric Probability 6. P(PQ or QR) = P(PQ) + P(QR) Example 1C: Using Length to Find Geometric Probability The point is on PQ or QR. 7. Use the figure below to find the probability that the point is on BD. Check It Out! Example 1 8. Example 2A: Transportation Application A pedestrian signal at a crosswalk has the following cycle: “WALK” for 45 seconds and “DON’T WALK” for 70 seconds. What is the probability the signal will show “WALK” when you arrive? To find the probability, draw a segment to represent the number of seconds that each signal is on. The signal is “WALK” for 45 out of every 115 seconds. 9. Example 3A: Using Angle Measures to Find Geometric Probability Use the spinner to find the probability of each event. the pointer landing on yellow The angle measure in the yellow region is 140°. 10. Example 3B: Using Angle Measures to Find Geometric Probability Use the spinner to find the probability of each event. the pointer landing on blue or red The angle measure in the blue region is 52°. The angle measure in the red region is 60°. 11. Example 3C: Using Angle Measures to Find Geometric Probability Use the spinner to find the probability of each event. the pointer not landing on green The angle measure in the green region is 108°. Subtract this angle measure from 360°. 12. Example 4: Using Area to find Geometric Probability Find the probability that a point chosen randomly inside the rectangle is in each shape. Round to the nearest hundredth. 13. ≈ 0.18. The probability is P = 254.5 1400 Example 4A: Using Area to find Geometric Probability the circle The area of the circle is A = r2 = (9)2 =81 ≈ 254.5 ft2. The area of the rectangle is A = bh = 50(28)=1400 ft2. 14. The area of the trapezoid is The probability is Example 4B: Using Area to find Geometric Probability the trapezoid The area of the rectangle is A = bh = 50(28)=1400 ft2. 15. The probability is Example 4C: Using Area to find Geometric Probability one of the two squares The area of the two squares is A = 2s2 = 2(10)2 =200 ft2. The area of the rectangle is A = bh = 50(28)=1400 ft2.
How Do You Calculate a Square Root? By Staff WriterLast Updated Apr 5, 2020 10:37:15 AM ET Mongkol Nitirojsakul / EyeEm/EyeEm/Getty Images In order to calculate the square root of a non-perfect square number, first find two perfect squares between which the number lies. Second, divide the number by one of the two square roots. Thirdly, take the average of the result and the root to discover the answer. For example, to find the square root of 10, first note that it lies between 3 and 4, the square roots of 9 and 16, respectively. Then, divide 10 by 3 to get 3.33. Average 3.33 and 3 to get the answer 3.1667. To come up with a more accurate answer, repeat the second and third steps. First, divide 10 by 3.1667 to get 3.1579. Then, average 3.1579 and 3.1667 to get 3.1623. Lastly, test the answer 3.1623 by squaring it. It should equal 10.0001. To calculate the square root of a number by hand, one can use a variety of approaches. If the goal is to simplify the square root, you could, for example, change the square root of 50 into 5 times the square root of 2. Reducing the number to its simplest terms is always a good solution to finding the square root of a number. Going further, by hand, involves estimation first. In the previous example, 5 times the square root of 2 can be estimated by multiplying 5 by 1.41, which is the square root of 2, to get approximately 7.1. For a larger number, you could find the square root using prime factorization. For example, to find the square root of 2,000, you could find the square roots of 500 and 4, which are multiplied together to equal 2,000. The square root of 4 is 2, and the square root of 500 is 5 times the square root of 20, which is 10 times the square root of 5. Overall, the square root of 2,000 is then 20 times the square root of 5, or approximately 40. More From Reference
# Suppose the acreage of forest is decreasing by 2% per year because of development. If there are currently 4,500,000 acres of forest, determine the amount of forest land after each of the following number of years? Feb 8, 2018 See below an explanation of how to do it, as cannot directly answer question as no number of years was given... But use: $A = 4 , 500 , 000 \times {\left(0.98\right)}^{N}$ Where $N$ is the years. #### Explanation: Even though there's no years, I will do a demonstration of how to do it for certain years Even though this isn't money related, I would use compound interest, where a certain percentage of a value is lost over a certain amount of time. It is repeated loss of money or other over a period of time. $A = P \times {\left(1 + \frac{R}{100}\right)}^{N}$ Where $A$ is the amount after the amount of time, $P$ is the original amount, $R$ is the rate and $N$ is the number of years. Plugging our values into the formula we get: $A = 4 , 500 , 000 \times {\left(1 - \frac{2}{100}\right)}^{N}$ As you did not state the number of years we will leave this blank for the moment. Notice that we minus as it is decreasing... $\frac{2}{100} = 0.02$ Therefore instead of $\frac{2}{100}$ minus this from $1$ and re-do the formula: $A = 4 , 500 , 000 \times {\left(0.98\right)}^{N}$ Let's just do an example: Someone puts £50,000 in a bank, he gets interest of 2.5% each year, calculate the amount he would have after $3$ years: (Focus on that it is addition as he is getting money) Using the formula $A = P \times {\left(1 + \frac{R}{100}\right)}^{N}$ we get... A=£50,000xx(1+2.5/100)^3 $\frac{2.5}{100} = 0.025$ Therefore we add this onto $1$ giving us $1.025$ This gets us... A=£50,000 xx (1.025)^3 Plug this in your calculator you get... =£53844.53125 which is rounded to £53844.53 Just do the exact same for your question, putting with the values I gave, just input the power as the amount of years that you want to work out.
Question Video: Dividing Complex Numbers \| Nagwa Question Video: Dividing Complex Numbers \| Nagwa # Question Video: Dividing Complex Numbers Mathematics • First Year of Secondary School ## Join Nagwa Classes Simplify (3 β 6π)/(1 β 5π). 02:09 ### Video Transcript Simplify three minus six π divided by one minus five π. When working with imaginary numbers and complex numbers, you cannot have an π on a denominator. So here, to get rid of the π on the denominator, we have to multiply by the complex conjugate. So the complex conjugate is where you keep the first number the same and you change the sign of the second number. So instead of one minus five π, the complex conjugate is one plus five π. And whatever we do to the denominator, we have to also do to the numerator. So now we need to FOIL. We need to use the distributive property. Three times one is three. Three times five π is 15π. Negative six π times one is negative six π. And negative six π times five π is equal to negative 30π squared. Now for the denominator, one times one is one, one times five π is five π, negative five π times one is negative five π, and negative five π times positive five π is negative 25π squared. However, π squared is equal to negative one. The reason why is because π squared is equal to π times π. And π is equal to square root negative one. So π times π is the same thing as square root negative one times square root negative one, which would be negative one. So again π squared is equal to negative one. So letβs replace these with negative one. And multiplying negative 30 times negative one is positive 30 and negative 25 times negative one is positive 25. So now we can combine the πs and we can combine the constant numbers. On the numerator three plus 30 is 33, and 15π minus six π is nine π. On the denominator, one plus 25 is 26 and then five π minus five π is zero π, which we do not have to write plus zero π. We can leave it out. And splitting this into two fractions would be thirty-three twenty-sixths plus nine π twenty-sixths. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
Wednesday , December 8 2021 # How to Find the Greatest Common Factors of Two Numbers? Learning to identify the greatest common factors of two numbers is a key math skill highly crucial for calculations particularly in maths. It is quite useful especially for simplifying fractions. Since determining the common denominators for two fractional numbers may seem difficult for students, learning two simple methods for determining the greatest common factor will enable them to understand and learn the process quickly. Although, they need to first have a clear understanding of the basics of factors to learn these two approaches of finding common factors. Eventually, they can apply their acquired knowledge to simplify fractions. ## What Is a Factor? The factors are numbers multiplied together to give other numbers. For instance, four and two are the factors of eight, because four multiplied by two equals eight. Similarly, three and three are factors of nine, since three multiplied by three is equal to nine. As you may know, prime numbers are numbers that have no factors other than themselves, and one. So 3 is a prime number because the only two whole numbers (integers) that can multiply together to give 3 as an answer are 3 and 1. In the same way, 7 is a prime number, and so is 13. Because of this, it’s often helpful to break down a number into “prime factors.” This means finding all of the prime number factors of another number. It basically breaks the number down into its fundamental “building blocks,” which is a useful step towards finding the greatest common factor of two numbers and is also invaluable when it comes to simplifying square roots. ## Finding the Greatest Common Factor: In algebra, the greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers have the same in a given set. This factor should divide all the numbers without any remainder. There are various methods involved in finding the GCF of a set of given numbers: • Listing the factors Method. • Prime Factorization Method. • Division Method ### Listing the factors Method: The simplest method for finding the greatest common factor of two numbers is to simply list all of the factors of each number and look for the highest number that both of them share. This method of finding GCF includes listing all the factors of any given number and then finding the highest common factor among them. Steps of listing the factors method: • Write a list of all the factors of each number. • Mark all the common factors. • Find the common factor with the greatest value. ### Prime Factorization Method: This process of determining GCF includes factorization of given numbers into their factors. It is also known as the highest common factor(HCF) or highest common divisor(HCD). Understanding the concept of GCF requires the knowledge of factors and common factors of numbers. The prime factorization method includes representing a number as a product of all its prime factors, starting from that number’s smallest prime factor. Steps of prime factorization method: • Write all the prime factors of any given set of numbers. • Mark all the common factors among the factors of numbers. • Multiply the common factors found in each number, and this number is the greatest common factor of the set of numbers. ### Conclusion Learning the process of finding HCF and LCM is highly important for every student as it forms the basis for understanding many advanced skills. Cuemath enables students to visualize math thus helping them gain a deep understanding of math topics. By gaining a basic understanding of these core concepts students can establish the right approach for learning advanced math with ease.
Related Topics: More Lessons for Algebra Math Worksheets A series of free, online Basic Algebra Lessons. In this lesson, we will learn • how to solve radical equations • how to graph radical equations using a table • how to graph radical equations using shifts • how to estimate square roots • how to find or simplify cube roots Radical equations are equations that have square root terms. When solving radical equations, try to isolate the radical expression on one side of the equation and then square both sides (it is the inverse operation of taking a square root). If there are two or more terms on the opposite side of the equation, remember to draw parenthesis around that expression before squaring ### Graphing Radical Equations using a Table One way to graph radical functions is to create a table of values and then plot the points. Before starting the table, first determine the domain of the function. Remember, the radical must be greater than or equal to zero. Once this lower limit for input (domain) values is established, create the table of values. When graphing radicals we plot the points in the coordinate plane. Graphing Radical Equations using a Table This video is a demonstration of graphing radical equations by first (1) finding the domain, and (2) using an x-y table. ### Graphing Radical Equations using Shifts When graphing radical equations using shifts, adding or subtracting a constant that is not in the radical will shift the graph up (adding) or down (subtracting). Adding or subtracting a constant that is in the radical will shift the graph left (adding) or right (subtracting). Multiplying a negative constant by the equation will reflect the graph over the x-axis. Multiplying by a number larger than one increases the y-values. How to graph radical functions without making a table of values. Graphing Radicals using Vertical/Horizontal shifts and scaling. ### Estimating Square Roots When a radical is not a perfect square (1, 4, 9, 16, …), estimating square roots is a valuable tool. When asked to estimate the value of a radical between two consecutive integers, find the two perfect squares that are slightly less and slightly more than the radicand. Also, remember that negative numbers do not have a real number square root. How to find square roots of integers. Estimating Square Roots to the Nearest Integer ### Simplify Cube Roots A square root is an exponent of one-half. A cube root is an exponent of one-third. Square roots of negative numbers do not have real number roots since the product of any real number and itself is positive. Cube roots do exist for negative numbers since the product of three negatives is a negative. Cube roots re-appear often in Geometry and in Algebra II. Finding Cube Roots Simplify cube root Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Converting Fractions to Decimals Steps for converting fraction into a decimal: 1. Find such integer number, that when multiplied by the denominator will give a power of 10 (10 or 100 or 1000 etc.) 2. Multiply both numerator and denominator by that number (this can be done because of equivalence of fractions) 3. As a result, we obtain a decimal fraction, that can be easily converted into a decimal. As can be seen we use ease of converting decimal fraction and equivalence of fractions to convert arbitrary fraction. Example 1. Convert $\frac{{4}}{{5}}$ into decimal. By what integer number should we multiply 5 to get 10? By 2. Now, multiply both numerator and denominator by 2: $\frac{{4}}{{5}}=\frac{{{4}\cdot{\color{red}{{{2}}}}}}{{{5}\cdot{\color{red}{{{2}}}}}}=\frac{{8}}{{10}}$. We obtained decimal fraction and it can be easy converted to decimal. Since ${10}={{10}}^{{{\color{blue}{{1}}}}}$, we move decimal point one position to the left: $\frac{{8}}{{10}}={0.8}$. Answer: $\frac{{4}}{{5}}={0.8}$. Let's do slightly harder example. Example 2. Convert $\frac{{3}}{{8}}$ into decimal. By what integer number should we multiply 8 to get ${{10}}^{{1}}={10}$? There is no such number. By what integer number should we multiply 8 to get ${{10}}^{{2}}={100}$? There is no such number. By what integer number should we multiply 8 to get ${{10}}^{{3}}={1000}$? By 125. Now, multiply both numerator and denominator by 125: $\frac{{3}}{{8}}=\frac{{{3}\cdot{\color{red}{{{125}}}}}}{{{8}\cdot{\color{red}{{{125}}}}}}=\frac{{375}}{{1000}}$. Since ${1000}={{10}}^{{{\color{blue}{{3}}}}}$, we move decimal point three places to the left: $\frac{{375}}{{1000}}={0.375}$. Answer: $\frac{{3}}{{8}}={0.375}$. We can convert improper fractions this way as well. Example 3. Convert $\frac{{26}}{{25}}$ into decimal. By what integer number should we multiply 25 to get ${{10}}^{{1}}={10}$? There is no such number. By what integer number should we multiply 25 to get ${{10}}^{{2}}={100}$? By 4. Now, multiply both numerator and denominator by 4: $\frac{{26}}{{25}}=\frac{{{26}\cdot{\color{red}{{{4}}}}}}{{{25}\cdot{\color{red}{{{4}}}}}}=\frac{{104}}{{100}}$. Since ${100}={{10}}^{{{\color{blue}{{2}}}}}$, we move decimal point two places to the left: $\frac{{104}}{{100}}={1.04}$. Answer: $\frac{{26}}{{25}}={1.04}$. Note, that not all fractions can be converted into decimal. This occurs when we can't find such number, that when multiplied by denominator will give power of 10. This is true for prime numbers and their multiples. For example, $\frac{{1}}{{3}}$, $\frac{{5}}{{70}}$ can' t be converted. Exercise 1. Convert $\frac{{7}}{{20}}$ into decimal. Answer: $\frac{{7}}{{20}}={0.35}$. Next exercise. Exercise 2. Convert $\frac{{1}}{{80}}$ into decimal. Answer: $\frac{{1}}{{80}}={0.0125}$. Next exercise. Exercise 3. Convert $-\frac{{19}}{{4}}$ into decimal. Answer: $-{4.75}$. Next exercise. Exercise 4. Convert $\frac{{5}}{{11}}$ into decimal. We can convert mixed numbers as well! Exercise 5. Convert mixed number ${3}\frac{{5}}{{16}}$ into decimal. Either convert $\frac{{5}}{{16}}$ into decimal and add to 3: ${3}\frac{{5}}{{16}}={3}+{0.3125}={3.3125}$. Or convert mixed number into improper fraction, and then convert result into decimal. Answer: ${3.3125}$. last one with mixed number. Exercise 6. Convert mixed number $-{2}\frac{{11}}{{20}}$ into decimal. Answer: $-{2.55}$. Hint: ignore minus sign, perform conversion, and then place minus sign back.
### GRAPHING OF FUNCTIONS USING FIRST AND SECOND DERIVATIVES The following problems illustrate detailed graphing of functions of one variable using the first and second derivatives. Problems range in difficulty from average to challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by carefully labeling critical points, intercepts, and inflection points. In addition, it is important to label the distinct sign charts for the first and second derivatives in order to avoid unnecessary confusion of the following well-known facts and definitions. Here are instruction for establishing sign charts (number line) for the first and second derivatives. To establish a sign chart (number lines) for f' , first set f' equal to zero and then solve for x . Mark these x-values underneath the sign chart, and write a zero above each of these x-values on the sign chart. In addition, mark x-values where the derivative does not exist (is not defined). For example, mark those x-values where division by zero occurs in f' . Above these x-values and the sign chart draw a dotted vertical line to indicate that the value of f' does not exist at this point. These designated x-values establish intervals along the sign chart. Next, pick points between these designated x-values and substitute them into the equation for f' to determine the sign ( + or - ) for each of these intervals. Beneath each designated x-value, write the corresponding y-value which is found by using the original equation y = f(x) . These ordered pairs (x, y) will be a starting point for the graph of f . This completes the sign chart for f' . Establish a sign chart (number line) for f'' in the exact same manner. To avoid overlooking zeroes in the denominators of f' and f'' , it is helpful to rewrite all negative exponents as positive exponents and then carefully manipulate and simplify the resulting fractions. FACTS and DEFINITIONS 1. If the first derivative f' is positive (+) , then the function f is increasing () . 2. If the first derivative f' is negative (-) , then the function f is decreasing ( ) . 3. If the second derivative f'' is positive (+) , then the function f is concave up () . 4. If the second derivative f'' is negative (-) , then the function f is concave down () . 5. The point x=a determines a relative maximum for function f if f is continuous at x=a , and the first derivative f' is positive (+) for x<a and negative (-) for x>a . The point x=a determines an absolute maximum for function f if it corresponds to the largest y-value in the range of f . 6. The point x=a determines a relative minimum for function f if f is continuous at x=a , and the first derivative f' is negative (-) for x<a and positive (+) for x>a . The point x=a determines an absolute minimum for function f if it corresponds to the smallest y-value in the range of f . 7. The point x=a determines an inflection point for function f if f is continuous at x=a , and the second derivative f'' is negative (-) for x<a and positive (+) for x>a , or if f'' is positive (+) for x<a and negative (-) for x>a . 8. THE SECOND DERIVATIVE TEST FOR EXTREMA (This can be used in place of statements 5. and 6.) : Assume that y=f(x) is a twice-differentiable function with f'(c)=0 . a.) If f''(c)<0 then f has a relative maximum value at x=c . b.) If f''(c)>0 then f has a relative minimum value at x=c . These are the directions for problems 1 through 10. For each function state the domain. Determine all relative and absolute maximum and minimum values and inflection points. State clearly the intervals on which the function is increasing () , decreasing ( ) , concave up () , and concave down () . Determine x- and y-intercepts and vertical and horizontal asymptotes when appropriate. Neatly sketch the graph. • PROBLEM 1 : Do detailed graphing for f(x) = x3 - 3x2 . Click HERE to see a detailed solution to problem 1. • PROBLEM 2 : Do detailed graphing for f(x) = x4 - 4x3 . Click HERE to see a detailed solution to problem 2. • PROBLEM 3 : Do detailed graphing for f(x) = x3 (x-2)2 . Click HERE to see a detailed solution to problem 3. • PROBLEM 4 : Do detailed graphing for . Click HERE to see a detailed solution to problem 4. • PROBLEM 5 : Do detailed graphing for . Click HERE to see a detailed solution to problem 5. • PROBLEM 6 : Do detailed graphing for . Click HERE to see a detailed solution to problem 6. • PROBLEM 7 : Do detailed graphing for f(x) = x - 3x1/3 . Click HERE to see a detailed solution to problem 7. • PROBLEM 8 : Do detailed graphing for . Click HERE to see a detailed solution to problem 8. • PROBLEM 9 : Do detailed graphing for for x in . Click HERE to see a detailed solution to problem 9. • PROBLEM 10 : Do detailed graphing for . Click HERE to see a detailed solution to problem 10. • PROBLEM 11 : Consider the cubic polynomial y = A x3 + 6x2 - Bx , where A and B are unknown constants. If possible, determine the values of A and B so that the graph of y has a maximum value at x= -1 and an inflection point at x=1 . Click HERE to see a detailed solution to problem 11. ### Click HERE to return to the original list of various types of calculus problems. Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address : Duane Kouba 1998-06-03
Seek Homework Help & learn solving the questions of grade 6 via practice test, chapter test, Cumulative Practice, etc. with the help of Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120. We have provided detailed explanations in simple methods on this page. Refer to Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 to learn the concepts of the chapter. ## Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120 Are you looking for the best material to score top in the exams? Then, you are in the right place. Here you can find the explanations for each and every question in different methods @ Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120. So, access the links and start your preparation for the exams. Vocabulary Lesson: 1 Count to 120 by Ones Lesson: 2 Count to 120 by Tens Lesson: 3 Compose Numbers 11 to 19 Lesson: 4 Tens Lesson: 5 Tens and Ones Lesson: 6 Make Quick Sketches Lesson: 7 Understand Place Value Lesson: 8 Write Numbers in Different Ways Lesson: 9 Count and Write Numbers to 120 Chapter: 6 – Count and Write Numbers to 120 ### Count and Write Numbers to 120 Vocabulary Organize It Review Words: column decade numbers hundred chart row Use the review words to complete the graphic organizer. Answer: Explanation: The given graphic organizer title is called as hundred chart, and numbers 41 to 50 are called as row, numbers 4,14,24 to 84, 94 are called as column and numbers 10,20,30 to 90,100 are called decade numbers. Define It Use your vocabulary cards to identify the words. Explanation: Given in the picture to identify words 2 – Two and 3 – Three. ### Lesson 6.1 Count to 120 by Ones Explore and Grow Point to each number as you count to 120. Color the first two rows and the last two rows. How are the rows the same? How are they different? Answer: Explanation: As we move from 1,2,3 count to 120 have pointed with an arrow for each number. Colored with green first two rows from 1 to 10, 11 to 20 and the last two rows from 101 to 110, 111 to 120. The rows are same – as we move from number 1 to 120 from first number to second number the number is increased by 1 as 1, 1 + 1 = 2, 2 + 1 = 3 and so on 119 + 1 = 120 means the next number is addition of 1 plus the previous number. The rows are different – as we see the numbers are not the same they differ its 1,2 ,3 to 119, 120 from 1 to 120. Show and Grow Count by ones to write the missing numbers. Question 1. 82, _____83_____, ____84____, _____85_____, ____86______, ____87______ Answer: The missing numbers after 82 are 83, 84, 85, 86 ,87 Explanation: Given the starting number as 82 we count by ones and write the next missing numbers as 82 + 1 = 83, 83 + 1 = 84, 84 + 1 = 85 , 85 +1 = 86, 86 + 1 = 87, now the missing numbers after 82 are 83, 84, 85, 86 ,87. Question 2. 103, ____104______, ____105______, ___106_______, ____107______, ____108______ Answer: The missing numbers after 103 are 104,105,106,107,108. Explanation: Given the starting number as 103 we count by ones and write the next missing numbers as 103 + 1 = 104, 104 + 1 =105, 105 + 1 = 106, 106 + 1 = 107,107 + 1 = 108, now the missing numbers after 103 are 104,105,106,107,108. Apply and Grow: Practice Count by ones to write the missing numbers. Question 3. 56, ____57______, ____58____, ____59______, ____60______, ____61______ Answer: The missing numbers after 56 are 57,58,59,60,61 Explanation: Given the starting number as 56 we count by ones and write the next missing numbers as 56 + 1 = 57, 57 + 1 = 58, 58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61, now the missing numbers after 56 are 57,58,59,60,61. Question 4. 98, _____99_____, ____100______, ____101______, ____102______, ____103______ Answer: The missing numbers after 98 are 99,100,101,102,103 Explanation: Given the starting number as 98 we count by ones and write the next missing numbers as 98 + 1 = 99,99 + 1 = 100, 100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103, So the missing numbers after 98 are 98,99,100,101,102,103. Question 5. 115, ____116______, ____117______, ____118______, ____119______, ___120_______ Answer: The missing numbers after 115 are 116,117,118,119,120 Explanation: Given the starting number as 115 we count by ones and write the next missing numbers as 115 + 1 = 116, 116 + 1 = 117, 117 + 1 = 118, 118 + 1 = 119, 119 + 1 = 120, So the missing numbers after 115 are 116,117,118,119,120. Question 6. ___40_______, ____41______, 42, ____43______, ____44______, ____45______ Answer: The missing numbers before and after 42 are 40,41,43,44,45 Explanation: Given number is 42 and 2 numbers before 42 are count 1 before 42 is 42- 1= 41 and count 1 before 41 is 41 – 1 = 40 and count 1 after 42 is 42 + 1 = 43, 43 + 1 = 44, 44 + 1 = 45, So the missing numbers before and after 42 are 40,41,43,44,45. Write the missing numbers in the chart. Question 7. Answer: The missing numbers in the chart are 33,35,42,44. Explanation: The missing number after 32 is 32 + 1 = 33, the number after 34 is 34 + 1 = 35, the number before 43 is 43 – 1 = 42 and number after 43 is 43 + 1 = 44, So the missing numbers in the chart are 33,35,42,44. Question 8. Answer: The missing numbers in the chart are 81,83,84,92,94. Explanation: The missing number before 82 is 82 – 1 = 81, the number after 82 is 82 + 1 = 83,now number after 83 is 83 + 1 = 84. The number after 91 is 91 + 1 = 92, the number after 93 is 93 + 1 = 94, therefore the missing numbers in the chart are 81,83,84,92,94. Question 9. MP Structure Structure Write a number between 95 and 105. Then count by ones to write the next 7 numbers. ____96___, _____97____, _____98_____, ____99____, ___ 100___, ____101___, ___102__, ___103___ Answer: Let the number be 96 which is between 95 and 105 and the next 7 numbers after 96 are 97,98,99,100,101,102,103. Explanation: Given to write a number between 95 and 105, Let us take the number as 96, now we will write the next 7 numbers are 96 + 1 = 97,97+ 1 = 98, 98 + 1 =99, 99 + 1 = 100,100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103, therefore the next 7 numbers after 96 are 97,98,99,100,101,102,103. Think and Grow: Modeling Real Life You hove 108 bouncy balls. You want 112. How many more bouncy bails do you need? Draw more bails to show 112: ______4______ more bouncy balls Answer: 4 more bouncy balls are needed, Explanation: Given I have 108 bouncy balls and I want 112, So I need 108 + 1 = 109, 109 + 1 = 110, 110 + 1 = 111, 111+ 1 = 112 or 112 – 108 = 4, so we need 4 more bouncy balls. Show and Grow Question 10. You have 66 rocks. You want 75. How many more rocks do you need? Draw more rocks to show 75: ______9_______ more rocks Answer: I need 9 more rocks. Explanation: Given I have 66 rocks and I want 75, So more rocks I needed are 75 – 66 = 9. ### Count to 120 by Ones Practice 6.1 Count by ones to write the missing numbers. Question 1. 57, _____58_____, _____59_____, ____60______, ____61______, ____62______ Answer: The missing numbers after 57 are 58,59,60,61,62. Explanation: Given the starting number as 57 we count by ones and write the next missing numbers as 57 + 1 = 58, 58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61, 61 + 1 = 62, So the missing numbers after 57 are 58,59,60,61,62. Question 2. 109, ____110___, ____111___, ____112___, ____113___, __114_____ Answer: The missing numbers after 109 are 110,111,112,113,114 Explanation: Given the starting number as 109 we count by ones and write the next missing numbers as 109 + 1 = 110, 110 + 1 =111,111 + 1 = 112, 112 + 1 = 113, 113 + 1 = 114, So the missing numbers after 109 are 110,111,112,113,114. Question 3. 40, ____41___, ____42___, ___43___, ___44___, ___45___ Answer: The missing numbers after 40 are 41,42,43,44,45 Explanation: Given the starting number as 40 we count by ones and write the next missing numbers as 40 + 1 = 41, 41 + 1 = 42,42 + 1 = 43, 43 + 1 = 44, 44 + 1 = 45, So the missing numbers after 40 are 40,41,42,43,44,45. Question 4. _____97_____, ____98______, ____99______, 100, ____101______, ____102______ Answer: The missing numbers before and after 100 are 97,98,99,101,102 Explanation: Given number is 100 and 3 numbers before 100 are count 1 before 100 is 100- 1= 99, count 1 before 99 is 99 – 1 = 98 and count 1 before 98 is 98 – 1 = 97, numbers after 100 are 100 + 1 = 101 and 101 + 1 = 102,therefore the missing numbers before and after 100 are 97,98,99,101,102. Write the missing numbers in the chart. Question 5. Answer: The missing numbers in the chart are 21,23,30,32. Explanation: The missing number after 20 is 20 + 1 = 21, the number after 22 is 22 + 1 = 23, the number before 31 is 31 – 1 = 30 and number before 33 is 33 – 1 = 32, So the missing numbers in the chart are 21,23,30,32. Question 6. Answer: The missing numbers in the chart are 102,104,105,113,115. Explanation: The missing number before 103 is 103 – 1 = 102, the number after 103 is 103 + 1 = 104, the number after 104 is 104 + 1 = 105, the number after 112 is 112 + 1 = 113 and the number after 114 is 114 + 1 = 115 So the missing numbers in the chart are 102,104,105,113,115. Question 7. MP Structure Write a number between 85 and 95. Then count by ones to write the next 7 numbers. __87__, _88___, _89___, __90___, _91__, _92_, _93_, _94__ Answer: Let the number be 87 which is between 85 and 95 and the next 7 numbers after 87 are 88,89,90,91,92,93,94. Explanation: Given to write a number between 85 and 95, Let us take the number as 87, now we will write the next 7 numbers are 87 + 1 = 88,88+ 1 = 89, 89 + 1 =90, 90 + 1 = 91,91 + 1 = 92, 92 + 1 = 93, 93 + 1 = 94, therefore the next 7 numbers after 87 are 88,89,90,91,92,93,94. Question 8. Modeling Real Life There are 110 tokens. You want 119. How many more tokens do you need? ____9_____ more tokens Answer: I need 9 more tokens. Explanation : Given there are 110 tokens. and I want 119, so I needed 119 – 110 = 9 more tokens. Review & Refresh Question 9. ? – 6 = 4 Think 6 + 4 = ___10_____ . So, ______10__ – 6 = 4. Answer: 6 + 4 = 10 so, 10 – 6 = 4 Explanation: Given ___ – 6 = 4, Let us take the missing number as X, therefore X – 6 = 4, X = 4 + 6 =10,the equation is 10 – 6 = 4. ### Lesson 6.2 Count to 120 by Tens Explore and Grow Count to 10. Circle the number. Count 10 more. Circle the number. Continue until you reach 120. Answer: Explanation: Start counting from 1 to 10 circled, Next from 11 to 20 counted circled, counted till 120 and circled. Show and Grow Count by tens to write the missing numbers. Question 1. 70, __80___, __90__, __100__, _110__, __120__ Answer: The missing numbers after 70 are 80,90,100,110,120 Explanation: Started from 70 counting by 10’s so 70 + 10 = 80, 80 + 10 = 90, 90 + 10 = 100, 100 + 10 = 110, 110 + 10 = 120, So the missing numbers after 70 are 80,90,100,110,120 Question 2. 31, __41___, __51__, _61__, __71__, _81__ Answer: The missing numbers after 31 are 41,51,61,71,81 Explanation: Started from 31 counting by 10’s so 31 + 10 = 41, 41 + 10 = 51, 51 + 10 =61, 61 + 10 = 71 ,71 +10 = 81, So the missing numbers after 31 are 41,51,61,71,81. Apply and Grow: Practice Count by tens to write the missing numbers Question 3. 62, __72__, __82__, _92__, _102__, _112___ Answer: The missing numbers after 62 are 72,82,92,102,112 Explanation: Started from 62 counting by 10’s are 62 + 10 = 72, 72 + 10 = 82, 82 + 10 =92, 92 + 10 = 102 ,102 +10 = 112, So the missing numbers after 62 are 72,82,92,102,112 Question 4. 43, __53__, _63__, __73__, __83__, _93__ Answer: The missing numbers after 43 are 53,63,73,83,93 Explanation: Started from 43 counting by 10’s are 43 + 10 = 53, 53 + 10 = 63, 63 + 10 =73, 73 + 10 = 83 ,83 +10 = 93, So the missing numbers after 43 are 53,63,73,83,93. Question 5. ___10__, __20___, 30, __40__, __50___, __60__ Answer: The missing numbers before and after 30 are 10,20,40,50,60 Explanation: Started from 30 counting by 10’s numbers before and after 30 are before 30 – 10 = 20, 20 – 10 = 10 and after 30 are 30 + 10 = 40, 40 + 10 = 50, 50 + 10 = 60, therefore numbers before and after 30 are 10,20,40,50,60. Question 6. Write the missing numbers from the chart. Then count on by tens to write the next three numbers. Answer: The missing numbers in the chart are 71,81 and count on 10’s the next three numbers after 81 are 91,101,111. Explanation: Given numbers in the chart, the number missing before 72 is 72 – 1 = 71 and number missing before 82 in the chart is 82 – 1 = 81,now counting by 10’s next three numbers after 81 are 81 + 10 = 91,91 + 10 = 101,101 + 10 = 111. Therefore the missing numbers in the chart are 71,81 and count on 10’s the next three numbers after 81 are 91,101,111. Question 7. YOU BE THE TEACHER Your friend counts by tens starting with 27. Is your friend correct? Show how you know. 27, 37, 47, 67, 77, 87 Answer: No, Friend is incorrect. As 57 is missing. Explanation: Given friend counts by tens starting with 27 and gets 27, 37, 47, 67, 77, 87 but after 47 we should get 47 + 10 = 57 as 57 is missing friend is incorrect. Think and Grow: Modeling Real Life You have 50 points. On your next turn, you knock over 6 cans. How many points do you have now? Write the numbers: ____110______ points Answer: I have 110 points now. Explanation: Given I have 50 points on my next turn I knock over 6 cans and 10 points for each knocked over can means I got 6 X 10 = 60 points, So in total I have 50 + 60 = 110 points now. Show and Grow Question 8. You have 21 points. On your next turn, 3 beanbags land in the circle. How many points do you have now? Write the numbers: _____51_____ points Answer: I have 51 points now. Explanation: Given I have 21 points on my next turn 3 beanbags land in the circle 10 points for each beanbag means I got 3 X 10 = 30 points, So in total I have 21 + 30 = 51 points now. ### Count to 120 by Tens Practice 6.2 Count by tens to write the missing numbers. Question 1. 69, __79_, _89__, _99_, _109_, _119__ Answer: The missing numbers after 69 are 79,89,99,109,119 Explanation: Started from 69 counting by 10’s numbers after 69 are 69 + 10 = 79, 79 + 10 = 89, 89 + 10 = 99, 99 + 10 = 109, 109 + 10 = 119, therefore the missing numbers after 69 are 79,89,99,109,119. Question 2. 41, __51__, _61___, _71__, _81__, _91__ Answer: The missing numbers after 41 are 51,61,71,81,91 Explanation: Started from 41 counting by 10’s numbers after 41 are 41 + 10 = 51,51 + 10 = 61, 61 + 10 = 71, 71 + 10 = 81, 81 + 10 =91, So the missing numbers after 41 are 51,61,71,81,91. Question 3. 16, __26__, _36__, _46__, _56__, __66__ Answer: The missing numbers after 16 are 26,36,46,56,66 Explanation: Started from 16 counting by 10’s numbers after 16 are 16 + 10 = 26, 26 + 10 = 36,36 + 10 = 46, 46 + 10 = 56, 56 + 10 = 66, So the missing numbers after 16 are 26,36,46,56,66. Question 4. _64__, __74__, __84__, 94, __104_, __114__ Answer: The missing numbers before 94 are 84,74 ,64 and after 94 are 104, 114. Explanation: Started from 94 counting by 10’s numbers before and after 94 are before 94 – 10 = 84, 84 – 10 = 74 , 74 – 10 = 64 and after are 94 + 10 = 104,104 + 10 = 114, So the missing numbers before 94 are 84,74 ,64 and after 94 are 104, 114. Question 5. Write the missing numbers from the chart. Then count on by tens to write the next three numbers. Answer: The missing numbers in the chart are 6,16 and count on 10’s the next three numbers after 16 are 26,36,46. Explanation: Given numbers in the chart, the number missing after 5 is 5 + 1 = 6 and number missing after 15 in the chart is 15 + 1 = 16,now counting by 10’s next three numbers after 16 are 16 + 10 = 26,26 + 10 = 36,36 + 10 = 46. Therefore the missing numbers in the chart are 6,16 and count on 10’s the next three numbers after 16 are 26,36,46. Question 6. DIG DEEPER! You count to 50. You only count 5 numbers. Did you count by ones or by tens? Show how you know. Answer: I count by tens. Explanation: Given I count to 50 and count only 5 numbers , If I count by ones I will get 50 numbers like 1,1+1=2, 2+ 1=3,…till I reach 49+ 1 = 50 as shown in the above picture, If I count by tens I get 5 numbers, Like 10,10 + 10 = 20,20 + 30 = 40, 40 + 10 = 50, as I count 5 numbers So I count by tens. Question 7. Modeling Real Life You have 30 points. On your next turn, 4 balls stick to the wall. How many points do you have now? ____70_____ points Answer: I have 70 points now. Explanation: Given I have 30 points. On my next turn, 4 balls stick to the wall and each ball we have 10 points, So 4 X 10 = 40 points means in total I have 30 + 40 = 70 points now. Review & Refresh Question 8. 3 + 1 = ___4_____ Answer: 3 + 1 = 4 Explanation: We are going to add 1 to 3 we get 4. Question 9. 5 – 1 = ____4_____ Answer: 5 – 1 = 4 Explanation: We will subtract 1 from 5 we get 4. ### Lesson 6.3 Compose Numbers 11 to 19 Explore and Grow Color to show 13 and 17. What is the same about the numbers? What is different? Answer: Both are same- 13, 17 are prime numbers. Both 13 and 17 are different as both are not same numbers, they both are not equal. Explanation: Taken numbers from 1 to 20 in the chart, colored numbers 13 and 17 with green. Both are the same- 13 and 17 are prime numbers- these are the numbers, which are only divisible by 1 or the number itself as 13 and 17 are divisible by 1 and only by them they both are prime. Both are different-As 13 and 17 are equal they differ. Show and Grow Question 1. Circle 10 feathers. Complete the sentence. ____One____ ten and ___six____ ones is __sixteen____ . Answer: One ten and six ones is sixteen. 1 ten and 6 ones is 16. Explanation: Circled 10 feathers, after counting we have total 16 feathers we write the sentence as one ten and six ones is sixteen means 10 + 6 = 16. 1 ten and 6 ones is 16. Apply and Grow: Practice Circle 10 feathers. Complete the sentence. Question 2. ___One_____ ten and __five___ ones is ____fifteen_____ . Answer: One ten and five ones is fifteen. 1 ten and 5 ones is 15. Explanation: Circled 10 fishes, after counting we have total 15 fishes we write the sentence as one ten and five ones is fifteen means 10 + 5 = 15. 1 ten and 5 ones is 15. Question 3. ____One____ ten and ___three__ ones is ____thirteen_____ . Answer: One ten and three ones is thirteen. 1 ten and 3 ones is 13. Explanation: Circled 10 sun flowers, after counting we have total 13 sunflowers we write the sentence as one ten and three ones is thirteen means 10 + 3 = 13. 1 ten and 3 ones is 13. Question 4. ___One_ ten and __seven__ ones is ___seventeen___ . Answer: One ten and seven ones is seventeen. 1 ten and 7 ones is 17. Explanation: Circled 10 bees, after counting we have total 17 bees we write the sentence as one ten and seven ones is seventeen means 10 + 7 = 17. 1 ten and 7 ones is 17. Question 5. MP Number Sense Color to show the number. Complete the sentence. ___one___ ten and __six___ ones is __sixteen____ . Answer: One ten and six ones is sixteen. 1 ten and 6 ones is 16. Explanation: Colored the shown number 16 with blue, we write the sentence as one ten and six ones is sixteen means 10 + 6 = 16. 1 ten and 6 ones is 16. Think and Grow: Modeling Real Life You have 15 footballs. A bag can hold 10. You fill a bag. How many footballs are not in the bag? Draw a picture: Write the missing numbers: __one_ ten and __five_ ones is __fifteen__ footballs Answer: One ten and five ones is fifteen footballs. 1 ten and 5 ones is 15. 5 footballs are not there in the bag, Zero ten and five ones or 0 ten and 5 ones is five or 5 footballs are missing. Explanation: Given I have 15 footballs means one ten and five ones is fifteen or 1 ten and 5 ones is 15 and a bag can hold 10 footballs, In the picture there are total 15 footballs in that I have taken 10 in the bag so footballs which are not there in the bag are 15 – 10 = 5, So missing number of footballs in the bag are zero ten and five ones is five or 0 ten and 5 ones is 5. Show and Grow Question 6. Your teacher has 18 calculators. A case can hold 10. Your teacher fills a case. How many calculators are not in the case? Draw a picture: Write the missing numbers: __one__ ten and __eights ones _ is eighteen_ calculators Answer: One ten and eight ones is eighteen calculators. 1 ten and 8 ones is 18. 8 calculators are not there in the case, Zero ten and eight ones is eight or 0 ten and 8 ones is 8 calculators are missing, Explanation: Given teacher ha 18 calculators means one ten and eight ones is eighteen or 1 ten and 8 ones is 18 and a case can hold 10, In the picture there are total 18 calculators in that I have taken 10 in the case so number of calculators which are not there in the case are 18 – 10 = 8, So missing calculators in the case are zero ten and eight ones is eight or 0 ten and 8 ones is 8. ### Compose Numbers 11 to 19 Practice 6.3 Circle 10 objects. Complete the sentence. Question 1. __one___ ten and ___one___ ones is eleven_ . Answer: One ten and one ones is eleven. 1 ten and 1 ones is 11. Explanation: Circled 10 balls, after counting we have total 11 balls we write the sentence as one ten and one ones is eleven means 10 + 1 = 11. 1 ten and 1 ones is 11. Question 2. __one____ ten and __nine____ ones is _nineteen_____ . Answer: One ten and nine ones is nineteen. 1 ten and 9 ones is 19. Explanation: Circled 10 fishes, after counting we have total 19 fishes we write the sentence as one ten and nine ones is nineteen means 10 + 9 = 19. 1 ten and 9 ones is 19. Question 3. ___one___ ten and __three____ ones __thirteen____ . Answer: One ten and three ones is thirteen. 1 ten and 3 ones is 13. Explanation: Circled 10 honeybees, after counting we have total 13 we write the sentence as one ten and three ones is thirteen means 10 + 3 = 13. 1 ten and 3 ones is 13. Question 4. MP Number Sense Color to show the number. Complete the sentence. __one____ ten and _eight___ ones is eighteen__ . Answer: One ten and eight ones is eighteen. 1 ten and 8 ones is 18. Explanation: Colored the shown number 18 with yellow, we write the sentence as one ten and eight ones is eighteen means 10 + 8 = 18. 1 ten and 8 ones is 18. Question 5. MP Number Sense Match. Answer: Explanation: Matched 1 ten and 3 ones with 13, 1 ten and 8 ones with 12 and 12 ones with 12 in the picture above. Question 6. Modeling Real Life You have 16 books. A backpack can hold 10. You fill a backpack. How many books are not in the backpack? _____6_____ books Answer: There are 6 books which are not in the backpack. Explanation: Given I have 16 books means and a backpack can hold 10 books, out of 16 books I have taken 10 in the backpack so books which are not there in the backpack are 16 – 10 = 6 books. Review & Refresh Question 7. 10 + 10 = ___20_____ Answer: 10 + 10 = 20 Explanation: We are going to add 10 with 10 we get 20. Question 8. 10 + 10 = ___20______ Answer: 10 + 10 = 20 Explanation: We are going to add 10 with 10 we get 20. ### Lesson 6.4 Tens Explore and Grow Circle groups of 10. Write the number of groups. Answer: How many counters are there in all? ____20______ counters Answer: There are 20 counters in all. Explanation: In the given picture circled groups of 10, There are 2 groups, and counted there are total 20 counters in all. Show and Grow Circle groups of 10. Complete the sentence. Question 1. ____6__ ten and _0__ ones is 60______ . Answer: 6 groups, 6 ten and 0 ones is 60. Explanation: In the given picture circled groups of 10, There are 6 groups, and counted there are total 60 in all. 6 ten and 0 ones is 60. Question 2. __5____ ten and ___0___ ones is 50_ . Answer: 5 groups, 5 ten and 0 ones is 50. Explanation: In the given picture circled groups of 10, There are 5 groups, and counted there are total 50 in all. 5 ten and 0 ones is 50. Apply and Grow: Practice Circle groups of 10. Complete the sentence. Question 3. ___7___ ten and ___0___ ones is 70______ . Answer: 7 groups, 7 ten and 0 ones is 70. Explanation: In the given picture circled groups of 10, There are 7 groups, and counted there are total 70 in all. 7 ten and 0 ones is 70. Question 4. ___9___ ten and ___0___ ones is _90__ . Answer: 9 groups, 9 ten and 0 ones is 90. Explanation: In the given picture circled groups of 10, There are 9 groups, and counted there are total 90 in all. 9 ten and 0 ones is 90. Question 5. ___1___ ten and ___0___ ones is __10____ . Answer: 1 group, 1 ten and 0 ones is 10. Explanation: In the given picture circled groups of 10, There are 1 group, and counted there are total 10 in all. 1 ten and 0 ones is 10. Question 6. MP Number Sense You have 4 groups of 10 linking cubes. How many linking cubes do you have? _____40_______ linking cubes Answer: I have 40 linking cubes. Explanation: Given I have 4 groups of 10 linking cubes, So I have 4 X 10 = 40 linking cubes. Think and Grow: Modeling Real Life You read 10 books every month. You 40 books. How many months does it take? Draw a picture: Write the missing numbers: ___4___ tens and ___0___ ones __4____ months Answer: 4 tens and 0 ones is 40 books. It will take 4 months. Explanation: Given I read 10 books every month and read 40 books, 4 tens and 0 ones is 40 books. So we will divide 40 by 10 we get 4, therefore it will take 4 months, Show and Grow Question 7. There are 10 dog bones in each box. You need 20 bones. How many boxes do you need? Draw a picture: Write the missing numbers: ___2___ tens and ___0___ ones __2____ boxes Answer: 2 tens and 0 ones is 20 boxes. We need 2 boxes. Explanation: Given there are 10 dog bones in each box and need 20 bones. 2 tens and 0 ones is 20 bones. So we will divide 20 by 10 we get 2, therefore we need 2 boxes. ### Tens Practice 6.4 Circle groups of 10. Complete the sentence. Question 1. ___8___ tens and __0____ ones is 80__ . Answer: 8 groups, 1 ten and 0 ones is 10. Explanation: In the given picture circled groups of 10, There are 8 groups, and counted there are total 80 in all. 8 ten and 0 ones is 80. Question 2. ___6___ tens and __0____ ones is __60____ . Answer: 6 groups, 6 ten and 0 ones is 60. Explanation: In the given picture circled groups of 10, There are 6 groups, and counted there are total 60 in all. 6 ten and 0 ones is 60. Question 3. ___3___ tens and __0____ ones _is 30_____ . Answer: 3 groups, 3 ten and 0 ones is 30. Explanation: In the given picture circled groups of 10, There are 3 groups, and counted there are total 30 in all. 3 ten and 0 ones is 30. Circle groups of 10. Complete the sentence. Question 4. ___2___ tens and __0____ ones is __20____ . Answer: 2 groups, 2 ten and 0 ones is 20. Explanation: In the given picture circled groups of 10, There are 2 groups, and counted there are total 20 in all. 2 ten and 0 ones is 20. Question 5. MP Number Sense You have 7 groups of 10 linking cubes. How many linking cubes do you have? _____70______ linking cubes Answer: I have 70 linking cubes. Explanation: Given I have 7 groups of 10 linking cubes, So I have 7 X 10 = 70 linking cubes. Question 6. Modeling Real Life You swim 10 laps at every practice. You want to swim 50 laps. How many practices will it take? _____5______ Practices Answer: 5 tens and 0 ones is 50 laps It will take 5 practices. Explanation: Given I swim 10 laps at every practice, I want to swim 50 laps 5 tens and 0 ones is 50 laps. So I will divide 50 by 10 I get 5, therefore I will take 5 practices. Review & Refresh Question 7. 4 + 3 + 4 = ____11________ Answer: 4 + 3 + 4 = 11 Explanation: First we add 4 and 3 then add 4 we get 11. Question 8. 1 + 5 + 9 = _____15______ Answer: 1 + 5 + 9 = 15 Explanation: First we add 1 and 5 then add 9 we get 15. Question 9. 2 + 2 + 1 = _____5______ Answer: 2 + 2 + 1 = 5 Explanation: First we add 2 and 2 then add 1 we get 5. Question 10. 7 + 3 + 6 = _____16______ Answer: 7 + 3 + 6 = 16 Explanation: First we add 7 and 3 then add 6 we get 16. ### Lesson 6.5 Tens and Ones Explore and Grow Model 2 tens and 3 ones. Write the number. Answer: 2 tens and 3 ones is 23. Explanation: Model shown in the picture 2 tens and 3 ones, So placed 2 ten blocks in tens place and 3 blocks in ones place, making 2 tens and 3 ones ,therefore the number is 23. Show and Grow Question 1. __2___ tens and ___1___ ones is ___21___ . Answer: 2 tens and 1 ones is 21 Explanation: Given in the picture there are 2 tens block in tens place and 1 block in ones make 2 tens and 1 ones is 21. Apply and Grow: Practice Question 2. ___3___ tens and __5____ ones is ___35___ . Answer: 3 tens and 5 ones is 35 Explanation: Given in the picture there are 3 ten blocks in tens place and 5 blocks in ones make 3 tens and 5 ones is 35. Question 3. ___6___ tens and ___6___ ones is ___6___ . Answer: 6 tens and 6 ones is 66 Explanation: Given in the picture there are 6 ten blocks in tens place and 6 blocks in ones make 6 tens and 6 ones is 66. Question 4. __8____ tens and __9____ ones is __89____ . Answer: 8 tens and 9 ones is 89 Explanation: Given in the picture there are 8 ten blocks in tens place and 9 blocks in ones make 8 tens and 9 ones is 89. Question 5. YOU BE THE TEACHER You have 92 linking cubes. Your friend says that there are 2 tens and 9 ones. Is your friend correct? Show how you know. Answer: No, friend is incorrect. Explanation: Given I have 92 linking cubes, friend says that there are 2 tens and 9 ones, but 92 means as shown in picture we get 9 tens and 2 ones, therefore friend is in correct. Think and Grow: Modeling Real Life Your teacher has 2 packages of dice and 3 extra dice. Each package has 10 dice. How many dice are there in all? Draw a picture: Write the missing numbers: ___2___ tens and __3____ ones _is__23___ dice Answer: 2 tens and 3 ones is 23 dice are there in all. Explanation: Given my teacher has 2 packages of dice and 3 extra dice. Each package has 10 dice, 2 packages means 2 X 10 = 20 dice and 3 extra dice making in all 2 tens and 3 ones is 23 dice as shown in picture. Show and Grow Question 6. You have 3 boxes of colored pencils and 4 extra colored pencils. Each box has 10 pencils. How many colored pencils are there in all? Draw a picture: Write the missing numbers: ___3___ tens and __4____ ones _is_34__ colored pencils Answer: 3 tens and 4 ones is 34 colored pencils. Explanation: Given I have 3 boxes of colored pencils and 4 extra colored pencils. Each box has 10 pencils, So 3 boxes have 3 X 10 = 30 pencils and 4 extra makes 30 + 4 = 34, 3 tens and 4 ones is 34 colored pencils are there in all. ### Tens and Ones Practice 6.5 Question 1. __7____ tens and ___9___ ones is __79____ . Answer: 7 tens and 9 ones is 79. Explanation: Given in the picture there are 7 ten blocks in tens place and 9 blocks in ones make 7 tens and 9 ones is 79. Question 2. __8____ tens and ___1___ ones _is 81_____ . Answer: 8 tens and 1 ones is 81. Explanation: Given in the picture there are 8 ten blocks in tens place and 1 blocks in ones make 8 tens and 1 ones is 81. Question 3. ___2___ tens and __4___ ones is 24_ . Answer: 2 tens and 4 ones is 24. Explanation: Given in the picture there are 2 ten blocks in tens place and 4 blocks in ones make 2 tens and 4 ones is 24. Question 4. YOU BE THE TEACHER You have 17 linking cubes. Your friend says that there is 1 ten and 7 ones. Is your friend correct? Show how you know. Answer: Yes, friend is correct. Explanation: Given I have 17 linking cubes, friend says that there is 1 ten and 7 ones, means as shown in picture we get 1 ten and 7 ones, similar to what friend says, therefore friend is correct. Question 5. Modeling Real Life You have 5 bags of apples and 1 extra apple. Each bag has 10 apples. How many apples are there in all? _____51______ apples Answer: 5 tens and 1 ones is 51 apples. Explanation: Given I have 5 bags of apples and 1 extra apple. Each bag has 10 apples, So 5 bags have 5 X 10 = 50 apples and 1 extra apple makes 50 + 1 = 51, 5 tens and 1 ones is 51 apples are there in all. Review & Refresh Question 6. _____4_____ + 6 = 10 Answer: 4 + 6 = 10 Explanation: Given ____ + 6 = 10, let us take the missing number as X , X + 6 = 10, X = 10 – 6 = 4 making the equation as 4 + 6 = 10. Question 7. _____6______ + 2 = 8 Answer: 6 + 2 = 8 Explanation: Given ____ + 2 = 8, let us take the missing number as X , X + 2 = 8, X = 8 – 2 = 6 making the equation as 6 + 2 = 8. ### Lesson 6.6 Make Quick Sketches Explore and Grow Model the number 2. Answer: 2 ones is 2. Explanation: Model shown in the picture are 2 ones, So 2 blocks in ones place, as shown in picture making 2 ones ,therefore the number is 2. Show and Grow Make a quick sketch. Complete the sentence. Question 1. 72 __72____ is ____7____ tens and __2____ ones. Answer: 72 is 7 tens and 2 ones. Explanation: The picture has 7 ten blocks in tens place and 2 blocks in ones makes 72 as 7 tens and 2 ones. Question 2. 36 __36____ is ___3__ tens and __6_ ones. Answer: 36 is 3 tens and 6 ones. Explanation: The picture has 3 ten blocks in tens place and 6 blocks in ones makes 36 as 3 tens and 6 ones. Apply and Grow: Practice Make a quick sketch. Complete the sentence. Question 3. 45 ___45___ is ___4_____ tens and ___5___ ones. Answer: 45 is 4 tens and 5 ones. Explanation: The picture has 4 ten blocks in tens place and 5 blocks in ones makes 45 as 4 tens and 5 ones. Question 4. 87 ___87___ is ___8___ tens and ___7___ ones. Answer: 87 is 8 tens and 7 ones. Explanation: The picture has 8 ten blocks in tens place and 7 blocks in ones makes 87 as 8 tens and 7 ones. Question 5. 64 __64____ is ___6____ tens and ___4___ ones. Answer: 64 is 6 tens and 4 ones. Explanation: The picture has 6 ten blocks in tens place and 4 blocks in ones makes 64 as 6 tens and 4 ones. Question 6. DIG DEEPER! Which sketch shows 54? Answer: First sketch shows 54. Explanation: Given 2 sketches in that first counts 54 and second counts 45, So first sketch shows 54, So selected first one for 54. Think and Grow: Modeling Real Life You need 58 plates for a party. You have 51. How many more plates do you need? Complete the model: ____7_______ more plates Answer: I need 7 more plates. Explanation: Given I need 58 plates for a party and I have 51, therefore more plates needed are 58 – 51 = 7 plates as shown in the picture above. Show and Grow Question 7. You need 80 tickets for a prize. You have 73. How many more tickets do you need? Complete the model: _____7______ more tickets Answer: I need 7 more tickets. Explanation: Given I need 80 tickets for a prize and I have 73, therefore more tickets needed are 80 – 73 = 7 plates as shown in the picture above. ### Make Quick Sketches Practice 6.6 Make a quick sketch. Complete the sentence. Question 1. 27 __27____ is ____2____ tens and ___7___ ones. Answer: 27 is 2 tens and 7 ones. Explanation: The picture has 2 ten blocks in tens place and 7 blocks in ones makes 27 as 2 tens and 7 ones. Question 2. 61 __61__ is ___6__ tens and ___1___ ones. Answer: 61 is 6 tens and 1 one. Explanation: The picture has 6 ten blocks in tens place and 1 blocks in one makes 61 as 6 tens and 1 one. Question 3. 92 ___92___ is ___9_____ tens and ___2___ ones. Answer: 92 is 9 tens and 2 ones. Explanation: The picture has 9 ten blocks in tens place and 2 blocks in ones makes 92 as 9 tens and 2 ones. Question 4. DIG DEEPER Which sketch shows 87? Answer: First sketch shows 87. Explanation: Given 2 sketches in that first counts 87 and second counts 78, So first sketch shows 87, So selected first one for 87. Question 5. Modeling Real Life You need 55 beads to make a necklace. You have 48. How many more beads do you need? ____7______ more beads Answer: I need 7 more beads. Explanation: Given I need 55 beads to make necklace and I have 48, therefore more beads needed are 55 – 48 = 7 beads. Review & Refresh Question 6. Color the shapes that have only 4 sides. Answer: Explanation: Given some shapes and to color that have 4 sides so selected shapes that has 4 sides and have colored as shown above in the picture. ### Lesson 6.7 Understand Place Value Explore and Grow Newton has 2 rods. Make a quick sketch. Write the number. Answer: Newton has 2 rods or 20 cubes Explanation: Given Newton has 2 rods as shown in the picture means 20 cubes. Descartes has 2 cubes. Make a quick sketch. Write the number. Answer: Descartes has 2 cubes Explanation: Given Descartes has 2 cubes as shown in picture 2 cubes. How are the models alike? How are they different? Answer: Both use cubes so alike, Both differ in number so different. Explanation: Models are alike means both use cubes. They are different Newton has 20 cubes and Descartes have 2 cubes, 20 and 2 are not the same and are not equal so they are different. Show and Grow Question 1. Make a quick sketch. Complete the sentences. 64 _____6_____ tens is ____60______ . ____4______ ones is ____4______ . _____6_____ tens and ____4______ ones is ___64_______. Answer: 6 tens is 60, 4 ones is 4, 6 tens and 4 ones is 64. Explanation: As shown in the picture above first we take 6 tens cubes as 60 and 4 ones cubes as 4, making 6 tens and 4 ones is 64. Apply and Grow: Practice Make a quick sketch. Complete the sentences. Question 2. 72 _____7_____ tens is ____70______ . _____2_____ ones is ____2______ . ______7____ tens and ____2______ ones is ___72__. Answer: 7 tens is 70, 2 ones is 2, 7 tens and 2 ones is 72. Explanation: As shown in the picture above first we take 7 tens cubes as 70 and 2 ones cubes as 2, making 7 tens and 2 ones is 72. Question 3. 98 __9____ tens is ___90____ . ___8____ ones is ___8_____ . ___9____ tens and ____8____ ones is ___98___. Answer: 9 tens is 90, 8 ones is 8, 9 tens and 8 ones is 98. Explanation: As shown in the picture above first we take 9 tens cubes as 90 and 8 ones cubes as 8, making 9 tens and 8 ones is 98. Question 4. 57 ____5______ tens is ___50_______ . _____7_____ ones is ____7______ . ______5____ tens and _____7_____ ones is ____57______. Answer: 5 tens is 50, 7 ones is 7, 5 tens and 7 ones is 57. Explanation: First we take 5 tens cubes as 50 and 7 ones cubes as 7, making 5 tens and 7 ones is 57. Think and Grow: Modeling Real Life You have 94 charms to make bracelets. There are lo charms on each bracelet. How many bracelets can you make? Model: Write the missing numbers: ___9___ tens and ___4___ ones ___9___ bracelets Answer: 9 tens and 4 ones is 94, I make 9 bracelets. Explanation: Given I have 94 charms to make bracelets. There are lo charms on each bracelet. So I divide 94 with 10 I get 9 as whole and 4 as remainder so I cannot make with 4 charms, I consider only whole so I take 9,therefore I can make 9 bracelets or as I have 9 tens and 4 ones I take only tens so 9 bracelets I can make. Show and Grow Question 5. You have 67 seeds. You plant 10 seeds in a row. How many rows can you plant? Model: Write the missing numbers: ___6___ tens and __7____ ones __6____ rows Answer: 6 tens and 7 ones is 67, I can plant 6 rows. Explanation: Given I have 67 seeds. I can plant 10 seeds in a row So I divide 67 with 10 I get 6 as whole and 7 as remainder so I cannot make with 7 seeds a row , I consider only whole so I take 6,therefore I can make 6 rows only, or as I have 6 tens and 7 ones I take only tens so 6 rows I can make. ### Understand Place Value Practice 6.7 Make a quick sketch. Complete the sentences. Question 1. 81 _____8_____ tens is ___80_______ . _____1_____ ones is ____1______ . _____8_____ tens and ____1______ ones is ____81______. Answer: 8 tens is 80, 1 one is 1, 8 tens and 1 one is 81. Explanation: As shown in the picture above first we take 8 tens cubes as 80 and 1 one cubes as 1, making 8 tens and 1 one is 81. Question 2. 53 _____5_____ tens is _____50_____ . ______3____ ones is ____3______ . ______5____ tens and _____3____ ones is ___53_____. Answer: 5 tens is 50, 3 ones is 3, 5 tens and 3 ones is 53. Explanation: As shown in the picture above first we take 5 tens cubes as 50 and 3 ones cubes as 3, making 5 tens and 3 ones is 53. Question 3. 49 ___4_______ tens is ____40______ . _____9_____ ones is ____9______ . ______4____ tens and ____9__ ones is __49__. Answer: 4 tens is 40, 9 ones is 9, 4 tens and 9 ones is 49. Explanation: As shown in the picture above first we take 4 tens cubes as 40 and 9 ones cubes as 9, making 4 tens and 9 ones is 49. Question 4. 76 ____7______ tens is ____70______ . ______6____ ones is ____6______ . ____7______ tens and ____6______ ones is ____76______. Answer: 7 tens is 70, 6 ones is 6, 7 tens and 6 ones is 76. Explanation: As shown in the picture above first we take 7 tens cubes as 70 and 6 ones cubes as 6, making 7 tens and 6 ones is 76. Question 5. Modeling Real Life You have 77 crayons. A box can hold 10 crayons. How many boxes can you fill? _____7____ boxes Answer: 7 tens and 7 ones is 77, I can fill 7 boxes. Explanation: Given I have 77 crayons, A box can hold 10 crayons, So I divide 77 with 10 I get 7 as whole and 7 as remainder I consider only whole so I take 7, therefore I need 7 boxes to fill or as I have 7 tens and 7 ones to fill I take only tens so 7 boxes I can fill. Review & Refresh Is the equation true or false? Question 6. Answer: 4 + 9 ≠ 2 + 3 + 5 Not equal, So false Equation: First we see 4 and 9 as 4 + 9 is 13 and if we add 2,3 and 5 we get 10 , therefore 4 + 9 ≠ 2 + 3 +5, So false. Question 7. Answer: 5 + 3 = 4 + 4 Is equal, So true Explanation: First we see 5 + 3 as we add 5 and 3 we get 8 and 4 + 4 is 8 both sides are equal, So equations are true. ### Lesson 6.8 Write Numbers in Different Ways Explore and Grow Model 27 two ways. _____2___ tens and __7_____ ones is ___27_____ . _____1___ ten and ___17____ ones is ___27_____ . Answer: one way is 2 tens and 7 ones is 27. Other way is 1 ten and 17 ones is 27. Explanation: One way is we add 2 tens and 7 ones equal to 27, 20 + 7 = 27 and other way is 1 ten and 17 ones is 27, 10 + 17 = 27. Show and Grow Question 1. Model 25 two ways. ____2____ tens and ____5___ ones is 25. Answer: 2 tens and 5 ones is 25. Explanation: we add 20 and 5 we get 25, So 2 tens and 5 ones is 25. ____1____ tens and __15_____ ones is 25. Answer: 1 ten and 15 ones is 25 Explanation: we add 10 and 25 we get 25, So 1 ten and 15 ones is 25. Apply and Grow: Practice Question 2. Model 52 two ways. ____5____ tens and ___2____ ones is 52. Answer: 5 tens and 2 ones is 52. Explanation: We add 50 and 2 we get 52, So 5 tens and 2 ones is 52. ______4__ tens and ____12___ ones is 52. Answer: 4 tens and 12 ones is 52. Explanation: First we take 40 and add to 12 we get 52, 40 + 12 = 52, therefore 4 tens and 12 ones is 52. Question 3. Model 14 two ways. ___1_____ tens and ___4____ ones is 14. Answer: 1 ten and 4 ones is 14 Explanation: we add 10 + 4 = 14, So 1 ten and 4 ones is 14. ___0_____ tens and ___14____ ones is 14. Answer: 14 ones is 14 Explanation: we add 0 + 14 = 14, 0 tens and 14 ones is 14. Question 4. DIG DEEPER! Circle all of the ways that show 39. 3 + 9 2 tens and 9 ones 9 tens and 3 ones 10 + 29 3 tens and 19 ones 39 ones Answer: Circled the ways that show 39 are 10 + 29 and 39 ones Explanation: As shown in picture circled all of the ways that show 39, 3 + 9 = 12 ≠ 39 so not circled, 2 tens and 9 ones is 29 ≠ 39 so not circled, 9 tens and 3 ones is 93 ≠ 39 so not circled, 10 + 29 = 39 = 39 so circled it, 3 tens and 19 ones is 30 + 19 is 49 ≠ 39 so not circled, 39 ones is 39 = 39 so circled it. Think and Grow: Modeling Real Life The models show how many seashells you and your friend have. Does your friend have the same number of seashells as you? Circle: Yes No Show how you know: Answer: No friend does not have the same number of seashells as much as I have. Explanation: After counting I have total 25 seashells and friend have 30, Both do not have same or equal number of seashells , So circled No as shown in the picture above. Show and Grow Question 5. The models show how many erasers you and your friend have. Does your friend have the same number of erasers as you? Circle: Yes No Show how you know: Answer: Yes, I and my friend have the same number of erasers. Explanation: After counting I have total 40 erasers and friend also have 40, Both have the same or equal number of erasers, So circled Yes as shown in the picture above. ### Write Numbers in Different Ways Practice 6.8 Question 1. Model 49 two ways. __4______ tens and ___9____ ones is 49. Answer: 4 tens and 9 ones is 49. Explanation: Adding 40 to 9 is 40 + 9 = 49, so 4 tens and 9 ones is 49. ___30_____ tens and ____19____ ones is 49. Answer: 30 tens and 19 ones is 49. Explanation: We add 30 and 19 we get 49, So 30 tens and 19 ones make 49. Question 2. DIG DEEPER! Circle all of the ways that show 45. 40 + 5 45 tens and 0 ones 4 tens and 5 ones 20 + 15 2 tens and 25 ones 54 ones Answer: Circled the ways that show 45 are 40 + 5, 4 tens and 5 ones and 2 tens and 25 ones. Explanation: As shown in picture circled all of the ways that show 45, 40 + 5 = 45 = 45 so circled it, 45 tens and 0 ones is 450 ≠ 45 so not circled, 4 tens and 5 ones is 45 = 45 so circled it, 20 + 15 = 35 ≠ 45 so not circled, 2 tens and 25 ones is 20 + 25 is 45 = 45 so circled it, 54 ones is 54 ≠ 45 so not circled. Question 3. Modeling Real Life The models show the number of toy cars you and your friend have. Does your friend have the same number of toy cars as you? Circle: Yes No Show how you know: Answer: Yes, friend have the same number of toy cars as I have. Explanation: After counting I have total 34 toy cars and friend also have 34 , Both have same or equal number of toy cars, So circled Yes as shown in the picture above. Review & Refresh Circle the heavier object. Question 4. Answer: Circled ball. Explanation: Given ball and balloon as ball is heavier object than balloon I circled it. Question 5. Answer: Circled 4 bananas Explanation: Given 4 bananas and 1 banana as 4 bananas are heavier object than 1 banana I circled it. ### Lesson 6.9 Count and Write Numbers to 120 Explore and Grow How many balls are there? How did you count? Answer: Number of balls are 106. Counted one after another. Explanation: There are total 106 balls, we start pointing and counting from first ball as one add next ball and so on counting till we reach the last ball ,we get total 106 balls. Show and Grow Question 1. Answer: 10 tens and 7 ones is 107. Explanation: Given in the picture if we count there are 100 + 7 = 107 , 10 tens and 7 ones is 107. Question 2. Answer: 11 tens and 5 ones is 115. Explanation: Given in the picture if we count there are 110 + 5 = 107 , 11 tens and 5 ones is 115. Apply and Grow: Practice Question 3. Answer: 6 tens and 4 ones is 64. Explanation: Given in the picture if we count there are 60 + 4 = 64 , 6 tens and 4 ones is 64. Question 4. Answer: 9 tens and 8 ones is 98. Explanation: Given in the picture if we count there are 90 + 8 = 98 , 9 tens and 8 ones is 98. Question 5. Answer: 10 tens and 1 one is 101. Explanation: Given in the picture if we count there are 100 + 1 = 101 , 10 tens and 1 one is 101. Question 6. Answer: 10 tens and 3 ones is 103. Explanation: Given in the picture if we count there are 100 + 3 = 103 , 10 tens and 3 ones is 103. Question 7. Answer: 11 tens and 0 ones is 110. Explanation: Given in the picture if we count there are 110 + 0 = 110 , 11 tens and 0 ones is 110. Question 8 DIG DEEPER! What number is equal to lo tens and 8 ones? Show how you know. Answer: The number equal to 10 tens and 8 ones is 108 Explanation: Given to find number equal to 10 tens and 8 ones, as shown in figure taken 10 tens and 8 ones , the number is equal to 100 + 8 = 108. Think and Grow: Modeling Real Life Your teacher has 12 bags of balloons. Each bag has 10 balloons. How many balloons are there in all? Model: ____120_______ balloons Answer: There are 120 balloons are there in all. Explanation: Given my teacher has 12 bags of balloons and each bag has 10 balloons so in all there are 12 X 10 = 120 balloons. Show and Grow Question 9. A dentist has 10 boxes of toothbrushes and 9 extra toothbrushes. Each box has 10 toothbrushes. How many toothbrushes are there in all? Model: ___109________ toothbrushes Answer: 10 tens and 9 ones is 109. There are 109 tooth brushes are there in all. Explanation: Given a dentist has 10 boxes of toothbrushes and 9 extra toothbrushes, each box has 10 toothbrushes means 10 x 10 = 100 and 9 extra makes 100 + 9 = 109 or 10 tens and 9 ones is 109 tooth brushes are there in all. ### Count and Write Numbers to 120 Practice 6.9 Question 1. Answer: 7 tens and 9 ones is 79. Explanation: Given in the picture if we count there are 70 + 9 = 79 , 7 tens and 9 ones is 79. Question 2. Answer: 11 tens and 5 ones is 115. Explanation: Given in the picture if we count there are 110 + 5 = 115 , 11 tens and 5 ones is 115. Question 3. Answer: 12 tens and 0 ones is 120. Explanation: Given in the picture if we count there are 120 + 0 = 120 , 12 tens and 0 ones is 120. Question 4. Answer: 9 tens and 2 ones is 92. Explanation: Given in the picture if we count there are 90 + 2 = 92 , 9 tens and 2 ones is 92. Question 5. DIG DEEPER! What number is equal to 11 tens and 2 ones? Show how you know. Answer: The number equal to 11 tens and 2 ones is 112. Explanation: We have 11 tens and 2 ones as shown in picture 11 tens is 110 and 2 ones is 2 so 110 + 2= 112. Question 6. Modeling Real Life You have 4 packs of baseball cards and 6 packs of football cards. Each pack has 10 cards. How many cards do you have in all? _____100_______ cards Answer: I have in total 100 cards in all. Explanation: Given I have 4 packs of baseball cards and 6 packs of football cards, Each pack has 10 cards, therefore 4 packs of baseball means 4 X 10 = 40 cards, 6 packs of football cards have 6 X 10 = 60 cards, In total I have 40 + 60 =100 cards in all. Review & Refresh Make a 10 to add. Question 7. Answer: Sum is 6 + 4 + 1 = 11 or 10 + 1 = 11, So 6 + 5 = 11 Explanation: First we make sum 10 , For 6 to make 10 we need 10-6=4, as we have to add 4 to 6 now we will minus 4 from 5 we get 5 – 4 = 1, Now we write the total sum as 6 + 4 + 1 = 11 or 10 + 1 = 11 or 6 + 5 = 11. Question 8. Answer: Sum is 7 + 3 + 5 = 15 or 10 + 5 = 15, So 7 + 8 = 15 Explanation: First we make sum 10 , For 7 to make 10 we need 10-7=3, as we have to add 3 to 7 now we will minus 3 from 8 we get 8 – 3 = 5, Now we write the total sum as 7 + 3 + 5 = 15 or 10 + 5 = 15 or 7 + 8 = 15. ### Count and Write Numbers to 120 Performance Task Question 1. Your class sells candles for a fundraiser. You earn 10 dollars for every large candle you sell and 1 dollar for every small candle. a. You sell 6 large candles. How much money do you raise? Answer: I raised 60 dollars. Explanation: Given my class sells candles for a fundraiser. I earn 10 dollars for every large candle I sell , I sold 6 large candles means 6 X 10 = 60 dollars, So I raised 60 dollars. b. You want to raise 72 dollars. How much more money do you need to raise? _____12_____ more dollars Answer: I need 12 more dollars to raise 72 dollars. Explanation: I know I raised 60 dollars to raise 72 dollars I need more is 72 – 60 = 12 dollars. c. You also sell 12 small candles. Do you reach your goal? Yes No Answer: Yes, I reached my goal. Explanation: Given I earn1 dollar for every small candle I sell, I sold 12 small candles means 12 X 1= 12 dollars, In total I sold 6 large candles and 12 small candles I raised total 60 + 12 = 72 dollars which is similar to which I want to raise, So yes I reached my goal. Question 2. Your friend wants to raise 54 dollars. What are two ways your friend can sell large and small candles to reach her goal? _____5____ large candles and ____4______ small candles _____4_____ large candles and ___14_______ small candles Answer: 5 large candles and 4 small candles, 4 large candles and 14 small candles. Explanation: I know if I sell 1 large candle I earn 10 dollars and if I sell 1 small candles I earn 1 dollar to raise 54 dollars my friend can sell 5 tens and 4 ones is 54, 5 X 10 + 4 X 1 = 50 + 4 = 54, So 5 large candles and 4 small candles makes 54 dollars She can even sell 4 tens and 14 ones is 54, 4 X 10 + 14 X 1 = 40 + 14 = 54, So 4 large candles and 14 small candles makes 54 dollars. ### Count to 120 by Ones Homework & Practice 6.1 Question 1. Count by ones to write the missing numbers. 99, __100___, ___101___, __102___, __103___, ___104___ Answer: The missing numbers after 99 are 100,101,102,103,104 Explanation: Started from 99 counting by 1’s numbers after 99 are 99 + 1 = 100, 100 + 1 = 101,101 + 1 = 102, 102 + 1 = 103, 104 + 1 = 105, So the missing numbers after 99 are 100,101,102,103,104. write the missing numbers. Question 2. Answer: The missing numbers in the chart are 69,71,78,80. Explanation: The missing number after 68 is 68 + 1 = 69, the number after 70 is 70 + 1 = 71, the number before 79 is 79 – 1 = 78 and number before 81 is 81 – 1 = 80, So the missing numbers in the chart are 69,71,78 and 80. Question 3. Answer: The missing numbers in the chart are 101,103,110,112,113 Explanation: The missing number after 100 is 100 + 1 = 101, the number after 102 is 102 + 1 = 103, the number before 111 is 111 – 1 = 110, the number after 111 is 111 + 1 = 112, the number after 112 is 112 + 1 = 113, So the missing numbers in the chart are 101,103,110,112,113. ### Count to 120 by Tens Homework & Practice 6.2 Question 4. Write the missing numbers from the chart. Then count on by tens to write the next two numbers. Answer: The missing numbers in the chart are 84,94 and count on 10’s the next two numbers after 94 are 104,114. Explanation: Given numbers in the chart, the number missing before 85 is 85 – 1 = 84 and number missing after 93 and before 94 in the chart is 93 + 1= 94 or 95 – 1 = 94,now counting by 10’s next two numbers after 94 are 94 + 10 = 104,104 + 10 = 114, Therefore the missing numbers in the chart are 84 and 94 count on 10’s the next two numbers after 94 are 104,114. Question 5. YOU BE THE TEACHER Your friend counts by tens starting with 53. Is your friend correct? Show how you know. 53, 63, 73, 83, 103 Answer: No friend is Incorrect. Explanation: Given friend counts by tens starting with 53 after 53 its is 53 + 10 = 63 then 63 + 10 = 73, 73 + 10 = 83 and after 83 it is 83 + 10 = 93 not 103 therefore friend is incorrect. ### Compose Numbers 11 to 19 Homework & Practice 6.3 Question 6. Circle 10 ducks. Complete the sentence. ____1______ ten and ____2______ ones is ___12____ . Answer: 1 ten and 2 ones is 12. Explanation: Circled 10 ducks, after counting we have total 12 ducks we write the sentence as one ten and 2 ones is twelve means 10 + 2 = 12. 1 ten and 2 ones is 12. Question 7. Modeling Real Life You have 19 tennis balls. A bag can hold 10. You fill a bag. How many tennis balls are not in the bag? _____9_______ tennis balls Answer: 9 tennis balls are not there in the bag. Explanation: Given I have 19 tennis balls and a bag can hold 10 and I have filled a bag so balls that are not there in the bag are 19 – 10 = 9 , So 9 tennis balls are not there in the bag. ### Tens Homework & Practice 6.4 Question 8. Circle groups of 10. Complete the sentence. ______5____ tens and ___0___ ones is ___50______ . Answer: fife tens and zero ones is fifty. 5 tens and 0 ones is 50. Explanation: Circled 5 tens ,five tens and zero ones is fifty or 5 tens and 0 ones is 50. ### Tens and Ones Homework & Practice 6.5 Question 9. ____9_____ tens and ____9_____ ones is ___99_____ . Answer: 9 tens and 9 ones is 99 Explanation: Given in the picture there are 9 ten blocks in tens place and 9 blocks in ones make 9 tens and 9 ones is 99. Question 10. Modeling Real Life You have 6 boxes of plastic cups and 3 extra cups. Each box has 10 cups. How many cups are there in all? _____63_____ plastic cups Answer: In all there are 63 plastic cups. Explanation: Given I have 6 boxes of plastic cups and 3 extra cups and each box has 10 cups means 6 X 10 = 60 and 3 extra cups make 60 + 3 = 63 or 6 tens and 3 ones is 63 Plastic cups. ### Make Quick Sketches Homework & Practice 6.6 Make a quick sketch. Complete the sentence. Question 11. 17 ____17____ is ____1___ ten and ____7____ ones. Answer: 17 is 1 tens and 7 ones. Explanation: Given in the picture there are 1 ten blocks in tens place and 7 blocks in ones make 17 as 1 tens and 7 ones. Question 12. 84 ___84____ is ____8___ tens and ___4__ ones. Answer: 84 is 8 tens and 4 ones is 84 Explanation: Given in the picture there are 8 ten blocks in tens place and 4 blocks in ones make 84 as 8 tens and 4 ones. ### Understand Place Value Homework & Practice 6.7 Question 13. 39 ______3____ tens is ____30______ . ______9____ ones is ____9______ . _____3____ tens and ____9______ ones is ____39______ . Answer: 3 tens is 30, 9 ones is 9, So 3 tens and 9 ones is 39. Explanation: As shown in the picture above first we take 3 tens cubes as 30 and 9 ones cubes as 9, making 3 tens and 9 ones is 39. ### Write Numbers in Different Ways Homework & Practice 6.8 Question 14. Model 59 two ways. _____5_____ tens and _____9_____ ones is 59. Answer: 5 tens and 9 ones is 59 Explanation: we add 5 X 10 = 50 and 9 X 1 = 9 we get 50 + 9 = 59, So 5 tens and 9 ones is 59. _____2_____ tens and ____39______ ones is 59. Answer: 2 tens and 39 ones is 59 Explanation: we add 2 X 10 = 20 and 39 X 1 = 39 we get 20 + 39 = 59, So 2 tens and 39 ones is 59. ### Count and Write Numbers to 120 Homework & Practice 6.9 Question 15. Answer: 11 tens and 4 ones is 114 Explanation: As shown in the picture there are 11 ten blocks in tens place and 4 blocks in ones make 11 tens and 4 ones is 114. Question 16. Answer: 10 tens and 8 ones is 108 Explanation: As shown in the picture there are 10 ten blocks in tens place and 8 blocks in ones make 10 tens and 8 ones is 108. Conclusion: Start solving the problems which are at the end of the chapter and check the solutions from here. Hope the information regarding Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120 is helpful for you to overcome the issues in maths. Check out the links and start solving all the questions.
## Precalculus (6th Edition) Blitzer Let us consider the left side of the given expression: $\cos \frac{\pi }{2}\cos \frac{\pi }{3}$ Now, put the values of the trigonometric functions to find the exact value as shown below: \begin{align} & \cos \frac{\pi }{2}\cos \frac{\pi }{3}=0\cdot \frac{1}{2} \\ & =0 \end{align} Also, consider the right side of the given expression: $\frac{1}{2}\left[ \cos \left( \frac{\pi }{2}-\frac{\pi }{3} \right)+\cos \left( \frac{\pi }{2}+\frac{\pi }{3} \right) \right]$ Then, put the values of the trigonometric functions to find the exact value as shown below: \begin{align} & \frac{1}{2}\left[ \cos \left( \frac{\pi }{2}-\frac{\pi }{3} \right)+\cos \left( \frac{\pi }{2}+\frac{\pi }{3} \right) \right]=\frac{1}{2}\left[ \cos \left( \frac{3\pi -2\pi }{6} \right)+\cos \left( \frac{3\pi +2\pi }{6} \right) \right] \\ & =\frac{1}{2}\left[ \cos \left( \frac{\pi }{6} \right)+\cos \left( \frac{5\pi }{6} \right) \right] \\ & =\frac{1}{2}\cdot \left[ \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \right] \\ & =0 \end{align} Thus, the left side of the expression is equal to the right side, which is $\cos \frac{\pi }{2}\cos \frac{\pi }{3}=\frac{1}{2}\left[ \cos \left( \frac{\pi }{2}-\frac{\pi }{3} \right)-\cos \left( \frac{\pi }{2}+\frac{\pi }{3} \right) \right]$.
### Learning objectives • Understand that derivatives are the instantaneous rate of change of a function • Understand how to calculate a derivative • Understand how to express taking a derivative at a given point, and evaluating a function at a given point mathematically ### Introduction In the last lesson, we saw that the derivative was the rate of change. We saw multiple ways of calculating this rate of change. • Essentially, the derivative is the rate of change of a function • Graphically this is rise over run • Which can be calculated by taking two points, $(x_1, y_1)$ and $(x_2, y_2)$ and calculating $\frac{y_2 - y_1}{x_2 - x_1}$ Finally, we said that when we have a function $f(x)$, we can calculate the derivative with knowing the starting point and the change in our input, $x$: $$\frac{f(x_1 + \Delta x) - f(x_1)}{\Delta x}$$ ### Derivatives of non-linear functions So we saw previously that the derivative is the rate of change of our function. We express this as $f'(x) = \frac{\Delta f}{\Delta x}$. So far we have only calculated the derivatives with linear functions. As we'll see, things becomes trickier when working with more complicated functions. For example, let's imagine that we are coaching our runner to perform in a track meet. We may want to know how well our track start does at one part of the race, say the starting point, versus another point later in the race. Then we will know what to focus on in practice. We can imagine the distance travelled by our track star's distance through time as represented by the function $f(x) = x^2$: import plotly from plotly.offline import iplot, init_notebook_mode init_notebook_mode(connected=True) from graph import plot, build_layout from calculus import function_values_trace x_squared = [(1, 2)] range_twenty_four = list(range(0, 25)) six = list(map(lambda x: x/4.0, range_twenty_four)) trace_x_squared = function_values_trace(x_squared, six) layout = build_layout(x_axis = {'title': 'number of seconds'}, y_axis = {'title': 'distance'}) plot([trace_x_squared], layout) The graph shows that from seconds zero through six, our track runner gets faster over time. #### Calculating speed at second two Now if we want to see how quickly our track star at second number two as opposed to some other second, what would we do? Well even if we knew nothing about derivatives, we would likely get a stop watch and at second 2 would use it to calculate the speed. Let's say that we start our stopwatch at second 2 and stop our stopwatch one second later. from calculus import delta_traces delta_layout = build_layout(x_axis = {'title': 'number of seconds'}, y_axis = {'title': 'distance'}) x_squared_delta_traces = delta_traces(x_squared, 2, line_length = 2, delta_x = 1) plot([trace_x_squared, x_squared_delta_traces], delta_layout) As the graph above shows, we measure the change at second two by starting our stopwatch at second 2 and stopping it one second later. So turning this into our formula for calculating a derivative of: $$f'(x) = \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ we do the following: • Set $x = 2$, as that's the point we want to calculate the rate of change at • Set $\Delta x = 1$, as that's the number of seconds that elapsed on our stopwatch and plugging in these values, we have: $$f'(2) = \frac{f(2 + 1) - f(2)}{ 1} = \frac{f(3) - f(2)}{1}$$ So our rate of change at second number 2, with a $\Delta x = 1$ is calculated by subtracting the function's output at second 2 from the function's output at second 3 and dividing by delta x, one. Simplifying our calculation of $f'(x)$ further by calculating the outputs at x = 2 and x = 3 we have: • $f(3) = (3)^2 = 9$ is the output at x = 3 and • $f(2) = (2)^2 = 4$ is the output at x = 2 so $$f'(2) = \frac{9 - 4}{1} = \frac{5}{1} = 5$$ from graph import plot x_squared = [(1, 2)] x_squared_delta_traces = delta_traces(x_squared, 2, line_length = 4, delta_x = 1) layout = build_layout(x_axis = {'title': 'number of seconds', 'range': [1.5, 4]}, y_axis = {'title': 'distance', 'range': [0, 12]}) plot([trace_x_squared, x_squared_delta_traces ], layout) ### The problem with our derivative formula Take a close look at the straight line in the graph above. That straight line is a supposed to be the rate of change of the function at the point $x = 2$. And it comes close. But it doesn't exactly line up. Our orange line quickly begins to move above the blue line, indicating that it has a faster rate of change than the blue line at $x = 2$. This means that our calculation that $f'(2) = 5$ is a little high. Here is the problem: • in our formula of $f'(x) = \frac{f(x_1 + \Delta x) - f(x_1)}{\Delta x}$, we are seeing the rate of change not just where x = 2, but from the period from $x = 2$ to $x = 3$. x_squared_delta_traces = delta_traces(x_squared, 2, line_length = 4, delta_x = 1) layout = build_layout(x_axis = {'title': 'number of seconds', 'range': [1.5, 3.5]}, y_axis = {'title': 'distance', 'range': [0, 12]}) plot([trace_x_squared, x_squared_delta_traces ], layout) In other words, the runner would tell us that we are not capturing their speed at precisely second two: This is because in between the clicks of our stopwatch from seconds two to three, our runner is getting faster and while we are supposed to be calculating his speed just at second 2, our calculation includes his increase in speed from seconds two to three. Therefore, the orange line has a larger rate of change than the blue line because we have included this increase in speed at second three. A mathematician would make the same point that we are not actually calculating the derivative: Our derivative means we are calculating how fast a function is changing at any given moment, and precisely at that moment. And unlike in where our functions were linear, here the rate of change of our function is always changing. The larger our value of $\Delta x$, the less our derivative reflects the rate of change at just that point. ### The solution: Decrease the change in x If you were holding a stopwatch and someone asked you to calculate their speed at second number 2, how could you be more accurate? Well, you would want decrease the change in seconds. Of course, our runner could continue to protest and say that we are still influenced by the speed at other times. However, the mathematician has a solution to this. To calculate the rate of change at precisely one point, the solution is to use our imagination. We calculate the derivative with a $\Delta$ of 1, then calculate it again with a $\Delta x$ of .1, then again with $\Delta x$ of .01, then again with $\Delta$ .001. Our derivative calculation should show convergence on a single number as our $\Delta$ approaches zero and that number is our derivative. * That is, the derivative of a function is a change in the function's output across $\Delta x$, as $\Delta x$ approaches zero *. In this example, by decreasing $\Delta x$ we can see a fairly clear pattern. $\Delta x$ $\frac{\Delta y}{\Delta x}$ 1 5 .1 4.1 .01 4.01 .001 4.001 ### Seeing this visually: tangent lines Another way to see how we approach the derivative is by seeing how a line becomes more tangent to the curve as delta x decreases. Tangent to the curve means that our line is just touching the curve. The more that a line is tangent to the curve at a point, the more it's slope matches the derivative. Ok, let's get a sense of what we mean by tangent to the curve. The orange line below is a line whose slope is calculated by using our derivative function, with delta x = 1. As you can see it is not tangent to our function, $f(x)$ * as it does not just touch the blue line, but rather touches it in two places. from calculus import derivative_trace tangent_x_squared_trace = derivative_trace(x_squared, 2, 2, 1) x_squared = [(1, 2)] range_twenty_four = list(range(0, 25)) six = list(map(lambda x: x/4.0, range_twenty_four)) trace_x_squared = function_values_trace(x_squared, six) layout_not_tangent = build_layout(x_axis = {'range': [0, 4]}, y_axis = {'range': [0, 10]}) plot([trace_x_squared, tangent_x_squared_trace], layout_not_tangent) If our orange line had the same slope, or rate of change, as our function at that x = 2, it would just touch the blue line. We know from above that we get closer to the rate of change of the function as we decrease delta x in our derivative formula. Let's look again using a smaller $\Delta x$. Below are the plots of our lines using our derivative formula for when $\Delta x = 1$ and when $\Delta x = .1$ from graph import plot_figure, make_subplots from calculus import derivative_trace range_twelve = list(range(0, 12)) three = list(map(lambda x: x/4.0, range_twelve)) trace_x_squared_to_three = function_values_trace(x_squared, three) tangent_x_squared = derivative_trace(x_squared, 2, 1, 1) tangent_x_squared_delta_tenth = derivative_trace(x_squared, 2, 1, .1) subplots = make_subplots([trace_x_squared_to_three, tangent_x_squared], [trace_x_squared_to_three, tangent_x_squared_delta_tenth]) subplots The graphs above illustrate when $\Delta x = 1$ and when $\Delta x = .1$ The graph to the left, with a smaller $\Delta x$ has a more tangent line. Let's keep decreasing $\Delta x$ and see if we can approach the derivative even further. tangent_x_squared_delta_hundredth = derivative_trace(x_squared, 2, 1, .01) tangent_x_squared_delta_thousandth = derivative_trace(x_squared, 2, 1, .001) subplots = make_subplots([trace_x_squared_to_three, tangent_x_squared_delta_hundredth], [trace_x_squared_to_three, tangent_x_squared_delta_thousandth]) $\Delta x = .01$ and $\Delta x = .001$ As you can see, as $\Delta x$ approaches zero, $f'(2)$ approaches $4$. This convergence around one number as we change another number, is the limit . ### Approaching our formula for a derivative So to describe the above, at the point $x = 2$, the limit of $\frac{\Delta y}{\Delta x}$ -- that is the number that $\frac{\Delta y}{\Delta x}$ settles upon as $\Delta x$ approaches zero -- is 4. We can abbreviate this into the following expression: When $x = 2,\lim_{\Delta x\to0} \frac{\Delta y}{\Delta x} = 4$. Or, better yet, we can update and correct our definition of derivative to be: $$f'(x) = \lim_{ \Delta x \to0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ So the derivative is the change in output as we just nudge our input. That is how we calculate instantaneous rate of change. We can determine the runners speed at precisely second number 2, by calculating the runner's speed over shorter and shorter periods of time, to see what that number approaches. One final definition before we go. Instead of $\Delta x$, mathematicians sometimes use the variable $h$ to describe the change in inputs. So replacing our $\Delta x$ symbols with $h$'s we have: $$f'(x) = \lim_{ h\to0} \frac{f(x + h) - f(x)}{h}$$ Above is the formula for the derivative for all types of functions linear and nonlinear. ### Summary In this section, we learned about derivatives. A derivative is the instantaneous rate of change of a function. To calculate the instantaneous rate of change of a function, we see the value that $\frac{\Delta y}{\Delta x}$ approaches as $\Delta x$ approaches zero. This way, we are not calculating the rate of change of a function across a given distance. Instead we are finding the rate of change at a specific moment. Learn to code.
# How do you find the perimeter or circumference, and the area of a circle with radius 4.2 in? Jun 13, 2018 See a solution process below: #### Explanation: The formula for the perimeter or circumference of a circle is: $C = 2 \pi r$ The formula for the area of a circle is: ${A}_{c} = \pi {r}^{2}$ In both formulas $r$ represents the radius of the circle. Substituting $4.2 \text{ in}$ for $r$ in both formulas gives: Circumference: $C = 2 \pi r$ becomes: $C = 2 \pi \cdot 4.2 \text{ in}$ $C = 8.4 \pi \text{ in}$ If you need a number we can substitute $3.14$ for $\pi$ giving: $C = 8.4 \cdot 3.14 \text{ in" = 26.376" in}$ Area: ${A}_{c} = \pi {r}^{2}$ becomes: ${A}_{c} = \pi \cdot {\left(4.2 \text{ in}\right)}^{2}$ ${A}_{c} = 17.64 \pi {\text{ in}}^{2}$ Again, if you need a number we can substitute $3.14$ for $\pi$ giving: ${A}_{c} = 17.64 \cdot 3.14 {\text{ in"^2 = 55.3896" in}}^{2}$
New Zealand Level 6 - NCEA Level 1 # Multiply and divide algebraic fractions Lesson ### Multiplying When it comes to working with algebraic fractions and applying the four operations, the process is exactly the same as when we worked with numeric fractions. Let's have a look at a simple example of multiplying two numerical fractions. ##### Example 1 Simplify $\frac{3}{4}\times\frac{5}{7}$34×57 $\frac{3}{4}\times\frac{2}{5}$34×25 $=$= $\frac{3\times5}{4\times7}$3×54×7 Multiplying numerators and denominators $=$= $\frac{15}{28}$1528 Simplifying the numerator Since $\frac{15}{28}$1528 doesn't have any common factors between the numerator and denominator, that is the most simplified form of our answer. Now let's apply the same process to multiplying algebraic fractions. ##### Example 2 Simplify $\frac{y}{5}\times\frac{3}{m}$y5×3m $\frac{y}{5}\times\frac{3}{m}$y5×3m $=$= $\frac{y\times3}{5m}$y×35m Multiplying numerator and denominators $=$= $\frac{3y}{5m}$3y5m Simplifying the numerator Again, since the numerator $3y$3y and the denominator $5m$5m don't have any common factors, $\frac{3y}{5m}$3y5m is the simplest form of our answer. #### Practice Questions ##### Question 1 Simplify the expression: $\frac{a}{7}\times\frac{a}{12}$a7×a12 ##### Question 2 Simplify the expression: $\frac{8u}{3v}\times\frac{2v}{7u}$8u3v×2v7u ### Dividing Again, the process for dividing is the same as when we divided numeric fractions. ##### Example 3 Simplify $\frac{2}{3}\div\frac{3}{5}$23÷35 $\frac{2}{3}\div\frac{3}{5}$23÷35 $=$= $\frac{2}{3}\times\frac{5}{3}$23×53 Dividing by a fraction is the same as multiplying by its reciprocal. So invert and multiply. $=$= $\frac{2\times5}{3\times3}$2×53×3 Multiply numerators and denominators respectively. $=$= $\frac{10}{9}$109 Since $\frac{10}{9}$109 doesn't have any common factors between the numerator and denominator, that is the most simplified form of our answer. Now let's apply the same process to dividing algebraic fractions. ##### Example 2 Simplify $\frac{m}{3}\div\frac{5}{x}$m3÷5x $\frac{m}{3}\div\frac{5}{x}$m3÷5x $=$= $\frac{m}{3}\times\frac{x}{5}$m3×x5 Dividing by a fraction is the same as multiplying by its reciprocal. So invert and multiply. $=$= $\frac{m\times x}{3\times5}$m×x3×5 Multiply numerators and denominators respectively. "=" $\frac{mx}{15}$mx15 Again, since the numerator $mx$mx and the denominator $15$15 don't have any common factors, $\frac{mx}{15}$mx15 is the simplest form of our answer. #### Practice Questions ##### Question 4 Simplify the expression: $\frac{m}{8}\div\frac{3}{n}$m8÷3n ##### Question 5 Simplify the following: $\frac{-2x}{11}\div\frac{7y}{5}$2x11÷7y5 ##### Question 6 Simplify $\frac{-2x}{11}\div\frac{2x}{3}$2x11÷2x3. ### Outcomes #### NA6-5 Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns #### NA6-6 Generalise the properties of operations with rational numbers, including the properties of exponents #### 91027 Apply algebraic procedures in solving problems
# NCERT Solutions and Notes for Class 6 Maths Chapter 10: Mensuration- Free PDF Download! In Chapter 10, Mensuration – Class 6, students will learn to analyze and compare the area of flat figures, i.e. figures on a 2D plane. The chapter is very important is it lays the foundation for understanding the measurements which will be useful for them even in their higher studies such as engineering, Maths, or any other core fields of study. In the Mensuration chapter of Class 6, students will learn to calculate the perimeter, perimeter formulas, area, and area formulas of various shapes. After completing the chapter Mensuration, Class 6, students will be able to differentiate between the area and perimeters of various shapes. Read through for CBSE NCERT Class 6 Maths Chapter 10 Mensuration Notes and Solutions. Below we have given topic-wise notes for Chapter 10, Mensuration, Class 6. We have also provided a downloadable free PDF at the end of these notes so you can download and take a printout to study later when you need quick revision before going to the exam hall. ### Topic 1: Perimeter Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once. ### Topic 2: Perimeter Formulas: Rectangle, Square, Triangles Below we have given the various perimeter formulas of some basic shapes. In the exercises, we will learn to use these perimeter formulas to calculate the perimeter of complex figures. 1. Perimeter of Rectangle Formula: The rectangle has 4 sides, having opposite sides of equal lengths. 1. Perimeter of Square Formula: A square has sides of equal lengths. 1. Perimeter of Triangle Formula: Here, we will determine the perimeter of an equilateral triangle. An equilateral triangle has all sides of equal lengths. We can determine the length of any regular figure in a similar way if we know its number of sides. Therefore, we can determine the perimeter of a regular pentagon in a similar way. ### Topic 3: Area The amount of surface enclosed by a closed figure is called its area. ### Topic 4: Area Formulas Below we have given the various area formulas of some basic shapes. In the exercises, we will learn to use these area formulas to calculate the area of complex figures. 1. Area of Rectangle Formula: The area of a rectangle can be calculated by multiplying its length and breadth. 1. Area of a Square: The area of a square can be calculated by multiplying its sides. ## Important Questions in NCERT Class 6 Maths Chapter 10: PDF Download Below we have provided some important exercise questions and their solutions from Chapter 10 of CBSE NCERT Maths for Class 6. ### Exercise 10.1 Q 1. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required? Ans. The length of the tape required to seal all the sides of the rectangular box’s lid will be equal to the perimeter of the rectangular box. We know, Perimeter of a Rectangle = 2 × (length + breadth) And, the length of the tape required = Perimeter of the rectangular box’s lid. Given: Length of the rectangular box’s lid = 40 cm The breadth of the rectangular box’s lid = 10 cm ∴ Length of tape required = 2 × (40 + 10) cm = 2 × 50 cm = 100 cm Q 2. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop? Ans. To find the perimeter of the tabletop, let’s find its measures in a single unit of length. Given: Length = 2 m 25cm and Breadth = 1 m 50 cm We know, 1 cm = 0.01 m, then, 25 cm = 0.25 m and 50 cm = 0.50 m Now, Length = 2 m + 0.25 m = 2.25 m And Breadth = 1 m + 0.50 m = 1.50 m Perimeter of tabletop = 2 × (2.25 + 1.50) m = 2 × 3.75 m = 7.5 m Q 3. What is the length of the wooden strip required to frame a photograph of length and breadth of 32 cm and 21 cm respectively? Ans. The length of the wooden strip required must be equal to the perimeter of the photograph. ∴ length of wooden strip = 2 × (32 + 21) cm = 2 × 53 cm = 106 cm Q 4. Find the perimeter of a regular hexagon with each side measuring 8 m. Ans. We know that regular shapes have all sides of equal lengths. Therefore, the given regular hexagon has 6 equal sides of 8 m. Perimeter of a Regular Hexagon = 6 × Length of Each Side = 6 × 8 m = 48 m ∴ The perimeter of the given regular hexagon is 48 m. Q 5. The perimeter of a regular pentagon is 100 cm. How long is its each side? Ans. From the above question, we know that the length of each side in a regular figure is equal. ∴ The perimeter of a regular pentagon = 5 × Length of Each Side Given: Perimeter of the regular pentagon = 100 cm ∴ the length of each side of the given regular pentagon is 20 cm. Q 6. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre. Ans. To determine the cost of fencing, we must first calculate the perimeter of the square park. Given: Length of each side of the square park = 250 m Perimeter of the park = 4 × 250 m = 1000 m Now, the cost of fencing for 1 metre = ₹ 20 ∴ cost of fencing 1000 m = ₹ 20 × 1000 = ₹ 20,000 Q 7. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square. 1. What is the perimeter of his arrangement? 2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement? 3. Which has a greater perimeter? Ans. The answers to each question are given below. 1. Length of each side of the square paving slabs = 0.5 m In the square arrangement, Avneet placed 3 square paving slabs in each row. ∴ Length of each side of the square = 3 × 0.5 m = 1.5 m Thus, the perimeter of the arrangement = 4 × 1.5 m = 6.0 m 1. Shari makes an arrangement in which 8 sides are made by placing two square paving slabs side to side and the remaining 4 sides are equal in length to 1 square paving slab. ∴ The perimeter of the arrangement made by Shari = 8 × (0.5 m + 0.5 m) + 4 × 0.5 m ⇒ 8 × 1 m + 2 m = 10 m Thus, the perimeter of the arrangement made by Shari is 10 m. 1. The arrangement made by Shari has a greater perimeter. ### Exercise 10.3 Q 1. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m? Ans. Let us find the area of the rectangular plot to determine the cost of tiling it. We know that the area of a rectangle = Length × Breadth So, area = 500 m × 200 m = 1,00,000 sq m Given, the cost of tiling 100 sq m of land = ₹ 8 ∴ the cost of tiling 1,00,000 sq m of plot = ₹ 8/100 × 1,00,000 = ₹ 8000 Q 2. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively 100 cm and 144 cm? Ans. We must divide the area of the rectangular region by the area of a tile to find the number of tiles that will fit in the given rectangular region. ∴ Area of a tile = 12 cm × 5 cm = 60 sq cm Area of the rectangular region = 100 cm × 144 cm = 14,400 sq cm So, the number of tiles that fit the rectangular region: So, 240 tiles will be needed to fit in a rectangular region. Also See: ## FAQs Q.1. What are regular closed figures? Ans: All the regular closed figures have all sides of equal lengths and all the angles of equal measure. Q.2. How to find the perimeter of a rectangle? Ans: The perimeter of a rectangle can be calculated by the formula: Perimeter = 2 × (length + breadth). Q.3. How to find the perimeter of a circle? Ans: The perimeter of a circle is referred to as its circumference. The circumference of a circle can be calculated by the formula: Circumference = 2πr. Q.4. How to find the perimeter of a regular triangle? Ans: The perimeter of a regular triangle can be determined by the formula: Perimeter = 3 × Length of each side. Q.5. How to find the area of a rectangle? Ans: The area of a rectangle can be found by multiplying each side, i.e. Length × Breadth. This was all about Mensuration Class 6 Maths in which we studied the perimeter, perimeter formulas, and area and area formulas of various shapes. Follow the CBSE Class 6 Maths Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 6 Maths.
# Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 361: 46 $(a+b)^{2}$ = $a^{2}$ + $b^{2}$ + 2ab #### Work Step by Step We know area of square A = side * side The given drawing consist of 4 rectangles The area of Square = (a+b)(a+b)= $(a+b)^{2}$ Lets find the area of all the 4 rectangles separately Rectangle I = ab Rectange II = $b^{2}$ Rectangle III = ab Rectangle IV = $a^{2}$ The total area of square = Area of I + Area of II + Area of III + Area of IV = ab+ $b^{2}$+ab+ $a^{2}$ =2ab+ $b^{2}$ + $a^{2}$ Therefore $(a+b)^{2}$ = $a^{2}$ + $b^{2}$ + 2ab After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
Middle Years 12.02 Theoretical probability Lesson Probability with numbers Now that we know how to describe events with language, we will now investigate using numbers to calculate probabilities. If we can split up the sample space into equally likely outcomes and can identify the favourable outcomes making up an event, we can use the formula: $\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Probability=Number of favourable outcomesTotal number of outcomes Worked example Example 1 Hunter has $11$11 black shirts, $9$9 white shirts, $4$4 yellow shirts and $3$3 red shirts. If he selects a shirt from his closet at random, what is the probability he selects either a yellow shirt or a red shirt? Think: Each individual shirt is an outcome. They are all equally likely to be picked, so we can use the equation: $\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Probability=Number of favourable outcomesTotal number of outcomes The event we are thinking about is "he selects either a yellow shirt or a red shirt", so the number of favourable outcomes is the number of red shirts plus the number of yellow shirts. The total number of outcomes is the total number of shirts. Do: The number of favourable outcomes is $4+3=7$4+3=7. The total number of outcomes is $11+9+4+3=27$11+9+4+3=27. The probability will be $\frac{7}{27}$727. We can also use a useful fact about complementary events - since exactly one of them must happen, their probabilities always add to $1$1. This means if we know the probability of an event, the probability of the complementary event will be one minus the probability of the original: $\text{Probability of complementary event}=1-\text{Probability of event}$Probability of complementary event=1Probability of event Worked example Example 2 Ursula selects a card at random from a standard deck. What is the probability she will select a card that isn't a king? Think: There are $13$13 different card values in a standard deck of cards. The event "selecting a card that isn't a king" is the complementary event to "selecting a card that is a king". We can calculate the probability of the complementary event and subtract it from $1$1 to find our probability. Do: The probability of selecting a king is $\frac{1}{13}$113, so the probability of selecting a card that isn't a king will be $1-\frac{1}{13}=\frac{12}{13}$1113=1213. Reflect: We could calculate this probability another way, by thinking about the $12$12 card values other than the king as "favourable outcomes", and dividing by the total number of card values ($13$13). Summary If a sample space can be split up into equally likely outcomes, then we can use the following formula: $\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Probability=Number of favourable outcomesTotal number of outcomes If two events are complementary then their probabilities will add to $1$1. This means: $\text{Probability of complementary event}=1-\text{Probability of event}$Probability of complementary event=1Probability of event Practice questions Question 1 Consider this list of numbers: $2,2,2,3,3,3,4,4,5,5,5,7,7,7,7,9,9$2,2,2,3,3,3,4,4,5,5,5,7,7,7,7,9,9 1. How many numbers are in the list? 2. A number is chosen from the list at random. What is the probability it is an odd number? 3. What number has the same probability of being picked as $4$4? 4. A number is chosen from the list at random. What number has the highest probability of being chosen? Question 2 The probability of the local football team winning their grand final is $0.36$0.36. 1. What is the probability that they won't win the grand final? Question 3 A standard deck of $52$52 cards is shown below. If a card is selected at random, what is the probability that it is: 1. a red card? 2. a card between $5$5 and $9$9 inclusive? 3. a card that is red and has a number between $5$5 and $9$9 inclusive? 4. a card that is red or a king?
4.5 Generalized Permutations and Combinations Presentation on theme: "4.5 Generalized Permutations and Combinations"— Presentation transcript: 4.5 Generalized Permutations and Combinations Lecture 6 4.5 Generalized Permutations and Combinations Recap r-permutations: The number of ways in which we can draw r balls from a collection of n different balls, where the order is important: P(n,r) = n! / (n-r)! r-combinations: The number of ways in which we can draw r balls from a collection of n different balls, where we do not care about the ordering: C(n,r) = n! / r! (n-r) ! Today: we study counting problems where repetitions are allowed, i.e. it is possible that the same ball is drawn multiple times. 4.5 X X X X r boxes n balls 1 2 3 4 1 2 3 4 we over-counted by r! n ways n-1 ways n-r+1 ways different because the slots have labels (distinguishable) n balls 1 2 3 4 X 1 2 3 4 X X r-permutation without repetition (order important) X r-combination without repetition (order not important) the balls are not replaced when they have been drawn the same because the slots have no labels (indistinguishable) we over-counted by r! 4.5 r boxes n balls 1 2 3 4 1 2 3 4 this one is tricky ! n ways n ways different because the slots have labels (distinguishable) n balls 1 2 3 4 1 2 3 4 r-permutation with repetition (order important) r-combination with repetition (order not important) the balls are replaced when they have been drawn. Or there is a very large stack of indistinguishable balls of each color. the same because the slots have no labels (indistinguishable) this one is tricky ! 4.5 Example: We want to draw 2 pieces of fruit from a bowl that contains many 2 apples, 2 pears, and 2 oranges. In how many ways can we do this such that: 1) The apples are different and the order matters. 2) The apples are different and the order does not matter. 3) The apples are indistinguishable but the order matters. 4) The apples are indistinguishable and the order does not matter.  1) 6 * 5 = P(6,2)  2) 6 * 5 / 2 = C(6,2)  3) Now there are 3 kinds of fruit that we draw with replacement (since there are enough of each kind to be able to pick any fruit at any draw). This is true because drawing apple 1 is no different than drawing apple 2. It’s like there are copies of the same apple present. Thus: 3 * 3 = 9 ( a a), (a p), (a o), (p a), (p p), (p o), (o a), (o p), (o o).  4) Since the order doesn’t matter (a p) = (p a), idem (a o)=(o a), (p o)=(o p). we over-counted 3 pieces: 9-3 = 6. 4.5 How to count the latter (r-combination with repetition)? One strategy could be to start from drawing where the order matters and try to count the number of ways we over-counted (last example). However, there is a much smarter way! Map the problem to a bit-string as follows: balls become indistinguishable indistinguishable slots become become distinguishable boxes. C(n+r-1,r) bit-strings ! 4.5 n is number of distinct classes of objects in the original bag! - r-permutation without repetition - order matters (r distinguishable slots) - without replacement (n distinguishable objects) - r-combination without repetition - order does not matter (r indistinguishable slots) - without replacement (n distinguishable objects) n! / r! (n-r)! n! / (n-r)! - r-permutation with repetition - order matters (r distinguishable slots) - with replacement (n distinct classes of indistinguishable objects) n^r - r-combination with repetition - order does not matter (r indistinguishable slots) - with replacement (n distinct classes of indistinguishable objects) (n+r-1)! / r! (n-1)! 4.5 Another problem: Assume we have exactly n objects in a bag, and we are going to draw all of them. The order in which we draw them is important, however n1 objects are indistinguishable, n2 different objects are indistinguishable etc. This is like: you can only replace the blue ball 2 times, the red ball 3 times etc.) (b b r r r) is different from (b r b r b r), but the blue balls are the same and the red balls are the same. (2 distinct classes of indistinguishable balls of size 2 and 3 and distinguishable slots). 1 2 3 4 5 Here we can first count the total number of permutations, pretending all balls are different  n! (n = number of balls, not kinds of balls) Now try to figure out which strings are equivalent: (b1 b2 r1 r2 r3) = (b2 b1 r1 r2 r3). There are 2! ways to permute the blue balls. (b1 b2 r2 r2 r3) = (b1 b2 r2 r1 r3). There are 3! ways to permute the red balls. 4.5 Therefore the total number of ways is: 5!/2! 3!. In general: n!/ n1! n2! ...nk! with n=n1+n2+...+nk Alternate derivation: -For the first kind we have n slots and n1 balls to place such that the order is unimportant: C(n,n1) -For the second kind there are then n-n1 slots still open to place n2 balls: C(n-n1,n2) ... etc. -For the last kind we have (n-n1-n2-...-n_(k-1)) slots for the remaining nk balls: C(n-n1-n2-...-n_(k-1),nk). Total thus: C(n,n1) C(n-n1,n2)..... = n!/ n1! n2! ...nk! (check it!) 4.5 Yet another problem: In how many ways can we distribute n different objects into k different boxes such that n1 objects go in box 1, n2 objects go in box 2,... nk objects go in box k. (k distinguishable boxes with n distinguishable balls) 4 1 2 6 5 3 (b r r b b g) Trick: the colors are now the boxes, so we assign boxes to balls instead of balls to boxes! From the previous problem we know there are n!/n1! n2!...nk! ways to assign the boxes with k colors to the n balls. 4.5 Examples: 1) How many ways are there to select five bills from a cash box containing \$1, \$2, \$5, \$10, \$20, \$50 and \$100 bills, such that the bills are indistinguishable and the order in which they are selected is unimportant. (there are also at least 5 bills of each kind).  This is like drawing colored balls with replacement. The colors correspond to the values. Since the order doesn’t matter we have: C(7+5-1,5)=462 2) A cookie shop has 4 kinds of cookies and we want to pick 6. We don’t care about the order and cookies from one kind are indistinguishable.  Again, drawing colored balls with replacement: colors are kind of cookies. C(6+4-1,6)=84. 3) How many solutions to x1+x2+x3=11 with xi nonnegative integers. This is like throwing 11 balls in 3 boxes. The balls inside each boxe are indistinguishable. C(11+3-1,11)=78. 4.5 Examples: 1) In how many ways can we place 10 indistinguishable items in 8 distinguishable boxes.  This is precisely the problem we saw to solve the r-combination with repetition: C(10+8-1,10) 2) How many different “words” can we create by reordering SUCCESS ? Total number of permutations is 7!. However permuting the 3 S’s does not create a new word, idem 2 C’s: 7!/3! 2! 3) In how many ways can we distribute a deck of 52 cards over 4 hands such that each hand gets 5 cards?  There are 4 boxes which receive 5 cards and 1 box which receives 32 cards. Assign people to cards instead: 52! / (5!)^5 32! 4.5 Exercise 52 p. 344: How many different cross terms will we generate when we multiply out: (x1+x2+...+xm)^n ?  How many different exponents are there of the sort x1^n1 x2^n2 ... xm^nm with n1+n2+...+nm=n. Equivalent to : how many different way are there to put n balls in m boxes: C(n+m-1,n) 38 p.342. Math teacher has 40 issues of a journal and packs them into 4 boxes, 10 issues each. a) How many ways if the boxes are numbered?  assign boxes to issues: 40! / (10!)^4 b) Now the boxes are indistinguishable.  There are 4! ways to label the boxes, once we have distributed them in unlabelled boxes. Since the number of ways to distribute them in labeled boxes is given by a) we get 40! / (10!)^4 4!.
The vertical line test to determine if a relation is a function. To use the vertical line test all that you have to do is ... # Vertical Line Test #### What is the Vertical Line Test for Functions? Answer: A method to distinguish functions from relations. The vertical Line test. • is a way to determine if a relation is a function. • states that if a vertical line intersects the graph of the relation more than once, then the relation is a NOT a function. If you think about it, the vertical line test is simply a restatement of the definition of a function. Definition of a function: Every x value has a unique y value. Think about it If any particular x value has 2 different y values, then a vertical line will intersect the at two different places. Let's examine the two relations below this These two relations differ by only 1 number! ### Practice Problems ##### Problem 1 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two time?" Answer: NO vertical lines only hit the graph once so this is a function. ##### Problem 2 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: YES a vertical line can hit the graph twice so this is not a function. ##### Problem 3 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: YES a vertical line hits the graph several times, so this is not is a function. ##### Problem 4 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: NO matter where you try to draw a vertical line, it only hits the graph once so this is a function. Be careful at x =2. The point (2, 1) is not filled in, indicating that the graph does not include the point (2, 1). However, notice that (2, 2) is completely filled in, because that point is included in the graph. ##### Problem 5 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: Yes, a vertical line can intersect this function more than once! . Unlike problem 3, in this case, the point (2, 1) is filled in and is,therefore, included in the graph. As you can see, that one point makes all the difference. This is not a function ##### Problem 6 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: YES This is not a function . In fact, circles, in general, are not functions. ##### Problem 7 Ask yourself: "Can I draw a vertical line (anywhere) that will hit the graph two times?" Answer: NO This is a function
1984 USAMO Problems/Problem 1 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) Problem In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$. Solution 1 (ingenious) Using Vieta's formulas, we have: \begin{align}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align} From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$. Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$, and so $ac+ad+bc+bd=k-30$. The key insight is now to factor the left-hand side as a product of two binomials: $(a+b)(c+d)=k-30$, so that we now only need to determine $a+b$ and $c+d$ rather than all four of $a,b,c,d$. Let $p=a+b$ and $q=c+d$. Plugging our known values for $ab$ and $cd$ into the third Vieta equation, $-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)$, we have $-200 = -32(c+d) + 62(a+b) = 62p-32q$. Moreover, the first Vieta equation, $a+b+c+d=18$, gives $p+q=18$. Thus we have two linear equations in $p$ and $q$, which we solve to obtain $p=4$ and $q=14$. Therefore, we have $(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30$, yielding $k=4\cdot 14+30 = \boxed{86}$. Solution 2 (cool) We start as before: $ab=-32$ and $cd=62$. We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$. Now \begin{align}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\ =& x^4+(r+s)x^3+(62-32+rs)x^2\\ &+(62s-32r)x-1984.\end{align} Equating the coefficients of $x^3$ and $x$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of $x^2$ and get $k=\boxed{86}.$ Solution 3 Let the roots of the equation be $a,b,c,$ and $d$. By Vieta's, \begin{align} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align} Since $abcd=-1984$ and $ab=-32$, then, $cd=62$. Notice that$$abc + abd + acd + bcd = -200$$can be factored into$$ab(c+d)+cd(a+b)=-32(c+d)+62(a+b).$$From the first equation, $c+d=18-a-b$. Substituting it back into the equation,$$-32(18-a-b)+62(a+b)=-200$$Expanding,$$-576+32a+32b+62a+62b=-200 \implies 94a+94b=376$$So, $a+b=4$ and $c+d=14$. Notice that$$ab+ac+ad+bc+bd+cd=ab+cd+(a+b)(c+d)$$Plugging all our values in,$$-32+62+4(14)=\boxed{86}.$$ ~ kante314 ~ pi_is_3.14
# $x^{4}-y^{4}=z^{2}$ has no solutions in positive integers We know (see example of Fermat’s Last Theorem) that the sum of two fourth powers can never be a square unless all are zero. This article shows that the difference of two fourth powers can never be a square unless at least one of the numbers is zero. Fermat proved this fact as part of his proof that the area of a right triangle with integral sides is never a square; see the corollary below. The proof of the main theorem is a great example of the method of infinite descent. ###### Theorem 1. $x^{4}-y^{4}=z^{2}$ has no solutions in positive integers. ###### Proof. Suppose the equation has a solution in positive integers, and choose a solution that minimizes $x^{2}+y^{2}$. Note that $x,y$, and $z$ are pairwise coprime, since otherwise we could divide out by their common divisor to get a smaller solution. Thus $z^{2}+(y^{2})^{2}=(x^{2})^{2}$ so that $z,y^{2},x^{2}$ form a pythagorean triple. There are thus positive integers $p,q$ of opposite parity (and coprime since $x,y$, and $z$ are) such that $x^{2}=p^{2}+q^{2}$ and either $y^{2}=p^{2}-q^{2}$ or $y^{2}=2pq$. Factoring the original equation, we get $(x^{2}-y^{2})(x^{2}+y^{2})=z^{2}$ If $y^{2}=p^{2}-q^{2}$, then $(xy)^{2}=p^{4}-q^{4}$, and clearly $p^{2}+q^{2}=x^{2}, so we have found a solution smaller than the assumed minimal solution. Assume therefore that $y^{2}=2pq$. Now, $x^{2}=p^{2}+q^{2}$; we may assume by relabeling if necessary that $q$ is even and $p$ odd. Then $p,q,x$ are pairwise coprime and form a pythagorean triple; thus there are $P>Q>0$ of opposite parity and coprime such that $q=2PQ,\quad p=P^{2}-Q^{2},\quad x=P^{2}+Q^{2}$ Then $PQ(P^{2}-Q^{2})=\frac{1}{2}pq=\frac{y^{2}}{4}$ is a square; it follows that $P,Q$, and $P^{2}-Q^{2}$ are all (nonzero) squares since they are pairwise coprime. Write $P=R^{2},\quad Q=S^{2},\quad P^{2}-Q^{2}=T^{2}$ for positive integers $R,S,T$. Then $T^{2}=R^{4}-S^{4}$, and $R^{2}+S^{2}=P+Q<(P+Q)(PQ)(P-Q)=\frac{1}{2}pq=\frac{y^{2}}{4}\leq y^{2} We have thus found a smaller solution in positive integers, contradicting the hypothesis. ∎ ###### Corollary 1. No right triangle with integral sides has area that is an integral square. ###### Proof. Suppose $x,y,z$ is a right triangle with $z$ the hypotenuse, and let $d=\gcd(x,y,z)$. Either $x/d$ or $y/d$ is even; by relabeling if necessary, assume $x/d$ is even. Then we can choose relatively prime integers $p,q$ with $p>q$ and of opposite parity such that $\displaystyle x=(2pq)d$ $\displaystyle y=(p^{2}-q^{2})d$ $\displaystyle z=(p^{2}+q^{2})d$ If the triangle’s area is to be a square, then $\frac{1}{2}xy=pq(p^{2}-q^{2})d^{2}$ must be a square, and thus $pq(p^{2}-q^{2})$ must be a square. Since $p$ and $q$ are coprime, it follows that $p$, $q$, and $p^{2}-q^{2}$ are all squares, and thus that $p^{2}-q^{2}$ is the difference of two fourth powers. But then $\frac{\frac{1}{2}xy}{pqd^{2}}=p^{2}-q^{2}$ must also be a square. Since both $p$ and $q$ are squares, this is impossible by the theorem. ∎ Title $x^{4}-y^{4}=z^{2}$ has no solutions in positive integers X4y4z2HasNoSolutionsInPositiveIntegers 2013-03-22 17:05:04 2013-03-22 17:05:04 rm50 (10146) rm50 (10146) 7 rm50 (10146) Theorem msc 11D41 msc 14H52 msc 11F80 ExampleOfFermatsLastTheorem IncircleRadiusDeterminedByPythagoreanTriple
# The Most Common Digital SAT Math Topics: Part 2 0 55 In the first installment of this series, we covered four of the eight most common digital SAT math topics: Finding an Equation to Fit the Data, Exponential Functions, Finding Intercepts, and Quadratics. This time we’ll look at the remaining four: Solving Equations, Systems of Equations, Linear Relationships, and Evaluating Functions at a Given Value Solving Equations Most of us are familiar with the process of solving an equation: find the value of the variable that makes the equation true. Most students can solve the most basic version of this problem in 10 seconds. We typically use inverse operations to isolate our variable. In this case, we’d subtract 30 from both sides to cancel out the + 30 next to the x: That was easy. But sometimes we’ll see trickier versions of this concept, like this. The basic strategy is still the same: isolate the variable, which happens to be w this time, solving for its value. One thing to note: we’re not asked to solve for w, but rather for 2w. So we’ll need to be careful about that. Let’s distribute the 3 into the parentheses, combine like terms, and find w, which we can then use to find the value of 2w: Since w is 7, 2w will be 2(7), which is 14, our final answer. One last thing: we can always test the value we get for the variable by plugging it back into the original equation to make sure the equation works. So we can plug in 7 to replace w in the original equation, and both sides of the equation will equal the same number, showing us that our answer is correct. Systems of Equations As we touched on in our post on the power of Desmos, a system of equations is just two equations that we’re looking at simultaneously; commonly, we’ll need to find that system’s solution, another name for the intersection points of their graphs. Here’s an example: Notice that they want not the (x, y) coordinates of the intersection point, but just the y-value there. The quickest way to do most systems of equations questions is to use Desmos, so let’s use it here: So the y-value at the intersection point is 2, our answer. You can certainly use algebra to solve this one, but doing so will take significantly longer for most students than simply using Desmos. Linear Relationships Any given situation can be modeled by a linear relationship if it consists of two variables, one of which changes at a constant rate for each unit change in the other variable. For example, if a loaf of bread costs \$4, there’s a linear relationship between the total cost and how many loaves of bread are purchased: each additional loaf adds \$4 to the total cost. However, there’s not a linear relationship between the population of the world and the number of years since 2000, for example, since population growth is exponential, not linear: the population doesn’t grow by the same number of people each year. How can we see this concept on the SAT? Here’s a common type of question: The given equation might remind us of the equation y = mx + b, where m is the slope of the line and b is its y-intercept. The only difference is that the order of mx and b is reversed, but that won’t matter. 18 would be the y-intercept of this line, the starting number of gallons of gas when x = 0 and no minutes have been driven since filling the gas tank. So the gas tank must have 18 gallons of gas in it when it’s full. Now, let’s consider the –0.04 part: notice that this is being multiplied by x, our time in minutes, so the total gallons of gas must drop by 0.04 gallons every minute; this is why this number is being multiplied by x. And 0.04 is negative because the amount of gas goes down as the car is driven. This matches up with the explanation given in C, which is our answer. Evaluating Functions at a Given Value Functions are usually given to us using function notation: f(x) = 2x + 1, for instance. That f(x), pronounced “f of x,” essentially means here’s the function with x plugged in. So here, 2x + 1 would be the function. If we’re asked to evaluate the function at a given value, we’ll just take out the x in the original function and replace it with whatever number they give us. So f(-2), for example, means replacing x in the function with -2, while f(0) means plugging in 0 for x. On the SAT, we could certainly see something like f(-2), which would make evaluating the given function simple. However, sometimes we aren’t given that same notation and will still need to know that we have to plug a value into the function: Notice that they don’t specifically use the notation of f(1) or f(2) or anything similar, which would make this one slightly more manageable. But they do give us a certain temperature, 75oF, and ask us how many cups of ice cream will be sold at this temperature. What’s the key here? 75 oF is the temperature, x, and the number of cups of ice cream sold is f(x). So they’re really asking us to find f(75); in other words, just plug 75, our x-value, in place of x in the function: So we get C: 280 cups of ice cream. As with anything, practice is the key to getting your goal score on the SAT, and it involves taking official practice tests, familiarizing yourself with the question types, and focusing on getting answers quickly. As we’ve shown, the eight most-tested topics are ones that are relatively easy to master and that can sometimes be solved using Desmos. On test day, if you can get all of the questions right for these eight most common topics, you’ll have already gotten about half of the entire math section correct! If you find yourself needing more material than just the College Board’s practice tests, or if you’d like detailed lessons on each SAT topic, check out our self-paced SAT course in Methodize, with guided lessons and hundreds of practice questions and explanations for finding each answer.
# Problems involving Two Unknowns In this section we will explore a method of solving equations in which there are two unknown values. In these types of problems, we will begin by assigning a variable to one of the unknown values, and expressing the other value in terms of the unknown. Doing this eliminates the need for multiple equations to describe the situation (we will do this later.) We saw this first in previous sections dealing with cutting a board into multiple pieces, or finding the values of the angles of a triangle. Let us begin with an example: Janet has 18 coins in her purse, all of them nickels and quarters. If she has $2.10, how many of each coin does she have? We do not know how many nickels Janet has, nor do we know how many quarters she has. These are our two unknowns. We can assign the variable to represent the number of nickels she has. To avoid using a second variable, we can note that if she has 18 coins total, and of them are nickels, then the difference between 18 and is the number of quarters she has. That is: the number of nickels the number of quarters And we know that the value of nickels is the number of nickels multiplied by the value of a nickel, and similarly for the value of the quarters: Value of nickels Value of quarters And if we were to add these two amounts together, they would equal the total amount Janet has,$2.10, we can create the following equation: And from here, we simplify and solve using our regular methods: So then we have 12 nickels. Plugging 12 into the expression for quarters, we then have 18-12=6 Quarters. Let’s view another example: Java Joe’s Coffee Shop sells two kinds of coffee. Mountain Aroma sells for $8 per pound, and Blue Ivy sells for$6.50 per pound. Joe wants to combine the two to sell at a rate of $7 per pound. How much of each coffee should he combine to make 30 pounds of the mix? Let represent the number of pounds of Mountain Aroma coffee. Then will represent the number of pounds of Blue ivy. We can use an equation similar to the previous example: Note that the right hand side of this equation is the total cost of the mixture, since it sells at$7 per pound, and there are thirty pounds of it, then the product of the two is the total sales price of the mixture. We solve this equation: So we have ten pounds of Mountain Aroma, and thus we have 20 pounds of Blue Ivy. Ginger has 60 coins, all of them nickels and dimes. If she has a total of \$4.45, how many nickels, and how many dimes does she have? We can call our variable , and note that is the number of nickels. If there are nickels, then there are dimes. And our equation becomes: And we solve: So we have 31 nickels, and therefore, we have dimes. Mixture problems are very similar to the previous problems: A scientist needs 42 gallons of a 15% mixture of saline for an experiment. He has ample 12% and 75% solutions in stock. How much of each should he mixed? We identify our variable , and can say that it is the amount of 12% mixture. Then we will also have gallons of 75% mixture. Our equation is: Then we need 40 gallons of 12% mixture, and 2 gallons of 75% A chemist needs to use 30 Liters of a 40% solution for an experiment. He notices that in his lab, however, that he only has 25% and 50% solutions. He decides to mix the two together to get his 40% mixture. How much of each should he use? The chemist wants 30 liters of the solution, but he doesn’t know how much of each to use. We can assign to represent the amount of the 25% solution to use. In doing so, then the rest of the solution is the 40% mixture. So the difference between 30 and is the amount of 40%. That is, we have: We can then create an equation using this information, and the fact that the product of a solutions volume and its strength (percentage) yields the actual amount of acid in the solution. We note that the sum of the two acids should equal the total amount of acid, and build the equation: And we can solve this equation through means discussed in earlier sections: So we see that we will use 12 liters of the 25% acid. We then will use Liters of the 50% solution.
In mathematics, we learn about quadratic functions. A graph that has a square and is a quadratic function, and a graph has a curve. In mathematics involving graphs, we first learn direct proportions. The graphs that apply direct proportions are linear functions. Further advancement of linear functions is quadratic functions. The concept of linear and quadratic functions is almost the same; the only difference is whether or not $x^2$ is included in the equation. However, the shape of the graph is quite different in quadratic functions. There are also some characteristics of quadratic functions that you need to remember. In addition, quadratic functions are often used with linear functions in problems. In this section, we will explain how to solve quadratic function problems. ## Equations with $y=ax^2$ Are Quadratic Functions When we substitute a number for $x$, the equation that clearly determines the value of $y$ is a function. In a linear function, for example, we have the following equation. • $y=x+2$ In such an equation, if $x$ is doubled, the increase in $y$ is also doubled. Therefore, the values of $y$ and $x$ are in a proportional relationship. On the other hand, in quadratic functions, for example, the equation is as follows. • $y=x^2$ In this case, if the value of $x$ is doubled, the value of $y$ is 22 times (four times). In other words, the value of $y$ is proportional to $x^2$. As the value of $x$ increases, the value of $y$ increases by the square. In quadratic equations, for example the following. • $y=5x^2$ • $y=-3x^2$ • $y=x^2$ • $y=\displaystyle\frac{1}{2}x^2$ Thus, the number in front of $x^2$ changes depending on the expression. Therefore, the quadratic function is represented by the following formula, • $y=ax^2$ The most basic quadratic function is expressed by $y=ax^2$. ### Many Natural Phenomena Involve Quadratic Functions Why do we need to learn quadratic functions? It’s because many of the phenomena around us involve quadratic functions. For example, when you place a ball on a slope and release it, what happens to the speed of the ball? Right after you take your hand off the ball, the speed of the ball is slow. However, over time, the speed of the ball will gradually increase. It is not proportional, which is always a constant speed, but a quadratic function, which increases in speed as time goes by. In other cases, when a stationary car speeds up, the speed increases in a quadratic function. Of course, we use quadratic functions not only for cars, but also for calculating the speed of a rocket launch. It is important to understand that quadratic functions are used in various situations. ## Table of Quadratic Functions Is Proportional to $x^2$ What does a quadratic function look like in a table? Unlike direct proportions, as mentioned above, quadratic functions are proportional to $x^2$. For example, in equation $y=x^2$, the table looks like this. If $x$ is doubled, $y$ is quadrupled. If $x$ is tripled, $y$ is multiplied by 9. If $x$ is quadrupled, $y$ is multiplied by 16. Since the equation is $y=x^2$, as the value of $x$ increases, the value of $y$ is $x$ squared. It is important to note that the value of $y$ is an expression that increases by the square. ### Quadratic Functions Become a Parabolic Graph What does a quadratic formula look like when you graph it? For example, the graph of $y=x^2$ looks like this. As you can see, it is a curved graph. To draw a graph, plot the coordinates of $x$ and $y$ on the graph. Then we can graph a quadratic function by connecting the coordinates with a smooth curve. The graph of $y=ax^2$ will always pass through the origin. Also, the quadratic function of $y=ax^2$ passes through the coordinates of $(0,0)$. Note that the curve on the graph of the quadratic function is the same as the trajectory of the ball when it is thrown. In fact, when you throw an object in the air, you can calculate its speed using the quadratic function. Earlier, we explained that the trajectory of a rocket is calculated by a quadratic function. The reason for this is that rockets fly in the air. Also, in the past, quadratic functions were used to calculate the trajectory of cannon fire. Quadratic functions are used in various situations, including science technology, and warfare. The line on the graph of a quadratic equation is called a parabola. Any object in the air will follow a parabola and will have the same curve as the quadratic function. ### The Value of $a$ Changes the Shape of the Graph When we draw a graph of $y=ax^2$, consider that the shape of the graph depends on the value of $a$. The larger the absolute value of $a$, the steeper the graph becomes. On the other hand, the smaller the absolute value of $a$, the looser the graph becomes. The shape of the graph depends on the value of $a$, as shown below. All graphs pass through the origin $(0,0)$. It is also the same that the curve is a parabola. The difference is that the shape of the graph changes depending on the value of $a$. ### Convex Downward for $a>0$ and Convex Upward for $a<0$ There is also another important point. The direction of the graph changes depending on whether the value of $a$ is a positive number or a negative number. In the graph of a quadratic function, if $a$ is a positive number ($a>0$), it will be convex downward. On the other hand, if $a$ is a negative number ($a<0$), it will be convex upward. Specifically, it looks like the following. In a quadratic function, the value of $x$ is squared. So whether the value of $x$ is positive or negative, the answer will always be positive. But in a quadratic function, you multiply the value of $a$. Therefore, if the value of $a$ is positive, the larger the value of $x$ is, the larger the value of $y$ will be. On the other hand, if the value of $a$ is negative, the larger the value of $x$ is, the smaller the value of $y$ will be. ## The Rate of Change in Quadratic Functions Varies In a quadratic function, we may need to find the rate of change. When calculating a linear function, the rate of change is very important. This is because the rate of change is the slope. By calculating the rate of change, we can get the slope $a$. However, the way of thinking about the rate of change is different between linear and quadratic functions. For a linear function, the rate of change is always the same. In quadratic functions, on the other hand, the rate of change varies from place to place. The following formula is used to calculate the rate of change. We don’t need to remember this formula for linear functions. As mentioned earlier, the rate of change and the slope have the same meaning. On the other hand, what about quadratic functions? Even for quadratic functions, there is no point in remembering the rate of change formula. The rate of change is, in essence, how much the value of $y$ increases when the value of $x$ increases by 1. If you understand this definition, you don’t need to remember the rate of change formula. We can use the properties of the ratio. It looks like the following. By using the properties of ratios, you can come up with a rate of change. For example, we have the following. • When the value of $x$ is increased by 1, the value of $y$ is increased by 2: the rate of change is 2 • When the value of $x$ is increased by 1, the value of $y$ is increased by -4: the rate of change is -4 • When the value of $x$ is increased by 7, the value of $y$ is increased by 3: the rate of change is $\displaystyle\frac{3}{7}$ For example, if the value of $x$ is increased by 7 and the value of $y$ is increased by 3, the ratio is $7:3=1:a$. Solving for this, we have $a=\displaystyle\frac{3}{7}$. If you remember a formula that is used infrequently and is not important, you are sure to forget it. Rather, you have to understand the principles of why it happens and come up with a value without a formula. In mathematics, there are situations where we have to remember formulas and situations where we don’t. For the rate of change, there is no point in remembering the formula. -The Rate of Change in Quadratic Equations As mentioned above, the rate of change in a quadratic function varies depending on the coordinates. How do we calculate the rate of change in a quadratic function? As an example, let’s calculate the rate of change for the graph of $y=x^2$ in the following case. • When the value of $x$ changes from 1 to 2 • When the value of $x$ changes from -3 to -1 When the value of $x$ changes from 1 to 2, the value of $y$ increases by 3. When $x$ increases by 1, $y$ increases by 3, so the rate of change is 3. On the other hand, when the value of $x$ changes from -3 to -1, the value of $y$ increases by -8. When $x$ increases by 2, $y$ increases by -8. The ratio of change is -4 because $2:-8=1:a$. ### Find a Linear Function Through Two Points Why do we need to find the rate of change in quadratic functions? It is because we have to solve a problem that combines a quadratic function and a linear function, as shown below. While the rate of change in quadratic functions changes, the rate of change in linear functions does not change. Therefore, for a quadratic function, the rate of change is always constant for the line connecting the two coordinates. This is because the slope is the rate of change. Finding the rate of change in a quadratic function is the same meaning as finding the slope of a line connecting two coordinates. For reference, in the previous graph, $(-1,1)$ and $(3,9)$ are the intersection points. In other words, an increase of $x$ by 4 results in an increase of $y$ by 8. Substituting “Increase in $x$: increase in $y$ = 1 : $a$,” we get $4:8=1:a$. Solving for this, we get $a=2$. Also, if we check the line passing through the coordinates of the two points, we can see that it passes through the coordinates of $(0,3)$. In other words, the intercept is 3. Therefore, the linear function passing through the intersection of the two points is the following. • $y=2x+3$ Thus, in a quadratic function, we can find the linear function of the line connecting the two coordinates. This kind of mathematical calculation problem is frequently given as an advanced problem in quadratic functions, so it is important to be able to solve it. ## Exercises: Quadratic Function Graph Problems Q1: Solve the following problems. We have the graph of $y=ax^2$ (1) and the graph of $y=ax+b$ (2). The origin is O, and the intersection of (1) and (2) is A and B. The $x$-coordinate of A is -2, and the $x$-coordinate of B is 6. Also, set C as the intersection of the line in (2) and the $x$-coordinate. 1. For the graph in (1), when the value of $x$ increased from -6 to -2, the value of $y$ increased by -16. Let’s find the value of $a$. 2. Find the equation for the line in (2). 3. Find the area of △AOB. 4. Take the point D on the graph in (1). If the area of △COD is 27, find the $x$-coordinates of point D. In quadratic function problems, it is not always the case that only problems related to quadratic functions are given. It is usually a complex of many problems, such as linear functions, simultaneous equations, graphs, etc., related to quadratic functions, as in this problem. (a) How can we get an expression for a quadratic function from the rate of change? To do this, consider the amount of increase in $x$ and $y$. When the value of $x$ is increased from -6 to -2, the value of $x$ is increased by 4. How much does the value of $y$ increase in this case? Substitute into the formula $y=ax^2$. If the value of $x$ is -6, $y=a×(-6)^2=36a$. If the value of $x$ is -2, $y=a×(-2)^2=4a$. $y$ has increased by -32a because it has changed from 36a to 4a. Since we already know that the change in $y$ is -16, we can calculate the following. $-32a=-16$ $a=\displaystyle\frac{-16}{-32}$ $a=\displaystyle\frac{1}{2}$ Thus, the function in (1) is found to be $y=\displaystyle\frac{1}{2}x^2$. (b) Since $y=\displaystyle\frac{1}{2}x^2$, we know the coordinates of A and B. By substituting the value of $x$, we can see that we get the following coordinates, respectively. • A$(-2,2)$ • B$(6,18)$ If we know the two coordinates, we can solve the linear equation using simultaneous equations. In other words, we can find equation (2). By substituting $y=ax+b$, we solve the following simultaneous equation. $\begin{eqnarray} \left\{\begin{array}{l}2=–2a+b\\18=6a+b\end{array}\right.\end{eqnarray}$ By solving this system of linear equations, we can get the equation for a linear function. By the way, it is also possible to get the equation (2) by another method. The other way is to use the rate of change. Since we learned about the rate of change in quadratic functions, let’s use the rate of change to get the equation (2). Comparing point A and point B, what is the rate of change: when moving from A $(-2,2)$ to B $(6,18)$, the value of $x$ increases by 8, and the value of $y$ increases by 16. The rate of change represents how much the value of $y$ increases when the value of $x$ increases by 1. Therefore, we have the following ratio. • $8:16=1:a$ Using the properties of the ratio, we can calculate the following. $16=8a$ $a=2$ Thus, we find that the slope (ratio of change) of the linear function in (2) is 2. Therefore, the formula for the linear function is $y=2x+b$. Then let’s get the intercept by substituting A $(-2,2)$ or B $(6,18)$. For example, substituting $(6,18)$ yields the following $18=2×6+b$ $18=12+b$ $b=6$ Thus, we find that the linear function in (2) is $y=2x+6$. (c) How do we calculate the area of △AOB? Let P be the intercept (intersection with the $y$-coordinate) in (2), and we will find the following. • $△AOB=△AOP+△BOP$ So, let’s find the areas of △AOP and △BOP. Since equation (2) is $y=2x+6$, the coordinate of P is $(0,6)$. In other words, the vertical length of the triangle (line segment PO) is 6. On the other hand, the $x$-coordinate of A is -2, so the horizontal length of △AOP is 2. The $x$-coordinate of B is 6, so the horizontal length of △BOP is 6. Therefore, we can get the area of each as follows. • The area of △AOP: $6×2×\displaystyle\frac{1}{2}=6$ • The area of △BOP: $6×6×\displaystyle\frac{1}{2}=18$ • $△AOB=6+18=24$ (d) If there is a point D on (1), △COD is as follows. The linear function in (2) is $y=2x+6$. Therefore, the intersection with the $x$-axis can be calculated by substituting $y=0$ as follows. $0=2x+6$ $-2x=6$ $x=-3$ The coordinate of point C is $(-3,0)$. In other words, the length of CO is 3. On the other hand, what is the vertical length of △COD? The quadratic function in (1) is $y=\displaystyle\frac{1}{2}x^2$. So the coordinate of point D is $(x,\displaystyle\frac{1}{2}x^2)$. In other words, the vertical length is $\displaystyle\frac{1}{2}x^2$. The area of the COD is 27. Substituting in the formula for the area of a triangle, we can calculate it as follows. $3×\displaystyle\frac{1}{2}x^2×\displaystyle\frac{1}{2}=27$ $x^2=27×\displaystyle\frac{4}{3}$ $x^2=36$ $x=-6,x=6$ Thus, when $x=-6$ or $x=6$, the area of △COD is 27. What is especially important in quadratic functions is the value of $a$ and the rate of change. Depending on the value of $a$, the shape of the graph varies. Also, by calculating the rate of change, we can get the slope of the line passing through the two coordinates.
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Chapter 3 Adding and Subtracting Fractions 2. Finding the Least Common Multiple Using Multiples Method 1: Find the LCM of a List of Numbers Using Multiples of the Largest Number Step 1: Write the multiples of the largest number (starting with the number itself) until a multiple common to all numbers in the list is found. Step 2: The multiple found in Step 1 is the LCM. 3. Example Find the LCM of 9 and 12. Write the multiples of 12 until we find a number that is also a multiple of 9. 12  1 = 12 Not a multiple of 9. 12  2 = 24 Not a multiple of 9. 12  3 = 36 A multiple of 9. The LCM of 9 and 12 is 36. 4. Finding the Least Common Multiple Using Multiples Method 2: Find the LCM of a List of Numbers Using Prime Factorization Step 1: Write the prime factorization of each number. Step 2: For each different prime factor in step 1, circle the greatest number of times that factor occurs in any one factorization. Step 3: The LCM is the product of the circle factors. 5. Example Find the LCM of 72 and 60. Circle the greatest number of prime factors found in either factorization. The LCM is the product of the circle factors. 6. Example Find the LCM of 15, 18, and 54. 7. Writing Equivalent Fractions To add or subtract unlike fractions, we first write equivalent fractions with the LCM as the denominator. To write an equivalent fraction, where a, b, and c are nonzero numbers. 8. Example Write an equivalent fraction with the indicated denominator. 9. Example Write an equivalent fraction with the indicated denominator.
# Algebra II : Hyperbolic Functions ## Example Questions ### Example Question #1 : Hyperbolic Functions Which of the following equations represents a vertical hyperbola with a center of and asymptotes at ? Explanation: First, we need to become familiar with the standard form of a hyperbolic equation: The center is always at . This means that for this problem, the numerators of each term will have to contain and . To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive term; a horizontal parabola has a positive term. In this case, we need a vertical parabola, so the term will have to be positive. (NOTE: If both terms are the same sign, you have an ellipse, not a parabola.) The asymptotes of a parabola are always found by the equation , where is found in the denominator of the term and is found in the denominator of the term. Since our asymptotes are , we know that must be 4 and must be 3. That means that the number underneath the term has to be 16, and the number underneath the term has to be 9. ### Example Question #1 : Hyperbolic Functions What is the shape of the graph depicted by the equation: Oval Hyperbola Circle Parabola Hyperbola Explanation: The standard equation of a hyperbola is: ### Example Question #2 : Hyperbolic Functions Express the following hyperbolic function in standard form: Explanation: In order to express the given hyperbolic function in standard form, we must write it in one of the following two ways: From our formulas for the standard form of a hyperbolic equation above, we can see that the term on the right side of the equation is always 1, so we must divide both sides of the given equation by 52, which gives us: Simplifying, we obtain our final answer in standard form: ### Example Question #1 : Hyperbolic Functions Which of the following answers best represent ? Explanation: The correct definition of hyperbolic sine is: Therefore, by multiplying 2 by both sides we get the following answer, ### Example Question #4 : Hyperbolic Functions Which of the following best represents , if the value of is zero? Explanation: Find the values of hyperbolic sine and cosine when x is zero. According to the properties: Therefore: ### Example Question #5 : Hyperbolic Functions What is the value of ? Explanation: The hyperbolic tangent will need to be rewritten in terms of hyperbolic sine and cosine. According to the properties: Therefore: ### Example Question #7 : Hyperbolic Functions Simplify: Explanation: The following is a property of hyperbolics that is closely similar to the problem. We will need to rewrite this equation by taking a negative one as the common factor, and divide the negative one on both sides. Substitute the value into the problem. ### Example Question #6 : Hyperbolic Functions Which of the following is the correct expression for a hyperbola that is shifted units up and to the right of ? Explanation: The parent function of a hyperbola is represented by the function where is the center of the hyperbola. To shift the original function up by simply add . To shift it to the right take away . ### Example Question #9 : Hyperbolic Functions Find the foci of the hyperbola: Explanation: Write the standard forms for a hyperbola. OR: The standard form is given in the second case, which will have different parameters compared to the first form. Center: Foci: , where Identify the coefficients and substitute to find the value of . ### Example Question #7 : Hyperbolic Functions Given the hyperbola , what is the value of the center? Explanation: In order to determine the center, we will first need to rewrite this equation in standard form. Isolate 41 on the right side. Subtract and add on both sides. The equation becomes: Group the x and y terms. Be careful of the negative signs. Pull out a common factor of 4 on the second parentheses. Complete the square twice. Divide the second term of each parentheses by two and square the quantity. Add the terms on both sides. This equation becomes: Factorize the left side and simplify the right. Divide both sides by nine. The equation is now in the standard form of a hyperbola. The center is at:
#### Logarithmic Functions and Their Graphs Logarithmic Functions and Their Graphs: Inverse of Exponential Functions: We stated in the part on exponential functions, which exponential functions were one-to-one. One-to-one functions had the unique property which they have inverses that are as well functions. And, as most of you said in class, one-to-one functions can applied to both the sides of an equation. They as well pass a horizontal line test. This part is about the inverse of exponential function. The inverse of an exponential function is the logarithmic function. Keep in mind that the inverse of a function is received by switching x and y coordinates. It reflects the graph regarding the line y = x. As you can state from the graph to the right, the logarithmic curve is a reflection of exponential curve. The table below illustrates how x and y values of points on an exponential curve can be switched to determine the coordinates of the points on logarithmic curve. Comparison of Exponential and Logarithmic Functions: Let us look at some of the properties of two functions. The standard form for logarithmic function is: y = loga x Note, if the ‘a’ in expression above is not a subscript (that is, lower than the ‘log’), then you require to update your web browser. Working Definition of Logarithm: In an exponential function, the x was exponent. The main aim of inverse of a function is to state you what x value were used when you already know the y value. Therefore, the aim of the logarithm is to state you the exponent. Therefore, our simple definition of a logarithm is that it is the exponent. The other way of looking at expression ‘loga x’ is ‘to what power (or exponent) should a be increased to get x?’ Equivalent Forms: The logarithmic form of equation y = loga x is equivalent to the exponential form x = ay. To rewrite one form in other, keep the base similar, and switch sides with other two values. Properties of Logarithms: loga 1 = 0 since a0 = 1 No matter what the base is, as long as it is legal, the log of 1 is for all time 0. That is as logarithmic curves always pass via (1,0) loga a = 1 since a1 = a Any of the value increased to the first power is that similar value. loga ax = x The log base a of x and a to x power are inverse functions. When inverse functions are applied to one other, they inverse out, and you are left with argument, in this case, x. loga x = loga y implies that x = y When two logs with similar base are equivalent, then the arguments should be equal. loga x = logb x implies that a = b When two logarithms with similar argument are equivalent, then the bases should be equal. Common Logs and Natural Logs: There are two logarithm buttons on your calculator. One is marked as ‘log’ and the other is marked as ‘ln’. Neither one of such consists of the base written in. The base can be recognized, though, by looking at inverse function that is written above the key and accessed by the 2nd key. Common Logarithm (base 10): Whenever you see ‘log’ written, with no base, suppose the base is 10. That is: log x = log10 x. Some of the applications which use common logarithms are in pH (to compute acidity), decibels (for sound intensity), the Richter scale (for earthquakes). The interesting (possibly) side note regarding pH. Sewers of the Village of Forsyth Code require forbids the release of waste with a pH of less than 5.5 or higher than 10.5 The common logs as well serve the other aim. Every rise of 1 in a common logarithm is the outcome of 10 times the argument. That is, an earthquake of 6.3 has 10-times the magnitude of 5.3 earthquakes. The decibel level of loud rock music or the chainsaw (115 decibels = 11.5 bels) is 10-times louder than the chickens within a building (105 decibels = 10.5 bels). Natural Logarithms (base e): Keep in mind that number e that we had from the prior part? You know the one that was around 2.718281828 (however does not repeat or terminate). That is the base for natural logarithm. ln x = loge x The exponential growth and decay models are one of the applications which use natural logarithms. This comprises continuous compounding, radioactive decay (that is, half-life), and population growth. Usually applications are procedure which is continually happening. Now, such applications were first illustrated in the exponential part, however you will be capable to solve for other variables included by using logarithms. In higher level of mathematics, the natural logarithm is a logarithm of choice. There are some special properties of natural logarithm function, and its inverse function, that make life much simpler in calculus. As ‘ln x’ and ‘ex’ are inverse functions of one other, any time an ‘ln’ and ‘e’ appear right next to one other, with absolutely nothing in between them (that is, whenever they are composed with one other), then they inverse out, and you are left with argument. Latest technology based Algebra Online Tutoring Assistance Tutors, at the www.tutorsglobe.com, take pledge to provide full satisfaction and assurance in Algebra help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Algebra, project ideas and tutorials. We provide email based Algebra help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Algebra. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Algebra Homework help and assignment help services. 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#### Combination circuits, circuits with series and parallel resistors Many circuits have resistors in both series and parallel. To find the total current through the circuit you generally want to replace the set of resistors with its equivalent resistor. This can be done by identifying a pair of resistors that is either in series or in parallel, replacing that pair by a single equivalent resistor, and iterating until you're left with one resistor in the circuit. This allows the total current in the circuit to be determined. The current flowing through each resistor can then be found by undoing the reduction process. Contract: Expand: #### A sample combination circuit In the circuit above, the resistors are: R1 = 6 W R2 = 36 W R3 = 12 W R4 = 3 W What is the potential difference across each resistor? How much current passes through each resistor? To solve this we need to find the equivalent resistance of the set of capacitors. We'll contract the circuit from 4 resistors to 1. Step 1 - R2 and R3 are in parallel. Replace this pair by a single capacitor R23: 1/R23 = 1/R2 + 1/R3 = 1/36 + 1/12 = 4/36. Therefore R23 = 36/4 = 9 W. Step 2 - R4 and R23 are in series. Replace that pair by a single resistor R234 = 3 + 9 = 12 W. Step 3 - R1 and R234 are in parallel. Replace that pair by a single resistor Req: 1/Req = 1/R1 + 1/R234 = 1/6 + 1/12 = 3/12 Req = 12/3 = 4 W. Step 4 - Determine the current through Req. I = DV / Req = 12/4 = 3A. Now we need to expand the circuit back to the original four resistors, and determine the current through, and potential difference across, each one as we go. Step 1 - Req represents R1 and R234 in parallel. Resistors in parallel have the same potential difference but split the current. The potential difference across each is 12 volts. The current is: I4 = DV / R4 = 12/6 = 2A. I234 = DV / R234 = 12/12 = 1A. Step 2 - R234 represents R4 and R23 in series. Devices in series have the same current through them (whatever current their series combination has), so they each have 1A. DV4 = I * R4 = 1 * 3 = 3 V. DV23 = I * R23 = 1 * 9 = 9 V. These add to the potential difference across the series combination. Step 3 - R23 represents R2 and R3 in parallel. The potential difference in both cases is 9 V. I2 = DV / R2 = 9/36 = 0.25 A. I3 = DV / R3 = 9/12 = 0.75 A. Step 4 - A good way to check for consistency is to label the potential at different points. Pick some point as a reference (say, 0 V at the negative terminal of the battery) and label other points relative to that. Check that the potential differences across the resistors are consistent with these potential values.
# Mathematics What Is Mathematics What Is The Problem Inside the School? Introduction The problem of the study of mathematics is one of the most difficult topics in the sciences. A study of the problems of mathematics in the later part of the last century created a huge amount of interest, and today, mathematics is the biggest single study problem in the sciences, largely because of its simplicity. The problem of the problem of the solution of a mathematical equation is, to a large extent, a problem of the form of equation. A problem is a necessary condition for solving a mathematical equation. In this regard, the problem of solving a differential equation is extremely difficult. A problem is a set of equations that are satisfied by some set of elements. The set of equations is a collection of equations. If we want to know the solution of the problem, we have to know the properties of the elements of the collection of equations, and they are not the same as the properties of equation. In the following, we will use the concept of the problem to study the problem of a set of elements, and we will be interested in the problem of finding the set of equations containing those elements. Problem 1: Find the set of elements containing the equation In this paper, we will be concerned with the problem of determining the set of equation containing the equation. The problem have a peek at this site a very complicated problem. In order to solve the problem, one has to know the elements of all the equations in the set. It is, to our knowledge, the first point of the problem. We will be concerned about the problem of computing a set of all the elements of a set. Using the fact that we have the set of all equations, we are able to compute the set of the elements containing the function. Let us first consider the problem of figuring out the set of functions that satisfy the equation. The set of functions is always a set. So we can write the set of function as a set of functions, and we can find the function with the given set of functions. Here is the problem of using the set of set of functions to find the set of sets of functions. In the first part, we will write down the problem of problem. ## Online Test Taker In the second part, we are interested in the set of problem. We can write the problem of having the set of given functions as a set, and we have to find the function that satisfies the given set. In order to solve this problem, one must define the function that satisfy the given sets, and find the set that satisfies the set of constructed functions. We will use the following definition. Definition 2: Let the set of unknowns be a set of unknown functions. We say that the set of defined functions is a set if it is a set that is a set. Recall that if the set of known functions is a subset of the set of non-unknown functions, then we can find a function that satisfies this set of functions; In the definition of a set, we will always say that the function is determined by its set of unknown, and we also will say that the functions that satisfy this set of unknown are determined by their set of unknown. There is a clear relationship between the set of possible functions and the set of a set that satisfies this definition. In this definition, we are going to use the following two terms: The sets ofMathematics What Is A Geometry? The concept of a geometry is a kind of mathematical term that refers to a geometric object, such as a mathematical object, a diagram or any other work of the mathematical world, something that is normally thought of as a single geometric object. Geometry is the artistry of mathematics. The concept of geometry is the art of mathematical analysis, this is the art that forms a part of mathematics. It is a science of physical analysis, from mathematical to physical sciences that we can classify the mathematical process of thought. From this we can look for the same mathematical objects as our physical objects, the objects that are used for analysis, the physical objects that are useful for study. If you are interested in this art, then you can have a look at this art by looking at the nature of mathematics. The geometric nature of a mathematical object is the nature of the mathematical process. It is the relationship between a mathematical object and a physical object. It is about the relationship between the physical object and the mathematical process that is the way a physical object is actually treated. A mathematical process is the act of thinking about this physical object. The geometric nature of the geometric process is the process of thinking about the mathematical process and it is the way that a physical object truly is. Mathematics is a science. ## What Difficulties Will Students Face Due To Online Exams? The mathematics is the art where the mathematical object and the physical object are the same. It is that way of thinking about mathematical objects and how they are manipulated and how they interact in the world. There are the ways that the mathematical process is manipulated, how they interact, where and how they get altered, how they are interpreted and what they are doing. A geometric field or field of mathematics is a field of mathematics. A field in mathematics is a way of thinking that one is thinking about the same physical object, the physical object. A field of mathematics also has a way of looking at the mathematical object that it is looking at. There are three types of geometric fields, an algebraic field, a geometry field and an algebraic geometry field. There is a geometry field where the mathematical process has been done by one of the types, the geometry field. The geometry field is the property of the mathematical object, the mathematical process, the mathematical object itself. It is where the mathematical objects and the physical objects have been manipulated. An algebraic field is a field that is a way to understand a physical object, and the way it is manipulated. It is how the physical objects are manipulated. There are two types of algebraic field. The algebraic field has a my sources to look at mathematical objects and its own properties. It is an algebraic way of thinking, thinking about mathematical processes and its own property. In mathematics, the mathematical processes are the ways in which the physical objects, their properties, and the mathematical object are changed. They are the processes that make the physical objects change. They are those processes that make a physical object change. As a result, the physical world becomes a geometry. If you want to study a geometry where you are just learning mathematics, it is possible to have a geometry that is the same as your physical object. ## Need Someone To Do My Statistics Homework In this geometry, there are two types, a geometry type and a geometry type. The geometry type is the way of thinking and thinking about mathematical things. It is also the way of studying physical objects. For this type of geometryMathematics What Is Mathematics? An algebraic description of the mathematics of algebraic geometry has been a subject of study for some time. The subject is now an active area of research. We have noticed that it is known that the mathematics of arithmetic is not different from the mathematics of geometry. The main task of this paper is to explain the mathematics of math and geometry in terms of this subject. Definition For a set of functions $f:\mathbb{N} \to \mathbb{R}$ that is not constant, we say that $f$ is [differentiable]{} iff $f$ has a derivative iff $\|f(x)-f(x)\|=\infty$, where $x\in \mathbb N$. In the case where $f$ does not have a derivative, we say $f$ [is differentiable at a point]{}. The following definition is a generalization of the definition of the linear functionals of a function (which can be seen as a generalization to the case of a function with Learn More derivative). Definition of the linear functions A linear function is called a More about the author function]{}, iff it is a linear function and the following conditions are satisfied: 1. It is not constant. 2. It does not vanish on a finite set. 3. It has a bounded inverse. 4. It depends only on the values of $f$. 5. It can be extended to the case where it does not depend on $f$. ## Can Someone Do My Homework For Me 6. It preserves the derivative property of $f$, iff $d_i f$ is differentiable at $x_i$ for $1\leq i\leq m$. The following lemma is an analogue of Lemma 2.1 in [@GrinLiv]. $lemma2.1$ Let $f : \mathbb R^n More Info \Pi$ be a linear function, then $f$ can be extended from $f((-\infty, \infty])$ to $f^{-1}((-\pi, \in \Pi))$ for some $\pi \in \mathcal P(f)$. Proof of Lemma ================ Lemma 2.2 in [@GasserLiv] gives an explicit representation of the linear power series \begin{aligned} \label{lemma2} \sum_{n=0}^\infty c_n \,f(x_n): \quad x_n= \sum_{i=0}^{n-1} a_i x_i,\end{aligned} when $a_i,b_i \in \{-1, 0\}$. $\bullet$ [Lipshitz linear function]\ A linear power series is called [Lipshritz]{ on $f$ iff it has a vanishing Fourier series for $f$ in $\mathbb{C}$ and $f$ satisfies the following condition: If $f$ was Lipschitz, then $d_1^2 f$ could be extended from $\mathbb C$ to $\mathbb official site which is a linear extension of the function$d_2^2 f$. When$f\$ could have a Lipschit function, then it is called [linear Lipschite*]{ as in [@ChenChen]. $$The same proof as in Lemma 2 has been given in [@Hou]. Let c_n be the n-th derivative of c_0, then it is easy to see that the following hold.$$\begin{array}{lcl} \displaystyle \displaystyle d_1^c c_0 & = & \displaystyle c_1 d_1 f(x_1) + d_2 c_2 f(x) + \displaystyle h(x) \\ \stackrel{\eqref{lemma1}}{
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Decimals as Percents ## Rewrite decimals as percents by moving the decimal point. Estimated4 minsto complete % Progress Practice Decimals as Percents Progress Estimated4 minsto complete % Decimals as Percents Have you ever eaten sour lemon candies? Have you ever counted them? At the candy store, .125 of the jar of sour lemon candies were purchased by a customer. Jake was working at the time and was amazed that only a small portion of the lemon candies went home with the customer. Can you write this decimal as a percent? What percent of the jar was purchased? To figure this out, you will need to know how to write decimals as percents. This Concept will teach you what you need to know. ### Guidance Just like we can write percents as decimals, we can also write decimals as percents. You saw a situation like this in the last Concept. In a way, you reverse the steps from turning a percent to a decimal to turn a decimal to a percent. Follow these steps to write a percent as a decimal Step 1: Move the decimal point two places to the right. Add zeros to the right of the decimal point as placeholders, if necessary. Step 2: Write a % symbol after the resulting number Take a few minutes to write these notes in your notebook. Then continue. Let’s apply these steps. Write .78 as a percent. This decimal is written in hundredths, so all we have to do is move the decimal two places to the right and add a percent sign. Write .6 as a percent. This decimal is written in tenths. When we move the decimal two places to the right, we will need to add a zero place holder. Write .345 as a percent. This decimal is written in thousandths. We only need to move the decimal two places to the right to make it a percent, so we will move it two places and add a percent sign. Notice that sometimes we can have a decimal in a percent. It means that we have 34 and one-half percent in this case. Don’t let that throw you off-not all percents are whole percents! Write each decimal as a percent. Solution: Solution: #### Example C Solution: Here is the original problem once again. At the candy store, .125 of the jar of sour lemon candies were purchased by a customer. Jake was working at the time and was amazed that only a small portion of the lemon candies went home with the customer. Can you write this decimal as a percent? What percent of the jar was purchased? To write this as a percent, we know that we need to move the decimal point two places, representing hundredths, and add a percent sign. The customer purchased 12.5% of the lemon candies. ### Vocabulary Here are the vocabulary words in this Concept. Decimal a number written according to place value. Numbers to the right of the decimal point represent parts of a whole. Numbers to the left of the decimal point represent whole numbers. Percent a part of a whole out of 100. Percents can be smaller than one represented by a decimal percent. They can also be greater than one hundred by having a decimal with a whole number and a part of a whole. ### Guided Practice Here is one for you to try on your own. Write 3.5 as a percent. Answer This decimal is written with a whole number and five tenths. We still move the decimal to the right two places. Notice that because we have a whole number that the percent will be greater than 100. This is our answer. ### Video Review Here is a video for review. ### Practice Directions: Write the following decimals as percents. 1. .45 2. .3 3. .675 4. .9 5. .08 6. .785 7. .22 8. .095 9. .54 10. .275 11. .04 12. .045 13. .112 14. 4.6 15. .672 ### Vocabulary Language: English Decimal Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths). ### Explore More Sign in to explore more, including practice questions and solutions for Decimals as Percents. Please wait... Please wait...
# Learning Place Value Views: Category: Education ## Presentation Description No description available. ## Presentation Transcript ### Learning Place Value: Learning Place Value ### Slide 2: Look at each of the numbers below and see if you can find the single numbers that are in all of them. 2,349 5,671 80 459 1,346,708 Hint: There are 10. ### Slide 3: What are the 10 numbers that are found in each number? Don’t repeat any. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 These DIGITS make up all numbers . ### Slide 4: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 What is the smallest number? What is the largest number? ### Slide 5: So what are we going to learn? M3N1. Students will further develop their understanding of whole numbers and decimals and ways of representing them and be able to i dentify place values from tenths through ten thousands. The State of Georgia says that… ### Slide 6: In kid terms please? I can tell what place a digit is in and will be able to tell the value of the digits in a number. ### Slide 7: So what is the difference between this… And this… ### The difference is place value!: The difference is place value! Ones Place Hundreds Place ### Slide 9: Place is where a digit sits in a number. Digit is any number 0-9. Value is how much a digit is worth because of the place it is in. Place Value ### Slide 10: Let’s Make Some Numbers What number is this? 2 4 3 ### Slide 11: Let’s Make Some Numbers What number is this? 6 3 5 2, ### Slide 12: Let’s Make Some Numbers What number is this? 1 5 0 7, ### Slide 13: Ones Tens Hundreds What are the names of the places in a number? Thousands ? ### Slide 14: Ones, tens, hundreds, Thousands, ten thousands, Hundred thousands, millions, That is place value! Here is a song to help you remember. Day 2 ### Slide 16: So what are we going to learn? M3N1. Students will further develop their understanding of whole numbers and decimals and ways of representing them and be able to i dentify place values from tenths through ten thousands. The State of Georgia says that… ### Slide 17: In kid terms please? I can tell what place a digit is in and will be able to tell the value of the digits in a number. ### Slide 18: Ones, tens, hundreds, Thousands, ten thousands, Hundred thousands, millions, That is place value! Here is a song to help you remember. ### Slide 19: Places in a Number (Going from left to right) Ones Tens Hundreds Thousands Ten Thousands Hundred Thousands Millions ### Slide 20: Can you draw a place value chart on your board? ### Slide 21: Can you put 451 on the chart correctly? 1 5 4 ### Slide 22: Can you put 9,473 on the chart correctly? 3 7 4 9, ### Slide 23: Can you put 8,205 on the chart correctly? 5 0 2 8, ### Slide 24: Can you identify the places without a chart? 5,023 Think about the song. Ones Tens Hundreds Thousands ### Slide 25: Try Another One 387 Think about the song. Ones Tens Hundreds ### Slide 26: Tricky, Tricky 24,821 Think about the song. Ones Tens Hundreds Thousands Ten Thousands Day 3 ### Slide 28: So what are we going to learn? M3N1. Students will further develop their understanding of whole numbers and decimals and ways of representing them and be able to i dentify place values from tenths through ten thousands. The State of Georgia says that… ### Slide 29: In kid terms please? I can tell what place a digit is in and will be able to tell the value of the digits in a number. ### Slide 30: Ones, tens, hundreds, Thousands, ten thousands, Hundred thousands, millions, That is place value! Here is a song to help you remember. ### Slide 31: Place is where a digit sits in a number. Digit is any number 0-9. Value is how much a digit is worth because of the place it is in. Place Value ### So how do we find the value of digits in numbers?: So how do we find the value of digits in numbers? ### Slide 34: Let’s Make Some Numbers the Place Value Mat What number is this? 2 4 3 Now how much is each digit worth? 2 40 300 ### Slide 35: Let’s Find Some Values What number is this? 6 3 5 2, Now how much is each digit worth? 6 30 500 2,000 ### Slide 36: Can you put 9,473 on the chart correctly? 3 7 4 9, 3 70 400 9,000 Day 4 ### Slide 39: Ones, tens, hundreds, Thousands, ten thousands, Hundred thousands, millions, That is place value! Here is a song to help you remember. ### Slide 40: Try This with Your Partner 47,906 Think about the song. Ones Tens Hundreds Thousands Ten Thousands What is the value of each digit? 40,000 7,000 900 6 ### How good are you?: How good are you? Which digit is in the ten-thousands place in the number 31,467? 3 1 7 4 ### How good are you?: How good are you? Which digit is in the thousands place in the numeral 5,427? 7 5 4 2 ### How good are you?: How good are you? What is the place value of the underlined digit in the number 7, 5 48? Ones Billions Ten-thousands Hundreds ### How good are you?: How good are you? What is the place value of the underlined digit in the number 1 2,098? Tens Thousands Ten-thousands Hundreds ### How good are you?: How good are you? What is the value of the underlined digit in the numeral 6,6 8 4? 8,000 10 800 80 ### How good are you?: How good are you? What is the value of the underlined digit in the number 535,85 2 ? 20 1 2 200 ### How good are you?: How good are you? What is the value of the underlined digit in the number 5 3 5,852? 30 10,000 3,000 30,000
Arithmetic Arithmetic Progression Associative Property Averages Brackets Closure Property Commutative Property Conversion of Measurement Units Cube Root Decimal Divisibility Principles Equality Exponents Factors Fractions Fundamental Operations H.C.F / G.C.D Integers L.C.M Multiples Multiplicative Identity Multiplicative Inverse Numbers Percentages Profit and Loss Ratio and Proportion Simple Interest Square Root Unitary Method Algebra Cartesian System Order Relation Polynomials Probability Standard Identities & their applications Transpose Geometry Basic Geometrical Terms Circle Curves Angles Define Line, Line Segment and Rays Non-Collinear Points Parallelogram Rectangle Rhombus Square Three dimensional object Trapezium Triangle Trigonometry Trigonometry Ratios Data-Handling Arithmetic Mean Frequency Distribution Table Graphs Median Mode Range Videos Solved Problems Home >> Commutative Property >> Addition of Integers >> Examples ## Commutative Property (Addition of Integers) : Solved Examples Addition of Integers Addition of Whole Numbers Division of integers Division of Whole Numbers Multiplication of Integers Multiplication of Whole Numbers Subtraction of Integers Subtraction of Whole Numbers We know that addition is commutative for whole numbers, does it apply same as on integers ? Let's check that if we change the order of integers in addition expression, the result remains same or not, and this can be checked as: Equation 1 : (-13) + 5 = 5 + (-13) On solving both side, we get: (-8) = (-8) Equation 2 : 27 +(-10) = (-10) + 27 On solving both side, we get: 17 = 17 In both the above equation, you can notice that L.H.S. = R.H.S., so we can that Addition is commutative for integers also. ### Related Question Examples We know that addition is commutative for whole numbers, does it apply same as on integers ? Explain commutative property for addition of integers, with variables x and y. Prove Addition is commutative for integers with the help of two negative integers i.e. (-86) & (-14). Prove Addition is commutative for integers with the help of two positive integers i.e. 67 & 49. Prove Addition is commutative for integers with the help of one positive integers & one negative integer i.e. 113 & (-247) A) If a = 58 and b = 72, prove a + b = b + a and also write what this property is known as ? B) If x = (-127) and y = 282, prove x + y = y + x and also write what this property is known as ? C) If p = (-369) and q = (-842), Prove p + q = q + p and also write what this property is known as ? Copyright@2022 Algebraden.com (Math, Algebra & Geometry tutorials for school and home education)
# Subtract and Simplify (0.04x3-0.03x2+0.02x) – (0.03x3+0.08x2-6) • Last Updated : 29 Oct, 2021 Mathematics is basically divided into different branches, out of which one branch is algebra. In algebra, people deal with numerals and variables, known value is termed as numerals and unknown value is termed as a variable. A variable can take any value. Arithmetic operations like addition, subtraction, multiplication, and division are also applied in algebra to find out the value of unknowns. ### Algebraic Expressions An algebraic expression is the combination of numerals and variables in systematic order. Numerals and variables are related by four fundamental mathematical operators, addition, subtraction, multiplication, or division. Example: By using two numerals (5 and 6), and one variable x, one can form an algebraic expression 6x + 5. In this expression, there are two terms. So on the basis of the number of terms algebraic expression is categorized into the following types, • Monomial: When an algebraic expression has only one term then that is known as a monomial. Example: 5t, 8y, etc • Binomial: When the number of terms in an algebraic expression is two then that expression is known as binomial. Example: 5t – 8k, 8t + 6, etc • Trinomial: An algebraic expression having three-term is called a trinomial. Example: 6x – 3y – 5z, 9t – 6u + 7w, etc • Polynomial: An algebraic expression having one or more than one term is known as a polynomial. To perform the fundamental arithmetic operation in algebraic expression, find out the like and unlike terms. Like and Unlike Terms: The term having same variables are known as like terms and the terms which does not have same variable is known as unlike terms. Example: In the algebraic expression, 5x +6y -7x² -4x +9, the terms which have same variables are 5x and 4x, so these two terms is called as like terms. ### Subtract and simplify: (0.04x3 – 0.03x2 + 0.02x) – (0.03x3 + 0.08x2 – 6) Solution: Steps to solve the problem Step 1: Find out the like terms of the given algebraic expression. Step 2: Apply the arithmetic operation on the numeral part of like terms. Step 3: No operation is applied on unlike terms. Step 4: By solving the like term, reduce the expression in the lowest term. In the given algebraic expression, like terms are 0.04x³ and 0.03x³, 0.03x² and 0.08x². The given expression can be written as, = 0.04x³ – 0.03x² + 0.02x – 0.03x³ – 0.08x² + 6 = 0.04x³ – 0.03x³ – 0.03x² – 0.08x² + 0.02x + 6 On solving like terms, = 0.01x³ – 0.11x² + 0.02x + 6 On simplifying the expression (0.04x³ – 0.03x² + 0.02x) – ( 0.03x³ + 0.08x² – 6), we got 0.01x³ – 0.11x² + 0.02x + 6. ### Similar Questions Question 1: Subtract and simplify: (5x³ – 12x² – 6x + 12) – (12x² +6x – 10) Solution: In the given expression like terms are 12x² and 12x², 6x and 6x, 12 and 10. The given expression can be written as = 5x³ – 12x² – 6x + 12 – 12x² – 6x +10 = 5x³ – 12x² – 12x² – 6x – 6x +12 +10 On solving the like terms, = 5x³ – 24x² – 12x + 22 So, on simplifying the expression (5x³ – 12x² – 6x + 12) – (12x² + 6x – 10), = 5x³ – 24x² – 12x + 22. Question 2: Subtract and simplify: (0.5y³ – 0.03x² – 0.3z + 12) – (0.4y³ – 0.36x² + 0.2z – 11) Solution: In the given expression, like terms are 0.5y³ and 0.4y³, 0.03x² and 0.36x², and 0.3z and 0.2z, 12 and 11 The given expression can be written as, = 0.5y³ – 0.03x² – 0.3z +12 – 0.4y³ + 0.36x² – 0.2z +11 = 0.5y³ – 0.4y³ – 0.03x² + 0.36x² – 0.3z – 0.2z + 12 + 11 On solving the like terms, = 0.1y³ + 0.06x² – 0.5z + 23 So, on solving the expression (0.5y³ – 0.03x² – 0.3z + 12) – (0.4y³ – 0.36x² + 0.2z -11), = 0.1y³ + 0.06x² – 0.5z + 23 My Personal Notes arrow_drop_up
Question Video: Using Addition to Check Subtraction Calculations \| Nagwa Question Video: Using Addition to Check Subtraction Calculations \| Nagwa # Question Video: Using Addition to Check Subtraction Calculations Mathematics Look at the expression 76 − 39. Choose a strategy and solve the expression. Hint: You can use the vertical column method. 04:44 ### Video Transcript Look at the expression 76 take away 39. Choose a strategy and solve the expression. Hint: you can use the vertical column method. In this problem, we’re given an expression, 76 take away 39. And we’re told to solve it by choosing a strategy. What are some of the strategies we could use? We could sketch a number line, start with 76 and count back 39. Another strategy is that we could partition 39 into 30 and nine and subtract each part separately. We could even recognise that 39 is near to 40 and use this to help by subtracting 40, and then adjusting our answer. These are just some of the strategies that we could choose. And each one, if we use them accurately, will give us the correct answer. But as part of our problem, we’re given a hint. So, let’s take the hint and use the vertical column method as it suggests. This method is called the vertical method because the numbers are written vertically, or on top of each other. This helps us to see the place value of each digit. The 10s are on top of each other. And the ones are on top of each other. And we start by looking at the ones column. In the ones column, we can see that we need to work out six take away nine. What are six ones subtract nine ones? Well, we can’t do this. Because the number of ones we want to take away is greater than the number of ones we already have. Is it possible? Well, it is possible because we have to remember we’re subtracting nine from a much larger number, 76. And we can subtract nine ones from 76. We just have to rearrange it a little. So, what can we do? Well, we can take one of the 10s from the tens column and exchange it for 10 ones. Instead of seven 10s, we now have six 10s. And let’s exchange it for 10 ones. Instead of six ones, we’ve now have got 16 ones. 60 and 16 is exactly the same number as 76. We’ve just partitioned it differently by exchanging one ten for 10 ones. So, now we have enough ones to take away nine ones. 16 take away nine. 16 take away nine equals seven. Now we subtract the tens column. What are six 10s take away three 10s? Remember, we’re starting with six 10s because we’ve already exchanged one ten for 10 ones. Six 10s take away three 10s leaves us with three 10s. And so, by using the vertical column method, we found that 76 take away 39 equals 37. There is a second part to this problem. Check your answer using the inverse operation. Remember, the inverse means the opposite. Which addition expression shows the opposite calculation you need to carry out? What this second part of the problem is reminding us is that we can always work backwards to check our answers. In our original calculation, we started with 76. We subtracted 39. And we were left with an answer of 37. So, to check our answer, we need to move in the opposite direction. Start with 37. Add 39 because we’re moving in the opposite direction. And if we calculated our first answer correctly, we should find that the answer to the addition is 76. And so, the addition expression we need to use is 37 plus 39. And we can use vertical addition this time to add those two numbers together and see that we do get 76. And so, our original calculation must have been correct. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
# How do I find the inverse of a 2xx2 matrix? Jun 9, 2018 ${\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)}^{- 1} = \frac{1}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$ #### Explanation: Let's have a go a this without simply plugging in a remembered formula. Given a matrix $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$, let's try multiplying it by $\left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$ and see what we get: $\left(\begin{matrix}a & b \\ c & d\end{matrix}\right) \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right) = \left(\begin{matrix}a d - b c & 0 \\ 0 & a d - b c\end{matrix}\right)$ So if we multiply by $\frac{1}{a d - b c} = \frac{1}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid}$ then we find the inverse matrix: $\frac{1}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right) = \left(\begin{matrix}\frac{d}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid} & - \frac{b}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid} \\ - \frac{c}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid} & \frac{a}{\left\mid \begin{matrix}a & b \\ c & d\end{matrix} \right\mid}\end{matrix}\right)$
## Taylor Series For any function $$f(x)$$, the Taylor series of $$f(x)$$ at $$a$$ is: $f(x) = f(a) + (x - a)f'(a) + \frac{(x - a)^2}{2!}f''(a) + \cdots + \frac{(x - a)^n}{n!}f^{(n)}(a) + \cdots$ Or, more compactly: $f(x) = \sum_{n = 0}^{\infty} \frac{(x - a)^n}{n!}f^{(n)}(a)$ In this course, we will often use two Maclaurin series, which are Taylor series with $$a = 0$$. ### Geometric Series Provided $$\|x\| < 1$$, $\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots.$ Or, more compactly: $\sum_{k = 0}^{\infty} x^k = \frac{1}{1 - x}$ We could also start from an arbitrary index: $\sum_{k = m}^{\infty} x^k = \frac{x^{m}}{1 - x}$ ### Exponential Function $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots$ Or, more compactly: $\sum_{k = 0}^{\infty} \frac{x^{k}}{k!} = e^{x}$ ## Binomial Expansion $(a + b)^{n} = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^{k}$ ## Sum of Power of Intergers $\sum_{k = 1}^{n} k = \frac{n(n+1)}{2}$ $\sum_{k = 1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
### Similar presentations 3 What You Should Learn Use multiple–angle formulas to rewrite and evaluate trigonometric functions Use power–reducing formulas to rewrite and evaluate trigonometric functions Use half–angle formulas to rewrite and evaluate trigonometric functions 4 Multiple–Angle Formulas 5 In this section, you will study four additional categories of trigonometric identities. 1. The first category involves functions of multiple angles such as sin ku and cos ku. 2. The second category involves squares of trigonometric functions such as sin 2 u. 3. The third category involves functions of half–angles such as 4. The fourth category involves products of trigonometric functions such as sin u cos v. 6 Use the sine sum formula to verify the identity : 7 Use the cosine sum formula to verify the identity: 8 Use the tangent sum formula to verify the identity: 9 Multiple – Angle Formulas You should learn the double–angle formulas below because they are used often in trigonometry and calculus. 10 Example 1 – Solving a Multiple –Angle Equation Solve 2 cos x + sin 2x = 0. Solution: Begin by rewriting the equation so that it involves functions of x (rather than 2x). Then factor and solve as usual. 2cos x + sin 2x = 0 2 cos x + 2 sin x cos x = 0 2 cosx(1 + sin x) = 0 Write original equation. Double–angle formula Factor. 11 2 cos x = 0 1 + sin x = 0 cos x = 0 sin x = –1 So, the general solution is x = + 2n  and x = + 2n  where n is an integer. Try verifying this solution graphically. Example 1 – Solution Set factors equal to zero. Isolate trigonometric functions. Solutions in [0, 2  ) General solution cont’d 12 Precalculus 5.5 Multiple Angle and Product- Sum Formulas 12 Solve the equation on : Example: 13 Precalculus 5.5 Multiple Angle and Product- Sum Formulas 13 You Try: Solve the equation on : 14 Example:Use the following to find,, : : 15 Example:Use the following to find,, : : 17 Precalculus 5.5 Multiple Angle and Product- Sum Formulas 17 You Try : Solve the equation: 19 Homework Day 1 5.5 pg. 382 1-25 odd 20 5.5 Day 2 HWQ 21 Derive Power-Reducing You should learn the double–angle formulas below because they are used often in trigonometry and calculus. Using the 2 nd two cosine double angle formulas, solve for sin 2 x and cos 2 x to derive power reducing formulas for sine, cosine, and tan. 22 Power–Reducing Formulas The double–angle formulas can be used to obtain the following power–reducing formulas. 23 Example 4 – Reducing a Power Rewrite sin 4 x as a sum of first powers of the cosines of multiple angles. Solution: sin 4 x = (sin 2 x) 2 = = (1 – 2 cos2x + cos 2 2x) Property of exponents Power–reducing formula Expand binomial. Power–reducing formula 24 Example 4 – Solution Distributive Property Simplify. Factor. cont’d 25 Half–Angle Formulas 26 Half–Angle Formulas You can derive some useful alternative forms of the power–reducing formulas by replacing u with u/2. The results are called half-angle formulas. 27 Example 5 – Using a Half–Angle Formula Find the exact value of sin 105 . Solution: Begin by noting that 105  is half of 210 . Then, using the half–angle formula for sin(u/2) and the fact that 105  lies in Quadrant II, you have The positive square root is chosen because sin  is positive in Quadrant II. 28 Example 6 – Solving a Trigonometric Equation Find all solutions of in the interval [0, 2  ). Solution: 1 + cos 2 x = 1 + cos x cos 2 x – cos x = 0 Write original equation. Half-angle formula Simplify. 29 Example 6 – Solution cos x(cos x – 1) = 0 By setting the factors cos x and cos x – 1 equal to zero, you find that the solutions in the interval [0,2  ) are x =, x =, and x = 0. Factor. cont’d 30 Product–to–Sum Formulas Each of the following product–to–sum formulas is easily verified using the sum and difference formulas. 31 Example 7 – Writing Products as Sums Rewrite the product as a sum or difference. cos 5x sin 4x Solution: Using the appropriate product-to-sum formula, you obtain cos 5x sin 4x = [sin(5x + 4x) – sin(5x – 4x)] = sin 9x – sin x. 32 Sum–to–Product Formulas 33 Example 8 – Using a Sum–to–Product Formula Find the exact value of cos 195° + cos 105°. Solution: Using the appropriate sum-to-product formula, you obtain cos 195° + cos 105° = = 2 cos 150° cos 45° 34 Homework Day 2 5.5 pg. 382 33-61 odd
# Question Video: Describing the Asymptote of a Function from Its Graph Mathematics • 10th Grade Consider the graph of the function π(π₯) = 1/(π₯ + 2). What happens to the function when the value of π₯ approaches β2? 03:13 ### Video Transcript Consider the graph of the function π of π₯ is equal to one divided by π₯ plus two. What happens to the function when the value of π₯ approaches negative two? Option (A) the value of π¦ approaches β when π₯ gets closer to negative two from the positive direction and approaches negative β when π₯ gets closer to negative two from the negative direction. Option (B) the value of π¦ approaches β when π₯ gets closer to negative two from the negative direction or from the positive direction. (C) The value of π¦ approaches negative β when π₯ gets closer to negative two from the negative direction or from the positive direction. Or is it option (D) the value of π¦ approaches negative β when π₯ gets closer to negative two from the positive direction and approaches β when π₯ gets closer to negative two from the negative direction? In this question, weβre given the graph of a function π of π₯ is equal to one divided by π₯ plus two. And since this function is the quotient of two polynomials, we can say π of π₯ is a rational function. We want to determine what happens to the graph of our function as our values of π₯ approach negative two. And thereβs a few different ways we could go about this. For example, we could determine what happens to the output of our functions around π₯ is equal to negative two directly by using the given function. For example, we could construct a function table with our values of π₯ getting closer and closer to negative two. However, this is not necessary because weβre given a graph of the function. And even if we werenβt given a graph of the function, we could just sketch this graph by noting itβs a translation of the graph of one over π₯ two units to the left. And in the graph of a function, the π₯-coordinate of any point on the curve tells us the input value and the corresponding π¦-coordinate of this point tells us the output value of the function. So we can determine what happens to the output of this function as the values of π₯ approach negative two by seeing what happens to the π¦-coordinates of points on the curve as the input values of π₯ approach negative two from either direction. To do this, letβs start by sketching the vertical line π₯ is equal to negative two onto the given diagram. We can see that the curve approaches this line. So this is a vertical asymptote of the function. Letβs now see what happens to the outputs of the function as π₯ approaches negative two from either side. Letβs start with the positive direction. And remember this is the side from the positive values of π₯. As our values of π₯ approach negative two, we can see that the output values, thatβs the π¦-coordinates of the points on the curve, are getting larger and larger. We can in fact see that the π¦-coordinates are growing without bound. So, as our values of π₯ get closer to negative two from the positive direction, the π¦-values are approaching β. We can do the exact same thing to determine what happens to the outputs of the function as the values of π₯ get closer to negative two from the negative direction. This time, the π¦-coordinates of the points on the curve are decreasing without bound. So theyβre approaching negative β. And of the four given options, we can see that this only matches option (A). The value of π¦ approaches β when π₯ gets closer to negative two from the positive direction and approaches negative β when π₯ gets closer to negative two from the negative direction.
# Past Papers’ Solutions \| Cambridge International Examinations (CIE) \| AS & A level \| Mathematics 9709 \| Pure Mathematics 1 (P1-9709/01) \| Year 2018 \| May-Jun \| (P1-9709/13) \| Q#6 Hits: 67 Question The coordinates of points A and B are (3k – 1, k + 3) and (k + 3, 3k + 5) respectively, where k is a constant (k 1). i. Find and simplify the gradient of AB, showing that it is independent of k. ii. Find and simplify the equation of the perpendicular bisector of AB. Solution i. We are given coordinates of two points A(3k – 1, k + 3) and B(k + 3, 3k + 5) and are required to find gradient of AB. Expression for slope of a line joining points and ; Therefore; ii. Next we are required to find the equation of the perpendicular bisector of AB. To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line). Therefore, to write equation of the perpendicular bisector of AB we need coordinates of a point on the perpendicular bisector of AB and slope of the perpendicular bisector of AB. Let us first find coordinates of a point on the perpendicular bisector of AB. Since it is the bisector of AB the mid-point of AB also lies on the perpendicular bisector of AB. Therefore, we find coordinates of the mid-point of AB and same will be also the coordinates of the point on the perpendicular bisector of AB. To find the mid-point of a line we must have the coordinates of the end-points of the line. Expressions for coordinates of mid-point of a line joining points and ; x-coordinate of mid-point of the line y-coordinate of mid-point of the line We are given coordinates of two points A(3k – 1, k + 3) and B(k + 3, 3k + 5). Hence; x-coordinate of mid-point of the line y-coordinate of mid-point of the line x-coordinate of mid-point of the line y-coordinate of mid-point of the line Hence, coordinates of mid-point of AB are and these are also coordinates of a point on the perpendicular bisector of AB. Next we need to find slope of the perpendicular bisector of AB. If two lines are perpendicular (normal) to each other, then product of their slopes and is; Therefore, if we can find slope of the AB, from this we can find slope of the perpendicular bisector of AB. As demonstrated in (i); Hence; Now we can write equation of the perpendicular bisector of AB. Point-Slope form of the equation of the line is; Hence;
Comparing Percentages Revision \| KS3 Maths Resources ## What you need to know Things to remember: • It doesn’t matter which number you choose as the percentage $x\% \text{ of } y = y\% \text{ of } x$ We have two methods for finding percentage amounts: To do use this method, you need to remember how to find some simple percentages. Percentage Division 100% $\div1$ 50% $\div2$ 25% $\div4$ 20% $\div5$ 10% $\div10$ 5% $\div20$ 4% $\div25$ 2% $\div50$ 1% $\div100$ To do this method, we need to remember that we can add and subtract percentages to make other ones. This method is best when you don’t have a calculator. $$9\%=5\%+4\%$$ $$28\%=20\%+10\%-2\%$$ Find 35% of 250 $$35\%=25\%+10\%$$ Step 2: Find the value of these percentages $$25\%\text{ of } 250 = 250\div4=62.5$$ $$10\%\text{ of } 250 = 250\div10=25$$ $$35\%=25\%+10\%$$ $$35\%\text{ of } 250=62.5+25=87.5$$ Multiplication works too. $$35\%=7\times5\%$$ Step 2: Find the value of these percentages $$5\%\text{ of } 250 = 250\div20=12.5$$ $$35\%=7\times5\%$$ $$35\%\text{ of } 250=7\times12.5=87.5$$ Our second method is the most powerful can we can find any percentage amount in two steps. This is useful for decimal amounts and you have a calculator. Method 2: Multiplying from 1% Find 12.56% of 300. Step 1: Divide the number you’re finding the percentage of by 100 to find 1%. $$300\div100=3$$ $$1\%=3$$ Step 2: Multiply the value of 1% found in Step 1 by your starting percentage. $$12.56\%=12.56\times1\%$$ $$12.56\%\text{ of }300=12.56\times3=37.68$$ We can now use these two methods to compare two percentage amounts. Now that we have the strategies, we can look at some questions. Which is bigger, 55% of 20 or 40% of 30? Find 55% of 20 Method 1 $$55\%=50\%+5\%$$ Step 2: Find the value of these percentages $$50\%\text{ of }20=20\div2=10$$ $$5\%\text{ of }20=20\div20=1$$ $$55\%=50\%+5\%$$ $$55\%\text{ of }20=10 + 1 =11$$ Find 40% of 30 Method 2 Step 1: Divide the number you’re finding the percentage of by 100 to find 1%. $$30\div100=0.3$$ $$1\%=0.3$$ Step 2: Multiply the value of 1% found in Step 1 by your starting percentage. $$40\%=40\times1\%$$ $$40\%\text{ of }30=40\times0.3=12$$ Which is bigger, 70% of 55 or 55% of 70? Find 70% of 55 Method 1 $$70\%=50\%+20\%$$ Step 2: Find the value of these percentages $$50\%\text{ of }55=55\div2=27.5$$ $$20\%\text{ of }55=55\div5= 11$$ $$70\%=50\%+20\%$$ $$70\%\text{ of }55= 27.5+11=38.5$$ Find 55% of 70 Method 2 Step 1: Divide the number you’re finding the percentage of by 100 to find 1%. $$70\div100=0.7$$ $$1\%=0.7$$ Step 2: Multiply the value of 1% found in Step 1 by your starting percentage. $$55\%=55\times1\%$$ $$55\%\text{ of }70=55\times0.7=38.5$$ So, these two percentages are the same. It is worth remembering that whenever you find a percentage of something, it is the same as using the number as the percent and the percent as the number. 55% of 20 = 20% of 55 78.5% of 24 = 24% of 78.5 13.35% of 72.8 = 72.8% of 13.35 ## KS3 Maths Revision Cards (78 Reviews) £8.99 ## Example Questions 28% of 110 Method 1 $$28\%=25\%+2\%+1\%$$ Step 2: Find the value of these percentages $$25\%\text{ of }110=110\div4=27.5$$ $$2\%\text{ of }110=110\div50=2.2$$ $$1\%\text{ of }110=110\div100=1.1$$ $$28\%=25\%+2\%+1\%$$ $$28\%\text{ of }110=27.5+2.2+1.1=30.8$$ 27% of 120 Method 2 Step 1: Divide the number you’re finding the percentage of by 100 to find 1%. $$120\div100=1.2$$ $$1\%=1.2$$ Step 2: Multiply the value of 1% found in Step 1 by your starting percentage. $$27\%=27\times1\%$$ $$27\%\text{ of }120=27\times1.2=32.4$$ So, 27% of 120 is bigger than 28% of 110. Neither is smaller, they’re equal! We have just swapped the numbers around. ## KS3 Maths Revision Cards (78 Reviews) £8.99 • All of the major KS2 Maths SATs topics covered • Practice questions and answers on every topic
Comment Share Q) # A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. This is a multi part question answered separately on Clay6.com Comment A) • In any problem that involves tossing a coin or a die and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes. • Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$ • Given P(E), P(F), P(E $\cap$ F), P(E $\cup$ F) = P(E) + P(F) - P(E $\cap$ F) Given one black die and one red die are rolled. The sample space of equally likely events = 6 $\times$ 6 = 36. $\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$ $\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$ For our set of events $E \cap F =(5,5), (6,6). Total number of outcomes = 2.$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{2}{36}$. Given P(E), P(F), P(E$\cap$F), P(E/F)$= \large \frac{P(E \;\cap \;F)}{P(F)}\Rightarrow P(E/F) = \Large\frac {\Large \frac{2}{36}}{\Large\frac{1}{6}}$=$\large\frac{1}{3}\$
# Incredible How To Solve Area Of Triangle Ideas Incredible How To Solve Area Of Triangle Ideas. That means we can solve for the height h. We will look at several types of triangles in this lesson. We already have rc k r c k ready to use, so let's try the formula on it: Area of an equilateral triangle. This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles. ### Any Side Of The Triangle Can Be A Base. There are several ways to find the area of a triangle. Start with the sas rule for area. The base refers to any side of the triangle where the height is represented by the length. See also Review Of How To Solve Problems In A Relationship Ideas ### The Base Of A Triangle Is 17 Meters. Area of an equilateral triangle. To calculate the area of an equilateral triangle you only need to have the side given: Area = aÂ² * âˆš3 / 4. ### To Find The Area Of A Triangle, Multiply The. How to find the area of a triangle: There are multiple different equations for calculating the area of a triangle, dependent on what information is known. Letâ€™s say we are asked to find the area of an acute triangle, given the base as 15 inches and the height as 4 inches. ### We Substitute The Known Values Into The Formula Then Solve For The Leftover Variable. Such a triangle can be solved by using angles of a triangle to find the other angle, and the law of sines to find each of the other two sides. Given one angle and one leg, find the area using e.g. To find the area of a triangle, we should know the base $(b)$ and height $(h)$ of it. ### We All Know The Usual Formula For The Area Of A Triangle: The area of a triangle can be defined as the total space or region occupied by the three sides of any triangle. To solve this, all we need to do is introduce our general formula which is a=1/2 b*h. Check out how this formula works in an actual problem. ## Incredible How To Solve Sudoku Tricks 2022 Incredible How To Solve Sudoku Tricks 2022. There are already many numbers filled in on the grid in the simple sudoku puzzles. The numbers...
## Engage NY Eureka Math 7th Grade Module 4 Lesson 6 Answer Key ### Eureka Math Grade 7 Module 4 Lesson 6 Example Answer Key Example 1: Mental Math and Percents a. 75% of the students in Jesse’s class are 60 inches or taller. If there are 20 students in her class, how many students are 60 inches or taller? → Is this question a comparison of two separate quantities, or is it part of the whole? How do you know? The problem says that the students make up 75% of Jesse’s class, which means they are part of the whole class; this is a part of the whole problem. → What numbers represent the part, whole, and percent? The part is the number of students that are 60 inches or taller, the whole is the 20 students that make up Jesse’s class, and the percent is 75%. Instruct students to discuss the problem with a partner; challenge them to solve it using mental math only. After 1–2 minutes of discussion, ask for students to share their mental strategies with the class. Possible strategies: 75% is the same as $$\frac{3}{4}$$ of 100%; 20 → 100% and 20 = 4(5), so 3(5) = 15, which means 15 is $$\frac{3}{4}$$ of 20. 100% → 20 25% → 5 75% → 15 Have students write a description of how to mentally solve the problem (including the math involved) in their student materials. The numbers involved in the problem shared factors with 100 that were easy to work with. b. Bobbie wants to leave a tip for her waitress equal to 15% of her bill. Bobbie’s bill for her lunch is $18. How much money represents 15% of the bill? Answer: → Is this question a comparison of two separate quantities, or is it part of a whole? How do you know? She is leaving a quantity that is equal to 15% of her bill, so this is a comparison of two separate quantities. → What numbers represent the part, the whole, and the percent? Is the part actually part of her lunch bill? The part is the amount that she plans to leave for her waitress and is not part of her lunch bill but is calculated as if it is a part of her bill; the whole is the$18 lunch bill, and the percent is 15%. Instruct students to discuss the problem with a partner; challenge them to solve it using mental math only. After 1–2 minutes of discussion, ask for students to share their mental strategies with the class. Possible strategy includes the following: 15% = 10%+5%; 10% of $18 is$1.80; half of 10% is 5%, so 5% → 1/2( $1.80) =$0.90; $1.80+$0.90 = $2.70. → Was this problem easy to solve mentally? Why? The numbers involved in the problem shared factors with 100 that were easy to work with. → Could you use this strategy to find 7% of Bobbie’s bill? Yes; 7% = 5%+2(1%); 1% of$18 is $0.18, so 2% →$0.36; $0.90+$0.36 = $1.26, so 7% →$1.26. Have students write a description of how to mentally solve the problem in their student materials including the math involved. ### Eureka Math Grade 7 Module 4 Lesson 6 Exercise Answer Key Exercise 1. Express 9 hours as a percentage of 3 days. 3 days is the equivalent of 72 hours since 3(24) = 72. 72 hours represents the whole. Quantity = Percent × Whole. Let p represent the unknown percent. 9 = p(72) $$\frac{1}{72}$$ (9) = p(72) ∙ $$\frac{1}{72}$$ $$\frac{9}{72}$$ = p(1) $$\frac{1}{8}$$ = p $$\frac{1}{8}$$ (100%) = 12.5% Exercise 2. Richard works from 11:00 a.m. to 3:00 a.m. His dinner break is 75% of the way through his work shift. What time is Richard’s dinner break? The total amount of time in Richard’s work shift is 16 hours since 1+12+3 = 16. 16 hours represents the whole. Quantity = Percent × Whole. Let b represent the number of hours until Richard’s dinner break. b = 0.75(16) b = 12 Richard’s dinner break is 12 hours after his shift begins. 12 hours after 11:00 a.m. is 11:00 p.m. Richard’s dinner break is at 11:00 p.m. Exercise 3. At a playoff basketball game, there were 370 fans cheering for school A and 555 fans cheering for school B. a. Express the number of fans cheering for school A as a percent of the number of fans cheering for school B. The number of fans for school B is the whole. Quantity = Percent × Whole. Let p represent the unknown percent. 370 = p(555) $$\frac{1}{555}$$ (370) = p(555)$$\frac{1}{555}$$ $$\frac{370}{555}$$ = p(1) $$\frac{2}{3}$$ = p $$\frac{2}{3}$$ (100%) = 66$$\frac{2}{3}$$% The number of fans cheering for school A is 66 2/3% of the number of fans cheering for school B. b. Express the number of fans cheering for school B as a percent of the number of fans cheering for school A. The number of fans cheering for school A is the whole. Quantity = Percent × Whole. Let p represent the unknown percent. 555 = p(370) $$\frac{1}{370}$$ (555) = p(370)$$\frac{1}{370}$$ $$\frac{555}{370}$$ = p(1) $$\frac{3}{2}$$ = p $$\frac{3}{2}$$ (100%) = 150% The number of fans cheering for school B is 150% of the number of fans cheering for school A. c. What percent more fans were there for school B than for school A? There were 50% more fans cheering for school B than for school A. Exercise 4. Rectangle A has a width of 8 cm and a length of 16 cm. Rectangle B has the same area as the first, but its width is 62.5% of the width of the first rectangle. Express the length of Rectangle B as a percent of the length of Rectangle A. What percent more or less is the length of Rectangle B than the length of Rectangle A? To find the width of Rectangle B: The width of Rectangle A is the whole. Quantity = Percent × Whole. Let w represent the unknown width of Rectangle B. w = 0.625(8) = 5 The width of Rectangle B is 5 cm. To find the length of Rectangle B: The area of Rectangle B is 100% of the area of Rectangle A because the problem says the areas are the same. Area = Width × Length. Let A represent the unknown area of Rectangle A. A = 8 cm(16 cm) = 128 cm2 Area = Width × Length. Let l represent the unknown length of Rectangle B. 128 cm2 = 5 cm (l) 25.6 cm = l The length of Rectangle B is 25.6 cm. To express the length of Rectangle B as a percent of the length of Rectangle A: The length of Rectangle A is the whole. Quantity = Percent × Whole. Let p represent the unknown percent. 25.6 cm = p(16 cm) 1.6 = p 1.6(100%) = 160%; The length of Rectangle B is 160% of the length of Rectangle A. Therefore, the length of Rectangle B is 60% more than the length of Rectangle A. Exercise 5. A plant in Mikayla’s garden was 40 inches tall one day and was 4 feet tall one week later. By what percent did the plant’s height increase over one week? 4 feet is equivalent to 48 inches since 4(12) = 48. 40 inches is the whole. Quantity = Percent × Whole. Let p represent the unknown percent. 8 = p(40) $$\frac{1}{5}$$ = p $$\frac{1}{5}$$ = 20/100 = 20% The plant’s height increased by 20% in one week. Exercise 6. Loren must obtain a minimum number of signatures on a petition before it can be submitted. She was able to obtain 672 signatures, which is 40% more than she needs. How many signatures does she need? The number of signatures needed represents the whole. Quantity = Percent × Whole. Let s represent the number of signatures needed. 672 = 1.4(s) 480 = s Loren needs to obtain 480 signatures on her petition. ### Eureka Math Grade 7 Module 4 Lesson 6 Problem Set Answer Key Question 1. Micah has 294 songs stored in his phone, which is 70% of the songs that Jorge has stored in his phone. How many songs are stored on Jorge’s phone? Quantity = Percent × Whole. Let s represent the number of songs on Jorge’s phone. 294 = $$\frac{70}{100}$$ ∙ s 294 = $$\frac{7}{10}$$ ∙ s 294 ∙ $$\frac{10}{7}$$ = $$\frac{7}{10}$$ ∙ $$\frac{10}{7}$$ ∙ s 42 ∙ 10 = 1 ∙ s 420 = s There are 420 songs stored on Jorge’s phone. Question 2. Lisa sold 81 magazine subscriptions, which is 27% of her class’s fundraising goal. How many magazine subscriptions does her class hope to sell? Quantity = Percent × Whole. Let s represent the number of magazine subscriptions Lisa’s class wants to sell. 81 = $$\frac{27}{100}$$ ∙ s 81 ∙ $$\frac{100}{27}$$ = $$\frac{27}{100}$$ ∙ $$\frac{100}{27}$$ ∙ s 3 ∙ 100 = 1 ∙ s 300 = s Lisa’s class hopes to sell 300 magazine subscriptions. Question 3. Theresa and Isaiah are comparing the number of pages that they read for pleasure over the summer. Theresa read 2,210 pages, which was 85% of the number of pages that Isaiah read. How many pages did Isaiah read? Quantity = Percent × Whole. Let p represent the number of pages that Isaiah read. 2,210 = $$\frac{85}{100}$$ ∙ p 2,210 = $$\frac{17}{20}$$ ∙ p 2,210 ∙ $$\frac{20}{17}$$ = $$\frac{17}{20}$$ ∙ $$\frac{20}{17}$$ ∙ p 130 ∙ 20 = 1 ∙ p 2,600 = p Isaiah read 2,600 pages over the summer. Question 4. In a parking garage, the number of SUVs is 40% greater than the number of non-SUVs. Gina counted 98 SUVs in the parking garage. How many vehicles were parked in the garage? 40% greater means 100% of the non-SUVs plus another 40% of that number, or 140%. Quantity = Percent × Whole. Let d represent the number of non-SUVs in the parking garage. 98 = $$\frac{140}{100}$$ ∙ d 98 = $$\frac{7}{5}$$ ∙ d 98 ∙ $$\frac{5}{7}$$ = $$\frac{7}{5}$$ ∙ $$\frac{5}{7}$$ ∙ d 14 ∙ 5 = 1 ∙ d 70 = d There are 70 non-SUVs in the parking garage. The total number of vehicles is the sum of the number of the SUVs and non-SUVs. 70+98 = 168. There is a total of 168 vehicles in the parking garage. Question 5. The price of a tent was decreased by 15% and sold for $76.49. What was the original price of the tent in dollars? Answer: If the price was decreased by 15%, then the sale price is 15% less than 100% of the original price, or 85%. Quantity = Percent × Whole. Let t represent the original price of the tent. 76.49 = $$\frac{85}{100}$$ ∙ t 76.49 = $$\frac{17}{20}$$ ∙ t 76.49 ∙ $$\frac{20}{17}$$ = $$\frac{17}{20}$$ ∙ $$\frac{20}{17}$$ ∙ t $$\frac{1,529.8}{17}$$ = 1 ∙ t 89.988 ≈ t Because this quantity represents money, the original price was$89.99 after rounding to the nearest hundredth. Question 6. 40% of the students at Rockledge Middle School are musicians. 75% of those musicians have to read sheet music when they play their instruments. If 38 of the students can play their instruments without reading sheet music, how many students are there at Rockledge Middle School? Let m represent the number of musicians at the school, and let s represent the total number of students. There are two whole quantities in this problem. The first whole quantity is the number of musicians. The 38 students who can play an instrument without reading sheet music represent 25% of the musicians. Quantity = Percent × Whole 38 = 25/100 ∙ m 38 = $$\frac{1}{4}$$ ∙ m 38 ∙ $$\frac{4}{1}$$ = $$\frac{1}{4}$$ ∙ $$\frac{4}{1}$$ ∙ m $$\frac{152}{1}$$ = 1 ∙ m 152 = m There are 152 musicians in the school. Quantity = Percent × Whole 152 = $$\frac{40}{100}$$ ∙ s 152 = $$\frac{2}{5}$$ ∙ s 152 ∙ $$\frac{5}{2}$$ = $$\frac{2}{5}$$ ∙ $$\frac{5}{2}$$ ∙ s $$\frac{760}{2}$$ = 1 ∙ s 380 = s There are 380 students at Rockledge Middle School. Question 7. At Longbridge Middle School, 240 students said that they are an only child, which is 48% of the school’s student enrollment. How many students attend Longbridge Middle School? Quantity → 100% 240 → 48% $$\frac{240}{48}$$ → 1% $$\frac{240}{48}$$ (100) → 100% 5(100) → 100% 500 → 100% There are 500 students attending Longbridge Middle School. Question 8. Grace and her father spent 4 1/2 hours over the weekend restoring their fishing boat. This time makes up 6% of the time needed to fully restore the boat. How much total time is needed to fully restore the boat? Quantity → % 4 $$\frac{1}{2}$$ → 6% $$\frac{\frac{9}{2}}{6}$$ → 6% $$\frac{\frac{9}{2}}{6}$$ → 1% $$\frac{\frac{9}{2}}{6}$$ (100) → 100% ($$\frac{9}{2}$$)($$\frac{1}{6}$$)100 → 100% ($$\frac{9}{12}$$)100 → 100% ($$\frac{3}{4}$$)100 → 100% 300/4 → 100% 75 → 100% The total amount of time to restore the boat is 75 hours. Question 9. Bethany’s mother was upset with her because Bethany’s text messages from the previous month were 218% of the amount allowed at no extra cost under her phone plan. Her mother had to pay for each text message over the allowance. Bethany had 5,450 text messages last month. How many text messages is she allowed under her phone plan at no extra cost? Quantity → % 5,450 → 218% $$\frac{5,450}{218}$$ → 1% $$\frac{5,450}{218}$$ (100) → 100% 25(100) → 100% 2,500 → 100% Bethany is allowed 2,500 text messages without extra cost. Question 10. Harry used 84% of the money in his savings account to buy a used dirt bike that cost him $1,050. How much money is left in Harry’s savings account? Answer: Quantity → % 1,050 → 84% $$\frac{1,050}{84}$$ → 1% $$\frac{1,050}{84}$$ (100) → 100% 12.5(100) → 100% 1,250 → 100% Harry started with$1,250 in his account but then spent $1,050 of it on the dirt bike. 1,250-1,050 = 200 Harry has$200 left in his savings account. Question 11. 15% of the students in Mr. Riley’s social studies classes watch the local news every night. Mr. Riley found that 136 of his students do not watch the local news. How many students are in Mr. Riley’s social studies classes? If 15% of his students do watch their local news, then 85% do not. Quantity → % 136 → 85% $$\frac{136}{85}$$ → 1% ($$\frac{136}{85}$$)(100) → 100% 1.6(100) → 100% 160 → 100% There are 160 total students in Mr. Riley’s social studies classes. Question 12. Grandma Bailey and her children represent about 9.1% of the Bailey family. If Grandma Bailey has 12 children, how many members are there in the Bailey family? Quantity → % 13 → 9.1% ($$\frac{1}{9.1}$$)(13) → 1% 100($$\frac{13}{9.1}$$) → 100% $$\frac{1,300}{9.1}$$ → 100% 142.857″…” → 100% The Bailey family has 143 members. Question 13. Shelley earned 20% more money in tips waitressing this week than last week. This week she earned $72.00 in tips waitressing. How much money did Shelley earn last week in tips? Answer: Quantity = Percent × Whole. Let m represent the number of dollars Shelley earned waitressing last week. 72 = $$\frac{120}{100}$$ m 72($$\frac{100}{120}$$) = $$\frac{120}{100}$$ ($$\frac{100}{120}$$)m 60 = m Shelley earned$60 waitressing last week. Question 14. Lucy’s savings account has 35% more money than her sister Edy’s. Together, the girls have saved a total of $206.80. How much money has each girl saved? Answer: The money in Edy’s account corresponds to 100%. Lucy has 35% more than Edy, so the money in Lucy’s account corresponds to 135%. Together, the girls have a total of$206.80, which is 235% of Edy’s account balance. Quantity = Pecent × Whole. Let b represent Edy’s savings account balance in dollars. 206.8 = $$\frac{235}{100}$$ ∙ b 206.8 = $$\frac{47}{20}$$ ∙ b 206.8 ∙ $$\frac{20}{47}$$ = $$\frac{47}{20}$$ ∙ $$\frac{20}{47}$$ ∙ b $$\frac{4,136}{47}$$ = 1 ∙ b 88 = b Edy has saved $88 in her account. Lucy has saved the remainder of the$206.80, so 206.8-88 = 118.8. Therefore, Lucy has $118.80 saved in her account. Question 15. Bella spent 15% of her paycheck at the mall, and 40% of that was spent at the movie theater. Bella spent a total of$13.74 at the movie theater for her movie ticket, popcorn, and a soft drink. How much money was in Bella’s paycheck? $13.74 → 40%$3.435 → 10% $34.35 → 100% Bella spent$34.35 at the mall. $34.35 → 15%$11.45 → 5% $229 → 100% Bella’s paycheck was$229. Question 16. On a road trip, Sara’s brother drove 47.5% of the trip, and Sara drove 80% of the remainder. If Sara drove for 4 hours and 12 minutes, how long was the road trip? There are two whole quantities in this problem. First, Sara drove 80% of the remainder of the trip; the remainder is the first whole quantity. 4 hr.12 min. is equivalent to 4 12/60 hr. = 4.2 hr. Quantity → % 4.2 → 80% $$\frac{4.2}{80}$$ → 1% $$\frac{4.2}{80}$$ (100) → 100% $$\frac{420}{80}$$ → 100% $$\frac{42}{8}$$ → 100% 5.25 → 100% The remainder of the trip that Sara’s brother did not drive was 5.25 hours. He drove 47.5% of the trip, so the remainder of the trip was 52.5% of the trip, and the whole quantity is the time for the whole road trip. Quantity → % 5.25 → 52.5% $$\frac{5.25}{52.5}$$ → 1% ($$\frac{5.25}{52.5}$$)(100) → 100% $$\frac{525}{52.5}$$ → 100% 10 → 100% The road trip was a total of 10 hours. ### Eureka Math Grade 7 Module 4 Lesson 6 Exit Ticket Answer Key Question 1. Parker was able to pay for 44% of his college tuition with his scholarship. The remaining $10,054.52 he paid for with a student loan. What was the cost of Parker’s tuition? Answer: Parker’s tuition is the whole; 56% represents the amount paid by a student loan. Quantity = Percent × Whole. Let t represent the cost of Parker’s tuition. 10,054.52 = 0.56(t) $$\frac{10,054.52}{0.56}$$ = t 17,954.50 = t Parker’s tuition was$17,954.50. Question 2. Two bags contain marbles. Bag A contains 112 marbles, and Bag B contains 140 marbles. What percent fewer marbles does Bag A have than Bag B? The number of marbles in Bag B is the whole. There are 28 fewer marbles in Bag A. Quantity = Percent × Whole. Let p represent the unknown percent. 28 = p(140) $$\frac{2}{10}$$ = p $$\frac{2}{10}$$ = $$\frac{20}{100}$$ = 20% Bag A contains 20% fewer marbles than Bag B. Question 3. There are 42 students on a large bus, and the rest are on a smaller bus. If 40% of the students are on the smaller bus, how many total students are on the two buses?
GED Mathematical Reasoning: Introduction To Polynomials \| Open Window Learning Algebraic Expressions and Polynomials # GED Mathematical Reasoning: Introduction To Polynomials • The prefix “poly” means “many” so it makes sense that a polynomial is a string of one or more terms being added or subtracted. • A polynomial may contain many more than three terms, but we only have special names for those having one, two or three terms. A polynomial containing only one term is called a “monomial.” For example: and A polynomial containing two terms is called a “binomial.” For example: and A polynomial containing three terms is called a “trinomial.” For example: • The number in front of the variable part is called “coefficient” and every term contains a coefficient. consider The coefficient of the first term is 7 The coefficient of the second term is 1 The coefficient of the third term, even though there appears to be no variable part, is -10 • When it comes to polynomials, the sign preceding a term goes with the term. What I mean is: If the sign is addition, the term is positive. If the sign is subtraction, the term is negative. • A term that appears to have no variable part is called a “constant” term since there is no variable to make it vary – meaning it will always remain the same. For example, -10 in • The degree of a polynomial containing only one variable is equal to the largest exponent. Example 1 Identify each term of the polynomial: The polynomial in example one consists of three terms: , and Notice that we state the middle term to be negative, since there is a subtraction sign preceding . This is because we can rewrite subtraction using the idea of “adding the opposite,” which makes the term negative. Example 2 Classify the polynomial in Example 1 as a monomial, binomial or trinomial. Since the polynomial in example one has three terms, it is classified as a trinomial. Example 3 State the degree of the polynomial: This polynomial contains 4 terms: -14 “x” cubed, -10 “x” squared, 9 “x” and -5. The coefficients in this polynomial are: -14, -10, 9, and -5. And the constant term is: -5. The degree of a polynomial containing only one variable is equal to the largest exponent. The polynomial shown in example 3 involves only one variable, which is “x,” and the largest power of “x” is 3. Therefore, the degree of the polynomial is 3. You have seen 1 out of 15 free pages this month. Get unlimited access, over 1000 practice questions for just \$29.99.
# Complement of 27 degrees ## Supplementary and Complementary Angles There are many special relationships that can be formed using angles. Here is a look at two of the relationships. Supplementaryangles are two angles that have a sum of 180°. Complementaryangles are two angles that have a sum of 90°. There is an easy way to try and remember these using the first letters of each word. The Sin supplementarycan be used to form the 8 in 180. The Cin complementarycan be used to form the 9 in 90. If we know that a set of angles form one of these special relationships, we can determine the measure of the other angle. Example #1: 43° To determine the supplement, subtract the given angle from 180. 180 - 43 = 137° The supplement of 43° is 137°. To determine the complement, subtract the given angle from 90. 90 - 43 = 47° The complement of 43° is 47°. Example #2: 61° 180 - 61 = 119° The supplement of 61° is 119°. 90 - 61 = 29° The complement of 61° is 29°. Example #3: 127° 180 - 127 = 53° The supplement of 127° is 53°. 127° is already greater than 90°. Therefore, there is no complement. Example #4: Determine the missing angle. Notice that the two angles for a right angle when together. This means that the angles are complementary and have a sum of 90°. 90 - 62 = 28° The missing angle measures 28 degrees. Example #5: Determine the missing angle. These two angles form a straight line. Straight lines measure 180°. That means that these two angles are supplementary. 180 - 77 = 103° The missing angle measures 103 degrees. Let's Review Complementary angles form a right angle (L shape) and have a sum of 90 degrees. Supplementary angles form a straight line and have a sum of 180 degrees. If the relationship is given, you can subtract the given angle from the sum to determine the measure of the missing angle. Sours: https://www.softschools.com/math/geometry/topics/supplementary_and_complementary_angles/ ## Complementary Angles Two angles are Complementary when they (a Right Angle ). These two angles (40° and 50°) are Complementary Angles, because they add up to 90°: Notice that together they make a right angle. But the angles don't have to be together. These two are complementary because 27° + 63° = 90° ### Play With It ... (Drag the points) ### Right Angled Triangle In a right angled triangle, the two non-right angles are complementary, because in a triangle the three angles add to 180°, and 90° has already been taken by the right angle. When two angles add to 90°, we say they "Complement" each other. Complementary comes from Latin completum meaning "completed" ... because the right angle is thought of as being a complete angle. Spelling: be careful, it is not"Complimentary Angle" (with an "i") ... that would be an angle you get for free! ### Complementary vs Supplementary A related idea is Supplementary Angles - those add up to 180° How to remember which is which? Well, alphabetically they are: You can also think: • "C" of Complementary is for "Corner" (a Right Angle), and • "S" of Supplementary is for "Straight" (180° is a straight line) Or you can think: • when you are right you get a compliment (sounds like complement) • "supplement" (like a vitamin supplement) is something extra, so is bigger Supplementary AnglesSymbols in GeometryDegrees (Angle)Angles On a Straight LineAngles Around a PointParallel Lines and Pairs of AnglesInterior AnglesGeometry Index Sours: https://www.mathsisfun.com/geometry/complementary-angles.html ## Complementary Angles When the sum of the measures of two angles is 90°, such angles are called complementary angles and each angle is called a complement of the other. The vertices of two angles may be same or different. In the given figure ∠AOB and ∠BOC are complementary as ∠AOB + ∠BOC = 30° + 60° = 90°. Again, ∠PQR and ∠QRP are complementary as ∠PQR + ∠QRP = 40° + 50° = 90°. Angles of measure 25° and 65° are complementary angles. The angle of 25° is the complement of the angle of 65° and the angle of 65° is the complement of the angle of 25°. The complement of an angle of measure 32° is the angle of 58°. And, the complement of the angle of measure 58° is the angle of 32°. Observations: (i) If two are complement of each other, then each is an acute angle. But any two acute angles need not be complementary. For example, angles of measure 30° and 50° are not complement of each other. (ii) Two obtuse angles cannot be complement of each other. (iii) Two right angles cannot be complement of each other. Worked-out Problems on Complementary Angles: 1. Find the complement of: (a) 68° Solution: 90° - 68° = 22° Therefore, the complement of 68° is 22° (b) 27°20' Solution: 90° - 27°20' = 89°60' - 27°20' = 62°40' Therefore, the complement of 27°20' is 62°40' (c) x + 52° Solution: 90° - (x + 52°) = 90° - x + 52° = 38° - x Therefore, the complement of x + 52° is 38° - x 2. Find the complement of the angle (10 + y)°. Solution: Complement of the angle (10 + y)° = 90° - (10 + y)° = 90° - 10° - y° = (80 - y)° 3. Find the measure of an angle which is 46° less than its complement. Solution: Let the unknown angle be x, then measure of its complement = 90 - x According to the question, (90 - x) - x = 46° 90 - x - x = 46° 90 - 2x = 46° 90 - 90 - 2x = 46° - 90 -2x = 46° - 90 -2x = 46° - 90 -2x = -44° 2x = 44° x = 44/2 x = 22° Therefore, 90 - x (Put the value of x = 22°) = 90 - 22° = 68° Therefore, the pair of complementary angles are 68° and 22° Lines and Angles Fundamental Geometrical Concepts Angles Classification of Angles Related Angles Some Geometric Terms and Results Complementary Angles Supplementary Angles Complementary and Supplementary Angles Linear Pair of Angles Vertically Opposite Angles Parallel Lines Transversal Line Parallel and Transversal Lines Didn't find what you were looking for? Or want to know more information aboutMath Only Math. Use this Google Search to find what you need. Sours: https://www.math-only-math.com/complementary-angles.html ### How do you find the supplement of 52 degrees? What is the complement and the supplement of 52 degrees 1. The complement is 38 degrees because you subtract 52 – 90 = 38. 2. The supplement is 128 because you do 180-52=128. 3. thank you so much. 4. I meant 90-52. 5. ok. 118° ### What is the complement of 27 degrees? The complement of 27° is the angle that when added to 27° forms a right angle (90° ). ### What is the supplement of an angle measuring 59 degrees? The supplement is 121∘ . 90° ### What is the measure of the complement of degree? When two angles add to 90°, we say they “Complement” each other. ### What is the measure of the complement of a 9 angle? X=81 DEGREES. IS THE COMPLEMENT ANGLE. 123degree ### What is the supplement of a 60 degree angle? The supplementary angle of 60∘ is 120∘ . ### What is the complement and supplement of 60 degrees? Finally complement of the supplement of twice the angle 60°=90°-60°=30°. ### What is the supplement of 63 degrees? Trigonometry Examples The supplement of 63° is the angle that when added to 63° forms a straight angle (180° ). ### What is the supplement of 58 degrees? Trigonometry Examples The supplement of 58° is the angle that when added to 58° forms a straight angle (180° ). ### Are the angles 63 degree and 27 degree complementary? (ii) 63°+27°=90°63°+27°=90° These are complementary angles. ### What is the complementary and supplementary angle of 63 degree? supplementary to the complement = 180°-90° = 90° which is complement. (90–63)° = 27°. ### Can two angles be supplementary if both of them are acute? Thus, the two acute angles cannot be supplementary angles. Thus, the two obtuses angles cannot be supplementary angles. ### What is the measure of an angle complementary to 70? 1. Complement of 70 is 90-70= 20 degrees. 2021-06-19Alex SmithMath Sours: https://rehabilitationrobotics.net/how-do-you-find-the-supplement-of-52-degrees/ ## Degrees 27 complement of Its a very simple one: The complement of an angle is what, when added to it, equals 90 degrees (90°). This means that the complement of 85° is 5°, since they add up to equal 90 degrees a right angle. The supplement of an angle is what, when added to it, equals 180 degrees. For example, in your problem, 180°-85°=95°. Correspondingly, what is the complement of a 39 degree angle? You are correct in setting up the equation, and your final answer is correct; you probably typoed when you wrote "180 = 219" when it should be "180 = 219 - x." And the complement of 39 degrees is 90 - 39 = 51 degrees. Additionally, how do you find the complement and supplement of an angle? To determine the supplement, subtract the given angle from 180. 180 - 43 = 137° The supplement of 43° is 137°. To determine the complement, subtract the given angle from 90. 90 - 43 = 47° The complement of 43° is 47°. In this manner, what is the complement of an angle? Two Angles are Complementary when they add up to 90 degrees (a Right Angle). They don't have to be next to each other, just so long as the total is 90 degrees. What is the supplement of the complement of 42 degree angle? For you to get the complement of 42 degrees, subtract 42 degrees from 90 degrees. 48 degrees is the complement of 42 degrees. Trigonometry - Find the complement and supplement of an angle Else, was also involved in this rhythm, this stream that carried her to no one knows where, but she could not, and did not want to resist him. Her gaze more and more often stopped at the top of the structure - the symbol of the deity, which symbolized the masculine principle. This shiny barrel seemed to change color. ### You will also like: Kolka twisted the tourniquet in case his dog fell off the leash. And if he wants to bite him again, he will crack her on the bite. While everything was smooth, his dog breathed quickly and licked its lips with its tongue, squealing softly. A member already accustomed to rubbing against the inner walls of the vagina stood firmly, not paying attention to the frequent frictions. 3017 3018 3019 3020 3021
# (solved)Question 9 SSC-CGL 2020 March 4 Shift 3 If 2x² + y² + 8z² - 2√2xy + 4√2yz - 8zx = (Ax + y + Bz)², then the value of A² + B² - AB =? (Rev. 04-Jun-2024) ## Categories \| About Hoven's Blog , ### Question 9SSC-CGL 2020 Mar 4 Shift 3 If 2x² + y² + 8z² - 2√2xy + 4√2yz - 8zx = (Ax + y + Bz)², then the value of A² + B² - AB =? ### Solution in Short Coefficient of x² on LHS is 2, and orally we can see that it is A² on the RHS. So A² = 2, and therefore A = √2. Similarly comparing coeff. of z², we can get B² = 8, and therefore B = 2√2. Finally, A² + B² - AB = 2 + 8 - 2√2 x √2 = 6 ans! ### Solution in Detail We shall use the well known algebraic principle that if two algebraic expressions are equal then the coefficients of like terms on both sides should be equal. [1] on LHS coeff. of $\displaystyle x^2 = 2$ To obtain the coefficient on the RHS we do not have to actually open the square. We can orally observe that the term on RHS is Ax². Hence [2] on RHS coeff. of $\displaystyle x^2 = A^2$ Equating, $\displaystyle A^2 = 2\text{. . . (1)}$ Similarly, equating for $\displaystyle z^2$ $\displaystyle B^2 = 8, \therefore B = 2\sqrt 2$ Both A and B are known. Substituting in A² + B² - AB, we can get $\displaystyle 2 + 8 - 2\sqrt 2 \cdot \sqrt 2$ $\displaystyle = 6\:\underline{Ans}$ This Blog Post/Article "(solved)Question 9 SSC-CGL 2020 March 4 Shift 3" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
# Descriptive statistics The mean, mode and median, are all ways of measuring “averages”. Depending on the distribution of the data, the values for the mean, mode and median can differ slightly or a lot. Therefore, the mean, mode and median are all useful for understanding your data set. Example data set: 6, 3, 6, 13, 7, 7 in a table: Mean the average value, Mode the value that occurs most often (highest frequency) e.g. The example data set has 2 modes: 6 and 7 Median the middle value when the data set is ordered low to high. Even number of values: the median is the average of the two middle values. Find for larger values as n + ½ . e.g. data set from low to high: 3, 6, 6, 7, 7, 13 Range largest x-value − smallest x-value e.g. range=13−3=10 Variance Standard deviation Grouped data data presented as an interval, e.g.10 < x ≤ 20 where: • lower boundary = 10 • upper boundary = 20 • interval width = 20 − 10 = 10 • mid-interval value (midpoint) = (20 + 10)/2 = 15 Use the midpoint as the x-value in all calculations with grouped data. Adding a constant to all the values in a data set or multiplying the entire data set by a constant influences the mean and standard deviation values in the following way: Table 7.1: Adding or multiplying by a constant Q1 the value for x so that 25% of all the data values are ≤ to it first quartile = 25th percentile Q2 median = 50th percentile Q3 third quartile = 75th percentile Q3 − Q1 interquartile range (IQR) = middle 50 percent Example: Snow depth is measured in centimetres: 30, 75, 125, 55, 60, 75, 65, 65, 45, 120, 70, 110. Find the range, the median, the lower quartile, the upper quartile and the interquartile range. First always rearrange data into ascending order: 30, 45, 55, 60, 65, 65, 70, 75, 75, 110, 120, 125 1. The range 125−30=95cm 2. The median: there are 12 values so the median is between the 6th and 7th value. (65 + 70) / 2 = 67.5 cm 3. The lower quartile: there are 12 values so the lower quartile is between the 3rd and 4th value. (55 + 60) / 2 = 57.5 cm 4. The upper quartile: there are 12 values so the lower quartile is between the 9th and 10th value. (75 + 110) / 2 = 92.5 cm 5. The IQR 92.5 − 57.5 = 35cm
Lesson Video: Rounding to the Nearest Ten \| Nagwa Lesson Video: Rounding to the Nearest Ten \| Nagwa # Lesson Video: Rounding to the Nearest Ten Mathematics • 3rd Grade In this video, we will learn how to round whole numbers within 1000 to the nearest ten by identifying the tens numbers it is between on a number line. 17:59 ### Video Transcript Rounding to the Nearest 10 In this video, we’re going to learn how to round three-digit whole numbers to the nearest 10. Now in maths, if we round a number, it’s a way of saying about how much it’s worth, not the exact amount, but about that amount. And if you keep your ears open, it’s the sort of thing that we hear a lot in everyday life. Someone might say we’ve got roughly 150 kilometers to go, when perhaps the exact distance on the sign you’ve just driven past says 153 kilometers. Or somebody might say, “It took me about 40 minutes to get to town,” when it actually took 39 minutes. We round numbers up all the time. And if we hear words like “roughly” or “about,” this sometimes means that our number has been rounded. And there’s something interesting about the numbers that we usually round to. If we look at the two numbers in these examples, 150 and 40, they both end in a zero. Numbers that end in a zero are multiples of 10 and they’re easy to deal with. And in this video, we’re going to be learning how to do exactly what these two people have done, round numbers to the nearest 10. And we’re going to be using three-digit numbers. So let’s start off with an example that’s all about rounding a three-digit number. Let’s imagine that someone’s playing a computer game and they’ve scored 763 points. And then a friend asks us, “About how many points have you scored?” Now by the way that they’ve asked this question, we know that they don’t want an exact number, just a rough idea. So we could round 763 to the nearest 10. One way we can do this is using a number line. Now eventually, we want to put the number 763 on this number line. But to begin with, we need to ask ourselves a question about this number. Which multiples of 10 does this number belong in between? For a clue, we could look at the tens digit of this number, which is a six. This tells us that the nearest multiple of 10 before we get to this number must be 760 and the nearest multiple of 10 after this number must be when this six turns to a seven, 770. Now to round this number, we need to think to ourselves, is 763 nearer to 760 or 770? Now it’s always helpful to mark the halfway point on a number line like this. Halfway between 760 and 770 is 765. Now we’ve got a better idea of where 763 belongs. We know that 763 is less than 765, the halfway point. And so we can now see whether we need to round our number up or down to the nearest 10. 763 is nearest to 760. And so in answer to our friend’s question, we could just say, “About 760.” 763 rounded to the nearest 10 is 760. Let’s imagine we carry on playing the game, and a new score flashes up and our friend asks us the same question. About how many points have you scored? This time, our score is 825, so we need to sketch a number line and think about where this number belongs. Remember, first of all, we can ask ourselves, which two multiples of 10 is this number in between? The tens digit is a two. This gives us a clue as to what the first multiple of 10 is. The nearest multiple of 10 less than 825 is 820. And the nearest multiple of 10 larger than 825 will be when those two tens turn into three tens. It’s 830. And you remember the third thing that we can do. We can think about the halfway point in our number line. What number is halfway between 820 and 830? It’s 825. This is interesting. The number that we want to round is exactly halfway between two multiples of 10. What are we gonna do? We’re going to round it down or up? Well, there’s a rule here that we can use to help us. And we need to remember it when it comes to rounding. If a number is exactly halfway between two multiples of 10, in other words, if it ends in a five, we always round it up. 825 rounded to the nearest 10 is 830. We’re doing a lot of looking at digits here, aren’t we? Perhaps there’s a way we could round numbers using place value to help. And being as we’re about to talk about place value, let’s use place value blocks to make a three-digit number. What’s this number rounded to the nearest 10? We can see that this number is made up of two hundreds, four tens, and six ones. This is the number 246. Now without drawing a number line, which part of this number should we be looking at first? Well, it’s probably a good idea to start looking at the tens digit. This is going to give us a clue as to which tens digits we either need to round down or up to. Which multiples of 10 this number lives in between? Our tens digit here is a four, and this tells us that we’re either going to have to around our number down to 240 or up to 250. Now which part of our number do we need to look at next? It’s the ones digit. This is going to tell us whether to round down or to round up. And you remember in our last example, we had a number that ended in a five. And we said there was a rule for this, and that number that end in a five in the middle of two multiples of 10 always round up. And if we know that numbers that end in a five round upwards, we can include some other digits too, any number that has a ones digit of five or more — so that’s five, six, seven, eight, or nine — we round up. And so we know that any number that has a ones digit of four or less — so that’s four, three, two, or one — we round down. Now there’s one digit we haven’t just mentioned, and that’s a zero. Can you spot why? A number that ends in a zero already is a multiple of 10. We can’t round it up or down; it is. So this is a really important rule to remember: four or less and we round down, five or more and we round up. And we can see in this particular number, we have six ones. And of course, six is greater than five. So we know we’re going to have to round this number up. So we can say 246 rounded to the nearest 10 is 250. Let’s see what you’re like at rounding three-digit numbers to the nearest 10. We’re going to try some questions where you need to practice these skills. To round three-digit numbers to the nearest 10, use a number line. Think about the number 437. We can add in the midpoint to help us. Which two multiples of 10 is 437 between? Hint: We use the digit in the tens place to guide us. What is 437 rounded to the nearest 10? Hint: Think about which side of the midpoint 437 will be on. This question is all about rounding three-digit numbers. When we round a number, we say about what it’s worth. And in this question, we’re rounding numbers to the nearest 10. One way we can do this and the way that this question gets us to do this is by using a number line. And we’re told to think about the number 437. Now, if we want to round the number 437 to the nearest 10, we’re going to need to draw a number line where this number belongs. So our number belongs somewhere on this line. Now, the first part of this question asks us, which two multiples of 10 is 437 between? The number 437 lives or belongs in between two multiples of 10, numbers that end in a zero. And the first thing we can do on our number line is to label these two multiples of 10. Now we’re given a hint here. We’re told that we can use the digit in the tens place to guide us. Now we know the number 437 is made up of four hundreds, three tens, and seven ones. And so the tens digit that we can use to guide us is three. This tells us that the number 437 come somewhere between the number 430 and then when these three tens turn to four tens, 440. So now that we know the number 437 is in between 430 and 440, we need to decide whether to round it up or down. What is 437 rounded to the nearest 10? Again, we’re given a hint to help us here. And we’re told to think about which side of the midpoint 437 will be on. Did you notice the word “midpoint” as we read the question through? If we look right at the beginning of the question when we were given that number line to help us, we can see that the middle of the line has been marked in pink. And it’s also been labeled. We can add in the midpoint to help us. So let’s do this on our number line that we’ve sketched. The midpoint is about here. Now what number is in the middle of 430 and 440? It’s 435. And labeling the midpoint like this can really help us work out whether our number is nearer to 430 or 440. Where would we put 437 on our number line? 437 comes after 435. It’s a greater number. And because we can see that it’s on this side of the midpoint, we can see which multiple of 10 it’s nearest to. We’re going to need to round this three-digit digit number up. We’ve used a number line to help us round a three-digit number to the nearest 10. The two multiples of 10 that 437 is in between are 430 and 440. And we used a midpoint to help us work out which one of these numbers our number was closest to. 437 rounded to the nearest 10 is 440. The teacher asked Michael and David to round the number 572 to the nearest 10. Michael said, “The answer is 570.” David said, “The answer is 580.” Who is correct? What’s happening in this question is the sort of thing that might happen in your maths lessons. A teacher has asked a question, but we’re getting two different answers from two different people. Who is correct? Now the task that both Michael and David have tried to do in this question is to around the number 572 to the nearest 10. And we’re given two possible answers. Either Michael’s right when he said the answer is 570 or David’s right when he said 580. One way we could find the answer is to look at the place value of the number that we’re rounding. Now we know that the number 572 is made up of five hundreds, seven tens, and two ones. Now the first thing we can do with the digits of this number is look at that tens digit. This gives us an idea of the two multiples of 10 that this number lives in between. We can take this digit as it is but change the ones digit for a zero. This gives us the multiple of 10 that is less than our number, 570. And then we can add one to this digit to make it eight tens. And this would give us the multiple of 10 that’s larger than our number, 580. Our number is in between 570 and 580. And you can see that this is where Michael and David have both got their answers from. Michael has thought to himself, I need to round this number down. And David has rounded the number up. And to find out who’s correct, we need to look at another of the digits in our number. This is the ones digit. Now there’s a rule for ones digits and rounding to the nearest 10. Do you remember what it is? If the ones digit is four or less — so that’s four, three, two, or one — we round down. And if the number ends in a five or more — so that’s five, six, seven, eight, or nine — we round up. Now that we’ve remembered this rule, we can apply it to this number. The number of ones in 572 is two. And this is definitely four or less, isn’t it? So we need to round down. 572 rounded to the nearest 10 must be 570. We’ve used place value and our knowledge of rounding to round 572 to the nearest 10. The answer is 570. So the person who is correct is Michael. Which of the following numbers can be 230 when rounded to the nearest 10? 237, 224, 228, or 220. Each of the possible answers to this question is a three-digit number. And this question is all about rounding numbers like these to the nearest 10. Now often, with questions like these where we have to round numbers, we’re given a number, and then we have to round that number to the nearest 10. But this question is slightly different. In this question, we’re given a multiple of 10. That’s the number 230 in our question. And we need to find the number that rounds to this multiple of 10. So in a way, we’re sort of working backwards here. If we were asked to round each of these numbers, which one would we round to 230? Well, there are only four answers to choose from, and one of them we can cross off straightaway. Can you spot which one? Our last answer is a multiple of 10 already; it’s 220. So we would never need to round this number to the nearest multiple of 10. It is a multiple of 10. This leaves us with only three possible answers. Now we know that the number 237 is in between two multiples of 10, 230 and 240. So do you think we need to round this number down to 230 or up to 240? Well, if we look at the ones digit of our number, we can see it’s a seven. And we know that any number that has a ones digit of five or more, we round up. 237 rounded to the nearest 10 is 240 not 230. Our next three-digit number is 224. We can draw this on the same number line, but we are going to have to extend it in one direction here because we know 224 is in between 220 and 230. And the ones digit of this number is a four. Now, whilst any number that ends in a five, six, seven, eight, or nine we round up, any number that ends in a four or less we round down. And so 224 rounded to the nearest 10 is going to be 220 not 230. So this only leaves us with one possible answer. We know that 228 is in between 220 and 230. But this time, the digit is five or more, so we can round up. The number that we write as 230 when we round it to the nearest 10 is 228. What have we learned in this video? We’ve learned how to round three-digit whole numbers to the nearest 10. We have used number lines and our knowledge of place value to help us.
# Calculate: {\text{begin}array l 5x+3y-5z=45 }-5x-y+z=-29 4x+3y+5z=0\text{end}array . ## Expression: $\left\{\begin{array} { l } 5x+3y-5z=45 \\ -5x-y+z=-29 \\ 4x+3y+5z=0\end{array} \right.$ Solve the equation for $x$ $\left\{\begin{array} { l } 5x+3y-5z=45 \\ -5x-y+z=-29 \\ x=-\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z\end{array} \right.$ Substitute the given value of $x$ into the equation $5x+3y-5z=45$ $\left\{\begin{array} { l } 5\left( -\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z \right)+3y-5z=45 \\ -5x-y+z=-29\end{array} \right.$ Substitute the given value of $x$ into the equation $-5x-y+z=-29$ $\left\{\begin{array} { l } 5\left( -\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z \right)+3y-5z=45 \\ -5\left( -\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z \right)-y+z=-29\end{array} \right.$ Simplify the expression $\left\{\begin{array} { l } -3y-45z=180 \\ -5\left( -\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z \right)-y+z=-29\end{array} \right.$ Simplify the expression $\left\{\begin{array} { l } -3y-45z=180 \\ 11y+29z=-116\end{array} \right.$ Multiply both sides of the equation by $-11$ $\left\{\begin{array} { l } 33y+495z=-1980 \\ 11y+29z=-116\end{array} \right.$ Multiply both sides of the equation by $-3$ $\left\{\begin{array} { l } 33y+495z=-1980 \\ -33y-87z=348\end{array} \right.$ Sum the equations vertically to eliminate at least one variable $408z=-1632$ Divide both sides of the equation by $408$ $z=-4$ Substitute the given value of $z$ into the equation $11y+29z=-116$ $11y+29 \times \left( -4 \right)=-116$ Solve the equation for $y$ $y=0$ Substitute the given values of $\begin{array} { l }y,& z\end{array}$ into the equation $x=-\frac{ 3 }{ 4 }y-\frac{ 5 }{ 4 }z$ $x=-\frac{ 3 }{ 4 } \times 0-\frac{ 5 }{ 4 } \times \left( -4 \right)$ Simplify the expression $x=5$ The possible solution of the system is the ordered triple $\left( x, y, z\right)$ $\left( x, y, z\right)=\left( 5, 0, -4\right)$ Check if the given ordered triple is a solution of the system of equations $\left\{\begin{array} { l } 5 \times 5+3 \times 0-5 \times \left( -4 \right)=45 \\ -5 \times 5-0+\left( -4 \right)=-29 \\ 4 \times 5+3 \times 0+5 \times \left( -4 \right)=0\end{array} \right.$ Simplify the equalities $\left\{\begin{array} { l } 45=45 \\ -29=-29 \\ 0=0\end{array} \right.$ Since all of the equalities are true, the ordered triple is the solution of the system $\left( x, y, z\right)=\left( 5, 0, -4\right)$ Random Posts Random Articles
# SOLUTION: begin by solving the linear equation for y. This will put the equation in slope-intercept form. Then find the slope and the y-intercept of the line with this equation. 3x - 4y = Algebra -> Algebra -> Coordinate Systems and Linear Equations -> Lessons -> SOLUTION: begin by solving the linear equation for y. This will put the equation in slope-intercept form. Then find the slope and the y-intercept of the line with this equation. 3x - 4y = Log On Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo . Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Linear Solvers Practice Answers archive Word Problems Lessons In depth Question 186056This question is from textbook : begin by solving the linear equation for y. This will put the equation in slope-intercept form. Then find the slope and the y-intercept of the line with this equation. 3x - 4y = 12This question is from textbook Answer by hkelson(7) (Show Source): You can put this solution on YOUR website!begin by solving the linear equation for y. This will put the equation in slope-intercept form. Then find the slope and the y-intercept of the line with this equation. 3x - 4y = 12 It says to solve the equation for y first. 3x - 4y = 12 : Start by subtracting 3x from both sides -4y = -3x + 12 : Then divide both sides by -4 y = 3/4x - 3 : Dividing the -3 by -4 it became a positive and 12/-4 is negative 3. The equation is now in slope-intercept form. The next part of the question said to figure out the slope and the y-intercept of the line using this equation. Slope-intercept form is y = mx + b m = slope (3/4) b = y-intercept (-3)
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 5.3: Complex Eigenvalues $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ In the last section, we found that if $\textbf{x}'=A\textbf{x}$ is a homogeneous linear system of differential equations, and $$r$$ is an eigenvalue with eigenvector z, then $\textbf{x}=\textbf{z}e^{rt}$ is a solution. (Note that x and z are vectors.) In this discussion we will consider the case where $$r$$ is a complex number $r = l + mi.$ First we know that if $$r = l + mi$$ is a complex eigenvalue with eigenvector z, then $r = l - mi$ the complex conjugate of $$r$$ is also an eigenvalue with eigenvector z. We can write the solution as $\textbf{x}=k_1\textbf{z}e^{(l+mi)t}+ k_2\textbf{z}e^{(l-mi)t}.$ We can use Euler's formula to get $\textbf{x}=k_1 \textbf{z} e^{lt}(\cos(mt)+i\sin(mt))+ k_2 \textbf{z}e^{lt}(\cos(mt)-i\sin(mt)).$ Writing $\textbf{z}=\textbf{a}+\textbf{a}i \;\;\; \text{and} \;\;\; \textbf{z}=\textbf{a}-\textbf{a}i$ we get $\textbf{x}=k_1 (\textbf{a}+\textbf{b}i)e^{lt}(\cos(mt)+i\sin(mt))+ k_2 (\textbf{a}-\textbf{b}i)e^{lt}(\cos(mt)-i\sin(mt)).$ Now multiplying and separating into real and imaginary parts, we get $\textbf{x}=e^{lt}[k_1(\textbf{a} \cos(mt)- \textbf{b} \sin(mt) + i(\textbf{a} \sin(mt)+ \textbf{b} \cos(mt)) \\ + k_2(\textbf{a} \cos(mt)- \textbf{b} \sin(mt) - i(\textbf{a} \sin(mt)+ \textbf{b} \cos(mt))].$ Now let $k_1+k_2=2c_1 \;\;\; \text{and} \;\;\; (k_1-k_2)i=2c_2 .$ Then we get $\textbf{x} =e^{lt}[c_1 (\textbf{a} \cos(mt) - \textbf{b} \sin(mt) + c_2 (\textbf{a} \sin(mt) + \textbf{b} \cos(mt))) ] .$ Example $$\PageIndex{1}$$ Solve the system of differential equations $x'=-2x+6y$ $y'=-3x+4y.$ Solution We have $A=\begin{pmatrix} -2 &6 \\ -3 &4 \end{pmatrix}$ To find the eigenvalues, we find the determinant of $A-rl = \begin{pmatrix} -2-r & 6 \\ -3 & 4-r \end{pmatrix} .$ We get $(-2-r)(4-r)+18 = r^2-2r+10=0.$ The quadratic formula gives the roots $r=1+3i \;\;\; \text{and} \;\;\; r=1-3i .$ Now we find and eigenvector corresponding to the eigenvalue $$1 + 3i$$. Plugging into $$A - rI$$, we get $A-(1+3i)l = \begin{pmatrix} -3-3i & 6 \\ -3 &3-3i \end{pmatrix}.$ The top row gives $(-3 - 3i)x + 6y = 0$ or $(1 + i)x - 2y = 0.$ An eigenvector is $z=\begin{pmatrix} 2 \\ 1+i \end{pmatrix}= \begin{pmatrix} 2 \\ 1 \end{pmatrix} + i\begin{pmatrix} 0 \\ 1 \end{pmatrix}.$ Hence the general solution is $x=e^t \left[ c_1\left( \begin{pmatrix}2\\1\end{pmatrix}\cos(3t)-\begin{pmatrix}0\\1\end{pmatrix}\sin(3t) \right) + c_2 \left(\begin{pmatrix}2\\1\end{pmatrix}\sin(3t) + \begin{pmatrix}0\\1\end{pmatrix} \cos(3t) \right) \right] .$ This can be written as $x=e^t [2c_1 \cos(3t) + 2c_2\sin(2t) ]$ $y=e^t[c_1 (\cos(3t)+\sin(3t)) + c_2 (\sin(3t)+ \cos(3t) )] .$ Below is the phase portrait Clearly the solutions spiral out from the origin, which is called a spiral node. The spiral occurs because of the complex eigenvalues and it goes outward because the real part of the eigenvalue is positive. If the real part of the eigenvalue had been negative, then the spiral would have been inward.
# Category Archives: Operation ## 3rd Edition – Time 4 Fractions – Problem #1 – Walking along a pond Welcome to our first problem ! This week will be a warm-up, as I want to make sure we are all aboard and comfortable with pursuing the journey from home. Bear with me with the length of this post, next week will be much shorter. The goal of this journey is to provide opportunities for children to explore word problems in “any way that they wish” (Carpenter et al, 2015, page 80), extend their reasoning skills, and gradually strengthen their foundation in fractions. Each problem is differentiated to target all elementary grades and is quite short. A child may be done within 5-10 min, or may decide to take more time to fully explore it with a visual representation and manipulatives. It is not a test, it is not a race. Week after week, problem after problem, children strengthen their reasoning skills by creating their own strategies to solve problems. When children receive their formal fraction instruction in class, they will have a stronger background to build upon. If you decide to take the journey with us, from home, I hope you will enjoy observing your child’s thinking as much as I do with mine. It is fascinating. They explore. We listen. So, here we go: Problem #1 – Walking along the pond • Level Yellow : Mr. Wood is walking along a pond. He sees 3 waterlily pads. On each pad, there are 2 frogs. How many frogs does Mr. Wood see ? • Level Orange: Mr. Wood is walking along a pond. He sees 4 giant waterlily pads. On each pad, there are 5 frogs. How many frogs does Mr. Wood see ? • Level Red : Complete the problem with the numbers of your choice. Mr. Wood is walking along a pond. He sees ____ giant waterlily pads. On each pad, there are ___ flies. How many flies does Mr. Wood see ? (e.g. 10 pads and 5 flies; 12 pads and 8 flies; 13 pads and 21 flies, etc.) What to do as a parent ? Invite your child to solve one of the problems, and listen to his/her way of solving it. He/she can make sense of the problem while using small objects (such as buttons, marbles, etc, and small containers) or drawing a picture. He/she may write an equation. I purposely stepped away from grade level. Each child should pick the problem that he/she feels like exploring. If your child calls out the answer right away, remind him/her that the answer is fine, but how it was obtained is even more important in this journey. How would he/she explain it to a younger child? Could he/she represent the problem with a drawing? a diagram? Using small objects ? If your child is not used to solving multiplication problems, you may have to read the problem again, and say things like “I am wondering if these cups and buttons could help us solve the problem” or “Do you think it would help to draw the situation? What should we draw?”. Level Yellow is great for that. Just resist to showing him/her how you would solve the problem. I am including a link to 2 videos that we did a while ago. Just bear with the French accent, the camera made me quite uncomfortable… : • Video Level Yellow : this short video (2 min) shows the material we use at home, and how a child may solve Level Yellow with a drawing • Video Level Orange : this one (3 min) is an example of a child solving Level Orange with manipulative These videos are just examples, but I hope they help you see what can be done at home. It is all about the exploration. Your child may not use the same approach, but as long as he/she solve the problem a way that makes sense to him/her, it is all that matters. One more thing: you are right, there is no fraction involved in this problem. Just remember that we are going to explore the concept gradually. We will start with 2 weeks on Multiplication problems (see problem #1) above. Then, we will continue with 2 weeks on Measurement Division problems (Carpenter et al, 2015). E.g. An elf has 10 berries and some bags. He wants to put 2 berries in each bag. How many bags can he fill? Finally, we will explore Partitive Division problems and Equal Sharing problems, the core of our fractions exploration (Epson & Levi, 2011). E.g. An elf has 15 berries. He puts the berries into 3 bags with the same number in each bag. How many berries are in each bag ? E.g. Two elves want to share 5 berries so that each of them gets the same amount. How many berries would each get? Please, feel free to comment or email at journey2helpchildrenwithmath(at)gmail(dot)com if you have any question about our journey. The more feedback I receive, the more complete the next post will be ! References: • Carpenter, T., Fennema, E., Franke, M., Levi, L. and Empson S. (2015). Children’s Mathematics, Second Edition: Cognitively Guided Instruction. Portsmouth, NH: Heinemann. ISBN-13:978-0325052878. • Empson, S. E., and Levi, L. (2011). Extending Children’s Mathematics: Fractions and Decimals. Portsmouth, NH : Heinemann. ISBN-13: 978-0325030531. ## Update Time 4 Fractions – Problem #1 – Walking along a pond My daughter and I went on a 12 week- journey last year to explore Fractions. We are doing it again this Fall. I am updating the posts from last year, in case you want to join us this year. I am adding videos this time :-) Click here if you want to know more about the journey. Welcome to our first problem ! This week will be a warm-up, as I want to make sure we are all aboard and comfortable with pursuing the journey from home. Bear with me with the length of this post, next week will be much shorter. The goal of this journey is to provide opportunities for children to explore word problems in “any way that they wish” (Carpenter et al, 2014, page 80), extend their reasoning skills, and gradually strengthen their foundation in fractions. Each problem is differentiated to target all elementary grades and is quite short. A child may be done within 5-10 min, or may decide to take more time to fully explore it with a visual representation and manipulatives. It is not a test, it is not a race. Week after week, problem after problem, children strengthen their reasoning skills by creating their own strategies to solve problems. When children receive their formal fraction instruction in class, they will have a stronger background to build upon. If you decide to take the journey with us, from home, I hope you will enjoy observing your child’s thinking as much as I do with mine. It is fascinating. They explore. We listen. So, here we go: Problem #1 – Walking along the pond • Level Yellow : Mr. Wood is walking along a pond. He sees 3 waterlily pads. On each pad, there are 2 frogs. How many frogs does Mr. Wood see ? • Level Orange: Mr. Wood is walking along a pond. He sees 4 giant waterlily pads. On each pad, there are 5 frogs. How many frogs does Mr. Wood see ? • Level Red : Complete the problem with the numbers of your choice. Mr. Wood is walking along a pond. He sees ____ giant waterlily pads. On each pad, there are ___ flies. How many flies does Mr. Wood see ? What to do as a parent ? 1. Invite your child to solve one of the problems. He/she can model the problem with some manipulatives ( such as buttons, marbles, etc, and small containers), represent the problem on a piece of paper, write an equation. 2. I purposely stepped away from grade level. Each child should pick the problem that he/she feels like exploring. In the coming weeks, some upper graders may decide to pick a Level orange to model, represent, and write the equation. Some lower graders may decide to pick a Level Red and model it only. 3. When your child is done, invite him/her to share his/her reasoning with you. If he/she writes only an equation, encourage him to share his/her strategy another way (with a visual representation, or with manipulatives). Enjoy following his/her way of thinking. Just resist teaching him/her symbols ! They will come in time ! 4. Our journey starts slowly, as children must explore a variety of problems, and build up strategies they can use with fractions later on. If you child solve Level Red, providing an equation and another way of representing his/her strategy, you can always invite him/her to invent his/her own multiplication problem. 5. One more thing, you may want to find an example on how fractions could be useful in your child’s life. For instance, my child loves working with me in the workshop, where we measure, add length, etc. That way, even if the problem is on a topic that may be less appealing to her, I can remind her why, in the long run, it can contribute to her learning all the same. I am including a link to 2 videos this week: • Video Level Yellow : this short video (2 min) shows the material we use at home, and how a child may solve Level Yellow with a drawing • Video Level Orange : this one (3 min) is an example of a child solving Level Orange with manipulative These videos are just examples, but I hope they help you see what we do at home. It is all about the exploration. Your child may not use the same approach, but as long as he/she solve the problem a way that makes sense to him/her, it is all that matters. Questions you may have ? I do not see any fraction in this problem ! Good point ! Just remember that we are going to explore the concept gradually. We will start with 2 weeks on Multiplication problems (see problem #1) above. Then, we will continue with 2 weeks on Measurement Division problems (Carpenter et al, 2014). E.g. An elf has 10 berries and some bags. He wants to put 2 berries in each bag. How many bags can he fill? Finally, we will explore Partitive Division problems and Equal Sharing problems, the core of our fractions exploration (Epson & Levi, 2011). When we reach that step, our children will have developed/reinforced the habit of modeling directly a problem with manipulative, or representing the problem on paper, and will naturally continue to do so with fractions. E.g. An elf has 15 berries. He puts the berries into 3 bags with the same number in each bag. How many berries are in each bag ? E.g. Two elves want to share 5 berries so that each of them gets the same amount. How many berries would each get? My child does not know how to start If your child is not used to solving multiplication problems, it is to be expected ! You may read the problem again, and say things like “I am wondering if these cups and buttons could help us solve the problem” or “Do you think it would help to draw the situation? What should we draw?”. Level Yellow is great for that. My child provides the answer straight away Remind him/her that the answer is important, but how it was obtained is even more important. That’s when learning happens ! Invite him/her to model and/or represent the problem with a drawing or a diagram. Invite him/her to write an equation and connect each part of the equation to his/her model and/or representation. Please, feel free to comment or email at journey2helpchildrenwithmath(at)gmail(dot)com if you have any question about our journey. The more feedback I receive, the more complete the next post will be ! Let’s build up a community of people supporting at home what our children learn during Math instruction ! References: • Carpenter, T., Fennema, E., Franke, M., Levi, L. and Empson S. (2014). Children’s Mathematics, Second Edition: Cognitively Guided Instruction. Portsmouth, NH: Heinemann. ISBN-13:978-0325052878. • Empson, S. E., and Levi, L. (2011). Extending Children’s Mathematics: Fractions and Decimals. Portsmouth, NH : Heinemann. ISBN-13: 978-0325030531. ## WedWoPro #14 – Last one ! Every Wednesday, I give a chance to my child to explore a word problem a way that makes sense to her. And every Wednesday, I share the word problem, and my experience with you, so you can do the same ! Click here to start from the beginning ! Hope you join us ! Here comes WedWoPro #14, that will end our journey for a little while. I don’t know if Rosie is getting tired with the end of the year but I think it is time to take a little break in word problems to come back even stronger once Summer break starts. WedWoPro #14 – Last one Today, you are the teacher ! Write a word problem, and I will solve it ! Sharing my experience: Here is what I had to solve : “Once upon a time, 6 butterflies came to have an ice-cream at a friend’s house because it was very hot outside. 4 more came at the friend’s house to have ice cream. How many butterflies were there in total? One butterfly left. Then, 3 more left. Then, 3 more left again. How many butterflies were still at the friend’s house?” I love doing this kind of task, and opening the door to creativity. It always lead to fun discussion. Today, it gave us a chance to discuss again how to select information that was necessary to solve the problem (e.g. 6 butterflies, 4 more, etc) vs the information that was not (e.g. it was very hot outside). It was also a fun way to model how I would solve the problem, and share my reasoning, by pretending to be the kid. Until next time ! ## WedWoPro #13 – Few more days ! Every Wednesday, I give a chance to my child to explore a word problem a way that makes sense to her. And every Wednesday, I share the word problem, and my experience with you, so you can do the same ! Click here to start from the beginning ! Hope you join us ! Here comes WedWoPro #13. We will be working with time for another week. Just to show Rosie that time is not only about minutes and hours. A one-step problem. WedWoPro #13 – Few more days ! Rosie was very excited. Soon, his grandma would come to visit. • “I can’t wait to see her tomorrow ! “ • “I can’t wait to see her ! Just 2 more days !” • “I can’t wait to see her ! Just 3 more weeks !” If today is Wednesday, April 13, when is Grandma coming ? Here is the .pdf if you want to print it out (WedWoPro13). • Level Green: Grandma is coming on Thursday, April 14. • Level Orange: Grandma is coming on Friday, April 15. • Level Red: Grandma is coming on Wednesday, May 4. As always, invite your child to solve the level of his/her choice a way that is meaningful to him/her (Carpenter et al., 2014) ! Sharing my experience: Well, Rosie’s grandma actually arrived TODAY from France. Too much excitement around here… Will have to update the post later ! A bientôt ! Reference: • Carpenter, T., Fennema, E., Franke, M., Levi, L. and Empson S. (2014). Children’s Mathematics, Second Edition: Cognitively Guided Instruction. Portsmouth, NH: Heinemann. ISBN-13:978-0325052878. ## Making sense of subtracting in column I thought I should complete my latest post, making sense of adding in column (here), with a quick post on using Base Ten blocks to make sense of subtracting in column. My daughter is not there yet, but your child may be. The picture above presents a concrete illustration of 43 – 15. It is quite helpful for kids to visualize that, when subtracting 15 to 43, they trade a Ten from the Ten column into Ones. Then, they can subtract 5 Ones to 13 Ones, 1 Ten to 3 Tens, and end up with 2 Tens and 8 Ones i.e. 28. I just spent the morning in my son’s classroom. He is 4 and attends a Montessori school. Their approach to teaching math is amazing. Indeed, using concrete objects, little 4-5 years old kiddos solve 4-digits addition without even thinking about it. I should write a special post on the Montessori approach to Math. Quite inspiring, indeed. ## Making sense of adding in column Adding in column 23 + 14 ? 3 +4 = 7, 2 + 1 = 3… so the answer is 37. Adding 37 + 44 ? 7 +4 = 11, 3 + 4 = 7… so the answer is 711. Wait, Mom. It does not make sense, doesn’t it? Nope, Rosie, 711 doesn’t seem to make sense. So let’s step back an inch, with the Base Ten Blocks (click here if you want to read more about these blocks). Here is an example of 32 +23, and the connection between the blocks, and the addition in column. While adding in column, you add the Ones, then the Tens, then the Hundreds, and so on, and the blocks provide a neat concrete representation of such process. Indeed, it shows why you have to “align” digits (because if you don’t, you end up adding Ones to Tens !). But what I like the most with these blocks is how they help children make sense of carrying an over to the next column. Here is an example with 37 +44. From the 11 Ones you get from the right column (i.e. the Ones column 7+4), you trade 10 Ones from 1 Ten that you carry over to the left column (i.e. the Tens column). Here you go, Rosie, 711 does not make sense, but 8 Tens 1 Ones aka 81 does. ## WedWoPro #12 – You have few more minutes ! Every Wednesday, I give a chance to my child to explore a word problem a way that makes sense to her. And every Wednesday, I share the word problem, and my experience with you, so you can do the same ! Click here to start from the beginning ! Hope you join us ! Here comes WedWoPro #12. We will be working with time this week ! WedWoPro #12 – You have few more minutes ! Rosie was reading a book about unicorns and fairies. Soon, it will be time to get ready to go to School. • “It is 7:00. In 30 minutes, it will be time to go to School.” • “It is 7:00. In 15 minutes, it will be time to go to School.” • “It is 6:50. If you read for 20 min, you will have 15 min left before it is time to go to School.” What time does Rosie need to go to School? Here is the .pdf if you want to print it out (WedWoPro12). • Level Green: Rosie needs to leave at 7:30. • Level Orange: Rosie needs to leave at 7:15 • Level Red: Rosie needs to leave at 7:25 As always, invite your child to solve the level of his/her choice a way that is meaningful to him/her (Carpenter et al, 2014)! Sharing my experience: Here came a loud “7:30!” as soon as Rosie finished reading the first level. OK, Rosie, but I want more… How did you figure it out? So she started drawing a clock, the 2 hands, writing 1 to 12 on it. What are these numbers for ? What about the hands? How would the big hand move in Level Green? How about the little hand? In Level Orange? How could you figure it out Level Red? And so on. And so on. Indeed, these 3 levels took us to a fun discussion on how clock works. So engaging that I will come up soon with a post dedicated to time…. Stay tuned, I should finally have more time to write on my blog very soon. Until next time ! Reference: • Carpenter, T., Fennema, E., Franke, M., Levi, L. and Empson S. (2014). Children’s Mathematics, Second Edition: Cognitively Guided Instruction. Portsmouth, NH: Heinemann. ISBN-13:978-0325052878.
LCM the 10 and also 15 is the smallest number amongst all usual multiples of 10 and also 15. The first few multiples the 10 and 15 are (10, 20, 30, 40, 50, 60, 70, . . . ) and (15, 30, 45, 60, 75, 90, . . . ) respectively. There are 3 generally used techniques to uncover LCM the 10 and 15 - by prime factorization, by department method, and by listing multiples. You are watching: Least common denominator of 10 and 15 1 LCM of 10 and also 15 2 List the Methods 3 Solved Examples 4 FAQs Answer: LCM of 10 and also 15 is 30. Explanation: The LCM of two non-zero integers, x(10) and y(15), is the smallest confident integer m(30) the is divisible through both x(10) and y(15) without any type of remainder. The approaches to discover the LCM of 10 and also 15 are explained below. By department MethodBy Listing MultiplesBy prime Factorization Method ### LCM of 10 and 15 by division Method To calculate the LCM that 10 and also 15 by the department method, we will certainly divide the numbers(10, 15) by their prime components (preferably common). The product of this divisors provides the LCM that 10 and 15. Step 3: proceed the measures until only 1s are left in the critical row. The LCM that 10 and 15 is the product of all prime numbers on the left, i.e. LCM(10, 15) by department method = 2 × 3 × 5 = 30. ### LCM the 10 and 15 by Listing Multiples To calculate the LCM of 10 and 15 through listing out the common multiples, we have the right to follow the given below steps: Step 1: perform a couple of multiples of 10 (10, 20, 30, 40, 50, 60, 70, . . . ) and 15 (15, 30, 45, 60, 75, 90, . . . . )Step 2: The common multiples indigenous the multiples the 10 and also 15 room 30, 60, . . .Step 3: The smallest common multiple that 10 and 15 is 30. ∴ The least common multiple of 10 and 15 = 30. See more: Where Do I Get Squirtle In Pokemon Yellow :: Getting The Starters ### LCM that 10 and also 15 by prime Factorization Prime administrate of 10 and also 15 is (2 × 5) = 21 × 51 and (3 × 5) = 31 × 51 respectively. LCM of 10 and 15 can be acquired by multiplying prime determinants raised to your respective highest possible power, i.e. 21 × 31 × 51 = 30.Hence, the LCM that 10 and also 15 by element factorization is 30.
# ACT Math : How to find the length of the side of a right triangle ## Example Questions ### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle Given a right triangle with a leg length of 6 and a hypotenuse length of 10, find the length of the other leg, x. 64 4 16 8 8 Explanation: Using Pythagorean Theorem, we can solve for the length of leg x: x2 + 62 = 102 Now we solve for x: x2 + 36 = 100 x2 = 100 – 36 x2 = 64 x = 8 Also note that this is proportionally a 3/4/5 right triangle, which is very common. Always look out for a side-to-hypoteneuse ratio of 3/5 or 4/5, or a side-to-side ratio of 3/4, in any right triangle, so that you may solve such triangles rapidly. ### Example Question #81 : Right Triangles In a right triangle a hypotenuse has a length of 8 and leg has a length of 7. What is the length of the third side to the nearest tenth? 2.4 1.0 3.6 3.9 Explanation: Using the pythagorean theorem, 82=72+x2. Solving for x yields the square root of 15, which is 3.9 ### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle Given a right triangle with a leg length of 2 and a hypotenuse length of √8, find the length of the other leg, x. 2 10 √8 6 4 2 Explanation: Using Pythagorean Theorem, we can solve for the length of leg x: x2 + 22 = (√8)2 = 8 Now we solve for x: x2 + 4 = 8 x2 = 8 – 4 x2 = 4 x = 2 ### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle The legs of a right triangle are and . Rounded to the nearest whole number, what is the length of the hypotenuse? Explanation: Use the Pythagorean Theorem. The sum of both legs squared equals the hypotenuse squared. ### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle Points , , and are collinear (they lie along the same line). , , Find the length of segment . Explanation: The length of segment is Note that triangles and are both special, 30-60-90 right triangles. Looking specifically at triangle , because we know that segment has a length of 4, we can determine that the length of segment is 2 using what we know about special right triangles. Then, looking at triangle now, we can use the same rules to determine that segment has a length of which simplifies to . ### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle A handicap ramp is long, and a person traveling the length of the ramp goes up vertically. What horizontal distance does the ramp cover?
Notes On Algebra of Continuous Functions - CBSE Class 12 Maths Let f(x) and g(x) be two real functions continuous at real number c. $\underset{xâ†’\text{c}}{\text{lim}}\text{f(x) = f(c)}$ . Now check the continuity of the sum, difference, product and quotient of these functions f + g, f - g, f . g and f/g Since f(x) and g(x) are continuous at c, f(x) and g(x) are defined at c. i) (f + g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) + g(x)]}$$\text{}$ = $\text{}$ = f(c) + g(c) = (f + g)(c) â‡’ Hence, f + g is continuous at x = c. ii) (f - g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) â€“ g(x)]}$$\text{}$ = $\text{}$ = f(c) - g(c) = (f - g)(c) $\underset{xâ†’\text{c}}{\text{lim}}\text{(fâ€“g)(x) =}$ (f - g)(c) Hence, f - g is continuous at x = c. iii) (f . g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) . g(x)]}$$\text{}$ = = f(c) . g(c) = (f . g)(c) â‡’ (f.g)(c) Hence, f . g is continuous at x = c. iv) f(x)/g(x), g(c) â‰ 0 is also defined at c. (f/g)(x) = f(x)/g(x) $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) / g(x)]}$$\text{}$ = = f(c)/g(c) = (f/g)(c) Hence, f/g and g â‰ 0 is continuous at x = c. Special case: f(x) = k, where k is a real number. â‡’ f(x). g(x) = (k . g)(x) = k . g(x) is also continuous. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) . g(x)]}$$\text{}$ $\text{}$ = k .$\underset{xâ†’\text{c}}{\text{lim}}\text{g(x)}$$\text{}$ If f(x) is continuous at c and k âˆˆ R, then k. f(x) is also continuous at c. k = -1 â‡’ -f is continuous. If f(x) is continuous at c the, then -f(x) is also continuous at c. â‡’ f(x)/g(x) = k/g(x) is also continuous. If g(x) is continuous at c and k âˆˆ R, then k/g(x) is also continuous at c provided g(c) â‰ 0. k = 1 â‡’ 1/g(x) is continuous. The reciprocal of a continuous function is also continuous #### Summary Let f(x) and g(x) be two real functions continuous at real number c. $\underset{xâ†’\text{c}}{\text{lim}}\text{f(x) = f(c)}$ . Now check the continuity of the sum, difference, product and quotient of these functions f + g, f - g, f . g and f/g Since f(x) and g(x) are continuous at c, f(x) and g(x) are defined at c. i) (f + g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) + g(x)]}$$\text{}$ = $\text{}$ = f(c) + g(c) = (f + g)(c) â‡’ Hence, f + g is continuous at x = c. ii) (f - g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) â€“ g(x)]}$$\text{}$ = $\text{}$ = f(c) - g(c) = (f - g)(c) $\underset{xâ†’\text{c}}{\text{lim}}\text{(fâ€“g)(x) =}$ (f - g)(c) Hence, f - g is continuous at x = c. iii) (f . g)(x) is also defined at c. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) . g(x)]}$$\text{}$ = = f(c) . g(c) = (f . g)(c) â‡’ (f.g)(c) Hence, f . g is continuous at x = c. iv) f(x)/g(x), g(c) â‰ 0 is also defined at c. (f/g)(x) = f(x)/g(x) $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) / g(x)]}$$\text{}$ = = f(c)/g(c) = (f/g)(c) Hence, f/g and g â‰ 0 is continuous at x = c. Special case: f(x) = k, where k is a real number. â‡’ f(x). g(x) = (k . g)(x) = k . g(x) is also continuous. $\underset{xâ†’\text{c}}{\text{lim}}\text{[f(x) . g(x)]}$$\text{}$ $\text{}$ = k .$\underset{xâ†’\text{c}}{\text{lim}}\text{g(x)}$$\text{}$ If f(x) is continuous at c and k âˆˆ R, then k. f(x) is also continuous at c. k = -1 â‡’ -f is continuous. If f(x) is continuous at c the, then -f(x) is also continuous at c. â‡’ f(x)/g(x) = k/g(x) is also continuous. If g(x) is continuous at c and k âˆˆ R, then k/g(x) is also continuous at c provided g(c) â‰ 0. k = 1 â‡’ 1/g(x) is continuous. The reciprocal of a continuous function is also continuous Previous Next âž¤
SHARE: In Geometry, transformation means changing the appearance of an object. There are many ways in which this can be done viz translation, reflection, rotation etc. Rotation in math definition Rotation is a transformation in which a geometric figure is turned around a certain point. Thus in rotation there is a central point that is fixed and the figure moves around that point in a circle. A complete rotation is when the object rotates by 360°. In rotation, the figure does not change its size or shape. The only change is in the direction. Terms related to rotation of shape Let us understand some of the terms that are frequently used along with the rotation function: Preimage - this term is used to refer to the original object Image - this term is used to refer to the object after it has been rotated. Angle of rotation - the amount by which the object is rotated is measured in degrees and is called the angle of rotation. Properties of a rotation • In rotation, the distance between any point on the shape and the center of rotation always remains the same. • The point around which the object is rotated can be a point inside the object itself or a point outside it. • When the object is rotated in a clockwise direction then the angle of rotation is negative and when the object is rotated in an anti-clockwise direction, the angle of rotation is positive. Did you know? When a triangle is rotated about X-axis, it forms a cone with its axis of symmetry as the same X-axis. Rotating 90° is the same as rotating 270° in the opposite direction. Practice questions: Question: Draw the new position of the point A after it has been rotated around the origin (0,0) Pre-image of point A is as shown in the figure below: Rotation 1. 90° clockwise rotation 2. 180° clockwise rotation 3. 270° clockwise rotation 4. 360° rotation 5. 90° counter clockwise rotation 6. 270° counter clockwise rotation Question: Draw the new position of the triangle ABC after it has been rotated around the origin (0,0) Pre-image of the triangle ABC is as shown in the figure below Rotation 1. 90° clockwise rotation 2. 180° clockwise rotation 3. 270° clockwise rotation 4. 360° rotation 5. 90° counter clockwise rotation 6. 270° counter clockwise rotation As you can see, a 360° rotation brings the shape back to its original position. SHARE: © Hozefa Arsiwala and teacherlookup.com, 2018-2019. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Hozefa Arsiwala and teacherlookup.com with appropriate and specific direction to the original content.
# NCERT Solutions for Class 12 Maths Chapter 13 - Probability - Exercise 13.3 ### Access Exercises of Class 12 Maths Chapter 13 – Probability Exercise 13.1 Solutions 17 Questions Exercise 13.2 Solutions 18 Questions Exercise 13.3 Solutions 14 Questions Exercise 13.4 Solutions 17 Questions Exercise 13.5 Solutions 15 Questions Miscellaneous Exercise on Chapter 13 Solutions 10 Questions Exercise 13.3 1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreoever, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red? Sol. The urn contains 5 red and 5 black balls. i.e., n(R) = 5, n(B) = 5 and n(S) = 10 Let a ball be drawn in the first attempt. $$\therefore\space\text{P(drawing a red ball)}\\=\frac{\text{n(R)}}{\text{n(S)}} = \frac{5}{10} =\frac{1}{2}$$ If two red balls are added to urn, then the urn contains 7 red and 5 black balls i.e., n(R) = 7, n(B) = 5 and n(S) = 12 $$\text{P(drawing a red ball)}\\=\frac{\text{n(R)}}{\text{n(S)}}= \frac{7}{12}$$ Let a black ball be drawn in the first attempt Then, n(R) = 5, n(B) = 5 and n(S) = 10 ∴ P(drawing a black ball in the first attempt) $$= \frac{\text{n(B)}}{\text{n(S)}} = \frac{5}{10} =\frac{1}{2}$$ If two black balls are added to the urn, then the urn contains 5 red and 7 black balls. i.e., n(R) = 5, n(B) = 7 and n(S) = 12 $$\therefore\space\text{P(drawing a red ball) =}\\\frac{\text{n(R)}}{\text{n(S)}} =\frac{5}{12}$$ Therefore, probability of drawing second ball as red is $$=\frac{1}{2}×\frac{7}{12} + \frac{1}{2}×\frac{5}{12}\\=\frac{1}{2}\bigg(\frac{7}{12} + \frac{5}{12}\bigg)\\=\frac{1}{2}×1 =\frac{1}{2}$$ 2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. Sol. Let E1 : first bag is selected, E2 : second bag is selected. Then, E1 and E2 are mutually exclusive and exhaustive events. Moreover, $$\text{P(E}_1) = \text{P(E}_2) = \frac{1}{2}$$ Let E : ball drawn is red. $$\therefore\space \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = \text{P}\\\text{(drawing a red ball} \\\text{from second bag)}\\=\frac{2}{8} =\frac{1}{4}$$ By Baye's theorem, $$\text{Required probability =}\\\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg)\\=\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P(E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}\\=\frac{\frac{1}{2}×\frac{1}{2}}{\frac{1}{2}×\frac{1}{2} + \frac{1}{2}×\frac{1}{4}}\\=\frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}}\\=\frac{\frac{1}{4}}{\frac{2+1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}}$$ $$= \frac{1}{4}×\frac{8}{3} =\frac{2}{3}$$ 3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier? Sol. Let E1 : the event that the students is residing in hostel and E2 : the event that the students in not residing in the hostel. Let E : a student attains A grade, Then, E1 and E2 are mutually exclusive and exhaustive events. Moreover, $$\text{P}(E_1) = 60\% =\\\frac{60}{100} = \frac{3}{5}\\\text{and}\space\text{P}(\text{E}_2) = 40\% = \\\frac{40}{100} =\frac{2}{5}\\\text{Then,}\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = 30\%\\=\frac{30}{100}=\frac{3}{10}$$ $$\text{and}\space \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = 20\%\\=\frac{20}{100} =\frac{2}{10}$$ By using Baye's theorme, we obtain $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) \\=\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}\\=\frac{\frac{3}{10}×\frac{3}{5}}{\frac{3}{10}×\frac{3}{5} + \frac{2}{10}×\frac{2}{5}}\\=\frac{9}{9 + 4}= \frac{9}{13}$$ 4. In answering a question on a mutliple choice test, a student either knows the answer or guesses. $$\textbf{Let\space}\frac{\textbf{3}}{\textbf{4}}\space\textbf{be the probability that he}\\\textbf{knows the answer and}\space\frac{\textbf{1}}{\textbf{4}}\textbf{be the}$$ probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $$\frac{\textbf{1}}{\textbf{4}}\textbf{.}\space\textbf{What is the probability}$$ that the student knows the answer given that he answered it correctly? Sol. Let E1 : the event that the students knows the answer and E2 : the event that the student guesses the answer. Therefore, E1 and E2 are mutually exclusive and exhaustive events. $$\therefore\space \text{P(E}_1) = \frac{3}{4}\space\text{and P}(\text{E}_2) = \frac{1}{4}$$ Let E : the answer is correct. The probability that the student answered correctly given that he knows the answer, is 1 i.e. $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = 1$$ Probability that the students answered correctly, given that the he guessed, is $$\frac{1}{4}\space\text{i.e.,}$$ $$\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)=\frac{1}{4}.$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) =\\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_2)}\\=\frac{1×\frac{3}{4}}{1×\frac{3}{4} + \frac{1}{4}×\frac{1}{4}}\\=\frac{\frac{3}{4}}{\frac{3}{4} +\frac{1}{16}} =\frac{\frac{3}{4}}{\frac{12+1}{6}}\\=\frac{3}{4}×\frac{16}{13} = \frac{12}{13}$$ Note : If two events are not mutually exclusive/exhaustive, then you do not use Baye's theorem. 5. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive? Sol. Let E1 : the event that the person has disease and E2 : the event that the person is healthy. Then, E1 and E2 are mutually exclusive and exhaustive events. $$\text{Moreever,}\space\text{P(E}_1) =0.1\% \\= \frac{0.1}{100} = 0.001$$ and P(E2) = 1 – 0.001 = 0.999 Let E : the event that test is positive. $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{(P(result is positive}\\\text{given the person has disease})\\ = 99\% = \frac{99}{100} = 0.99$$ Probability that a person does not have disease and test result is positive. $$\therefore\space \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = 0.5\% \\= \frac{0.5}{100} = 0.005$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg)= \\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}$$ $$=\frac{0.001×0.99}{0.001×0.99×0.999×0.005}\\=\frac{0.00099}{0.00099 + 0.004995}\\ = \frac{0.00099}{0.005985} =\frac{990}{5985}\\=\frac{110}{665} = \frac{22}{133}$$ 6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin? Sol. Let E1 : the event that the coin chosen is two headed, E2 : the event that the coin chosen is biased and E3 : the event that the coin chosen is unbiased. ⇒ E1, E2, E3 are mutually exclusive and exhausitive events. Morever, $$\text{P(E}_1) =\text{P(E}_2) =\text{P(E}_3) = \frac{1}{3}$$ Let E : tossed coin shows up a head, $$\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{P (coin showing}\\\text{heads, given that it is a two}\\\text{headed coin)}=1\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\text{P}(\text{coin showing heads,}\\\text{given that it is a biased coin})\\=75\% =\frac{75}{100} =\frac{3}{4}$$ $$\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\text{P}(\text{coin showing head,}\\\text{given that it is an unbiased coin})\\=\frac{1}{2}$$ The probability that the coin is two headed, given that is shows head, is given by $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg)$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) =\\\frac{\text{P}\bigg(\frac{E}{E_1}\bigg)\text{P(\text{E}}_1)}{\text{P}\bigg(\frac{E}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}\text{(E}_2) + \text{p}\bigg(\frac{\text{E}}{\text{E}_3}\bigg)\text{P}(\text{E}_3)}\\=\frac{1×\frac{1}{3}}{1×\frac{1}{3} + \frac{3}{4}×\frac{1}{3} + \frac{1}{2}×\frac{1}{3}}\\=\frac{1}{1 + \frac{3}{4} + \frac{1}{2}}\\=\frac{1}{\frac{4+3+2}{4}} =\frac{4}{9}$$ 7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? Sol. There are 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total number of drivers = 2000 + 4000 + 6000 = 12000 Let E1 : the event that insured person is a scooter driver, E2 : the event that insured person is a car driver and E3 : the event that insured person is a truck driver. Then, E1, E2, E3 are mutually exclusive and exhaustive events. Moreover, $$\text{P(E}_1) =\\\frac{\text{Number of scooter drivers}}{\text{Total number of drivers}}\\=\frac{2000}{12000} = \frac{1}{6}\\\text{P(E}_2) =\\ \frac{\text{Number of car drivers}}{\text{Total number of drivers}}\\=\frac{4000}{12000} =\frac{1}{3}\\\text{and}\space\text{P}(\text{E}_3) =\\\frac{\text{Number of truck drivers}}{\text{Total number of drivers}}\\=\frac{6000}{12000}= \frac{1}{2}$$ Let E : the events that insured person meets with an accident. $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{P(scooter driver met}\\\text{with an accident)}\\= 0.01 = \frac{1}{100}\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\text{P}(\text{car driver met}\\\text{ with an accident})\\=0.03 = \frac{3}{100}\\\text{P}\bigg(\frac{\text{E}}{\text{E}_3}\bigg)= \text{P}\text{(truck driver met}\\\text{ with an accident)}\\=0.15 = \frac{15}{100}$$ Probability that the driver is a scooter driver, given he met with an accident is given by $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg).$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) =\\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2) +\text{P}\bigg(\frac{E}{\text{E}_3}\bigg)\text{P}(\text{E}_3)}$$ $$= \frac{\frac{1}{6}×\frac{1}{100}}{\frac{1}{6}×\frac{1}{100} + \frac{1}{3}×\frac{3}{100} + \frac{1}{2}×\frac{15}{100}}\\=\frac{\frac{1}{6}×\frac{1}{100}}{\frac{1}{6}×\frac{1}{100} + \frac{1}{3}×\frac{3}{100} + \frac{1}{2}×\frac{15}{100}}\\=\frac{\frac{1}{6}}{\frac{1}{6} + 1+\frac{15}{2}}\\=\frac{1}{1 + 6 + 45} =\frac{1}{52}$$ 8. A factory has two machines A and B. Past recored shows that machine A produced 60% of the item of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B? Sol. Let E1 : the event that the item is produced by machine A and E2 : the event that the item is produced by machine B. Then, E1 and E2 are mutually exclusive and exhaustive events. Moreover, $$\text{P}(\text{E}_1) = 60\% = \frac{60}{100} =\frac{3}{5}\\\text{and}\space\text{P(E}_2) = 40\%\\=\frac{40}{100} = \frac{2}{5}$$ Let E : the event that the item chosen is defective, $$\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) =\text{P}(\text{machine A produced}\\\text{ defective items})$$ $$= 2\% = \frac{2}{100}\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\text{P(machine B produced}\\\text{defective items)}\\ = 1\% = \frac{1}{100}$$ The probability that the randomly selected item was from machine B, given that is is defective, is given by $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg).$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg) = \\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}{\text{P}\bigg(\frac{E}{E_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{E}{E_2}\bigg)\text{P}(\text{E}_2)}\\=\frac{\frac{1}{100}×\frac{2}{5}}{\frac{2}{100}×\frac{3}{5} + \frac{1}{100}×\frac{2}{5}}\\=\frac{\frac{2}{500}}{\frac{6}{500} + \frac{2}{500}}\\=\frac{2}{6 +2} = \frac{1}{4}$$ 9. Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3, if the second group wins. Find the probability that the new product introduced was by the second group. Sol. Let E1 : the event that the first group wins and E2 : the event that the second groups wins. Then, E1 and E2 are mutually exclusive and exhaustive events. $$\therefore\space\text{P(E}_1) = 0.6 = \frac{6}{10}\\\text{and}\space \text{P}(\text{E}_2) = 0.4 = \frac{4}{10}$$ Let E : the event that the new product is introduced. $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{P}(\text{introducing a new ,}\\\text{product if the first group wins})\\ =0.7 = \frac{7}{10}\\\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = \text{P(introducing a new},\\\text{product, if the second group wins)}\\=0.3 =\frac{3}{10}$$ Probability that the new product is introduced by the second group is given by $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg).$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg) =\\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}\text{(E}_2)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}\\=\frac{\frac{3}{10}×\frac{4}{10}}{\frac{7}{10}×\frac{6}{10} + \frac{3}{10}×\frac{4}{10}}\\=\frac{12}{42 + 12} = \frac{12}{54} = \frac{2}{9}.$$ 10. Suppose, a girl throws a dice. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die? Sol. Let E1 : the event that 5 or 6 is shown on die and E2 : the event that 1, 2, 3 or 4 is shown on die. Then, E1 and E2 are mutually exclusive and exhaustive events. and n(E1) = 2, n(E2) = 4 Also, n(S) = 6 $$\therefore\space\text{P}(\text{E}_1) = \frac{2}{6} =\frac{1}{3}\\\text{and}\space \text{P}(\text{E}_2) = \frac{4}{6} =\frac{2}{3}$$ Let E : Event that exactly one head shows up, $$\therefore\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{P(exactly one head}\\\text{shows up when coin is tossed thrice)}\\ =\text{P}\lbrace\text{HTT, THT, TTH}\rbrace= \frac{3}{8}\\(\because\space \text{Total number of events} = 2^{3} = 8)\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\text{P}\text{(head shows up when)}\\\text{coin is tossed once)}\\=\frac{1}{2}$$ Probability that the girls threw 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg).$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_2}{\text{E}}\bigg) = \\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}\text{(E}_2)}\\=\frac{\frac{1}{2}×\frac{2}{3}}{\frac{3}{8}×\frac{1}{3} + \frac{1}{2} × \frac{2}{3}}\\=\frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{8}}\\=\frac{8}{8+3} =\frac{8}{11}$$ 11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items, respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A? Sol. Let E1 : the event that item is produced by machine A, E2 : the event that item is produced by machine B and E3 : the event that item is produced by machine C. Here, E1, E2 and E3 are mutually exclusive and exhaustive events. Moreover, $$\text{P}\text{(E}_1) = 50\% = \frac{50}{100}\\\text{P}(\text{E}_2) = 30\% =\frac{30}{100}\\\text{and}\space\text{P}(\text{E}_3) = 20\% =\frac{20}{100}$$ Let E : Event that item chosen is found to be defective, $$\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)= \frac{1}{100}\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) =\frac{5}{100}\\\text{and\space}\text{P}\bigg(\frac{\text{E}}{\text{E}_3}\bigg) = \frac{7}{100}$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) = \\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P(E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}(\text{E}_2) + \text{P}\bigg(\frac{\text{E}}{\text{E}_3}\bigg)\text{P}(\text{E}_3)}\\=\\\frac{\frac{1}{100}×\frac{50}{100}}{\frac{1}{100}×\frac{50}{100} + \frac{5}{100}×\frac{30}{100} + \frac{7}{100}×\frac{20}{100}}\\= \frac{50}{50 + 150 +140} = \frac{50}{340}\\=\frac{5}{34}$$ 12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond. Sol. Let E1 : the event that the lost cards is a diamond ⇒ n(E1) = 13 E2 : lost cards is not a diamond ⇒ n(E2) = 52 – 13 = 39 and n(S) = 52 Then, E1 and E2 are mutually exclusive and exhaustive events. $$\therefore\space\text{P(E}_1) = \frac{13}{52}= \frac{1}{4}\\\text{and}\space\text{P(E}_2) =\frac{39}{52} = \frac{3}{4}$$ Let E : the event that two cards drawn from the remaining pack are diamonds, When one diamond card is lost, there are 12 diamond cards out of 51 cards. The cards can be drawn out of 12 diamond cards is 12C2 ways. Similarly, 2 diamond cards can be drawn out of 51 cards in 51C2 ways. The probability of getting two cards, when one diamond card is lost, is given by $$\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg).\\\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)= \frac{^{12}\text{C}_2}{^{51}\text{C}_2}\\=\frac{\frac{12×11}{1×2}}{\frac{51×50}{1×2}}= \frac{12×11}{51×50}\\\text{and}\space\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = \frac{^{13}\text{C}_{2}}{^{51}\text{C}_{2}}\\ =\frac{\frac{13×12}{1×2}}{\frac{51×50}{1×2}} = \frac{13×12}{51×50}$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) = \\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P(\text{E}}_2)}\\=\frac{\frac{12×11}{51×50}×\frac{1}{4}}{\frac{12×11}{51×50}×\frac{1}{4} + \frac{13×12}{51×50}×\frac{3}{4}}\\=\frac{12×11}{12×11 + 13×12×3}\\=\frac{132}{132 + 468} = \frac{132}{600}\\=\frac{11}{50}$$ 13. Probability that A speaks truth is $$\frac{\textbf{4}}{\textbf{5}}\textbf{.}\space\textbf{A coin is tossed.}$$ A reports that a head appears. The probability that actually there was head is : $$\textbf{(A)\space}\frac{\textbf{4}}{\textbf{5}}\\\textbf{(B)\space}\frac{\textbf{1}}{\textbf{2}}\\\textbf{(C)\space}\frac{\textbf{1}}{\textbf{5}}\\\textbf{(D)\space}\frac{\textbf{2}}{\textbf{5}}\\\textbf{Sol.}\space\text{(A)\space}\frac{4}{5}$$ Let E1 : the event that coins comes up with a head, E2 : the event that coin comes up with a tail. Then, E1 and E2 are mutually exclusive and exhaustive events. Moreover, $$\text{P}(\text{E}_1) = \text{P}(\text{E}_2)= \frac{1}{2}$$ E : the events that A reports that a head appears, $$\therefore\space\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg) = \text{P}(\text{head comes up and}\\\text{A speaks truth})\\=\frac{4}{5}\\\text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg) = \text{P}(\text{tails comes up and}\\\text{speaks lie})\\=\frac{1}{5}$$ Probability that there is actually a head is given by $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg).$$ By using Baye's theorem, we obtain $$\text{P}\bigg(\frac{\text{E}_1}{\text{E}}\bigg) =\\\frac{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1)}{\text{P}\bigg(\frac{\text{E}}{\text{E}_1}\bigg)\text{P}(\text{E}_1) + \text{P}\bigg(\frac{\text{E}}{\text{E}_2}\bigg)\text{P}\text{(E}_2)}\\=\frac{\frac{4}{5}×\frac{1}{2}}{\frac{4}{5}×\frac{1}{2} + \frac{1}{5}×\frac{1}{2}}\\=\frac{4}{4+1} = \frac{4}{5}$$ 14. If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct? $$\textbf{(A)\space P(A/B)} = \frac{\textbf{P(A)}}{\textbf{P(B)}}\\\textbf{(B)\space}\textbf{P(A/B)}\lt \textbf{P}\textbf{(A)}\\\textbf{(C) P(A/B)}\geq\textbf{P(A)}$$ (D) None of these Sol. (C) P(A/B) ≥ P(A) If A ⊂ B, then A ∩ B = A ∴ P(A ∩ B) = P(A) We know that $$\text{P}(A/B) =\frac{\text{P}(A\cap B)}{\text{P(B)}}\\=\frac{\text{P(A)}}{\text{P(B)}}$$ But P(B) ≤ 1 ∴ P(A/B) ≥ P(A)
Sie sind auf Seite 1von 34 # CHAPTER Deductive Geometry Deductive geometry is the art of deriving new geometric facts from previously-known facts by using logical reasoning. In elementary school, many geometric facts are introduced by folding, cutting, or measuring exercises, not by logical deduction. But as we have seen, fifth and sixth grade students are already practicing and enjoying deductive reasoning as they solve unknown angle problems. In geometry, a written logical argument is called a proof. Section 4.1 introduces one type of proof: unknown angle proofs. Unknown angle proofs are natural continuations of students experience in solving unknown angle problems; the transition is a small step that requires no new concepts. Indeed, as you will see, unknown angle proofs are almost identical to the Teachers Solutions that you wrote in the previous chapter! Section 4.2 describes how congruent triangles are introduced in middle school. Congruence is a powerful geometric tool that opens a door to new aspects of geometry; some of this is covered in Sections 4.3 and 4.4. These sections also describe how the facts about triangles and quadrilaterals that students learned in grades 5 and 6 are revisited at a higher level in middle school. In this chapter we reach the last stage in the preparation of students for high school geometry. As you read and do problems, think about how these problems are part of a story line that goes back to learning to measure angles in grade 4 and learning to measure lengths in grade 1. You will be teaching part of this story, and it is important to know how it unfolds. Background Knowledge Here is a list of the geometric facts at our disposal at this point. These facts will be used in the examples and homework problems in this chapter. Several additional facts will be added to this list in Section 4.2. (c = a + b.) c b a 73 ## The sum of adjacent angles on a straight line is 180 . (If L is a line then a + b = 180 .) Abbreviation: s on a line. b L ## The sum of adjacent angles around a point is 360 . (a + b + c + d = 360 .) Abbreviation: s at a pt. ## Vertically opposite angles are equal. (At the intersection of two straight lines, a = c and b = d). Abbreviation: vert. s. ## When a transversal intersects parallel lines, corresponding angles are equal. (If ABCD then a = b.) b A Abbreviation: corr. s, ABCD. Conversely, if a = b then ABCD. Abbreviation: corr. s converse. When a transversal intersects parallel lines, alternate interior angles are equal. (If ABCD then a = c.) Abbreviation: alt. s, ABCD. Conversely, if a = c then ABCD. Abbreviation: alt. s converse. c a When a transversal intersects parallel lines, interior angles on the same side of the transversal are supplementary. (If ABCD then a + d = 180 .) Abbreviation: int. s, ABCD. Conversely, if a + d = 180 then ABCD. Abbreviation: int. s converse. The angle sum of any triangle is 180 . () (a + b + c = 180 .) Abbreviation: sum of . A a Each exterior angle of a triangle is the sum of the opposite interior angles. () (e = a + b). Abbreviation: ext. of . a b ## Base angles of an isosceles triangle are equal. () (If AC = BC then a = b.) Abbreviation: base s of isos. . A ## Each interior angle of an equilateral triangle is 60 . () Abbreviation: equilat. . 60 60 60 ## Opposite angles in a parallelogram are equal. () (a = b). Abbreviation: opp. s-ogram. b a ## The sum of the interior angles of an n-gon is (n 2) 180 . Abbreviation: sum of n-gon. The sum of the exterior angles of a convex n-gon is 360 . Abbreviation: ext. s of cx. n-gon. () The starred facts were established by fifth grade classroom demonstrations. Later in this chapter we will give deductive proofs for them. ## 4.1 Unknown Angle Proofs The previous chapter introduced the idea of a Teachers Solution, which you then used for homework solutions. This specific format is designed to help make you aware of all the aspects of a solution that must be communicated to students and to emphasize that this communication requires only a few words. Many problems in the New Elementary Math textbooks lead students to write teacher solutions themselves. The Teachers Solution format serves another purpose: it helps pave the way for proofs. In fact, a Teachers Solution can be made into a proof by simply changing one specific measurement (such as 31 ) into a measurement specified by a letter (such as x ). Beyond that, there are only stylistic differences between unknown angle problems and unknown angle proofs. Examples 1.1 and 1.2 illustrate the transition from unknown angle problems to unknown angle proofs. This section also introduces a format for writing simple proofs in a manner that is almost identical to the Teachers Solutions you have done already. ## 76 CHAPTER 4. DEDUCTIVE GEOMETRY EXAMPLE 1.1. In the figure, angles A and C are right angles and angle B is 78 . Find d. Teachers Solution: 78 d D 90 + 78 + 90 + d = 360 180 + 78 + d = 360 78 + d = 180 d = 102. sum in 4-gon Example 1.1 is a fact about one particular shape. But if we replace the specific measurement 78 by an unspecified angle measure b , then the identical reasoning yields a general fact about quadrilaterals with two 90 interior angles. EXAMPLE 1.2. In the figure, angles A and C are right angles. Prove that d = 180 b. Proof. b d D 90 + b + 90 + d = 360 180 + b + d = 360 b + d = 180 d = 180 b. sum in 4-gon Notice the distinction between the above examples. Example 1.1 is an unknown angle problem because its answer is a number: d = 102 is the number of degrees for the unknown angle. We call Example 1.2 an unknown angle proof because the conclusion d = 180 b is a relationship between angles whose size is not specified. A D C EXAMPLE 1.3. In the figure, AB EC and BD EF. Find b. B D 43 43 ## Teachers Solution: Extend the lines as shown. Mark angle c as shown. c b c = 43 b=c b = 43. alt. s, BC EF alt. s, BA ED ## There is nothing special about the number 43. The same reasoning shows that b = a in the picture on the right. The proof below is a Teachers Solution with two embellishments. First, it is launched by a preamble that states, in very few words, what we are assuming as known and what we wish to show. Second, the solution involves two auxiliary lines, as we explain to the reader on a line labeled construction. To prove: b = a. D ## Construction: Extend sides BC and ED. Mark angle c as shown. A F c b Proof. c=a b=c b = a. alt. s, BC EF alt. s, BA ED We have just turned Example 1.3 into a proof. The proof requires no additional effort! x B A EXAMPLE 1.4. In the figure, ABCD. Prove that z = x + y. You have seen this problem before: it is almost identical to Example 2.3 on page 65. The previous version was an unknown angle problem: two of the three angles x, y, and z were given (one was 37 and another was 75 ) and the problem was to find the third. The version above leaves x, y, and z unspecified and the problem is to prove that they are related. Just as before, the proof requires a construction, and several different constructions will work. Here is one proof. Given: ABCD. To prove: z = x + y. B ## Construction: Extend line as shown. Mark angles a and b as shown. z C Proof. b y a=x b=y z=a+b z = x + y. corr. s, ABCD vert. s ext. s of a ## A polygon whose vertices lie on a circle is said to be inscribed in the circle. example proves a famous fact about inscribed triangles. B EXAMPLE 1.5. Any inscribed triangle with a side passing through the center of the circle is a right triangle. The next ## Given: ABC is inscribed in a circle. AC contains center O. To prove: B = 90 . Construction: Draw segment OB. Mark angles as shown. B Proof. x y A def. of circle OAB and OBC are isosceles s a=x b=y x + y + a + b = 180 x + y + x + y = 180 2(x + y) = 180 B = x + y = 90 OA = OB = OC base s of isos. base s of isos. sum of ABC There are two standard ways for writing angles in geometric figures. One possibility is to label angles in figures with a letter and a degree sign, as in x . Then x stands for a number, for example, x might be 42. The alternative is to label angles in figures with a letter only, as was done in Example 1.5. Then x stands for a quantity, x might be 42 degrees. Both are correct provided the notation is consistent within each problem. In particular, equations should be written so that every term is a number, or every term is a quantity. EXERCISE 1.6. In which of the solved examples in this section do letters stand for numbers? In ## which to they stand for quantities? For elementary students, it is best to make the units the degree signs plainly visible in the figure. But for unknown angle proofs, the presentation is often clearer when letters stand for quantities. Accordingly, we scrupulously included degree signs in Chapters 1, 2 and 3, but we will now begin using both notations freely. ## The Elementary Proof Format Proofs are exercises for students! The challenge and the fun lies in figuring out the sequence of steps that take you to the desired conclusion. But, as with multi-step problems, it is easy for students to get confused unless they systematically record their steps in writing. Students who have learned a specific format for recording steps can devote more attention to geometric thinking and less to organizing their writing. A uniform format also facilitates classroom discussions and makes it easier for teachers to read and evaluate student work. Which format is best? There is no single answer, but several principles are important. Writing proofs shouldnt be a chore, so the chosen format should be simple, minimal, and natural. It is useful to include a preamble stating the hypotheses and the goal, because this helps students clearly understand the task before starting. A clearly-marked picture obviously helps students work out the solution. Finally, the completed proof should make the reasoning clear to And who is this mysterious reader? At all levels of K-12 mathematics, students write for two readers: themselves and their teacher. For students, clear writing aids clear thinking. For ## SECTION 4.1 UNKNOWN ANGLE PROOFS 79 teachers, asking students to use a simple, clear format is a matter of self-interest: it makes student work easier to read. In fact, this applies to everything written by mathematics students. In this book, we will adopt the format used in the examples in this section. We will call a proof written in this format an Elementary Proof. You will be expected to use this in all homework problems that ask for an Elementary Proof. The following template shows the features of an Elementary Proof. Preamble states the given information and the goal. Elementary Proof ## Given: AD CE and AB BC. To Prove: x+ y = 90. y ## Any needed construction is explained before the proof starts. E C Proof: x=z z+y = 90 ext. of ## Diagram shows all points, lines and angles used. ## Facts used are recorded using our abbreviations. Only 1 fact per line. ## Hints. Clear proofs are short and simple. To that end, Do not label the two columns statement and reason (everyone already knows this!). Do not include reasons for simple arithmetic and algebra steps. To avoid cluttering the picture, label only those points, lines and angles used in the proof. ## Establishing Facts Using Proofs The next two examples show how two familiar facts about triangles follow from the properties of parallel lines. In fifth grade, students justified the angle sum of triangle fact by cutting and re-arranging paper triangles. Middle school students can give a purely geometric argument for this fact. These are very important proofs: learn them. THEOREM 1.7. In any triangle, the sum of the interior angles is 180 . Given: ABC. To prove: a + b + c = 180 . B L ## Construction: Draw line L through C parallel to AB. Mark angles x and y as shown. Proof. a A y c C a=x b=y c + x + y = 180 a + b + c = 180 . corr. s, L AB alt. int. s, L AB s on a line ## 80 CHAPTER 4. DEDUCTIVE GEOMETRY THEOREM 1.8. Each exterior angle of a triangle is the sum of the opposite interior angles. Given: ABC. To prove: e = a + b. b e Proof. e = 180 c a + b = 180 c e = a + b. s on line sum of EXERCISE 1.9. Compare both theorems with the picture proof of the same fact described in Section 2.2 of Chapter 2. Are they compatible? Do these proofs elaborate on the picture proofs? The next proof is different from the fifth-grade paper-folding explanation, but is still easy to understand. THEOREM 1.10. Opposite angles in a parallelogram are equal. To prove: a = c. x B c C a A ## Construction: Extend sides AB, DC, and BC. Mark angle x as shown. Proof. a=x x=c a = c. alt. int. s, CD AB ## Theorems and Proofs in the Classroom Students often acquire misconceptions about the meaning of the words theorem and proof. Many believe that a theorem is a mathematical fact and a proof is an explanation of why a fact is true. This viewpoint embodies a subtle misunderstanding that teachers should try to prevent from taking root. Theorems are not statements of universal truths. Rather, they are if-then statements: If certain assumptions are true then a stated conclusion is true. The assumptions are of two types: those explicitly stated as hypotheses (after the word Given in our format), and a collection of background knowledge facts (often not explicitly mentioned) that have already been accepted or proven true. A proof is a sequence of deductive steps that explain how the conclusion follows from the hypothesis. This can be said more compactly as follows: DEFINITION 1.11. A proof of a mathematical statement is a detailed explanation of how that ## statement follows logically from other statements already accepted as true. A theorem is a mathematical statement with a proof. Theorems are the building blocks of geometry. Once a theorem has been proved, it can be added to the list of background facts and used in subsequent proofs. For example, after proving ## SECTION 4.1 UNKNOWN ANGLE PROOFS 81 Theorem 1.7 we were able to use it as a step in the proof of Theorem 1.8. In this way one builds, bit by bit, a large body of knowledge. This structure puts a burden on textbooks and teachers. A proof only makes sense in the context of a particular collection of accepted background facts, and that collection changes grade by grade. Teachers should devote some time to clarifying the already-accepted background facts Homework Set 13 Following the format described in this section, give an Elementary Proof of each statement. 1. In the figure, prove that a + b = 90. R C Q Y B b D S B b D d e B c D b d D C c a B ## 8. In the figure, prove that d = a + b + c. A 3. In the figure, B = 90 . Prove that a + c = 90. B a z ## 9. In the figure, ABCD and CB ED. Prove that ABC = CDE. D B A X C B ## 4. In the figure, prove that z = x + y. E x F a d D A 10. In the figure, ABCD and BC ED. Prove that b + d = 180. C d Z Y b A XYZ. X A q R ## 4.2 Congruent Triangles Two line segments are called congruent if they have equal lengths. Two angles are congruent if they have equal measures. Similarly, two triangles are called congruent if their sides have the same lengths and their angles have the same measures. The first part of this section explores how curricula build up to this notion of congruent triangles. The second part focuses on the criteria that ensure that two triangles are congruent. A Curriculum Sequence Early Grades. In geometry, figures that are duplicates exact copies of one another are called congruent. Duplicate figures can be drawn in different positions and orientations. Thus, to check whether two figures are congruent one must realign them to see if they match. Young children are taught to match shapes, first with figures made of cardboard or thin plastic, then visually with pictures. The phrase same size and same shape is often used to mean congruent. The grade 1 Primary Mathematics books contain exercises like this: Do the figures have the same size and same shape? a) b) c) d) These are exercises in visualization only; students are not expected to make measurements. The goal is to convey the idea that matching requires first sliding and rotating, and then comparing lengths and angles. The term congruence is not used, but the seed of the idea is planted. Teaching comment: Children may misinterpret the term same shape to mean shapes with the same name. But the word shape refers to the angles and proportions in a figure, not just its name. Thus, the figures in b) above are both triangles, but do not have the same shape. ## Matching exercises also introduce ideas that are used later for defining area and for modeling fractions. For example, a regular hexagon, decomposed into 6 congruent triangles, can be used to illustrate fractions whose denominator is 6. whole unit fractional unit 5 units 6 ## Middle School Introduction. The Primary Mathematics curriculum introduces congruence in grade 7 (some curricula start as early as grade 5). One approach starts with see-through tracing paper or overhead transparency sheets as in the following exercise. EXERCISE 2.1. Trace triangle A on a transparent sheet and lay it over figures B and C. Can you make them match? Flip the sheet over and try again. This exercise makes the idea of same size, same shape more precise, yet it still relies on visual matching. But notice that matching can be described purely in terms of the 3 side lengths and 3 angle measures: two triangles match if they can be aligned so that all six of these measurements exactly agree. In diagrams, such matching can be shown by marking the triangles as in the figure below. C But it is more efficient to use symbols. This requires (i) a symbolic way of describing how to align two triangles, and (ii) a definition for what is meant by exactly match. To align triangles, one pairs up their vertices. Such a pairing is called a correspondence. More precisely, a correspondence between two triangles pairs each vertex of one triangle with one and only one vertex of the other. The figure shows triangles PQR and XYZ aligned in a way that suggests the correspondence P X, Q Y, and R Z. X Z Once a correspondence is chosen, we can talk about corresponding sides and corresponding angles. For the correspondence P X, Q Y, and R Z, P corresponds to X, side QR corresponds to side YZ, etc. In this way we can compare the six measurements of two triangles, even if they dont have the same shape or size. When the corresponding measurements are equal, the correspondence gives us a precise way of stating that the triangles are exact copies of each other. DEFINITION 2.2. Two triangles are congruent if, under some correspondence, ## all pairs of corresponding sides are equal, and all pairs of corresponding angles are equal. ## 84 CHAPTER 4. DEDUCTIVE GEOMETRY To indicate that two triangles are congruent we use the symbol " and write the vertices of the triangles in corresponding order. Thus ABC " PQR means that ABC and PQR are congruent under the correspondence A P, B Q, C R. To see the correspondence, one can visualize arrows pairing up the vertices in order. ABC PQR Here is a grade 7 exercise that helps students move from checking congruence visually to thinking about the measurements of corresponding parts: EXERCISE 2.3. These triangles are congruent. Fill in the blanks by matching side lengths and ## estimating angle measures. AB corresponds to BC corresponds to corresponds to XY m 8c 1. 3 cm A corresponds to B corresponds to corresponds to Y. ABC " 3 cm 1.8 cm .4 cm 2.4 cm X C Hint: In such a congruence, one can associate each letter with an interior angle, and then the smallest angle of one triangle is paired with the smallest angle of the other, the largest is paired with the largest, and the mid-sized with the mid-sized. Symmetry can be expressed as a congruence of a triangle with itself. For example, the isosceles triangle below has a line of symmetry. Reflecting across the line of symmetry interchanges A and C and does not move B. Thus this reflection is described by the correspondence B A C, C A, and B B. The reflection takes the 3 sides of the triangle to sides of equal length (since AB = BC) and takes the 3 angles to angles of equal measure (since base angles are equal). Consequently, the reflection is a congruence, which we write as ABC " CBA. EXERCISE 2.4. An equilateral triangle ABC is congruent to itself in six ways. Write down all six, beginning with ABC " ABC. Write R next to the congruences that are rotations and LS next to those that are reflections across a line of symmetry. It will help to draw pictures. ## Congruence Tests for Triangles: a Teaching Sequence Congruent triangles have six pairs of equal measurements (3 pairs of angles and 3 pairs of side lengths). However, to show that triangles are congruent, it isnt necessary to check all six pairs. Four congruence criteria make the task easier. The criteria are often called tests and are named by triples: the Side-Side-Side Test, Angle-Side-Angle Test, Side-Angle-Side Test and the Right-Hypotenuse-Side Test. Each test states that two triangles are congruent whenever the named measurements match under some correspondence. Congruence tests are often introduced in middle school by activities that ask students to construct triangles with specified measurements. The aim is to make the meaning of the tests clear and convince students of their validity. Side-Side-Side Test. Construction 6 on page 53 showed how to use the 3 side-lengths of a triangle to create a duplicate triangle with the same side-lengths. To verify that the two triangles are congruent, one must also know that corresponding angles have equal measure. C EXERCISE 2.5. Using a compass and a straightedge, copy ABC onto a blank paper as in Construction 6 on page 53. Then make the following checks: First Check: Measure the angles in the copied triangle with a protractor. Are they equal to the corresponding angles in ABC? Are the two triangles congruent? Second Check: Cut out your triangle and place it on top of the original triangle in this book. Align the sides. Are all angle measures equal? Are the triangles congruent? In Exercise 2.5, both checking methods refer directly to the definition of congruence both require verifying that all corresponding angles and sides are equal. The teaching goal at this stage is to make students explicitly aware of the definition: congruence means 6 matching measurements. Side-Side-Side Test C ## Two triangles are congruent if corresponding sides are equal under some correspondence, i.e., A ## If AB = PQ, BC = QR, CA = RP, then ABC " PQR. (Abbreviation: SSS.) B R ## 86 CHAPTER 4. DEDUCTIVE GEOMETRY Angle-Side-Angle Test. In the next example, students duplicate a triangle by copying two angles and the side between them. C EXERCISE 2.6. Using a straightedge and protractor, copy ABC onto a blank paper as described below. Then make the two checks stated beneath the picture. 64 35 4 cm T ## Draw PQ of length AB. R ? ? 64 P Draw ray PS so S PQ = 64 . 35 4 cm ## Mark R where the two rays intersect and draw PR and QR. First Check: Use a protractor to measure R; is it equal to C? Use your compass to compare lengths QR and PR; are they equal to BC and AC? Are the two triangles congruent? Second Check: Cut out your triangle and place it on top of ABC. Align the sides. Are all angles and side lengths equal? Are the two triangles congruent? Angle-Side-Angle Test C ## Two triangles are congruent if two pairs of corresponding angles and their included sides are equal, i.e., A ## If A = P, B = Q, AB = PQ, then ABC " PQR. (Abbreviation: ASA.) B R The ASA Test should really be called the two angles and a side test. After all, if two pairs of angles are equal, then the third pair is also equal (because the angles of a triangle total 180 ). Consequently, each side lies between two pairs of corresponding angles, and we can apply the ASA Test. Some books distinguish between the ASA and AAS conditions, but we will the term ASA Test for both. ## SECTION 4.2 CONGRUENT TRIANGLES 87 Side-Angle-Side Test. In the next exercise, students copy a side, the included angle and side of a triangle to create a copy of ABC above. C EXERCISE 2.7. Using a ruler and protractor, copy 3 cm ## ABC onto a blank paper using the construction below and make the check described. A 40 4 cm S R 3 cm 40 P ## Draw ray PS with S PQ = 40 . 4 cm Draw RQ. First Check: Measure angles Q and R; are they equal to B and C? Use your compass to compare the lengths QR and BC; are they equal? What do you conclude about the two triangles? Second Check: Cut out your triangle and place it on top of ABC. Align the sides. Are the two triangles congruent? Side-Angle-Side Test C ## Two triangles are congruent if two pairs of corresponding sides and their included angle are equal, i.e., A ## If AB = PQ, AC = PR, A = P, then ABC " PQR. B R (Abbreviation: SAS.) ## Some sets of three equal corresponding measurements do not guarantee congruence. In particular, triangles with three matching angles neednt be congruent. F ## These triangles have the same Angle-Angle-Angle data but are not congruent. D A ## 88 CHAPTER 4. DEDUCTIVE GEOMETRY Likewise, triangles with one pair of equal corresponding angles and two pairs of equal corresponding sides neednt be congruent, as the following swinging girl picture shows. ## The shaded triangle (base 11 cm) and the large triangle (base 21 cm) have the same Angle-Side-Side data but are not congruent. 13 cm cm 13 c 20 x 11 cm 10 cm The fact that Angle-Side-Side measurements do not determine a unique triangle is easily said: there is no A.S.S. test. This phrasing is not recommended for classroom use. Typically, textbooks tell students that A.A.A. and S.S.A. do not confirm congruence of triangles. Right-Hypotenuse-Leg Test. For pairs of right triangles, it is enough to compare two pairs of sides. In a right triangle, the side opposite the 90 angle is called the hypotenuse and the other sides are called legs. The following exercise shows that two right triangles are congruent if they have equal hypotenuses are equal and one leg of equal length. A EXERCISE 2.8. Copy the right triangle ABC below. 2 cm 3 cm Construction: P ## Draw a segment PQ of length 2 cm. Draw line L perpendicular to PQ passing through Q. 3 cm 2 cm S ## Draw circle, center P, radius 3 cm. Mark as R one of the two points where the circle intersects L. Measure angles P and R; are they equal to A and C? Use your compass to compare the lengths QR and BC; are they equal? What do you conclude about the two triangles? Right-Hypotenuse-Leg Test If two right triangles have hypotenuses of the equal length and a pair of legs with equal length, then the triangles are congruent, i.e., ## If B = Q = 90 , AC = PR, AB = PQ, then ABC " PQR. (Abbreviation: RHL.) ## SECTION 4.2 CONGRUENT TRIANGLES 89 The RHL Test uses two sides and a non-included angle Angle-Side-Side measurements. In general, for such measurements, the swinging girl picture produces two non-congruent triangles with the same Angle-Side-Side measurements. But the RHL Test is true because, when the angle x is a right angle, the swinging girl picture becomes the diagram in Exercise 2.8, and the triangles are actually congruent. Homework Set 14 ## 1. Do Exercise 2.3 in this section (just fill in the blanks in your textbook). Then fill in all of the blanks in the exercises below. (On your homework sheet, just state the correspondence for each part.) a) B A = AB = ## 3. Is there an SSSS Test for quadrilaterals? That is, is it true that two quadrilateral with four pairs of corresponding sides of equal length are necessarily congruent? Explain and illustrate. ## 4. Do Exercise 2.6 in this section. First do the construction (in the first step, use your compass to carry the 4 cm the First Check and Second Check questions. C = CA = 5. Do Exercise 2.7 on page 87, this time using ruler and protractor. Then answer the Check questions. Y B = BC = ## 2. Do Exercise 2.5 on page 85 of this section. Answer the First Check and Second Check questions beneath the picture. 6. For each figure, name the congruent triangles and state the reason they are congruent (i.e., SSS, ASA, SAS, RHL). b) a) B F D = DE = L E = EF = M F = FD = E b) ## Correspondence: DEF " G H c) R c) W V T R = S = T = RS = ST = TR = Correspondence: RS T " d) ## 8. In the figure below, which triangles are congruent? Write down each congruence. e) B R 70 60 80 W f) 50 50 The figure is a circle with center Z. Z Y 70 E X 7. For each figure, find the values of the unknowns. a) ABC " DCB ## 9. Given any three positive real numbers, a, b, and c, with b less than 180 and c less than 180, can you construct a triangle having two interior angles whose measures are b and c, respectively, and the included side having measure a? Illustrate. b) RS T " VUT 40 3.7 cm m 5.8 c b S ## 10. Given any three positive real numbers, a, b, and c, with c less than 180, can you construct a triangle having two sides whose measures are a and b with the included angle having measure c? Illustrate. 11. Prove or give a counterexample: If two triangles both have interior angle measures equal to 32 , 100 and 48 , and both have a side of length 10 cm, then the triangles are congruent. ## 4.3 Applying Congruences Once students have learned the triangle tests, they can move beyond unknown angles problems to the next stage in the curriculum: proving facts about side-lengths and angles within figures. This section is an introduction to the simplest proofs. At this point, students appreciate and perhaps enjoy unknown angle problems and short proofs. They are prepared for high school geometry. B ## with AB = BC. Let X and Y be distinct points on AC such that AX = YC. Prove that XBY is also isosceles. ## Given: AB = BC and AX = YC. To Prove: BX = BY. B Proof. A AB = BC A = C AX = YC AXB " CY B given base s of isos. given SAS. BX = BY ## The triangle tests can also be used in unknown angle proofs. B EXAMPLE 3.2. In the figure, A = D and AE = DE. ## Prove ECB = EBC. E C To prove this, mark angles as shown and concentrate your attention on the two shaded triangles. Given: a = d and AE = DE. To Prove: b = c. B A D a d x y Proof. E c C a=d AE = DE x=y AEB " DEC given given vert. s ASA. EB = EC b=c ## corr. sides of " s. base s of isos. . Study Examples 3.1 and 3.2 for a moment. In each proof, Line 4 states that two triangles are congruent by a congruence test, Lines 1-3 are the facts needed to apply the test, and Line 5 is a conclusion based on this congruence. In this chapter, almost all of the proofs will have this format. It is simple, but it has many applications! false starts in proofs As students learn to construct proofs, it is easy for them to make a false start their first approach doesnt work. This is completely normal! Proofs are like puzzles: the fun lies in trying different strategies to find a solution. The reward, like the reward in solving a tricky puzzle, is a feeling of accomplishment. In fact, geometric proofs were a common amusement of educated people in the 19th century, just as crossword and sudoku puzzles are today. The next example is a problem in which it is easy to make a false start. The figure contains several pairs of congruent triangles; which pair should be used in the proof? Try to find a strategy before you look at the proof written below. Here is a strategy that helps: color, highlight or shade the segments that appear in the Given and the To Prove statements. Look for a pair of congruent triangles that contain these sides. ## EXAMPLE 3.3. In the figure, CA = CB. Prove that AS = BR. Given: CA = CB. To Prove: AS = BR. A R Proof. CA = CB C = C R = S = 90 ACS " BCR given common angle given AAS. AS = BR ## corr. sides of " s. EXERCISE 3.4. This proof used the congruence ACS " BCR. Name (without proof) two other pairs of congruent triangles in this figure (use the letter T to label the intersection point in the middle of the figure). These examples indicate that there are two levels of congruent triangle proofs: ones that are especially simple because the figure contains only one pair of congruent triangles, and ones in which students must find the appropriate congruent triangles from among several pairs. Careful textbooks (and careful teachers!) provide plenty of practice at the first level before challenging students with problems at the second level. The problems in the homework for this section (HW Set 15) are similar to a well-written eighth grade textbook. As you do these proofs, notice how they are arranged so that they slowly increase in difficulty. ## Proofs for Symmetry Explanations Congruence tests can be used to prove many of the facts that were introduced in fifth and sixth grade using symmetry arguments. We will give three examples. As you will see, the elementary school folding proofs contain the key ideas of a complete mathematical proof. For example, the following proof is just the detailed explanation of the fifth grade picture proof shown on page 62 of Primary Math 5B and on page 44 of this book. To prove: a = b. C Construction: Proof. a A b D ## Let D be the midpoint of AB. Draw CD. AC = BC DC = DC given D is the midpoint common. a=b corr. s of " s. When we first introduced compass and straightedge constructions in Section 2.5, we focused on the steps required and used symmetry to argue that the construction accomplished its intended purpose. We can now replace those symmetry arguments with proofs based on congruent triangles. For example, the construction below shows how to bisect an angle. A EXAMPLE 3.6. Bisect ABC. C ## Draw circle, center B, any radius. Mark R and S where the circle intersects the sides of the angle. ## Draw circle, center R, radius RS . Draw circle, center S , radius RS . R X ## Draw BX, RX and S X. We claim that this construction works: that BX bisects ABC. Symmetry suggests that BRX and BS X are congruent. Here is a proof: C Given: BR = BS , RX = S X. To prove: a = b. R X B a b S Proof. BR = BS RX = S X BX = BX BRX " BS X given given common SSS. a=b corr. s of " s. All constructions are theorems. When written down in complete form, each entails a construction followed by a proof that the construction works. We can therefore use constructions ## as reasons in proofs. Here is an example: THEOREM 3.7. If two angles of a triangle are congruent, then the triangle is isosceles. ## (Abbreviation: base s converse.) Given: ABC with b = c. To Prove: AB = AC. A ## Construction: Bisect A and mark angles x and y as shown. x y Proof. b B c D b=c x=y given bisect constr. common AAS. AB = AC ## corr. sides of " s. Homework Set 15 This homework set gives more practice with proofs using congruent triangles. Please write Elementary Proofs to the following problems. Most of these problems are similar to homework problems found in Japan and Hong Kong in 1. In the figure, prove that ABC = EDC. B D A P A Q R A A 130 C R T A = T B. 130 C HK = HC. A H O B B K AB = AC. A a) APV = BPV. b) AQ = BQ. A x B ## 12. In the figure, AB = CD, BC = AD. Prove that a) ABE " ACE. b) AC and BD bisect each other. b) AB = AC and AD BC. a1 a2 e1 e2 D A C ## 13. In the figure, P is equidistant from the lines OA and OB (i.e. PA = PB). Prove that OP bisects AOB. B b) e = f . ## 14. In the figure, OX and OY are the perpendicular bisectors of AB and AC respectively. Prove that H A b c d f K ## a) OAX " OBX. b) OA = OB = OC. c) What can you say about the circle with center O and E C Y D a1 A a2 O b2 b1 B ## 96 CHAPTER 4. DEDUCTIVE GEOMETRY You have already seen how the grade 5 (page 68) and grade 6 (page 6464) Primary Mathematics textbooks present facts about parallelograms. Middle school students can dig deeper and see that the facts they learned about parallelograms and other quadrilaterals are actually consequences of basic facts, namely those listed as Background Knowledge at the beginning of this chapter. This section shows how the background knowledge facts can be used to prove various properties of quadrilaterals. It ends with a review of the K-8 curriculum sequence that Parallelograms By definition, a parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Using this definition and congruent triangles, we can prove additional properties of parallelograms, as in the following proofs. (1) ## In ABC and CDA, s=x r=y AC = AC ABC " CDA alt. s, ABCD common ASA AB = CD and AD = BC x y (2) B = D A = r + s = x+y = C corr. s of " s s = x, r = y r s A (3) r=y b=d ## from part (1) corr. s of " s corr. sides of " s ASA AP = PC and BP = PD C y d P r A b B ## These three properties can also be seen by symmetry but it is not a folding symmetry. Instead, put a pin at the center of parallelogram (the point where the diagonals intersect) and rotate 180 around that point. Do you see why this symmetry gives all three properties (opposite sides equal, opposite angles equal, diagonals bisect)? ## SECTION 4.4 CONGRUENCES IN QUADRILATERALS 97 EXERCISE 4.1. Try the activity suggested above using transparency sheet. The converse of each of these three properties is also true: If a quadrilateral satisfies any one of the conditions (1), (2), (3) above, then that quadrilateral is a parallelogram. Students can be led to this conclusion by working through the following eighth-grade activity. EXERCISE 4.2. The chart below gives four different criteria for recognizing parallelograms. Fill in the blanks to complete these sketches of proofs: If both pairs of opposite angles are equal, then the figure is a parallelogram. sum n-gon x + y = 180 x+y+x+y= ABCD int. s converse. x A ABCD is a parallelogram. ## If both pairs of opposite sides are equal, then the figure is a parallelogram. BAC = DCA ABCD alt. s converse alt. s converse ABCD is a ## If the diagonals bisect each other, then the figure is a parallelogram. Similarly, APD " CPB ABCD is a P A If two opposite sides are equal and parallel, then the figure is a parallelogram. PAB = PCD DC = AB given DCA = BAC AC = AC ACD " CAB alt. s, AB DC common SAS DAC = corr. s, " ABCD is a ## 98 CHAPTER 4. DEDUCTIVE GEOMETRY These criteria justify the school definitions of rhombus and rectangle (see page 46 and the comment on page 47). The standard definitions used in high school geometry are: A rhombus is a quadrilateral with all four sides equal in length. A rectangle is a quadrilateral with four right angles. With these definitions, the second boxed criterion on the previous page shows that every rhombus is a parallelogram and the first criterion shows that every rectangle is a parallelogram. Thus we lose nothing by defining a rhombus to be a parallelogram with four equal sides and a rectangle to be a parallelogram with four right angles, as we did on page 46. Kites By definition, a kite is a quadrilateral in which two adjacent sides have equal length and the remaining two sides also have equal length. A kite has the following properties: 1. There is a pair of equal opposite angles. 2. One of its diagonals bisects a pair of opposite angles. 3. The two diagonals are perpendicular. The first two of these properties are obvious from symmetry. In the kite on the right, the line of symmetry is BD; it splits the kite into two triangles that are congruent by SSS (or by symmetry). Therefore, A = C, which shows the first statement. Furthermore, BD bisects B and D (by symmetry or because these are corresponding angles of congruent triangles), which shows the second statement. A D ## To see that the diagonals are perpendicular, note that the two shaded triangles pictured on the left are congruent by SAS. Angles x and y are therefore equal, so both are 90 . Thus the diagonals are perpendicular. We have seen that rectangles, rhombuses, and squares are also parallelograms. These figures therefore have all the properties of parallelograms. Likewise, rhombuses and squares are kites, so the properties of a kite also hold for rhombuses and squares. Below is a summary of the ## By definition, a parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. For parallelograms, 1. both pairs of opposite sides are equal, ## 2. both pairs of opposite angles are equal, 3. the diagonals bisect each other, and 4. two opposite sides are equal and parallel. ## Furthermore, any quadrilateral satisfying (1), (2), (3), or (4) is a parallelogram. By definition, a kite is a quadrilateral with two consecutive sides of equal length and the other two sides also of equal length. For kites, ## 1. at least one pair of opposite angles are equal, 2. there is a diagonal that bisects a pair of opposite angles, and ## Furthermore, any quadrilateral satisfying (2) is a kite. B By definition, a rhombus is a quadrilateral with all sides of equal length. A rhombus has all the properties of a parallelogram and of a kite, and the diagonals bisect the interior angles. ## By definition, a rectangle is a quadrilateral all of whose angles are right angles. A rectangle has all of the properties of a parallelogram, and its diagonals are equal. Furthermore, any quadrilateral whose diagonals are equal and bisect each other is a rectangle. C D ## By definition, a square is a rectangle with all sides of equal length. A square has the properties of a rectangle and of a rhombus. Curriculum Overview Look how far weve come! In four chapters, we have followed the main thread of the geometry curriculum starting in Kindergarten and now we are reaching the level of beginning high school geometry. This upward path is built on two themes: solving word problems involving measurements, and solving unknown angle problems. ## 100 CHAPTER 4. DEDUCTIVE GEOMETRY The part of the curriculum we have studied so far can be organized into four phases: Grades 14. Students learn to measure and calculate with lengths and angles. Grades 46. Students learn angle, triangle, and parallelogram facts (convincingly introduced by brief paper-folding, cutting, and measurement exercises). They use those facts to solve many unknown angle problems. Grades 67. Students learn parallel line facts and use them to give geometric proofs of some of the angle and triangle facts learned in grades 46. Students continue to solve many unknown angle problems, which now involve simple algebra. Grades 78. Congruence and similarity tests are introduced using constructions and measurement, and the Pythagorean Theorem is introduced with a simple proof. Students continue working on unknown angle problems and begin doing simple proofs. Students with this background are well-prepared for a 9th or 10th grade geometry course that concentrates on the logical structure of geometry. By that time they have spent five years developing knowledge and intuition for geometry and logical deduction. This built-up intuition and knowledge make learning the axiomatic approach to geometry in high school courses enormously easier. We have reached the limits of elementary school mathematics. The next section briefly describes some aspects of the transformation approach to geometry which are included in area is developed in elementary school. Homework Set 16 1. Do Problem 2 on page 282 of NEM1. Copy only the letters for the properties (you dont need to write the sentence). For example, the first line of the table should start: (a) T F T . . .. 2. Do all parts of Problem 3 on pages 283-284 of NEM1. Do not give proofs or Teachers Solutions just find the unknowns x and y in each figure. 3. Prove that ABCD is a parallelogram. D C 110 70 E ## 4. In the figure, S R PQ, PS = S R, and PQ = QR. Prove that PQRS is a rhombus, i.e., prove that PS = S R = PQ = QR. S R ## 5. In the figure, ABCD is a parallelogram, and AE and CF are perpendicular to BD. Prove that AE = CF. D C E ## SECTION 4.5 TRANSFORMATIONS AND TESSELLATIONS 101 6. In the figure, ACDF is a rectangle and BCEF is a parallelogram. Prove that ABF " DEC. A B C ## 8. Give an Elementary Proof: the diagonals of a square are perpendicular. Start by drawing a picture and writing: Given: Square ABCD with diagonals AC and BD To Prove: AC BD. ## 7. In the figure, ABCD is a rectangle, and AP = CR and AS = CQ. Prove that PQRS is a parallelogram. R C D S Q A ## 4.5 Transformations and Tessellations This section describes two topics often used for enrichment: transformations (which sometimes appear under the title motion geometry) and tessellations (also called tilings). Both are serious subjects whose study is college-level mathematics. Both are also frequently taught in K-8 mathematics classes. At the K-8 level, these topics are rich in classroom activities, but the mathematical content sometimes gets lost. This section explains the mathematics that is involved and includes some additional teacher-knowledge information. Here is one way to introduce the idea of transformations in the classroom. Begin by drawing a figure on a piece of cardboard. Lay an overhead projector transparency on it and trace the figure. rotated figure P original figure translated figure original ## (1) At some point P, insert a pin through the transparency into the cardboard (this is the reason for using cardboard). Rotate the transparency. The traced figure moves to a new position. This motion is called a rotation about P and the moved figure is called a rotation of the original figure. (In many elementary school illustrations, the rotation is around a point P that lies inside the figure and is often the center of the figure.) (2) Instead of inserting a pin, draw a line L on the cardboard and trace the line on the transparency. Slide the transparency along the line, keeping the traced line exactly on top of the line L while moving the figure to a new position. This motion is called a translation parallel to L and the moved figure is called a translation of the original figure. ## (3) Now mark a point P on L and the corresponding point on the traced copy of L. Turn over the transparency and align the traced line exactly on top of L with the two marked points aligned. This produces a mirror image of the original figure. This motion is called a reflection across L, and the moved figure is called a reflection of the original figure. reflected figure P L original A composite motion is any succession of rotations, translations, and reflections. For example, reflecting across a line L and then translating in the direction of L is a composite motion called a glide reflection. Transformations move the entire plane, including all figures within it. The transformed figure is called the image of the figure under the transformation. Notice that if you trace a segment or angle, moving the transparency does not affect the length or angle. Thus: For rotations, translations, reflections and their composites: the image of a segment is a segment of equal length, the image of an angle is an angle of equal measure, the image of a figure is a congruent figure. rigid motion There is a more elementary version of these ideas that involves moving figures (rather than moving the entire plane). If students cut a figure out of cardboard and lay it on a sheet of paper, they are able to slide the figure around. The various ways of sliding and turning are called the motions of the figure. These are sometimes called rigid motions because the cardboard figure is rigid it does not stretch or distort, and hence does not change lengths and angles. In classroom language, a translation moves the figure parallel to a line, a rotation spins the figure, and a reflection turns the figure over and places it on the other side of a given line. These descriptions are not precise, but they are enough for students to recognize two key points: A motion takes a figure to a figure with the same size and shape. Every motion can be obtained as a succession of rotations, translations, and reflections. These two points are the basis for the informal elementary school definition of congruence. Two figures are congruent if one is obtained from the other by a rigid motion. The required motion can be thought of as a way of picking up the first figure and laying it down ## SECTION 4.5 TRANSFORMATIONS AND TESSELLATIONS 103 so that it exactly covers the second figure. Alternatively, one can think of sliding and flipping the first figure through a succession of rotations, translations, and reflections until it exactly matches the second figure. Middle school students learn that congruence is defined by the requirement that corresponding sides and angles are equal. At that point a mathematical question arises: is it true that any two congruent figures are related by a succession of rotations, translations, and reflections? The answer is Yes, that is a mathematical theorem, but the proof is beyond K-8 mathematics. Nevertheless, the basic ideas, discussed next, are important teacher knowledge. Isometries transformation One unifying idea in geometry is the notion of an isometry. The mathematical study of isometries begins with the precise definition of a (general) transformation of the plane, which is defined in college-level mathematics as an invertible mapping from the plane to the plane. For our purposes, it is enough to say that a transformation is any succession of rotations, translations, reflections and any kind of stretching and pulling that might be possible with a transparency DEFINITION 5.1. A transformation that takes every line segment to a line segment of equal length is called an isometry. Thus isometries are length-preserving transformations. They also preserve distances because the distance between two points is the length of a segment. The congruence tests for triangles show that isometries also preserve angles. THEOREM 5.2. Isometries preserve angles. A B Proof. A' B' ## Given: A& , B& , and C & are obtained from A, B and C by an isometry. To prove: B& = B Constr.: Draw AC and A&C & . C' A& B& = AB B&C & = BC A&C & = AC A& B&C & " ABD ## isometries preserve lengths SSS B& = B. Theorem 5.2, together with Definition 5.1, shows that isometries preserve both lengths and angles. Consequently, they take triangles to congruent triangles. In fact, isometries move any figure to one that is congruent to the original. identity fixed points Rotations, translations and reflections preserve lengths, so are isometries. What other kinds of isometries are there? The following theorem, which is proved in college geometry courses, provides a complete list. The statement involves two new terms. The isometry that does nothing that does not move points, lines, and figures at all is called the identity. In general, an isometry will move most points of the plane; the points that are not moved are called the fixed points of the isometry. ## 104 CHAPTER 4. DEDUCTIVE GEOMETRY THEOREM 5.3. All isometries are composites of rotations, translations and reflections. In fact, every isometry is one of four types: a rotation, a translation, a reflection or a glide reflection. Specifically, ## a) The only isometry that fixes 3 non-collinear points is the identity. b) Any isometry that fixes 2 points is a reflection across the line through those points or is the identity. c) Any isometry that fixes exactly 1 point P is either a rotation around P, or the composite of a rotation around a point P and a reflection across a line through P. d) Any isometry that fixes no points is either a translation or a glide reflection. To get a sense of a), put 3 pins through your tracing paper and try to move it. You cant! Inserting 2 pins, then 1 pin, will similarly provide intuition (but not a mathematical argument) for b) and c). Statement d) is less intuitive. It includes the fact, which is not at all evident, that the composite of a rotation and a glide reflection along a line L is a glide reflection along a different line. The details are complicated. Understanding the reasoning requires mathematical knowledge and maturity beyond the K-8 level. In other countries, transformations are not introduced until after students have begun using coordinate systems in the plane. At that point, students have the geometric tools and experience to study transformations as mathematics. In contrast, many U.S. curricula and state standards introduce motions at the level of cutting and folding paper in elementary or middle school, and then abandon it. This is fine for enrichment and building intuition, but teachers should be aware that the study of transformations is not part of the main story-line of geometry in the lower grades, which focuses on measurement and deduction. Tessellations One often sees floor tiles and wall decorations arranged in geometric patterns formed by joining together copies of geometric figures without gaps or overlaps. These patterns are called tessellations (the ancient Romans made mosaics from clay and glass tiles called tessella). regular tessellation ## More precisely, a tessellation or tiling of the plane is a collection of polygonal regions whose union is the entire plane and whose interiors do not intersect. A tessellation made of congruent regular polygons is called a regular tessellation. It is a useful exercise to give students examples of tessellations and have them shade exactly one of the congruent polygons in the tiling. They can then visually test whether the tessellation is regular by checking the two requirements: a) is the tile a regular polygon? and b) are all the tiles in the tessellation congruent to the shaded one? There are only three regular tessellations. The three possibilities are formed by equilateral triangles, squares, and regular hexagons as shown. ## SECTION 4.5 TRANSFORMATIONS AND TESSELLATIONS 105 90 120 60 The proof that these are the only regular tessellations is an interesting classroom exercise in grades 5-7. The proof is based on the fact, which is evident from the pictures, that each vertex is surrounded by congruent angles. Thus, in a regular tessellation by n-gons (i) every interior angle of each tile is a factor of 360 and we know (from Section 3.3) that (ii) interior angles have equal measure, and (iii) the sum of interior angles is 180 (n 2). With these facts students can show, as you will in a homework problem, that the only regular tessellations are the three pictured above. symmetry of a tessellation A symmetry of a tessellation is a transformation that moves the tessellation onto itself. For example, a tessellation by squares can be moved onto itself by translating one or more units up, down, left, or right, or by rotating 90 around the center point of one square. Fancy tessellations can be created by modifying simple tessellations by such transformations. Here is an example. ## Choose part of one rectangle and translate it to all rectangles. ## Translate another part. visual appeal. A semi-regular tessellation is a tiling using two or more tile shapes, each a regular polygon, so that the arrangement of polygons at every vertex point is identical (i.e., the same tiles surround each vertex in the same order). It is a mathematical theorem (too difficult for school mathematics) that there are exactly 11 such semi-regular tessellations (including the 3 regular tessellations). There are also non-periodic tessellations called Penrose tilings. Elementary students can have fun with tessellations. Making and coloring tessellations is an excellent project for the end of the school year when summer is coming and students need a break from learning mathematics. Homework Set 17 ## 1. In this problem you will determine which regular n-gons tessellate. The steps outline a typical classroom exercise. a) Recall that the sum of the interior angles of any ngon is (n 2)180 . Use this fact and a sketch to explain why the measure of each interior angle for any n-gon is 360 . n b) List all the factors of 360 greater than or equal to 60. c) Fill in the following table. In = 180 # of sides 3 4 5 6 7 8 9 10 2. 3. 4. 5. Interior angle 60 Factor of 360 ? # Tessellate? # ## d) Use the formula for In and some algebra to show that whenever n is bigger than 6, In is between 120 and 180 . regular n-gons tessellate only for n = 3, 4 and 6. Study the last picture of this section. Follow the same procedure to make two simple tessellations. Draw at least 6 tiles for each. Coloring may make your picture clearer. Do the following constructions. ## move the triangle ABC 4 cm in the direction DE using only a compass, ruler, and a set-square. b) Draw a triangle ABC and a line l. Show how to reflect the triangle ABC across the line l using a compass and straightedge. c) Draw a triangle ABC and mark a point D. Show how to rotate the triangle ABC about the point D an angle measure of 45 , using a compass and straightedge. (Study the textbook!) Math 5B. Use graph paper to show how Shape B on page 78 can tessellate (graph paper can be obtained from www.printfreegraphpaper.com). (Study the textbook!) Assuming that 1 to 2 pages can be covered in a days lesson, estimate how many class days fifth grade Primary Math teachers spend on tessellations. ## 6. Explain, with pictures, why every parallelogram tessellates. 7. Explain why every triangle can tessellate (this can be done with one sentence and one picture using your answer to Problem 6). 8. This problem takes you through a classroom explanation of the fact that every quadrilateral tessellates. 3 inches across. Make the sides straight (use a ruler) and make the four interior angles have clearly different measures (your quadrilateral neednt be convex). Label the interior angles w, x, y, z in clockwise order. a) What is the sum w + x + y + z of the interior angles? Cut out your quadrilateral with scissors; this is your template. On a large blank sheet of paper, make a copy of this figure Tile 1 and label its interior angles w, x, y, z as on the template. Draw additional tiles by repeatedly applying the following step (which is named to indicate the edge we fill across). STEP xy: Without flipping the template, align the template along the edge between interior angles x and y of one of the previously-drawn tiles so that y on the template matches with x on the tile and vice versa. Trace around the template to create a new tile. Label the interior angles of the new tile w, x, y, z as on the template. Do STEP xy to draw Tile 2. Then do STEP wx on Tile 1 to draw Tile 3. (b) Look at the vertex of Tile 1 labeled by angle x. It is surrounded by 3 tiled angles and an un-tiled angle. The un-tiled angle has the same measure as which interior angle of your template? Why? (c) Fill in the blanks: one can fill in this un-tiled angle on Tile or by doeither by doing STEP on Tile . Do these give the same ing STEP result? (d) After drawing Tile 4 by your first answer to (c), explain why the paired sides of Tiles 2 and 4 have equal length. (e) Similarly, explain why the paired sides of Tiles 3 and 4 have equal length. (f) Draw at least 6 tiles. Can this tiling procedure be continued indefinitely?
Algebra 1 assist » Functions and Lines » order » exactly how to find the following term in one arithmetic succession In the adhering to arithmetic sequence, what is ? Explanation: The inquiry states the the succession is arithmetic, which method we discover the following number in the sequence by including (or subtracting) a continuous term. We understand two of the values, be separate by one unknown value. You are watching: What is the next number in the sequence below 10 13 9, 14 8 We recognize that is equally far from -1 and also from 13; therefore is equal to fifty percent the distance between these 2 values. The distance between them can be found by adding the pure values. The continuous in the sequence is 7. From over there we have the right to go forward or behind to uncover out that . Given the sequence below, what is the amount of the next three number in the sequence? Explanation: By taking the difference in between two adjacent numbers in the sequence, we have the right to see that the typical difference rises by one every time. Our following term will certainly fit the equation , an interpretation that the next term have to be . After , the next term will certainly be , an interpretation that the following term need to be . Finally, after , the following term will certainly be , definition that the following term need to be The question asks for the amount of the next three terms, so currently we require to include them together. Explanation: First, find a sample in the sequence. Friend will notice that each time you relocate from one number to the an extremely next one, it boosts by 7. The is, the difference in between one number and the following is 7. Therefore, we can add 7 to 36 and the result will it is in 43. Thus . Explanation: Determine what type of sequence you have, i.e. Even if it is the sequence transforms by a consistent difference or a consistent ratio. You have the right to test this by spring at pairs of numbers, however this sequence has actually a consistent difference (arithmetic sequence). So the sequence developments by individually 16 every time. Use this come the last provided term. Explanation: First, find the typical difference for the sequence. Subtract the first term from the 2nd term. Subtract the 2nd term from the 3rd term. Subtract the third term indigenous the fourth term. To find the following value, add to the last given number. Explanation: First, uncover the common difference because that the sequence. Subtract the an initial term native the 2nd term. Subtract the second term indigenous the 3rd term. To uncover the next value, add to the last offered number. Explanation: First, discover the typical difference because that the sequence. Subtract the very first term native the 2nd term. Subtract the 2nd term from the third term. To find the following value, add to the last offered number. Explanation: First, discover the typical difference because that the sequence. Subtract the first term indigenous the 2nd term. Subtract the 2nd term indigenous the 3rd term. To uncover the following value, add to the last given number. Explanation: First, uncover the common difference because that the sequence. Subtract the first term indigenous the 2nd term. Subtract the 2nd term from the third term. Subtract the third term native the fourth term. To uncover the next value, add to the last given number. See more: What Does The Word Canoodle Mean Ing, Origin Of Canoodle University the Richmond, Bachelor the Science, Chemistry. University of California-Los Angeles, master of Science, Chemistry. Imperial university England, Bachelor the Engineering, mechanical Engineering. Columbia college in the City of new York, Docto... 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# How To Simplify Negative Fractions Contents ## How To Simplify Negative Fractions? 0:21 1:36 So that’s one way of reducing an improper negative fraction. Now another way if you have negative 32MoreSo that’s one way of reducing an improper negative fraction. Now another way if you have negative 32 over 6 same kind of idea. What you can do is reduce. Both numbers at the same time. ## When you simplify a negative fraction does it turn positive? If both the numerator and denominator are negative then the fraction itself is positive because we are dividing a negative by a negative. ## How do you simplify fractions? How to Reduce Fractions 1. Write down the factors for the numerator and the denominator. 2. Determine the largest factor that is common between the two. 3. Divide the numerator and denominator by the greatest common factor. 4. Write down the reduced fraction. ## Can you have negative fractions? A negative fraction could indicate the absence removal of the indicated quantity. For example adding −1/2 of a pizza would mean removing 1/2 of that pizza. So yes fractions can be negative. ## How do you bring down a negative exponent? Dividing negative exponents is almost the same as multiplying them except you’re doing the opposite: subtracting where you would have added and dividing where you would have multiplied. If the bases are the same subtract the exponents. Remember to flip the exponent and make it positive if needed. ## How do you change a negative fraction to a positive? Realize that negative signs represent multiplying the number by −1. Any fraction can be multiplied by aa without changing the value since it equals one. So multiply your fraction by −1−1 to clear negative signs (or to move them between the top and bottom). ## When a fraction has a negative exponent? A negative exponent helps to show that a base is on the denominator side of the fraction line. In other words the negative exponent rule tells us that a number with a negative exponent should be put to the denominator and vice versa. For example when you see x^-3 it actually stands for 1/x^3. ## How do you simplify polynomial fractions with exponents? To simplify a fraction with a factorable polynomial in the numerator and the denominator factor the polynomial in the numerator and the denominator. Then reduce the fraction to lowest terms by canceling out any monomials or polynomials that exist in both the numerator and denominator. ## How do I teach my 10 year old fractions? When they are confident with numbers up to 10 you can introduce the concept of parts of the whole. Show examples to children and ask them to identify the fraction involved – the part of a whole circle that is coloured for example. Get them to enunciate the fractions involved e.g. Half Quarter Third Five-eighths. ## Why do we simplify fractions? Simplifying fractions is the process of reducing the numerator and denominator to their smallest whole numbers so the fraction is in its simplest form. … This helps other mathematicians or scientists to easily interpret data and can also help avoid confusion when numbers and equations become large and complex. ## How do you simplify 2 fractions? The first step when multiplying fractions is to multiply the two numerators. The second step is to multiply the two denominators. Finally simplify the new fractions. The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator. ## Can polynomials have negative exponents? A polynomial cannot have a variable in the denominator or a negative exponent since monomials must have only whole number exponents. Polynomials are generally written so that the powers of one variable are in descending order. ## How do you do negative exponents in math? A positive exponent tells us how many times to multiply a base number and a negative exponent tells us how many times to divide a base number. We can rewrite negative exponents like x⁻ⁿ as 1 / xⁿ. For example 2⁻⁴ = 1 / (2⁴) = 1/16. Categories FAQ
# Bring the general equation of the straight line 3x-4y + 12 = 0 to the equation in segments and calculate Bring the general equation of the straight line 3x-4y + 12 = 0 to the equation in segments and calculate the length of the segment that is cut off from this straight line by the corresponding coordinate angle. As you know, the equation of a straight line in segments on a plane in a rectangular coordinate system Oxy has the form x / a + y / b = 1, where a and b are some nonzero real numbers. In order to bring this general equation of the straight line 3 * x – 4 * y + 12 = 0 to an equation in segments, move the number 12 from the left side of the equation to the right side. We have: 3 * x – 4 * y = –12. Divide both sides of the resulting equation by (–12). Then, we get: x / (–4) + y / 3 = 1. The resulting equation is an equation in the segments of this straight line with the values: a = –4 and b = 3. The resulting equation allows us to assert that this straight line passes through points (in other words, intersects the coordinate axes) with coordinates: A (–4; 0) and B (0; 3). Now let’s calculate the length of the segment that is cut off from this straight line by the corresponding coordinate angle, that is, we will find the length of the segment AB. To do this, use the formula for calculating the distance between two points A (xa; ya) and B (xb; yb) on the plane: AB = √ ((xb – xa) ² + (yb – ya) ²). We have: AB = √ ((0 – (–4)) ² + (3 – 0) ²) = √ (16 + 9) = √ (25) = 5. Answer: x / (–4) + y / 3 = 1 and 5. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
# Adding sum of leftmost digits Starting with a number of at least $$9$$ digits, every minute we add the sum of the leftmost $$9$$ digits of the current number to the number itself. Will we always, at some point, see $$9$$ consecutive numbers that are all not divisible by $$9$$? The sum of the leftmost $$9$$ digits is at most $$9\times 9=81$$, so we never add more than $$81$$ each time. Consequently, when the number is large enough, the leftmost $$9$$ digits will stay the same for more than $$9$$ minutes. We can wait until the leftmost $$9$$ digits is divisible by $$9$$, to make sure that we have $$9$$ consecutive numbers that leave the same remainder when divided by $$9$$. If we can ensure that at some point this remainder is not $$0$$, we would be done. Let $$a_n$$ be the $$n$$th number. Clearly, the sequence $$\{a_n\}_n$$ is strictly increasing and hence unbounded. Let $$M=10^m$$ for some $$m\ge 11$$ and such that $$M>a_1$$. Then there is a minimal $$n_0$$ with $$a_{n_0}\ge M$$. As noted by the OP, $$a_{n_0}\le a_{n_0-1}+ 81$$, hence certainly $$a_{n_0}$$ has leading digits $$100000000$$. This means that the sequence will grow in steps of $$1$$ for a while, namely until we hit $$10^m+10^{m-8}$$ exactly. From here on, we will advance in steps of $$2$$ until we hit $$10^m+2\cdot 10^{m-8}$$ exactly. Now follow steps of $$3$$ until we hit $$10^m+3\cdot 10^{m-8}+2$$, then steps of size $$4$$ until we hit $$10^m+4\cdot 10^{m-8}+2$$, then steps of size $$5$$ until we hit $$10^m+5\cdot 10^{m-8}+2$$, then steps of size $$6$$ until $$10^m+6\cdot 10^{m-8}+4$$. Note that $$10^m+5\cdot 10^{m-8}+2+k\cdot 6$$ is not a multiple of $$9$$ (or even a multiple of $$3$$). Thus we have found $$\approx \frac{10^{m-8}}6\gg 9$$ consecutive terms that are not multiples of $$9$$
Courses Courses for Kids Free study material Offline Centres More Store # Solve for $x$: $\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0$ Last updated date: 15th Jul 2024 Total views: 453k Views today: 9.53k Verified 453k+ views Hint- Here, we will be using a discriminant method to solve the given quadratic equation. Given, equation is $\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0$ As we know that for any general quadratic equation $a{x^2} + bx + c = 0$, the solution is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $d = \sqrt {{b^2} - 4ac}$ is the discriminant of the quadratic equation. On comparing the given quadratic equation with the general quadratic equation, we get $a = \sqrt 3$ ,$b = - 2\sqrt 2$ and $c = - 2\sqrt 3$ Now substitute these values in the formula, we get $x = \dfrac{{ - \left( { - 2\sqrt 2 } \right) \pm \sqrt {{{\left( { - 2\sqrt 2 } \right)}^2} - 4\left( {\sqrt 3 } \right)\left( { - 2\sqrt 3 } \right)} }}{{2\left( {\sqrt 3 } \right)}} = \dfrac{{2\sqrt 2 \pm \sqrt {8 + 24} }}{{2\sqrt 3 }} = \dfrac{{2\sqrt 2 \pm \sqrt {32} }}{{2\sqrt 3 }} \\ \Rightarrow x = = \dfrac{{2\sqrt 2 \pm 4\sqrt 2 }}{{2\sqrt 3 }} = \dfrac{{\sqrt 2 \pm 2\sqrt 2 }}{{\sqrt 3 }} \\$ $\Rightarrow {x_1} = \dfrac{{\sqrt 2 + 2\sqrt 2 }}{{\sqrt 3 }} = \dfrac{{3\sqrt 2 }}{{\sqrt 3 }} = \left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right) = \sqrt 6$ and $\Rightarrow {x_2} = \dfrac{{\sqrt 2 - 2\sqrt 2 }}{{\sqrt 3 }} = \dfrac{{ - \sqrt 2 }}{{\sqrt 3 }} = - \sqrt {\dfrac{2}{3}}$ . i.e., The two roots of the given quadratic equation are ${x_1} = \sqrt 6$ and ${x_2} = - \sqrt {\dfrac{2}{3}}$. Therefore, the two values of $x$ possible in order to satisfy the given quadratic equations are $\sqrt 6$ and $- \sqrt {\dfrac{2}{3}}$. Note- For any quadratic equation, $a{x^2} + bx + c = 0$, according to the value of $d = \sqrt {{b^2} - 4ac}$ we have three possible cases: i. If it is positive, then the quadratic equation will have two different real roots. ii. If it is equal to zero, then the quadratic equation will have real and equal roots. iii. If it is negative, then the quadratic equation will have two different imaginary roots.
# Multiplication Chart Color Coded Learning multiplication right after counting, addition, and subtraction is perfect. Kids learn arithmetic via a all-natural progression. This growth of understanding arithmetic is truly the subsequent: counting, addition, subtraction, multiplication, lastly department. This statement results in the issue why learn arithmetic in this particular sequence? More importantly, why understand multiplication soon after counting, addition, and subtraction just before department? ## The subsequent information respond to these concerns: 1. Children discover counting very first by associating visual things with their hands. A tangible illustration: Just how many apples are there any within the basket? A lot more abstract instance is just how aged are you currently? 2. From counting figures, the following logical phase is addition combined with subtraction. Addition and subtraction tables can be very useful training helps for the kids as they are visual instruments making the move from counting simpler. 3. That ought to be learned after that, multiplication or department? Multiplication is shorthand for addition. At this point, kids possess a organization understanding of addition. As a result, multiplication is the after that logical form of arithmetic to learn. ## Assess fundamentals of multiplication. Also, review the basics how to use a multiplication table. We will review a multiplication illustration. By using a Multiplication Table, flourish several times about three and obtain an answer 12: 4 x 3 = 12. The intersection of row about three and line 4 of the Multiplication Table is a dozen; a dozen will be the solution. For youngsters starting to learn multiplication, this can be simple. They are able to use addition to solve the situation hence affirming that multiplication is shorthand for addition. Illustration: 4 x 3 = 4 4 4 = 12. It is an exceptional introduction to the Multiplication Table. An added gain, the Multiplication Table is visible and reflects straight back to learning addition. ## In which should we commence learning multiplication utilizing the Multiplication Table? 1. Initial, get knowledgeable about the table. 2. Start with multiplying by one particular. Start at row number 1. Proceed to column number one. The intersection of row 1 and line one is the best solution: a single. 3. Recurring these methods for multiplying by one particular. Increase row a single by posts a single by means of twelve. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively. 4. Repeat these methods for multiplying by two. Multiply row two by posts one through 5. The replies are 2, 4, 6, 8, and 10 correspondingly. 5. We will hop ahead of time. Repeat these actions for multiplying by 5 various. Flourish row several by posts one particular via a dozen. The answers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively. 6. Now let us boost the degree of problems. Replicate these techniques for multiplying by about three. Flourish row 3 by posts one particular through a dozen. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly. 7. Should you be confident with multiplication to date, consider using a examination. Remedy the next multiplication troubles in your mind and then evaluate your answers towards the Multiplication Table: flourish half a dozen and 2, multiply 9 and a few, increase a single and eleven, multiply a number of and a number of, and multiply six and 2. The situation solutions are 12, 27, 11, 16, and 14 correspondingly. When you got 4 away from five issues proper, design your very own multiplication exams. Compute the replies in your mind, and check them while using Multiplication Table.
# Use the binomial series to find the Maclaurin series for the function f(x)=sqrt{1+x^3} Question Series Use the binomial series to find the Maclaurin series for the function $$f(x)=\sqrt{1+x^3}$$ 2021-03-08 Here we will use the standard binomial series Representation for the factorial power in order to determine the power series representation for the given function.First of all we convert the given function into the standard form then we will go for series representation an the interval of convergence. $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...$$ The function for which we have to determine the power series representation is given as $$f(x)=\sqrt{1+x^3}$$ Let us assume $$a=x^3$$, now $$\sqrt{1+a}=(1+a)^{\frac{1}{2}}=1+\frac{a}{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}a^2+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}a^3+...$$ the above series is valid only when $$\|a\|<1$$</span> $$\Rightarrow(1+a)^{\frac{1}{2}}=1+\frac{a}{2}-\frac{a^2}{8}+\frac{a^3}{16}+...$$ The power series representation is given as $$f(x)=\sqrt{1+x^3}=1+\frac{x^3}{2}-\frac{x^6}{8}+\frac{x^9}{16}+...$$ The interval of convergence is given as $$\|x^3\|<1\Rightarrow\|x\|<1\Rightarrow R=1$$</span> ### Relevant Questions Use the binomial series to find the Maclaurin series for the function. $$f(x)=\frac{1}{(1+x)^4} asked 2020-10-19 a) Find the Maclaurin series for the function \(f(x)=\frac11+x$$ b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function $$g(x)=\frac{1}{(x+1)^2}$$ c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function $$h(x)=\frac{1}{(x+1)^3}$$ d) Find the sum of the series $$\sum_{n=3}^\infty \frac{n(n-1)}{2n}$$ This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is $$\frac72$$ Binomial series a. Find the first four nonzero terms of the binomial series centered at 0 for the given function. b. Use the first four terms of the series to approximate the given quantity. $$f(x)=(1+x)^{\frac{2}{3}}$$, approximate $$(1.02)^{\frac{2}{3}}$$ Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\log_3(x+1),a=0$$ Find the Maclaurin series for the function $$f(x)=2\sin x^3$$ Use the table of power series for elementary functions Find the Maclaurin series for the function $$\displaystyle{f{{\left({x}\right)}}}={2}{{\sin{{x}}}^{{3}}}$$ $$(1+x)^{-2}=1-2x+3x^2-4x^3+...,for\ -1 \((1+4x)^{-2}$$ Find the Maclaurin series for the function $$f(x)=\cos4x$$. Use the table of power series for elementary functions $$f(x)=\frac{e^{x^4}}{4}$$ Find the Maclaurin series for the function $$f(x)=\sin5x$$. Use the table of power series for elementary functions
Select Page The ALU is the core of the computer – it performs arithmetic and logic operations on data that not only realize the goals of various applications (e.g., scientific and engineering programs), but also manipulate addresses (e.g., pointer arithmetic). In this section, we will overview algorithms used for the basic arithmetic and logical operations. A key assumption is that twos complement representation will be employed, unless otherwise noted. When adding two numbers, if the sum of the digits in a given position equals or exceeds the modulus, then a carry is propagated. For example, in Boolean addition, if two ones are added, the sum is obviously two (base 10), which exceeds the modulus of 2 for Boolean numbers (B = Z2 = {0,1}, the integers modulo 2). Thus, we record a zero for the sum and propagate a carry valued at one into the next more significant digit, as shown in Figure 3.1. Figure 3.1. Example of Boolean addition with carry propagation, adapted from [Maf01]. #### 3.1.2. Boolean Subtraction When subtracting two numbers, two alternatives present themselves. First, one can formulate a subtraction algorithm, which is distinct from addition. Second, one can negate the subtrahend (i.e., in a – b, the subtrahend is b) then perform addition. Since we already know how to perform addition as well as twos complement negation, the second alternative is more practical. Figure 3.2 illustrates both processes, using the decimal subtraction 12 – 5 = 7 as an example. Figure 3.2. Example of Boolean subtraction using (a) unsigned binary representation, and (b) addition with twos complement negation – adapted from [Maf01]. Just as we have a carry in addition, the subtraction of Boolean numbers uses a borrow. For example, in Figure 3.2a, in the first (least significant) digit position, the difference 0 – 1 in the one’s place is realized by borrowing a one from the two’s place (next more significant digit). The borrow is propagated upward (toward the most significant digit) until it is zeroed (i.e., until we encounter a difference of 1 – 0). #### 3.1.3. Overflow Overflow occurs when there are insufficient bits in a binary number representation to portray the result of an arithmetic operation. Overflow occurs because computer arithmetic is not closed with respect to addition, subtraction, multiplication, or division. Overflow cannot occur in addition (subtraction), if the operands have different (resp. identical) signs. To detect and compensate for overflow, one needs n+1 bits if an n-bit number representation is employed. For example, in 32-bit arithmetic, 33 bits are required to detect or compensate for overflow. This can be implemented in addition (subtraction) by letting a carry (borrow) occur into (from) the sign bit. To make a pictorial example of convenient size, Figure 3.3 illustrates the four possible sign combinations of differencing 7 and 6 using a number representation that is four bits long (i.e., can represent integers in the interval [-8,7]). Figure 3.3. Example of overflow in Boolean arithmetic, adapted from [Maf01]. #### 3.1.4. MIPS Overflow Handling MIPS raises an exception when overflow occurs. Exceptions (or interrupts) act like procedure calls. The register `\$epc` stores the address of the instruction that caused the interrupt, and the instruction `mfc register, \$epc` moves the contents of `\$epc` to register. For example, register could be `\$t1`. This is an efficient approach, since no conditional branch is needed to test for overflow. Two’s complement arithmetic operations (`add`, `addi`, and `sub` instructions) raise exceptions on overflow. In contrast, unsigned arithmetic (`addu` and `addiu`) instructions do not raise an exception on overflow, since they are used for arithmetic operations on addresses (recall our discussion of pointer arithmetic in Section 2.6). In terms of high-level languages, C ignores overflows (always uses`addu`, `addiu`, and `subu`), while FORTRAN uses the appropriate instruction to detect overflow. Figure 3.4 illustrates the use of conditional branch on overflow for signed and unsigned addition operations. Figure 3.4. Example of overflow in Boolean arithmetic, adapted from [Maf01].
# 2000 AMC 12 Problems/Problem 25 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) $[asy] import three; import math; unitsize(1.5cm); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); [/asy]$ $\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680$ ## Solution 1 Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red. $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);[/asy]$ There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$. $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)shift(2right)r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)shift(4right)s); [/asy]$ ## Solution 2 We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube. Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}$. ## Solution 3 There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 64 = 24 ways to orient an octahedron, and $8!/24 = 1680 \Rightarrow \mathrm{(E)}$ ## Solution 4 If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0$^{\circ}$ rotation, 120$^{\circ}$ rotation, and 240$^{\circ}$ rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is $8 \cdot 3 = 24$. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be $8!/24 \Rightarrow \mathrm{(E)}$ ~kempwood ## Solution 5 (Graph Theory) This problem can be approached by Graph Theory. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge. Now the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other. Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs. There are $8$ colors and $4$ outer vertices, therefore there are $\binom{8}{4}$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }$ ## Solution 6 Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$. Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have $3!$ ways to put the three remaining colors in three spaces. In total this is $7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }$ -mathfan2020
# The learning nerd! On a mission to learn everything! # Binary addition and subtraction — February 15, 2015 ## Intro This is a continuation of the learning binary series. Be sure to read learning binary before reading this post. The basic values column 1000 \| 100 \| 10 \| 1 Example 1 1 1 5 9 5 + 3 5 6 4 _ _ _ _ 5 1 5 9 We take the first digit and carry that. While the second digit remains below. The digit you carried is added to the number. 5 + 4 = 9 > 9 Is less then 10 so we don’t carry over. 9 + 5 = 15 > 15 is higher then 10. We need to carry a 1 over into the 100’s column. 1 + 5 + 5 = 11 > 11 is higher then 10. We need to carry a 1 over into the 1000’s column. 1 + 1 + 3 = 5 > 5 is less then 10 so we don’t carry over. Binary addition is the same concept. Except for your using binary numbers 1 and 0. #### The five basic addition rules 0 + 0 = 0 1 + 0 = 1 10 + 0 = 10 1 + 1 = 10 > 1 + 1 = 2 > 10 = 2 in binary 10 + 1 = 11 Example 1 1 1 1 0 1 1 + 1 0 1 1 _ _ _ _ _ 1 0 1 1 0 1 + 1 = 10 > We carry the 1 1 + 1 + 1 = 10 + 1 = 11 > We carry the 1 1 + 0 = 1 1 + 1 = 10 > Since there is no more columns to carry. We put 10. 1011 = 11 1011 = 11 11 + 11 = 22 # Binary subtraction Example 3 4 16 3 3 5 6 4 − 2 3 4 9 0 _ _ _ _ _ _ 1 0 0 7 4 4 – 0 = 4 Borrow 1 to make 16 – 9 = 7 Borrow to make 4 – 4 = 0 Borrow to make 3 – 3 = 0 3 – 2 = 1 10074 Onto binary subtraction #### The four basic subtraction rules 0 – 0 = 0 1 – 0 = 1 1 – 1 = 0 10 – 1 = 1 0 1 10 10 1 0 1 0 1 − 1 1 1 0 _ _ _ _ _ _ 0 0 1 1 1 1 – 0 = 1 Borrow 1 to make 10 – 1 = 1 Borrow 1 to make 10 – 1 = 1 Borrow 1 to make 1 – 1 = 0 Borrow 0 to make 0 #### You can double check 10101 = 21 1110 = 14 21 – 14 = 7 I hope you enjoy my tutorial on how to subtract and add binary numbers. The next guide in this series will be how to multiply and divide binary numbers. Categorised as: Binary
A right triangle, whose sides are $3 \mathrm{~cm}$ and $4 \mathrm{~cm}$ (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate.) Given: A right triangle, whose sides are $3 \mathrm{~cm}$ and $4 \mathrm{~cm}$ (other than hypotenuse) is made to revolve about its hypotenuse. To do: We have to find the volume and surface area of the double cone so formed. Solution: Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of the right triangle $ABC$. This implies, using Pythagoras theorem, $AC^2=AB^2+BC^2$ $AC^2=3^2+4^2$ $=9+16$ $=25$ $\Rightarrow AC=\sqrt{25}$ $=5\ cm$ When the triangle $ABC$ is revolved about the hypotenuse $AC$, the following double cone is formed. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BDC}$, $\angle \mathrm{ABC}=\angle \mathrm{CDB}=90^{\circ}$ ($\mathrm{BD} \perp \mathrm{AC})$) $\angle \mathrm{BCA}=\angle \mathrm{BCD}$ (Common) Therefore, by AA similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{BDC}$ This implies, $\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$ (Corresponding sides of similar triangles are in proportion) $\mathrm{BD}=\frac{\mathrm{AB} \times \mathrm{BC}}{\mathrm{AC}}$ $=\frac{3 \mathrm{~cm} \times 4 \mathrm{~cm}}{5 \mathrm{~cm}}$ $=\frac{12}{5} \mathrm{~cm}$ $=2.4 \mathrm{~cm}$ We know that, Volume of a cone of height $h$ and radius $r$ is $\frac{1}{3} \pi r^{2} h$ The volume of the double cone in the figure $=$ Volume of cone ABB' $+$ Volume of cone BCB' $=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AD}+\frac{1}{3} \pi(\mathrm{BD})^{2} \times \mathrm{DC}$ $=\frac{1}{3}\times \pi(\mathrm{BD})^{2}(\mathrm{AD}+\mathrm{DC})$ $=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AC})$ $=\frac{1}{3}\times 3.14 \times (2.4)^2 \mathrm{~cm} \times 5 \mathrm{~cm}$ $=30.144 \mathrm{~cm}^{3}$ The curved surface area of a cone $=\pi rl$ The curved surface area of the double cone $=$ curved surface area of the cone ABB' $+$ curved surface area of the cone BCB' $=\pi \times \mathrm{BD} \times \mathrm{AB}+\pi \times \mathrm{BD} \times \mathrm{BC}$ $=\pi \times \mathrm{BD}(\mathrm{AB}+\mathrm{BC})$ $=3.14 \times 2.4 \mathrm{~cm} \times(3 \mathrm{~cm}+4 \mathrm{~cm})$ $=3.14 \times 2.4 \mathrm{~cm} \times 7 \mathrm{~cm}$ $=52.752 \mathrm{~cm}{ }^{2}$ $=52.75 \mathrm{~cm}{ }^{2}$ The volume and surface area of the double cone so formed are $30.144 \mathrm{~cm}^{3}$ and $52.75 \mathrm{~cm}{ }^{2}$ respectively. Tutorialspoint Simply Easy Learning
NCERT Solutions: Rational Numbers # NCERT Solutions: Rational Numbers Notes \| Study Mathematics (Maths) Class 7 - Class 7 ## Document Description: NCERT Solutions: Rational Numbers for Class 7 2022 is part of Class 7 Mathematics for Mathematics (Maths) Class 7 preparation. The notes and questions for NCERT Solutions: Rational Numbers have been prepared according to the Class 7 exam syllabus. Information about NCERT Solutions: Rational Numbers covers topics like Exercise 9.1, Exercise 9.2  and NCERT Solutions: Rational Numbers Example, for Class 7 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Rational Numbers. Introduction of NCERT Solutions: Rational Numbers in English is available as part of our Mathematics (Maths) Class 7 for Class 7 & NCERT Solutions: Rational Numbers in Hindi for Mathematics (Maths) Class 7 course. Download more important topics related with Class 7 Mathematics, notes, lectures and mock test series for Class 7 Exam by signing up for free. Class 7: NCERT Solutions: Rational Numbers Notes \| Study Mathematics (Maths) Class 7 - Class 7 1 Crore+ students have signed up on EduRev. Have you? Exercise 9.1 Q1. List five rational numbers between: (i) -1 and 0 Ans: Let us write -1 and 0 as rational numbers with denominator 6. Therefore, five rational numbers between -1 and 0 would be (ii) -2 and -1 Ans: Let us write -2 and -1 as rational numbers with denominator 6. Therefore, Five rational numbers between -2 and -1 would be (iii) Ans: Let us write as rational numbers with the same denominators. Therefore, five rational numbers between would be (iv) Ans: Let us write as rational numbers with the same denominators. Therefore, five rational numbers between would be Q2. Write four more rational numbers in each of the following patterns: (i) Ans: Therefore, the next four rational numbers of this pattern would be (ii) Ans: Therefore, the next four rational numbers of this pattern would be (iii) Ans: Therefore, the next four rational numbers of this pattern would be (iv) Ans: Therefore, the next four rational numbers of tins pattern would be Q3. Give four rational numbers equivalent to: (i) -2/7 Ans: The four rational numbers equivalent to -2/7 are, Therefore, four equivalent rational numbers are (ii) 5/-3 Ans: The four rational numbers equivalent to 5/-3 are, Therefore, four equivalent rational numbers are (iii) 4/9 Ans: The four rational numbers equivalent to 5/-3 are, Therefore, four equivalent rational numbers are Q4. Draw the number line and represent the following rational numbers on it: (i) 3/4 Ans: We know that 3/4 is greater than 0 and less than 1. ∴ it lies between 0 and 1. It can be represented on number line as, (ii) -5/8 Ans: We know that -5/8 is less than 0 and greater than -1. ∴ it lies between 0 and -1. It can be represented on number line as, (iii) -7/4 Ans: Now above question can be written as, = (-7/4) = We know that (-7/4) is Less than -1 and greater than -2. ∴ it lies between -1 and -2. It can be represented on number line as, (iv) 7/8 Ans: We know that 7/8 is greater than 0 and less than 1. ∴ it lies between 0 and 1. It can be represented on number line as, Q5. The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S. Ans: Therefore, Similarly Thus, the rational numbers represented P, Q, R and S are respectively. Q6. Which of the following pairs represent the same rational numbers: (i) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is not represents the same rational number. (ii) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is represents the same rational number. (iii) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is represents the same rational number. (iv) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is represents the same rational number. (v) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is represents the same rational number. (vi) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest term] So, the given pair is not represents the same rational number. (vii) Ans: We have to check the given pair represents the same rational number. Then, [Converting into lowest terra] So, the given pair is not represents the same rational number. Q7. Rewrite the following rational numbers in the simplest form: (i) -8/6 Ans: The given rational numbers can be simplified further, Then, = -4/3 … [∵ Divide both numerator and denominator by 2] (ii) 25/45 Ans: The given rational numbers can be simplified further, Then, = 5/9 … [∵ Divide both numerator and denominator by 5] (iii) -44/72 Ans: The given rational numbers can be simplified further, Then, = -11/18 … [∵ Divide both numerator and denominator by 4] (iv) -8/10 Ans: The given rational numbers can be simplified further, Then, = -4/5 … [∵ Divide both numerator and denominator by 2] Q8. Fill in the boxes with the correct symbol out of <, > and =: (i) Ans: The LCM of the denominators 7 and 3 is 21 ∴ (-5/7) = [(-5 × 3)/ (7 × 3)] = (-15/21) And (2/3) = [(2 × 7)/ (3 × 7)] = (14/21) Now, -15 < 14 So, (-15/21) < (14/21) Hence, -5/7 [<] 2/3 (ii) Ans: The LCM of the denominators 5 and 7 is 35 ∴ (-4/5) = [(-4 × 7)/ (5 × 7)] = (-28/35) And (-5/7) = [(-5 × 5)/ (7 × 5)] = (-25/35) Now, -28 < -25 So, (-28/35) < (- 25/35) Hence, -4/5 [<] -5/7 (iii) Ans: 14/-16 can be simplified further, Then, 7/-8 … [∵ Divide both numerator and denominator by 2] So, (-7/8) = (-7/8) Hence, -7/8 [=] 14/-16 (iv) Ans: The LCM of the denominators 5 and 4 is 20 ∴ (-8/5) = [(-8 × 4)/ (5 × 4)] = (-32/20) And (-7/4) = [(-7 × 5)/ (4 × 5)] = (-35/20) Now, -32 > – 35 So, (-32/20) > (- 35/20) Hence, -8/5 [>] -7/4 (v) Ans: The LCM of the denominators 3 and 4 is 12 ∴ (-1/3) = [(-1 × 4)/ (3 × 4)] = (-4/12) And (-1/4) = [(-1 × 3)/ (4 × 3)] = (-3/12) Now, -4 < – 3 So, (-4/12) < (- 3/12) Hence, 1/-3 [<] -1/4 (vi) Ans: Since, (-5/11) = (-5/11) Hence, 5/-11 [=] -5/11 (vii) Ans: Since every negative rational number is less than 0. We have: = 0 [>] -7/6 Q9. Which is greater in each of the following: (i) Ans: The LCM of the denominators 3 and 2 is 6 (2/3) = [(2 × 2)/ (3 × 2)] = (4/6) And (5/2) = [(5 × 3)/ (2 × 3)] = (15/6) Now, 4 < 15 So, (4/6) < (15/6) ∴ 2/3 < 5/2 Hence, 5/2 is greater. (ii) Ans: The LCM of the denominators 6 and 3 is 6 ∴ (-5/6) = [(-5 × 1)/ (6 × 1)] = (-5/6) And (-4/3) = [(-4 × 2)/ (3 × 2)] = (-12/6) Now, -5 > -12 So, (-5/6) > (- 12/6) ∴ -5/6 > -12/6 Hence, – 5/6 is greater. (iii) Ans: The LCM of the denominators 4 and 3 is 12 ∴ (-3/4) = [(-3 × 3)/ (4 × 3)] = (-9/12) And (-2/3) = [(-2 × 4)/ (3 × 4)] = (-8/12) Now, -9 < -8 So, (-9/12) < (- 8/12) ∴ -3/4 < 2/-3 Hence, 2/-3 is greater. (iv) Ans: The given fraction is like friction, So, -¼ < ¼ Hence ¼ is greater, (v) Ans: First we have to convert mixed fraction into improper fraction, = -23/7 = -19/5 Then, The LCM of the denominators 7 and 5 is 35 ∴ (-23/7) = [(-23 × 5)/ (7 × 5)] = (-115/35) And (-19/5) = [(-19 × 7)/ (5 × 7)] = (-133/35) Now, -115 > -133 So, (-115/35) > (- 133/35) > Hence, is greater. Q10. Write the following rational numbers in ascending order: (i) -3/5, -2/5, -1/5 Ans: The given rational numbers are in form of like fraction, Hence, (-3/5)< (-2/5) < (-1/5) (ii) -1/3, -2/9, -4/3 Ans: To convert the given rational numbers into like fraction we have to find LCM, LCM of 3, 9, and 3 is 9 Now, (-1/3)= [(-1 × 3)/ (3 × 9)] = (-3/9) (-2/9)= [(-2 × 1)/ (9 × 1)] = (-2/9) (-4/3)= [(-4 × 3)/ (3 × 3)] = (-12/9) Clearly, (-12/9) < (-3/9) < (-2/9) Hence, (-4/3) < (-1/3) < (-2/9) (iii) -3/7, -3/2, -3/4 Ans: To convert the given rational numbers into like fraction we have to find LCM, LCM of 7, 2, and 4 is 28 Now, (-3/7)= [(-3 × 4)/ (7 × 4)] = (-12/28) (-3/2)= [(-3 × 14)/ (2 × 14)] = (-42/28) (-3/4)= [(-3 × 7)/ (4 × 7)] = (-21/28) Clearly, (-42/28) < (-21/28) < (-12/28) Hence, (-3/2) < (-3/4) < (-3/7) Exercise 9.2 Q1. Find the sum: (i) Ans: We have: = (5/4) – (11/4) = [(5 – 11)/4] … [∵ denominator is same in both the rational numbers] = (-6/4) = -3/2 … [∵ Divide both numerator and denominator by 3] (ii) Ans: Take the LCM of the denominators of the given rational numbers. LCM of 3 and 5 is 15 Express each of the given rational numbers with the above LCM as the common denominator. Now, (5/3) = [(5 × 5)/ (3 × 5)] = (25/15) (3/5) = [(3 × 3)/ (5 × 3)] = (9/15) Then, = (25/15) + (9/15) … [∵ denominator is same in both the rational numbers] = (25 + 9)/15 = 34/15 (iii) Ans: Take the LCM of the denominators of the given rational numbers. LCM of 10 and 15 is 30 Express each of the given rational numbers with the above LCM as the common denominator. Now, (-9/10)= [(-9 × 3)/ (10 × 3)] = (-27/30) (22/15)= [(22 × 2) / (15 × 2)] = (44/30) Then, = (-27/30) + (44/30) … [∵ denominator is same in both the rational numbers] = (-27 + 44)/30 = (17/30) (iv) Ans: We have, = 3/11 + 5/9 Take the LCM of the denominators of the given rational numbers. LCM of 11 and 9 is 99 Express each of the given rational numbers with the above LCM as the common denominator. Now, (3/11) = [(3 × 9)/ (11 × 9)] = (27/99) (5/9) = [(5 × 11)/ (9 × 11)] = (55/99) Then, = (27/99) + (55/99) … [∵ denominator is same in both the rational numbers] = (27 + 55)/99 = (82/99) (v) Ans: We have = -8/19 – 2/57 Take the LCM of the denominators of the given rational numbers. LCM of 19 and 57 is 57 Express each of the given rational numbers with the above LCM as the common denominator. Now, (-8/19)= [(-8 × 3)/ (19 × 3)] = (-24/57) (-2/57)= [(-2 × 1)/ (57 × 1)] = (-2/57) Then, = (-24/57) – (2/57) … [∵ denominator is same in both the rational numbers] = (-24 – 2)/57 = (-26/57) (vi) Ans: We know that any number or fraction is added to zero the answer will be the same number or fraction. Hence, = -2/3 + 0 = -2/3 (vi) Ans: [L.C.M. of 3 and 5 is 15] Q2. Find: (i) Ans: Take the LCM of the denominators of the given rational numbers. LCM of 24 and 36 is 72 Express each of the given rational numbers with the above LCM as the common denominator. Now, (7/24)= [(7 × 3)/ (24 × 3)] = (21/72) (17/36)= [(17 × 2)/ (36 × 2)] = (34/72) Then, = (21/72) – (34/72) … [∵ denominator is same in both the rational numbers] = (21 – 34)/72 = (-13/72) (ii) Ans: We can also write -6/21 = -2/7 = 5/63 – (-2/7) We have, = 5/63 + 2/7 Take the LCM of the denominators of the given rational numbers. LCM of 63 and 7 is 63 Express each of the given rational numbers with the above LCM as the common denominator. Now, (5/63)= [(5 × 1)/ (63 × 1)] = (5/63) (2/7)= [(2 × 9)/ (7 × 9)] = (18/63) Then, = (5/63) + (18/63) … [∵ denominator is same in both the rational numbers] = (5 + 18)/63 = 23/63 (iii) Ans: We have, = -6/13 + 7/15 LCM of 13 and 15 is 195 Express each of the given rational numbers with the above LCM as the common denominator. Now, (-6/13)= [(-6 × 15)/ (13 × 15)] = (-90/195) (7/15)= [(7 × 13)/ (15 × 13)] = (91/195) Then, = (-90/195) + (91/195) … [∵ denominator is same in both the rational numbers] = (-90 + 91)/195 = (1/195) (iv) Ans: Take the LCM of the denominators of the given rational numbers. LCM of 8 and 11 is 88 Express each of the given rational numbers with the above LCM as the common denominator. Now, (-3/8)= [(-3 × 11)/ (8 × 11)] = (-33/88) (7/11)= [(7 × 8)/ (11 × 8)] = (56/88) Then, = (-33/88) – (56/88) … [∵ denominator is same in both the rational numbers] = (-33 – 56)/88 = (-89/88) (v) Ans: First we have to convert the mixed fraction into improper fraction, = -19/9 We have, -19/9 – 6 Take the LCM of the denominators of the given rational numbers. LCM of 9 and 1 is 9 Express each of the given rational numbers with the above LCM as the common denominator. Now, (-19/9)= [(-19 × 1)/ (9 × 1)] = (-19/9) (6/1)= [(6 × 9)/ (1 × 9)] = (54/9) Then, = (-19/9) – (54/9) … [∵ denominator is same in both the rational numbers] = (-19 – 54)/9 = (-73/9) Q3. Find the product: (i) (9/2) × (-7/4) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) The above question can be written as (9/2) × (-7/4) We have, = (9 × -7) / (2 × 4) = -63/8 (ii) (3/10) × (-9) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) The above question can be written as (3/10) × (-9/1) We have, = (3 × -9)/ (10×1) = -27/10 (iii) (-6/5) × (9/11) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) We have, = (-6 × 9) / (5 × 11) = -54/55 (iv) (3/7) × (-2/5) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) We have, = (3 × -2) / (7 × 5) = -6/35 (v) (3/11) × (2/5) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) We have, = (3 × 2) / (11 × 5) = 6/55 (vi) (3/-5) × (-5/3) Ans: The product of two rational numbers = (product of their numerator)/ (product of their denominator) We have, = (3 × -5) / (-5 × 3) On simplifying, = (1 × -1)/ (-1 × 1) = -1/-1 = 1 Q4. Find the value of: (i) (-4) ÷ (2/3) Ans: We have, = (-4/1) × (3/2) … [∵ reciprocal of (2/3) is (3/2)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-4 × 3) / (1 × 2) = (-2 × 3) / (1 × 1) = -6 (ii) (-3/5) ÷ 2 Ans: We have, = (-3/5) × (1/2) … [∵ reciprocal of (2/1) is (1/2)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-3 × 1) / (5 × 2) = -3/10 (iii) (-4/5) ÷ (-3) Ans: We have, = (-4/5) × (1/-3) … [∵ reciprocal of (-3) is (1/-3)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-4× (1)) / (5× (-3)) = -4/-15 = 4/15 (iv) (-1/8) ÷ 3/4 Ans: We have, = (-1/8) × (4/3) … [∵ reciprocal of (3/4) is (4/3)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-1 × 4) / (8 × 3) = (-1 × 1) / (2 × 3) = -1/6 (v) (-2/13) ÷ 1/7 Ans: We have, = (-2/13) × (7/1) … [∵ reciprocal of (1/7) is (7/1)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-2 × 7) / (13 × 1) = -14/13 (vi) (-7/12) ÷ (-2/13) Ans: We have, = (-7/12) × (13/-2) … [∵ reciprocal of (-2/13) is (13/-2)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (-7× 13) / (12× (-2)) = -91/-24 = 91/24 (vii) (3/13) ÷ (-4/65) Ans: We have, = (3/13) × (65/-4) … [∵ reciprocal of (-4/65) is (65/-4)] The product of two rational numbers = (product of their numerator)/ (product of their denominator) = (3 × 65) / (13 × (-4)) = 195/-52 = -15/4 The document NCERT Solutions: Rational Numbers Notes \| Study Mathematics (Maths) Class 7 - Class 7 is a part of the Class 7 Course Mathematics (Maths) Class 7. 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## Thursday, February 10, 2011 ### Math problems of the week: 6th grade Connected Math vs. Singapore Math Initial 6th grade fractions problems: I. The second fractions problem set in 6th Grade Connected Math (Bits and Pieces I, p. 7): A. 1. Use strips of paper that are 8 1/2 inches long. Fold the strips to show halves, thirds, fourths, fifths, sixths, eighths, ninths, tenths, and twelfths. Mark the folds so you can see them better. 2. What strategies did you use to fold your strips? B. 1. How could you use the halves strip to fold eighths? 2. How could you use the halves strip to fold twelfths? C. What fractions strips can you make if you start with a thirds strip? D. Which of the fraction strips you folded have at least one mark with the arks on the twelfths strip? E. 1. Sketch a picture of a fifths strip and mark 1/5/, 2/5, 3/5, 3/5, and 5/5 on the strip. 2. Show 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10, and 10/10 on the fifths strip that you sketched. F. What do the numerator and the denominator of a fraction tell you? II. The second fractions problems set in 6th Grade Singapore Math (Primary Mathematics 6B, p. 7): 1. Divide (a) 1/3 ÷ 3 = 1/3 × 1/3 = (b) 1/6 ÷ 6 = 1/2 × ___ = (c) 1/6 ÷ 4 = (d) 4/5 ÷ 2 = (e) 2/5 ÷ 4 = (f) 8/9 ÷ 4 = (g) 3/4 ÷ 2 = (h) 2/3 ÷ 6 = III. Extra Credit: What is your preferred way of discovering things about fractions: fraction strips or repeated calculation of carefully contrived fractions problems? #### 1 comment: Cynthia said... I have to say that I would have enjoyed doing the fraction strips (first time I've felt that way about the problems you've posted from them) but I would have HATED having to write about it in what would have had to be either an incredibly tedious or frustratingly imprecise way.
# How do you solve 10/x - 12/(x-3) + 4 = 0? Feb 12, 2016 You must put on a common denominator. #### Explanation: The LCD (Least Common Denominator) is $x \left(x - 3\right)$ $\frac{10 \left(x - 3\right)}{x \left(x - 3\right)} - \frac{12 \left(x\right)}{x \left(x - 3\right)} + \frac{4 \left({x}^{2} - 3 x\right)}{x \times x - 3} = 0$ We can now eliminate the denominators: $10 x - 30 - 12 x + 4 {x}^{2} - 12 x = 0$ $4 {x}^{2} - 14 x - 30 = 0$ Solve by factoring. Two numbers that multiply to $\left(- 30 \times 4\right) = - 120$ and that add to -14 are -20 and 6. $4 {x}^{2} - 20 x + 6 x - 30 = 0$ $4 x \left(x - 5\right) + 6 \left(x - 5\right) = 0$ $\left(4 x + 6\right) \left(x - 5\right) = 0$ $x = - \frac{6}{4} \mathmr{and} 5$
# Class 9 Maths Chapter 10 Circles NCERT Solutions for Class 9 Maths Chapter 10 Circles. CBSE Online Eductional Study Material for India Schools, Students. Exercise 10.1 Question 1. Fill in the blanks. (i) The centre of a circle lies in ___ of the circle. (exterior/interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior) (iii) The longest chord of a circle is a ____ of the circle. (iv) An arc is a ____ when its ends are the ends of a diameter. (v) Segment of a circle is the region between an arc and ____ of the circle. (vi) A circle divides the plane, on which it lies, in ____ parts. Solution: (i) interior (ii) exterior (iii) diameter (iv) semicircle (v) the chord (vi) three Question 2. Write True or False. Give reason for your answers. (i) Line segment joining the centre to any point on the circle is a , radius of the circle. (ii) A circle has only finite number of equal chords. (iii) If a circle is divided into three equal arcs, each is a major arc. (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle. (v) Sector is the region between the chord and its corresponding arc. (vi) A circle is a plane figure. Solution: (i) True [∵ All points on the circle are equidistant from the centre] (ii) False [ ∵ A circle can have an infinite number of equal chords] (iii) False [∵ Each part will be less than a semicircle] (iv) True [ ∵ Diameter = 2 x Radius] (v) False [ ∵ The region between the chord and its corresponding arc is a segment] (vi) True [ ∵ A circle is drawn on a plane] Exercise 10.2 Question 1. Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres Solution: Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD. To Prove: ∠AOB = ∠CO’D Proof: In ∆AOB and ∆CO’D, we have AB = CD [Given] OA = O’C [Each equal to r] OB = O’D [Each equal to r] ∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria] ⇒ ∠AOB = ∠CO’D [C.P.C.T.] Question 2. Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal. Solution: Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D. To Prove: AB = CD Proof: In ∆AOB and ∆CO’D, we have OA = O’C [Each equal to r] OB = O’D [Each equal to r] ∠AOB = ∠CO’D [Given] ∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria] Hence, AB = CD [C.P.C.T.] Question 2. Suppose you are given a circle. Give a construction to find its centre. Solution: Steps of construction : Step I : Take any three points on the given circle. Let these points be A, B and C. Step II : Join AB and BC. Step III : Draw the perpendicular bisector, PQ of AB. Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O. Thus, ‘O’ is the required centre of the given drcle. Question 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. Solution: We have two circles with centres O and O’, intersecting at A and B. ∴ AB is the common chord of two circles and OO’ is the line segment joining their centres. Let OO’ and AB intersect each other at M. ∴ To prove that OO’ is the perpendicular bisector of AB, we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’, we have OA = OB [Radii of the same circle] O’A = O’B [Radii of the same circle] OO’ = OO’ [Common] ∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria] ⇒ ∠1 = ∠2 , [C.P.C.T.] Now, in ∆AOM and ∆BOM, we have OA = OB [Radii of the same circle] OM = OM [Common] ∠1 = ∠2 [Proved above] ∴ ∆AOM = ∆BOM [By SAS congruence criteria] ⇒ ∠3 = ∠4 [C.P.C.T.] But ∠3 + ∠4 = 180° [Linear pair] ∴∠3=∠4 = 90° ⇒ AM ⊥ OO’ Also, AM = BM [C.P.C.T.] ⇒ M is the mid-point of AB. Thus, OO’ is the perpendicular bisector of AB. Exercise 10.4 Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Solution: We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord. ∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord. ∴∠OLP = ∠OLQ = 90° and PL = LQ Now, in right ∆OLP, we have PL2 + OL2 = 2 ⇒ PL2 + (4 – x)2 = 52 ⇒ PL2 = 52 – (4 – x)2 ⇒ PL2 = 25 -16 – x2 + 8x ⇒ PL2 = 9 – x2 + 8x …(i) Again, in right ∆O’LP, PL2 = PO‘2 – LO‘2 = 32 – x2 = 9 – x2 …(ii) From (i) and (ii), we have 9 – x2 + 8x = 9 – x2 ⇒ 8x = 0 ⇒ x = 0 ⇒ L and O’ coincide. ∴ PQ is a diameter of the smaller circle. ⇒ PL = 3 cm But PL = LQ ∴ LQ = 3 cm ∴ PQ = PL + LQ = 3cm + 3cm = 6cm Thus, the required length of the common chord = 6 cm. Question 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. Solution: Given: A circle with centre O and equal chords AB and CD intersect at E. To Prove: AE = DE and CE = BE Construction : Draw OM ⊥ AB and ON ⊥ CD. Join OE. Proof: Since AB = CD [Given] ∴ OM = ON [Equal chords are equidistant from the centre] Now, in ∆OME and ∆ONE, we have ∠OME = ∠ONE [Each equal to 90°] OM = ON [Proved above] OE = OE [Common hypotenuse] ∴ ∆OME ≅ ∆ONE [By RHS congruence criteria] ⇒ ME = NE [C.P.C.T.] Adding AM on both sides, we get ⇒ AM + ME = AM + NE ⇒ AE = DN + NE = DE ∵ AB = CD ⇒ = ⇒ AM = DN ⇒ AE = DE …(i) Now, AB – AE = CD – DE ⇒ BE = CE …….(ii) From (i) and (ii), we have AE = DE and CE = BE Question 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. Solution: Given: A circle with centre O and equal chords AB and CD are intersecting at E. To Prove : ∠OEA = ∠OED Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OE. Proof: In ∆OME and ∆ONE, OM = ON [Equal chords are equidistant from the centre] OE = OE [Common hypotenuse] ∠OME = ∠ONE [Each equal to 90°] ∴ ∆OME ≅ ∆ONE [By RHS congruence criteria] ⇒ ∠OEM = ∠OEN [C.P.C.T.] ⇒ ∠OEA = ∠OED Question 4. If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure). Solution: Given : Two circles with the common centre O. A line D intersects the outer circle at A and D and the inner circle at B and C. To Prove : AB = CD. Construction: Draw OM ⊥ l. Proof: For the outer circle, OM ⊥ l [By construction] ∴ AM = MD …(i) [Perpendicular from the centre to the chord bisects the chord] For the inner circle, OM ⊥ l [By construction] ∴ BM = MC …(ii) [Perpendicular from the centre to the chord bisects the chord] Subtracting (ii) from (i), we have AM – BM = MD – MC ⇒ AB = CD Question 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? Solution: Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that RS = SM = 6 m [Given] Equal chords of a circle subtend equal angles at the centre. ∴ ∠1 = ∠2 In ∆POR and ∆POM, we have OP = OP [Common] OR = OM [Radii of the same circle] ∠1 = ∠2 [Proved above] ∴ ∆POR ≅ ∆POM [By SAS congruence criteria] ∴ PR = PM and ∠OPR = ∠OPM [C.P.C.T.] ∵∠OPR + ∠OPM = 180° [Linear pair] ∴∠OPR = ∠OPM = 90° ⇒ OP ⊥ RM Now, in ∆RSP and ∆MSP, we have RS = MS [Each 6 cm] SP = SP [Common] PR = PM [Proved above] ∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria] ⇒ ∠RPS = ∠MPS [C.P.C.T.] But ∠RPS + ∠MPS = 180° [Linear pair] ⇒ ∠RPS = ∠MPS = 90° SP passes through O. Let OP = x m ∴ SP = (5 – x)m Now, in right ∆OPR, we have x2 + RP2 = 52 RP2 = 52 – x2 In right ∆SPR, we have (5 – x)2 + RP2 = 62 ⇒ RP2 = 62 – (5 – x)2 ……..(ii) From (i) and (ii), we get ⇒ 52 – x2 = 62 – (5 – x)2 ⇒ 25 – x2 = 36 – [25 – 10x + x2] ⇒ – 10x + 14 = 0 ⇒ 10x = 14 ⇒ x = = 1.4 Now, RP2 = 52 – x2 ⇒ RP2 = 25 – (1.4)2 ⇒ RP2 = 25 – 1.96 = 23.04 ∴ RP = = 4.8 ∴ RM = 2RP = 2 x 4.8 = 9.6 Thus, distance between Reshma and Mandip is 9.6 m. Question 6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Solution: Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA i. e., ∆ASD is an equilateral triangle. Let the length of each side of the equilateral triangle be 2x. Draw AM ⊥ SD. Since ∆ASD is an equilateral triangle. ∴ AM passes through O. ⇒ SM = SD = (2x) ⇒ SM = x Now, in right ∆ASM, we have AM2 + SM2 = AS2 [Using Pythagoras theorem] ⇒ AM2= AS2 – SM2= (2x)2 – x2 = 4x2 – x2 = 3x2 ⇒ AM = m Now, OM = AM – OA= ( – 20)m Again, in right ∆OSM, we have OS2 = SM2 + OM2 [using Pythagoras theorem] 202 = x2 + ( – 20)2 ⇒ 400 = x2 + 3x2 – 40 + 400 ⇒ 4x2 = 40 ⇒ x = 10√3 m Now, SD = 2x = 2 x 10√3 m = 20√3 m Thus, the length of the string of each phone = 20√3 m Exercise 10.5 Question 1. In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC. Solution: We have a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30° ∵∠AOB + ∠BOC = ∠AOC ∴ ∠AOC = 60° + 30° = 90° The angle subtended by an arc at the circle is half the angle subtended by it at the centre. ∴ ∠ ADC = (∠AOC) = (90°) = 45° Question 2. A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. Solution: We have a circle having a chord AB equal to radius of the circle. ∴ AO = BO = AB ⇒ ∆AOB is an equilateral triangle. Since, each angle of an equilateral triangle is 60°. ⇒ ∠AOB = 60° Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle. Hence, the angle subtended by the chord on the minor arc = 150°. Similarly, ∠ADB = [∠AOB] = x 60° = 30° Hence, the angle subtended by the chord on the major arc = 30° Question 3. In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. Solution: The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference. ∴ reflex ∠POR = 2∠PQR But ∠PQR = 100° ∴ reflex ∠POR = 2 x 100° = 200° Since, ∠POR + reflex ∠POR = 360° ⇒ ∠POR = 360° – 200° ⇒ ∠POR = 160° Since, OP = OR [Radii of the same circle] ∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal] Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle is 180°] ⇒ ∠OPR + ∠ORP + 160° = 180° ⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP] ⇒ ∠OPR = = 10° Question 4. In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC. Solution: In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180° ⇒ 69° + 31° + ∠BAC = 180° ⇒ ∠BAC = 180° – 100° = 80° Since, angles in the same segment are equal. ∴∠BDC = ∠BAC ⇒ ∠BDC = 80° Question 5. In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC. Solution: ∠BEC = ∠EDC + ∠ECD [Sum of interior opposite angles is equal to exterior angle] ⇒ 130° = ∠EDC + 20° ⇒ ∠EDC = 130° – 20° = 110° ⇒ ∠BDC = 110° Since, angles in the same segment are equal. ∴ ∠BAC = ∠BDC ⇒ ∠BAC = 110° Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD. Solution: Since angles in the same segment of a circle are equal. ∴ ∠BAC = ∠BDC ⇒ ∠BDC = 30° lso, ∠DBC = 70° [Given] In ∆BCD, we have ∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°] ⇒ ∠BCD + 70° + 30° = 180° ⇒ ∠BCD = 180° -100° = 80° Now, in ∆ABC, AB = BC [Given] ∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal] ⇒ ∠BCA = 30° [∵ ∠B AC = 30°] Now, ∠BCA + ∠BCD = ∠BCD ⇒ 30° + ∠ECD = 80° ⇒ ∠BCD = 80° – 30° = 50° Question 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. Solution: Since AC and BD are diameters. ⇒ AC = BD …(i) [All diameters of a circle are equal] Also, ∠BAD = 90° [Angle formed in a semicircle is 90°] Similarly, ∠ABC = 90°, ∠BCD = 90° and ∠CDA = 90° Now, in ∆ABC and ∆BAD, we have AC = BD [From (i)] AB = BA [Common hypotenuse] ∠ABC = ∠BAD [Each equal to 90°] ∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria] ⇒ BC = AD [C.P.C.T.] Similarly, AB = DC Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle. ∴ ABCD is a rectangle. Question 8. If the non – parallel sides of a trapezium are equal, prove that it is cyclic. Solution: We have a trapezium ABCD such that AB ॥ CD and AD = BC. Let us draw BE ॥ AD such that ABED is a parallelogram. ∵ The opposite angles and opposite sides of a parallelogram are equal. ∴ ∠BAD = ∠BED …(i) and AD = BE …(ii) But AD = BC [Given] …(iii) ∴ From (ii) and (iii), we have BE = BC ⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal] Now, ∠BED + ∠BEC = 180° [Linear pair] ⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)] i.e., A pair of opposite angles of a quadrilateral ABCD is 180°. ∴ ABCD is cyclic. ⇒ The trapezium ABCD is cyclic. Question 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A,D and P, Q respectively (see figure). Prove that ∠ ACP = ∠QCD. Solution: Since, angles in the same segment of a circle are equal. ∴ ∠ACP = ∠ABP …(i) Similarly, ∠QCD = ∠QBD …(ii) Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles] ∴ From (i), (ii) and (iii), we have ∠ACP = ∠QCD Question 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. Solution: We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A. Let us join A and D. ∵ AB is a diameter. ∴∠ADB is an angle formed in a semicircle. ⇒ ∠ADB = 90° ……(i) Similarly, ∠ADC = 90° ….(ii) Adding (i) and (ii), we have ∠ADB + ∠ADC = 90° + 90° = 180° i. e., B, D and C are collinear points. ⇒ BC is a straight line. Thus, D lies on BC. Question 11. ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. Solution: We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC ∴ Both the triangles are in semi-circle. Case – I: If both the triangles are in the same semi-circle. ⇒ A, B, C and D are concyclic. Join BD. DC is a chord. ∴ ∠CAD and ∠CBD are formed in the same segment. ⇒ ∠CAD = ∠CBD Case – II : If both the triangles are not in the same semi-circle. ⇒ A,B,C and D are concyclic. Join BD. DC is a chord. ∴ ∠CAD and ∠CBD are formed in the same segment. ⇒ ∠CAD = ∠CBD Question 12. Prove that a cyclic parallelogram is a rectangle. Solution: We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral. ∴ Sum of its opposite angles is 180°. ⇒ ∠A + ∠C = 180° …(i) But ∠A = ∠C …(ii) [Opposite angles of a parallelogram are equal] From (i) and (ii), we have ∠A = ∠C = 90° Similarly, ∠B = ∠D = 90° ⇒ Each angle of the parallelogram ABCD is 90°. Thus, ABCD is a rectangle. Exercise 10.6 Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Solution: Given : Two circles with centres O and O’ respectively such that they intersect each other at P and Q. To Prove: ∠OPO’ = ∠OQO’. Construction : Join OP, O’P, OQ, O’Q and OO’. Proof: In ∆OPO’ and ∆OQO’, we have OP = OQ [Radii of the same circle] O’P = O’Q [Radii of the same circle] OO’ = OO’ [Common] ∴ AOPO’ = AOQO’ [By SSS congruence criteria] ⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.] Question 2. Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. Solution: We have a circle with centre O. AB \|\| CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm. Let r cm be the radius of the circle. Let us draw OP ⊥ AB and OQ ⊥ CD such that PQ = 6 cm Join OA and OC. Let OQ = x cm ∴ OP = (6 – x) cm ∵ The perpendicular drawn from the centre of a circle to chord bisects the chord. Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ? Solution: We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre. Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC. ∵ OP ⊥ AB ∴ P is the mid-point of AB. ⇒ AP = = = 3 cm Similarly, CQ = = = 4 cm Now in ∆OPA, we have OA2 = OP2 + AP2 ⇒ r2 = 42 + 32 ⇒ r2 = 16 + 9 = 25 ⇒ r = =5 Again, in ∆CQO, we have OC2 = OQ2 + CQ2 ⇒ r2 = OQ2 + 42 ⇒ OQ2 = r2 – 42 ⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5] ⇒ OQ ⇒ √9 = 3 The distance of the other chord (CD) from the centre is 3 cm. Note: In case if we take the two parallel chords on either side of the centre, then In ∆POA, OA2 = OP2 + PA2 ⇒ r2 = 42 + 32 = 52 ⇒ r = 5 In ∆QOC, OC2 = CQ2 + OQ2 ⇒ OQ2 = 42 + OQ2 ⇒ OQ2 = 52 – 42 = 9 ⇒ OQ = 3 Question 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. Solution: Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE. To prove: ∠ABC = [∠DOE – ∠AOC] Construction: Join AE. Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles. ∴ In ∆BAE, we have ∠DAE = ∠ABC + ∠AEC ……(i) The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle. ⇒ ∠ABC = [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)] ⇒ ∠ABC = [Difference of the angles subtended by the chords DE and AC at the centre] Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. Solution: We have a rhombus ABCD such that its diagonals AC and BD intersect at O. Taking AB as diameter, a circle is drawn. Let us draw PQ \|\| DA and RS \|\| AB, both are passing through O. P, Q, R and S are the mid-points of DC, AB, AD and BC respectively, ∵ Q is the mid-point of AB. ⇒ AQ = QB …(i) Since AD = BC [ ∵ ABCD is a rhombus] AD = BC ⇒ RA = SB ⇒ RA = OQ …(ii) [ ∵ PQ is drawn parallel to AD and AD = BC] We have, AB = AD [Sides of rhombus are equal] AB = AD ⇒ AQ = AR …(iii) From (i), (ii) and (iii), we have AQ = QB = OQ i.e. A circle drawn with Q as centre, will pass through A, B and O. Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD. Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD. Solution: We have a circle passing through A, B and C is drawn such that it intersects CD at E. ABCE is a cyclic quadrilateral. ∴∠AEC + ∠B = 180° …(i) [Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given] ∴∠D = ∠B …(ii) [Opposite angles of a parallelogram are equal] From (i) and (ii), we have ∠AEC + ∠D = 180° …(iii) But ∠AEC + ∠AED = 180° [Linear pair] …(iv) From (iii) and (iv), we have ∠D = ∠AED i.e., The base angles of AADE are equal. ∴ Opposite sides must be equal. ⇒ AE = AD Question 7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle. Solution: Given: A circle in which two chords AC and BD are such that they bisect each other. Let their point’of intersection be O. To Prove: (i) AC and BD are diameters. (ii) ABCD is a rectangle. Construction: Join AB, BC, CD and DA. Proof: (i) In ∆AOB and ∆COD, we have AO = CO [O is the mid-point of AC] BO = DO [O is the mid-point of BD] ∠AOB = ∠COD [Vertically opposite angles] ∴ Using the SAS criterion of congruence, ∆AOB ≅ ∆COD ⇒ AB = CD [C.P.C.T.] ⇒ arc AB = arc CD …(1) Similarly, arc AD = arc BC …(2) Adding (1) and (2), we get arc AB + arc AD = arc CD + arc BC ⇒ BD divides the circle into two equal parts. ∴ BD is a diameter. Similarly, AC is a diameter. (ii) We know that ∆AOB ≅ ∆COD ⇒ ∠OAB = ∠OCD [C.P.C.T] ⇒ ∠CAB = ∠ACD AB \|\| DC Similarly, AD \|\| BC ∴ ABCD is a parallelogram. Since, opposite angles of a parallelogram are equal. ∴ ∠DAB = ∠DCB But ∠DAB + ∠DCB = 180° [Sum of the opposite angles of a cyclic quadrilateral is 180°] ⇒ ∠DAB = 90° = ∠DCB Thus, ABCD is a rectangle. Question 8. Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – A, 90° – B and 90° – C. Solution: Given : A triangle ABC inscribed in a drcle, such that bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, £ and F respectively. Construction: Join DE, EF and FD. Proof: ∵ Angles in the same segment are equal. ∴ ∠ED A = ∠FCA …(i) ∠EDA = ∠EBA …(ii) Adding (i) and (ii), we have ∠FDA + ∠EDA = ∠FCA + ∠EBA ⇒ ∠FDE = ∠FCA + ∠EBA Question 9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ. Solution: We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q. Let us draw the common chord AB. Since angles subtended by equal chords in the congruent circles are equal. ⇒ ∠APB = ∠AQB Now, in ∆PBQ, we have ∠AQB = ∠APB So, their opposite sides must be equal. ⇒ BP = BQ. Question 10. In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC. Solution: ∆ABC with O as centre of its circumcirde. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcirde at P. In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of ∠A of ∆ABC. Arc BC makes angle θ at the circumference ∴ ∠BOC = 2θ [Angle at centre is double the angle made by an arc at circumference] Also, in ∆BOC, OB=OC and OP is perpendicular bisector of BC. So, ∠BOP = ∠COP = θ Arc CP makes angle θ at O, so it will make angle [/latex]\frac { \theta }{ 2 }[/latex] at circumference. So, ∠COP = [/latex]\frac { \theta }{ 2 }[/latex] Hence, AP is the angle bisedor of ∠A of ∆ABC.

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