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# How to Convert Decimal to Octal: Simple Techniques and Examples Published: Converting decimal numbers to octal numbers can be a bit tricky for those who are new to the process. Decimal numbers are base 10 numbers, meaning they use 10 digits (0-9), while octal numbers are base 8 numbers, meaning they use 8 digits (0-7). In this article, we will discuss some simple techniques and examples to help you convert decimal numbers to octal numbers with ease. ## Technique 1: Division Method The division method is one of the simplest techniques for converting decimal numbers to octal numbers. The process involves dividing the decimal number by 8 and writing down the remainder. Then, the quotient is divided by 8 again, and the remainder is written down. This process is repeated until the quotient becomes 0. Let's take an example to better understand this technique: Example 1: Convert the decimal number 158 to octal. Solution: Step 1: Divide 158 by 8, and write down the remainder. The quotient is 19. 158 ÷ 8 = 19 remainder 6 Step 2: Divide 19 by 8, and write down the remainder. The quotient is 2. 19 ÷ 8 = 2 remainder 3 Step 3: Divide 2 by 8, and write down the remainder. The quotient is 0. 2 ÷ 8 = 0 remainder 2 Therefore, the octal equivalent of the decimal number 158 is 236. ## Technique 2: Binary Method The binary method involves converting the decimal number to binary and then converting the binary number to octal. The binary number is divided into groups of three digits, starting from the right, and each group is converted to its octal equivalent. Let's take an example to better understand this technique: Example 2: Convert the decimal number 247 to octal. Solution: Step 1: Convert the decimal number 247 to binary. 24710 = 111101112 Step 2: Divide the binary number into groups of three digits, starting from the right. 111 101 112 Step 3: Convert each group to its octal equivalent. 1112 = 78 1012 = 58 112 = 38 Therefore, the octal equivalent of the decimal number 247 is 375. ## FAQs #### What is a decimal number? A decimal number is a base 10 number, meaning it uses 10 digits (0-9) to represent numbers. #### What is an octal number? An octal number is a base 8 number, meaning it uses 8 digits (0-7) to represent numbers. #### What is the easiest way to convert decimal to octal? The easiest way to convert decimal to octal is by using the division method. Divide the decimal number by 8 and write down the remainder. Then, divide the quotient by 8 again and write down the remainder. Repeat this process until the quotient becomes 0. The octal equivalent is the remainders read from bottom to top. #### Can I convert decimal to octal using the binary method? Yes, you can convert decimal to octal using the binary method. Convert the decimal number to binary and then divide the binary number into groups of three digits, starting from the right. Each group is then converted to its octal equivalent. The final answer is the combination of the octal equivalents of each group. ## Conclusion Converting decimal numbers to octal numbers is a simple process once you understand the techniques involved. The division method is the easiest and quickest technique, while the binary method is useful for those who prefer working with binary numbers. By following the examples given in this article, you should be able to convert decimal numbers to octal numbers with ease.
# Fermat’s little Theorem Fermat’s little theorem: Let p be a prime. Then a^p-a is a multiple of p. In other words a^p is congruent to a mod p. Proof: We will use Induction. For a=1 we have a^p-a=1^p-1=1-1=0. 0 is a multiple of p. Now we assume, that the theorem works for a. Last but not least, we have to show that a+1 works. We have (a+1)^p-(a+1)=(a+1)^p-a-1. Now we use the binomial theorem: We know that a^p-a is a multiple of p. Every summand of the sum is a multiple of p as well. It follows, that the theorem works for a+1 as well. This proves the theorem. # Egyptian Multiplikation In this post I will introduce egyptian multiplikation to you. Lets say you want to multiply 15 with 22. You make 2 coulums: If the right number is even, you simply half it and double the first number. If the second number is odd, you subtract one from it and then half it. The other number gets doubled. You keep on doing this until the second number has reached 1. Now you take all the rows, where the second number is odd. If you add the first numbers of these rows you get the product: 1522=30+60+240=330. Proof: Let our product be xy. x0 is always zero. If y is even, we have x2y/2 which is equal to xy. If y is odd we have x+2x(y-1)/2=x+x(y-1)=xy. So basically what we did, is that we just took 1 x out of the product. When we continue this algorithm, we eventually get to y=0 since we make y smaller with an integer amount in each step. If y is zero the product is zero, which means that we took the whole number out. Since we only took out numbers when y was odd, we have to add all the x’s from the rows where the y is odd. That proves the method.
# What is 7/16 as a decimal? ## Want to practice? Fill in the missing parts in the long division below and complete the answer. $$\frac7?=0.4\_7\_$$ ### Solution • As we can see that the quotient starts with 0…., we can say that the divisor is greater than the dividend. Hence 0 is added at each step to make the value divisible by the divisor. • Start by finding the divisor. As we have the first digit of the quotient and the product 64, we can conclude that 16 x 4=64. Hence the divisor is 16. • The blank in the second step is the difference between 70 and 64. 70 – 64 = 6 • Find the blank in the quotient by finding the number of times 60 is divisible by 16. 16 x 3 = 48. Hence the first blank in the quotient is 3. • In the third step, we can see that 16 divides 120 seven times. Hence 16 x 7 = 112 • The last blank in the quotient is the number of times 16 exactly divides 80. 16 x 5 = 80 Find below the correct long division with the answers. The long division method includes the conversion of a fraction into a decimal by division. The numerator of the fraction is the dividend and the denominator of the fraction is the divisor. $$\frac{\mathbf7}{\mathbf{16}}\boldsymbol=\mathbf0\boldsymbol.\mathbf{4375}$$ ## Remember Here are some common terms you should be familiar with. • In the fraction $$\frac{7}{16}$$, the number 7 is the dividend (our numerator) • The number 16 is our divisor (our denominator) ## Find More Fractions to Decimals We invite you to read on our blog, about what is 5 /16 as a decimal fraction.
The Square Root of a Negative Number # THE SQUARE ROOT OF A NEGATIVE NUMBER by Dr. Carol JVF Burns (website creator) Follow along with the highlighted text while you listen! • PRACTICE (online exercises and printable worksheets) Square roots of negative numbers were introduced in a prior lesson: Arithmetic with Complex Numbers in the Algebra II materials. For your convenience, that material is repeated (and extended) here. Firstly, recall some information from beginning algebra: ## The Square Root of a Nonnegative Real Number For a nonnegative real number $\,k\,$, $$\cssId{s7}{\overbrace{\ \ \ \sqrt{k}\ \ \ }^{\text{the square root of \,k\,}}} \cssId{s8}{:=} \ \ \cssId{s9}{\text{the unique nonnegative real number which, when squared, equals \,k\,}}$$ (Note: when the distinction becomes important in higher-level mathematics, then $\,\sqrt{k}\,$ is called the principal square root of $\,k\,$.) Thus, the number that gets to be called the square root of $\,k\,$ satisfies two properties: • it is nonnegative (not negative; i.e., greater than or equal to zero) • when squared, it gives $\,k\,$ For example, $\,\sqrt{9} = 3\,$ since the number $\,3\,$ satisfies these two properties: • $\,3\,$ is nonnegative • $\,3\,$, when squared, gives $\,9\,$: $\,3^2 = 9\,$ So, what is (say) $\,\sqrt{-4}\$? There does not exist a nonnegative real number which, when squared, equals $\,-4\$. Why not? Every real number, when squared, is nonnegative: for all real numbers $\,x\,$, $\,x^2 \ge 0\,$. Complex numbers to the rescue! ## The Square Root of a Negative Number Complex numbers allow us to compute the square root of negative numbers, like $\,\sqrt{-4}\$. Remember the key fact: $\,i:=\sqrt{-1}\,$, so that $\,i^2=-1\,$ THE SQUARE ROOT OF A NEGATIVE NUMBER Let $\,p\,$ be a positive real number, so that $\,-p\,$ is a negative real number. Then: $$\cssId{s31}{\overbrace{\ \ \ \sqrt{-p}\ \ \ }^{\text{the square root of negative \,p\,}}} \ \cssId{s32}{:=}\ \cssId{s33}{i\,\sqrt{\vphantom{h}p}}$$ (Note: when the distinction becomes important in higher-level mathematics, then $\,\sqrt{-p}\,$ is called the principal square root of $\,-p\,$.) Observe that $\,i\sqrt{\vphantom{h}p}\,$, when squared, does indeed give $\,-p\$: $$\cssId{s36}{(i\sqrt{\vphantom{h}p})^2} \ \cssId{s37}{=\ i^2(\sqrt{\vphantom{h}p})^2} \ \cssId{s38}{=\ (-1)(p)} \ \cssId{s39}{=\ -p}$$ Some of my students like to think of it this way: You can slide a minus sign out of a square root, and in the process, it turns into the imaginary number $\,i\,$! Here are some examples: • $\sqrt{-4} = i\sqrt{4} = i2 = 2i\,$ It is conventional to write $\,2i\,$, not $\,i2\,$. When there's no possible misinterpretation (see below), write a real number multiplier before the $\,i\,$. • $\sqrt{-5} = i\sqrt{5}$ In this situation, it's actually better to leave the $\,\sqrt{5}\,$ after the $\,i\,$. Why? If you pull the $\,\sqrt{5}\,$ to the front, it can lead to a misinterpretation, as follows: The numbers $\,\sqrt{5}i\,$ and $\,\sqrt{5i}\,$ are different numbers—look carefully! In $\ \sqrt{5}i\$, the $\,i\,$ is outside the square root. In $\ \sqrt{5i}\$, the $\,i\,$ is inside the square root. Unless you look carefully, however, you might mistake one of these for the other. By writing $\,i\sqrt{5}\,$, you eliminate any possible confusion. TWO DIFFERENT QUESTIONS; TWO DIFFERENT ANSWERS Recall these two different questions, with two different answers: • What is $\,\sqrt{4}\,$? Answer: By definition, $\,\sqrt{4} = 2\,$. The number $\ \sqrt{4}\$ is the nonnegative number which, when squared, gives $\,4\,$. • What are the solutions to the equation $\,x^2 = 4\,$? Answer: $\,x= \pm\ 2\,$ There are two different real numbers which, when squared, give $\,4\,$. Similarly, there are two different questions involving complex numbers, with two different answers: • What is $\,\sqrt{-4}\,$? Answer: $\sqrt{-4} = 2i\,$ • What are the (complex) solutions to the equation $\,x^2 = -4\,$? Answer: $\,x=\pm \sqrt{-4} = \pm\ 2i\,$ There are two different complex numbers which, when squared, give $\,-4\,$. Verifying the second solution: $\, \cssId{s74}{(-2i)(-2i)} \cssId{s75}{= 4i^2} \cssId{s76}{= 4(-1)} \cssId{s77}{= -4}$ ## Square Roots of Products The following statement is true: for all nonnegative real numbers $\,a\,$ and $\,b\,$, $\,\sqrt{ab} = \sqrt{a}\sqrt{b}\$. For nonnegative numbers, the square root of a product is the product of the square roots. Does this property work for negative numbers, too? The answer is NO, as shown next. Certainly, anything called ‘the square root of $\,-4\,$’ must have the property that, when squared, it equals $\,-4\$. Unfortunately, the following incorrect reasoning gives the square as $\,4\,$, not $\,-4\,$: $$\cssId{s87}{\text{? ? ? ? }}\ \ \ \ \cssId{s88}{(\sqrt{-4})^2} \ \ \cssId{s89}{=\ \ \sqrt{-4}\sqrt{-4}} \cssId{s90}{\overbrace{\ \ =\ \ }^{\text{this is the mistake}}}\ \ \cssId{s91}{\sqrt{(-4)(-4)}}\ \ \cssId{s92}{=\ \ \sqrt{16}}\ \ \cssId{s93}{=\ \ 4} \ \ \ \ \cssId{s94}{\text{? ? ? ? }}$$ Here is the correct approach: $$\cssId{s96}{(\sqrt{-4})^2} \ \ \cssId{s97}{=\ \ \sqrt{-4}\sqrt{-4}}\ \ \cssId{s98}{=\ \ i\sqrt{4}\ i\sqrt{4}} \ \ \cssId{s99}{=\ \ i^2(\sqrt{4})^2} \ \ \cssId{s100}{=\ \ (-1)(4)} \ \ \cssId{s101}{=\ \ -4}$$ Similarly, $\,\sqrt{-5}\sqrt{-3}\,$ is NOT equal to $\,\sqrt{(-5)(-3)}\,$. Instead, here is the correct simplication: $$\cssId{s104}{\sqrt{-5}\sqrt{-3}} \ \ \cssId{s105}{=\ \ i\sqrt{5}\ i\sqrt{3}} \ \ \cssId{s106}{=\ \ i^2\sqrt{5}\sqrt{3}} \ \ \cssId{s107}{=\ \ (-1)\sqrt{(5)(3)}} \ \ \cssId{s108}{=\ \ -\sqrt{15}}$$ Be careful about this! Master the ideas from this section
# Lesson 3 Let’s solve more problems involving proportional relationships using tables. ### 3.1: Equal Measures Use the numbers and units from the list to find as many equivalent measurements as you can. For example, you might write “30 minutes is $$\frac12$$ hour.” You can use the numbers and units more than once. 1 12 0.4 60 50 $$\frac{1}{2}$$ 40 0.01 $$3\frac{1}{3}$$ 30 0.3 24 $$\frac{1}{10}$$ 6 2 $$\frac{2}{5}$$ centimeter meter hour feet minute inch ### 3.2: Centimeters and Millimeters There is a proportional relationship between any length measured in centimeters and the same length measured in millimeters. 1. If you know the length of something in centimeters, you can calculate its length in millimeters. 1. Complete the table. 2. What is the constant of proportionality? length (cm) length (mm) 9 12.5 50 88.49 2. If you know the length of something in millimeters, you can calculate its length in centimeters. 1. Complete the table. 2. What is the constant of proportionality? length (mm) length (cm) 70 245 4 699.1 3. How are these two constants of proportionality related to each other? 4. Complete each sentence: 1. To convert from centimeters to millimeters, you can multiply by ________. 2. To convert from millimeters to centimeters, you can divide by ________ or multiply by ________. 1. How many square millimeters are there in a square centimeter? 2. How do you convert square centimeters to square millimeters? How do you convert the other way? ### 3.3: Pittsburgh to Phoenix On its way from New York to San Diego, a plane flew over Pittsburgh, Saint Louis, Albuquerque, and Phoenix traveling at a constant speed. Complete the table as you answer the questions. Be prepared to explain your reasoning. segment time distance speed Pittsburgh to Saint Louis 1 hour 550 miles Saint Louis to Albuquerque 1 hour 42 minutes Albuquerque to Phoenix 330 miles 1. What is the distance between Saint Louis and Albuquerque? 2. How many minutes did it take to fly between Albuquerque and Phoenix? 3. What is the proportional relationship represented by this table? 4. Diego says the constant of proportionality is 550. Andre says the constant of proportionality is $$9 \frac16$$. Do you agree with either of them? Explain your reasoning. ### Summary When something is traveling at a constant speed, there is a proportional relationship between the time it takes and the distance traveled. The table shows the distance traveled and elapsed time for a bug crawling on a sidewalk. We can multiply any number in the first column by $$\frac23$$ to get the corresponding number in the second column. We can say that the elapsed time is proportional to the distance traveled, and the constant of proportionality is $$\frac23$$. This means that the bug’s pace is $$\frac23$$ seconds per centimeter. This table represents the same situation, except the columns are switched. We can multiply any number in the first column by $$\frac32$$ to get the corresponding number in the second column. We can say that the distance traveled is proportional to the elapsed time, and the constant of proportionality is $$\frac32$$. This means that the bug’s speed is $$\frac32$$ centimeters per second. Notice that $$\frac32$$ is the reciprocal of $$\frac23$$. When two quantities are in a proportional relationship, there are two constants of proportionality, and they are always reciprocals of each other. When we represent a proportional relationship with a table, we say the quantity in the second column is proportional to the quantity in the first column, and the corresponding constant of proportionality is the number we multiply values in the first column to get the values in the second. ### Glossary Entries • constant of proportionality In a proportional relationship, the values for one quantity are each multiplied by the same number to get the values for the other quantity. This number is called the constant of proportionality. In this example, the constant of proportionality is 3, because $$2 \boldcdot 3 = 6$$, $$3 \boldcdot 3 = 9$$, and $$5 \boldcdot 3 = 15$$. This means that there are 3 apples for every 1 orange in the fruit salad. number of oranges number of apples 2 6 3 9 5 15 • equivalent ratios Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, $$8:6$$ is equivalent to $$4:3$$, because $$8\boldcdot\frac12 = 4$$ and $$6\boldcdot\frac12 = 3$$. A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because $$8:6$$ and $$4:3$$ are equivalent ratios. cups of water number of lemons 8 6 4 3 • proportional relationship In a proportional relationship, the values for one quantity are each multiplied by the same number to get the values for the other quantity. For example, in this table every value of $$p$$ is equal to 4 times the value of $$s$$ on the same row. We can write this relationship as $$p = 4s$$. This equation shows that $$s$$ is proportional to $$p$$. $$s$$ $$p$$ 2 8 3 12 5 20 10 40
# Trigonometry : Ambiguous Triangles ## Example Questions ← Previous 1 ### Example Question #85 : Triangles Given and determine to the nearest degree the measure of . Explanation: We are given three sides and our desire is to find an angle, this means we must utilize the Law of Cosines. Since the angle desired is the equation must be rewritten as such: Substituting the given values: Rearranging: Solving the right hand side and taking the inverse cosine we obtain: ### Example Question #86 : Triangles If and , determine the measure of to the nearest degree. Explanation: This is a straightforward Law of Cosines problem since we are given three sides and desire one of the corresponding angles in the triangle. We write down the Law of Cosines to start: Substituting the given values: Isolating the angle: The final step is to take the inverse cosine of both sides: ### Example Question #87 : Triangles If , and determine the length of side , round to the nearest whole number. Explanation: This is a straightforward Law of Sines problem as we are given two angles and a corresponding side: Substituting the known values: Solving for the unknown side: ### Example Question #4 : Ambiguous Triangles If , and determine the measure of , round to the nearest degree. Explanation: This is a straightforward Law of Sines problem since we are given one angle and two sides and are asked to determine the corresponding angle. Substituting the given values: Now rearranging the equation: The final step is to take the inverse sine of both sides: ### Example Question #1 : Ambiguous Triangles If , , and find to the nearest degree. Explanation: The problem gives the lengths of three sides and asks to find an angle. We can use the Law of Cosines to solve for the angle. Because we are solving for , we use the equation: Substituting the values from the problem gives Isolating by itself gives ### Example Question #51 : Law Of Cosines And Law Of Sines If , find to the nearest degree. Explanation: We are given the lengths of the three sides to a triangle. Therefore, we can use the Law of Cosines to find the angle being asked for. Since we are looking for we use the equation, Inputting the values we are given, Next we isolate by itself to solve for it ### Example Question #52 : Law Of Cosines And Law Of Sines If , and , find to the nearest degree. Explanation: Because the problem provides all three sides of the triangle, we can use the Law of Cosines to solve this problem. Since we are solving for , we use the equation Substitute in the given values Isolate ### Example Question #53 : Law Of Cosines And Law Of Sines If = , and find to the nearest degree. Explanation: Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. Because we are looking for we use Inputting the lengths of the triangle into this equation Isolating ### Example Question #54 : Law Of Cosines And Law Of Sines If , and = find to the nearest degree. Explanation: Because we are given the length of two sides of a triangle and the corresponding angle of one of the sides, we can find the angle being asked for with the Law of Sines. Inputting the values of the problem Rearranging the equation to isolate ### Example Question #55 : Law Of Cosines And Law Of Sines If = , , find to the nearest tenth of a degree.
# 8.02 Experimental probability Lesson In order to predict the future, we sometimes need to determine the probability by running experiments, or looking at data that has already been collected. This is called experimental probability, since we determine the probability of each outcome by looking at past events. #### Exploration Imagine we have a "loaded" die, where a weight is placed inside the die opposite the face that the cheater wants to come up the most (in this case, the $6$6): If the die is made like this, the probability of each outcome is no longer equal, and we cannot say that the probability of rolling any particular face is $\frac{1}{6}$16. Instead we will need to roll the die many times and record our results, and use these results to predict the future. Here are the results of an experiment where the die was rolled $200$200 times: Result Number of rolls $1$1 $11$11 $2$2 $19$19 $3$3 $18$18 $4$4 $18$18 $5$5 $20$20 $6$6 $114$114 We can now try to predict the future using this experimental data, and the following formula: $\text{Experimental probability of event}=\frac{\text{Number of times event occurred in experiments}}{\text{Total number of experiments}}$Experimental probability of event=Number of times event occurred in experimentsTotal number of experiments Here is the table again, with the experimental probability of each face listed as a percentage: Result Number of rolls Experimental probability $1$1 $11$11 $5.5%$5.5% $2$2 $19$19 $9.5%$9.5% $3$3 $18$18 $9%$9% $4$4 $18$18 $9%$9% $5$5 $20$20 $10%$10% $6$6 $114$114 $57%$57% A normal die has around $17%$17% chance of rolling a $6$6, but this die rolls a $6$6 more than half the time! Sometimes our "experiments" involve looking at historical data instead. For example, we can't run hundreds of Eurovision Song Contests to test out who would win, so instead we look at past performance when trying to predict the future. The following table shows the winner of the Eurovision Song Contest from 1999 to 2018: Year Winning country Year Winning country 1999 Sweden 2009 Norway 2000 Denmark 2010 Germany 2001 Estonia 2011 Azerbaijan 2002 Latvia 2012 Sweden 2003 Turkey 2013 Denmark 2004 Ukraine 2014 Austria 2005 Greece 2015 Sweden 2006 Finland 2016 Ukraine 2007 Serbia 2017 Portugal 2008 Russia 2018 Israel What is the experimental probability that Sweden will win the next Eurovision Song Contest? We think of each contest as an "experiment", and there are $20$20 in total. The winning country is the event, and we can tell that $3$3 of the contests were won by Sweden. So using the same formula as above, $\text{Experimental probability of event}=\frac{\text{Number of times event occurred in experiments}}{\text{Total number of experiments}}$Experimental probability of event=Number of times event occurred in experimentsTotal number of experiments the experimental probability is $\frac{3}{20}$320, which is $15%$15%. How many of the next $50$50 contests can Sweden expect to win? Just like in the last chapter, we can calculate this by multiplying the experimental probability of an event by the number of trials. In this case Sweden can expect to win $\frac{3}{20}\times50=\frac{150}{20}$320×50=15020 contests This rounds to $8$8 contests out of the next $50$50. Summary $\text{Experimental probability of event}=\frac{\text{Number of times event occurred in experiments}}{\text{Total number of experiments}}$Experimental probability of event=Number of times event occurred in experimentsTotal number of experiments You may also see the term relative frequency which is the same as the experimental probability. #### Practice questions ##### Question 1 A retail store served $773$773 customers in October, and there were $44$44 complaints during that month. Determine, as a percentage, the experimental probability that a customer submits a complaint. 1. Round your answer to the nearest whole percent. ##### Question 2 An insurance company found that in the past year, of the $2558$2558 claims made, $1493$1493 of them were from drivers under the age of 25. Give your answers to the following questions as percentages, rounded to the nearest whole percent. 1. What is the experimental probability that a claim is filed by someone under the age of 25? 2. What is the experimental probability that a claim is filed by someone 25 or older? ##### Question 3 The experimental probability that a commuter uses public transport is $50%$50%. Out of $500$500 commuters, how many would you expect to use public transport? ### Outcomes #### 7.SP.6 Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Staging content lifeboatΒ >Β Unit 14 Lesson 50: Multiplying and dividing decimals by 10, 100, and 1000 # Multiplying decimals by 10, 100, and 1000 Practice multiplying decimals by 10, 100, and 1000. ## Part 1: Multiplying decimals by 10 Key idea: Multiplying by 10 moves every digit one place to the left. Let's visualize it! Drag the dot all the way to the right. $5.14\Gamma 10=$ ## Let's try a few more. $4.006\Gamma 10=$ $583.2\Gamma 10=$ $0.7761\Gamma 10=$ ## Part 2: Multiplying decimals by 100 Key idea: Multiplying by 100 moves every digit two places to the left. Let's visualize it! Drag the dot all the way to the right. Notice that we add a zero to fill the empty place value. $23.8\Gamma 100=$ ## Let's try a few more. $90.5\Gamma 100=$ $6.33\Gamma 100=$ $0.0047\Gamma 100=$ ## Part 3: Multiplying decimals by 1,000 Key idea: Multiplying by 1,000 moves every digit three places to the left. Let's visualize it! Drag the dot all the way to the right. Notice that we add zeros to fill the empty place values. $0.6\Gamma 1000=$ ## Let's try a few more. $3.4\Gamma 1000=$ $62.11\Gamma 1000=$ $0.0577\Gamma 1000=$ ## Part 4: Let's look at the pattern. Multiplying by $10$ moves every digit $1$ place to the left. Multiplying by $100$ moves every digit $2$ places to the left. Multiplying by $1,000$ moves every digit $3$ places to the left. Multiplying by $10,000$ moves every digit places to the left. Multiplying by $10,000,000$ moves every digit places to the left. ## Part 5: Challenge time! Use the pattern above to answer the following questions. $71.55\Gamma 10,000=$ $0.8\Gamma 1,000,000=$ ## Want to join the conversation? • this is so easy • it might be hard for other people • why is it so hard for me • Another way to think of it is, shift the decimal point to the right. 23 = 23.0 so 23x10 = 230 • the 100's were kinda hard but the 10's were good • think about like this: if you want to multiply shift the decimal point to the right based on the numbers of zeros. for example if you want to multiply 4.25 X 10 you will shift the decimal point one time to the right (because there is one zero in the ten). if you want to multiply 4.35 X 100 you will shift the decimal point two times to the right because there is two zeros in the 100. • it's okay I mean hard but it's okay • I got 1 incorrect but I got the other correct • Can you just divide by a fraction? Ex: instead of 1.2 x 100, 1.2 / 1/100. If yes, why aren't they showing that? If no, why not? • You can, but the reason they are not showing that is because this method is slightly easier to do mentally and visually.
Test: Number System- 3 - CLAT MCQ # Test: Number System- 3 - CLAT MCQ Test Description ## 20 Questions MCQ Test Quantitative Techniques for CLAT - Test: Number System- 3 Test: Number System- 3 for CLAT 2024 is part of Quantitative Techniques for CLAT preparation. The Test: Number System- 3 questions and answers have been prepared according to the CLAT exam syllabus.The Test: Number System- 3 MCQs are made for CLAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Number System- 3 below. Solutions of Test: Number System- 3 questions in English are available as part of our Quantitative Techniques for CLAT for CLAT & Test: Number System- 3 solutions in Hindi for Quantitative Techniques for CLAT course. Download more important topics, notes, lectures and mock test series for CLAT Exam by signing up for free. Attempt Test: Number System- 3 \| 20 questions in 15 minutes \| Mock test for CLAT preparation \| Free important questions MCQ to study Quantitative Techniques for CLAT for CLAT Exam \| Download free PDF with solutions Test: Number System- 3 - Question 1 ### In the examination a candidate must get 2/5 marks to pass, out of total marks.Shiyam appeared in the exam ang got 198 marks and still failed by 36 marks.The maximum mark is Detailed Solution for Test: Number System- 3 - Question 1 Let the maximum mark be M. Since Shiyam got 198 marks and still failed by 36 marks, this means that Shiyam needed 2/5 of the maximum mark to pass. Therefore, 2/5 of M is equal to 198 + 36 = 234. To find M, we can multiply both sides of the equation by 5/2: M = 234 * 5/2 = 585. So, the maximum mark is 585. Test: Number System- 3 - Question 2 ### If 4 is added to the numerator of a fraction it becomes 1/3 and if 3 is added to the denominator it becomes 1/6 then find the numerator and denominator is Detailed Solution for Test: Number System- 3 - Question 2 Let the numerator be x and denominator be y Fraction will be x/y According to the question (x + 4)/y = 1/3 ⇒ 3x + 12 = y ......(1) Now, x/(y + 3) = 1/6 ⇒ 6x = y + 3 ......(2) Solving both the equations we get x = 5 and y = 27 So , fraction x/y = 5/27 1 Crore+ students have signed up on EduRev. Have you? Test: Number System- 3 - Question 3 ### One-third of a 2 digit number exceeds its one-fifth by 8.What is the 2 digit number ? Detailed Solution for Test: Number System- 3 - Question 3 1/3(10x+y) -1/5(10x+y) = 8 (10x+y)[5/15 – 3/15] = 8 10x+y = 815/2 = 60 Test: Number System- 3 - Question 4 12 +22 +……………………………………+242 Detailed Solution for Test: Number System- 3 - Question 4 Sum of square number = n(n+1)(2n+1)/6 N = 24 Sum = 242549/6 = 4900 Test: Number System- 3 - Question 5 Micael gets 3 marks for each correct questions and loses 2 marks for each wrong answers.He attempts 30 sum and obtain 30 marks.Find the no of Questions he answered correctly ? Detailed Solution for Test: Number System- 3 - Question 5 let x be the no of correct answer x+y=30 3x-2y=30 2x+2y=60 5x=90 x=18 Test: Number System- 3 - Question 6 What half of a number is added to 234, it becomes five time of itself. What is the number ? Detailed Solution for Test: Number System- 3 - Question 6 x/2 + 234 = 5x 5x – x/2 = 234 9x/2 = 234 X = 2342/9= 52 Test: Number System- 3 - Question 7 Twenty times a positive integer is less than its square by 96. What is the Integer ? Detailed Solution for Test: Number System- 3 - Question 7 Let the positive integer be 'x' Then, 20 times integer = 20x Square of integer = x2 Twenty times a positive integer is less than its square by 96 x2 = 20x + 96 ⇒ x2 - 20x - 96 = 0 ⇒ x- 24x + 4x - 96 = 0 ⇒ x ( x -24) + 4( x - 24) = 0 ⇒ (x + 4) × (x -24) = 0 ⇒ x = -4, x = 24 ∴ The positive integer is 24 Test: Number System- 3 - Question 8 There are some Parrots and Some Lions in a forest.If the total number of animals head in forest are 840 and total no of animal legs are 1760.What is the number of Parots in the forest ? Detailed Solution for Test: Number System- 3 - Question 8 2x+4(840-x) = 1760 2x+3360-4x = 1760 -2x = 1600 X = 1600/2 = 800 Test: Number System- 3 - Question 9 When all the students in a school are made to stand in row of 68, 40 such rows are formed.If the students are made to stand in the row of 20, how many such rows can be formed ? Detailed Solution for Test: Number System- 3 - Question 9 No of row = 6840/20 = 136 Test: Number System- 3 - Question 10 A banana costs Rs.1.25 and Orange costs Rs2.75. What will be the total cost of 10 bananas and 2 dozens Orange ? Detailed Solution for Test: Number System- 3 - Question 10 1.2510 +2.7524 = 12.5 +66 = 78.50 Test: Number System- 3 - Question 11 What will be the Unit digit of 27! Detailed Solution for Test: Number System- 3 - Question 11 123………………………………2627 Factorial after 6 should end with 0 Test: Number System- 3 - Question 12 xp × xq × xr = 1 then p3 + q3 + r3 is equal to Detailed Solution for Test: Number System- 3 - Question 12 so p + q + r = 0 Test: Number System- 3 - Question 13 If the divisor is five times the quotient and six times the remainder, if the remainder is 5 then the dividend is Detailed Solution for Test: Number System- 3 - Question 13 Q=6, R = 5 Dividend = divisorQ + R Dividend = 306 + 5 = 180+5 = 185 Test: Number System- 3 - Question 14 If the sum and the product of 2 numbers are 25 and 144 respectively then the difference of the number should be Detailed Solution for Test: Number System- 3 - Question 14 A+B = 25 AB = 24 (a-b)2 = (a+b)2 – 4ab (a-b)2 = 25 – 4(144) (a-b)2 = 625 – 576 (a-b)2 = 49 a-b = 7 Test: Number System- 3 - Question 15 The sum of the two digit number is 11. The number obtained by reversing the digit is 27 less than the original number.Then the original number is Detailed Solution for Test: Number System- 3 - Question 15 4+7 = 11 74-47 = 27 Test: Number System- 3 - Question 16 The product of 2 consecutive even number is 1224, then one of the number is Detailed Solution for Test: Number System- 3 - Question 16 3436 = 1224 Test: Number System- 3 - Question 17 212 +222 +232 +…………………302 = ? Detailed Solution for Test: Number System- 3 - Question 17 12 +22 +32 +…………………n2 = 1/6[n(n+1)(2n+1)] = 303161/6 – 202141/6 = 9455 – 2870 = 6585 Test: Number System- 3 - Question 18 A number when divided by 361 gives a remainder 47. If the Same number is divided by 19 ,the remainder is Detailed Solution for Test: Number System- 3 - Question 18 47÷19 ⇒ remainder 9 Ex : 769 ÷ 361 ⇒ remainder 47 769 ÷ 19 ⇒ remainder 9 Test: Number System- 3 - Question 19 The respective ratio between the two digit of a number is 5:3, if the digit at the one’s place is twice the digit at ten’s place.What is the number ? Detailed Solution for Test: Number System- 3 - Question 19 2x/x = 5/3 6x = 5x Test: Number System- 3 - Question 20 N is the smallest number which when added to the 2000 makes the resulting number divisible by 12, 16, 18 and 21, Then the N is Detailed Solution for Test: Number System- 3 - Question 20 LCM of 12, 16, 18, 21 = 1008 1008*2 = 2016 N = 16 ## Quantitative Techniques for CLAT 56 videos\|104 docs\|95 tests Information about Test: Number System- 3 Page In this test you can find the Exam questions for Test: Number System- 3 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Number System- 3, EduRev gives you an ample number of Online tests for practice ## Quantitative Techniques for CLAT 56 videos\|104 docs\|95 tests
# How do you solve division problems in 4th grade It is important to understand the basics of division before attempting to solve division problems in 4th grade. Division is the process of dividing one number by another to find the answer. In 4th grade, children are expected to know the basics of division and be able to solve basic division problems. The first step in solving a division problem is to break down the problem into smaller parts. This will help make the problem easier to understand and makes for easier math. For example, if given a problem such as 6 divided by 2, it can be broken down into two simpler problems: 6 divided by 1 = 6, and 6 divided by 2 = 3. This makes the problem easier to understand and calculate. Once the problem is broken down into smaller parts, it is time to calculate the answer. In 4th grade, children should know their times tables up to 10, so they will be able to use those facts to help solve their division problems. The most common method of solving a division problem is to divide the dividend (the number being divided) by the divisor (the number it is being divided by). By using this method, 6 divided by 2 becomes 6 divided by 2 = 3. In addition to using times tables and long division, there are other methods for solving division problems in 4th grade. Children can also use repeated subtraction or multiplication facts. For example, if given 8 divided by 4, students could subtract 4 from 8 four times (4 + 4 + 4 + 4 = 16) or multiply 4 x 2 = 8. Using these methods can help students quickly solve their division problems without having to use long division or memorize all their times tables. Finally, it is important that students practice solving division problems. By practicing on a regular basis, students will become more confident with their math skills and be able to easily solve more complex division problems in 4th grade. ## How do you do simple division for 4th grade When it comes to doing simple division in the fourth grade, it can be a bit tricky at first. But with some practice and the right strategies, it’s something that your child can learn quickly and easily. The basic concept of division is to break a larger number (the “dividend”) into groups of a smaller number (the “divisor”). For example, if you wanted to divide 10 by 2, you would break 10 into two groups of five. We call this the quotient. To solve a division problem, you must use a few key steps. The first step is to determine the dividend and the divisor. Once you know these numbers, you can move on to the next step. Next, you need to figure out how many times the divisor goes into the dividend. To do this, you will need to use multiplication or repeated addition. For example, if you were dividing 10 by 2, you would multiply 2 by 5 or add 2 five times to get 10. This tells us that 2 goes into 10 five times. Now that we know how many times the divisor goes into the dividend, we can start solving for the answer. The answer is called the quotient and it is written as a fraction or decimal. In our example above, we determined that 2 goes into 10 five times, so our answer is 5/2 or 2.5 (five halves or two and a half). Once your child has mastered this process for simple division problems, they can move on to more complex ones that involve three-digit numbers and larger divisors. But no matter what type of problem they are working on, it’s important that they understand each step of the process so that they can accurately solve more difficult problems in the future. ## How do I help my child with struggling with division If your child is having difficulty with division, the first thing you should do is determine the source of the problem. Is it a lack of understanding of the concept? Is it an inability to remember the multiplication tables? Or is it something else altogether? Once you have pinpointed what is causing the difficulty, you can take steps to help your child overcome it. One way to help your child with division is to provide them with hands-on practice. You can find a variety of activities online that will help your child work through different types of problems. For example, you can have them use candy or other small objects to divide into groups to practice simple division problems. You can also break down more complicated equations into simpler parts and have them work through each step one at a time. Another helpful strategy is to use visual aids to illustrate the process of division. You can draw pictures or diagrams on paper or use apps or software programs that allow for interactive demonstrations. Seeing how the equation works can often make it easier for children to understand. You can also try breaking down the process of division into smaller parts and focusing on each one separately. For example, have your child practice memorizing basic multiplication facts before attempting to tackle more complex equations that involve multiple steps. Finally, be sure to provide plenty of encouragement and praise when your child makes progress with their division skills. Let them know that you believe in them and are proud of their efforts even if they don’t get the answer right every time. Celebrate successes, no matter how small, and make sure they know that mistakes are part of learning and everyone has room for improvement. With patience and support, you can help your child gain confidence in their division skills and succeed in math. ## What is an example of a division problem A division problem is a type of mathematical equation in which one number is divided by another. Division problems usually come in the form of a fraction or ratio, with the numerator (the top number) representing the dividend (the number being divided) and the denominator (the bottom number) representing the divisor (the number used to divide the dividend). An example of a division problem would be: 7 ÷ 3 = 2 In this example, 7 is the dividend and 3 is the divisor. The answer to this division problem is 2. Another example of a division problem is: 42 ÷ 6 = 7 In this example, 42 is the dividend and 6 is the divisor. The answer to this division problem is 7. It is important to remember that when solving division problems, it is necessary to divide each number by the same divisor. For example, if you are asked to divide 8 by 4, you cannot divide 8 by 2 and expect it to work out correctly. The answer would be incorrect because you are not dividing each number by the same divisor. ## What is a easy division problem Division is one of the four basic operations in mathematics. It is the process of dividing a number into two or more parts. Division can be used to solve problems involving fractions, decimals, and other types of numbers. A “easy” division problem means that it is not too difficult to solve and is not too long. It may involve numbers that are not too large and the answer can be found quickly. A simple division problem could look something like this: 30 ÷ 6 = To solve this problem, you would divide 30 by 6. The answer is 5, so the answer to this easy division problem is 5. Other examples of easy division problems include: 24 ÷ 4 = 45 ÷ 5 = 100 ÷ 10 = As you can see, all of these problems involve dividing a number by another number and then finding the answer. In each case, the answer was found quickly and easily. Division does not have to be difficult! With a bit of practice and some basic knowledge of how division works, you can be sure to solve even the most challenging division problems with ease. ## What are good division word problems Division word problems can be a great way for students to practice their division skills in an engaging and meaningful way. A good division word problem should have a clear question that students can easily understand and a set of data that the student can use to solve the problem. Here are some examples of good division word problems: 1. At a restaurant, there were 18 customers who ordered a total of 72 slices of pizza. If each customer ordered an equal number of slices, how many slices did each customer order? 2. A farmer has 180 apples that need to be divided equally among 6 baskets. How many apples will go into each basket? 3. Dave had 45 marbles in his collection. He wanted to divide them equally between his three friends. How many marbles did each friend get? 4. A basketball team has 12 players, and they need to divide up 24 basketballs for practice. How many basketballs will each player get? 5. There are 30 books in a library, and they need to be divided equally between 5 shelves. How many books will go on each shelf? These are just a few examples of good division word problems that can help students practice their division skills. By providing students with clear and engaging questions, they will be able to apply their knowledge of division and hone their problem-solving abilities. ## What is the example of division Division is one of the four basic operations in mathematics, along with addition, subtraction, and multiplication. It is the process of breaking a number up into smaller parts or groups. For example, if you have 10 cookies and want to divide them among five people, each person will get two cookies. This is an example of division because you are breaking the total number of cookies (10) into smaller groups (2). Another example of division can be seen when you divide a number by another number. For instance, if you divide 10 by 5, the answer is 2. This is because 5 goes into 10 two times. The answer of 2 is the result of dividing 10 by 5. The last example on division involves fractions. When two fractions are divided, you find the quotient. The quotient is found by dividing the numerator (the top number in a fraction) by the denominator (the bottom number). For example, if you divide 1/2 by 3/4, the answer is 4/6 because 1 divided by 3 equals 1/3 and 2 divided by 4 equals 1/2. Therefore, 1/3 divided by 1/2 equals 4/6. Division can be used in many real-world situations such as dividing money among family members or splitting a pizza among friends. No matter what kind of problem you are solving, division can help you break it down into smaller parts for easier solution. ## What is a division sentence example A division sentence is an equation that shows how to divide a number into two or more parts. For example, the equation “72 ÷ 8 = 9” is a division sentence. Here, 72 is the dividend (the number being divided), 8 is the divisor (the number it’s being divided by), and 9 is the quotient (the result of the division). In this case, the equation tells us that if you divide 72 by 8, you get 9. Another example of a division sentence is “60 ÷ 5 = 12.” Here, 60 is the dividend, 5 is the divisor, and 12 is the quotient. This equation tells us that when we divide 60 by 5, we get 12. Division sentences can also include remainders. For example, “45 ÷ 6 = 7 R3” is a division sentence with a remainder. Here, 45 is the dividend, 6 is the divisor, 7 is the quotient, and 3 is the remainder. This equation tells us that if we divide 45 by 6, we get 7 with a remainder of 3. These are just a few examples of division sentences. With these examples in mind, you can easily create your own equations to practice your division skills!
# Unit 6: Exponents and Exponential Functions ## 6.3: Graphs of Exponential Functions Graphs of Exponential Functions $f\left(x\right)=b^x$ Consider the following graphs: $f\left(x\right)=2^x$ $f\left(x\right)=\left(\frac{1}{3}\right)^x$ • This function has base $2$, and it is increasing. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,1\right)$. • It has another identifiable point at $\left(1,2\right)$. • Its domain is all real numbers. • Its range is $y>0$. • This function has base $\frac{1}{3}$, and it is decreasing. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,1\right)$. • It has another identifiable point at $\left(1,\frac{1}{3}\right)$. • Its domain is all real numbers. • Its range is $y>0$. We could look at more examples, and end up with the following rules, for the graph of an exponential function $f\left(x\right)=b^x$: • If $0 \lt b \lt 1$, the function is decreasing. • If $b>1$, the function is increasing. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,1\right)$. • It has another identifiable point at $\left(1,b\right)$. • Its domain is all real numbers. • Its range is $y>0$. Graphs of Exponential Functions $f\left(x\right)=a \cdot b^x$ Consider the following graphs: $f\left(x\right)=3 \cdot 2^x$ $f\left(x\right)=2 \cdot \left(\frac{1}{3}\right)^x$ • This function has base $2$, and it is increasing. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,3\right)$. • It has another identifiable point at $\left(1,6\right)$. • Its domain is all real numbers. • Its range is $y>0$. • This function has base $\frac{1}{3}$, and it is decreasing. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,2\right)$. • It has another identifiable point at $\left(1,\frac{2}{3}\right)$. • Its domain is all real numbers. • Its range is $y>0$. So far so good, but what if $a<0$? $f\left(x\right)=-2 \cdot 3^x$ • This function has base $3$, but it is decreasing because $a$ is negative. • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,-2\right)$. • It has another identifiable point at $\left(1,-6\right)$. • Its domain is all real numbers. • Its range is $y<0$ (again, because $a$ is negative). Graphs of Exponential Functions $f\left(x\right)=a \cdot b^x$ $a$ positive $a$ negative decreasing increasing increasing decreasing • It has a horizontal asymptote at the $x$-axis. • It has a $y$-intercept at $\left(0,a\right)$. • It has another identifiable point at $\left(1,ab\right)$. • Its domain is all real numbers. • Its range is $y>0$ if $a$ is positive, and $y \lt 0$ if $a$ is negative.
# How do you evaluate the limit (abs(x+2)-2)/absx as x approaches 0? Feb 27, 2017 The limit: ${\lim}_{x \to 0} \frac{\left\mid x + 2 \right\mid - 2}{\left\mid x \right\mid}$ does not exist since the right and left limit are different. #### Explanation: Around $x = 0$ we have that: $x + 2 > 0$, so $\left\mid x + 2 \right\mid = x + 2$, and then: ${\lim}_{x \to 0} \frac{\left\mid x + 2 \right\mid - 2}{\left\mid x \right\mid} = {\lim}_{x \to 0} \frac{x + 2 - 2}{\left\mid x \right\mid} = {\lim}_{x \to 0} \frac{x}{\left\mid x \right\mid}$ Evaluate separately: ${\lim}_{x \to {0}^{+}} \frac{x}{\left\mid x \right\mid} = {\lim}_{x \to {0}^{+}} \frac{x}{x} = 1$ As when $x \to {0}^{+}$ we have $x > 0$ so $\left\mid x \right\mid = x$, while: ${\lim}_{x \to {0}^{-}} \frac{x}{\left\mid x \right\mid} = - 1$ We can conclude that: ${\lim}_{x \to 0} \frac{\left\mid x + 2 \right\mid - 2}{\left\mid x \right\mid}$ does not exist since the right and left limit are different.
# Introduction to statistics: Part 2 ## Nonparametric and qualitative data In a lot of cases, we don’t only have numbers, but categories as well. This can happen when we place measurable values into buckets, like [low, medium, high] or when we have distinct possibilities which are not necessarily ordered, like [red, green, blue, yellow]. These are qualitative parameters, while in part 1 we only worked with quantitative parameters. ### Spearman’s rank correlation coefficient When the data we use is nonparametric, but still has a certain order, we can use Spearman’s rank correlation coefficient instead of the Pearson correlation coefficient. #### Calculating the rank The Spearman’s rank correlation coefficient is obtained by calculating the rank of every value, and then using these values to calculate the Pearson correlation coefficient. There are several methods to calculate the rank of a list of values. The simplest method is to sort the values in ascending or descending order and to assign the values numbers from 1 until n, the amount of values. However if a value appears more than once, all these duplicates are assigned the rank of the first duplicate, while the numbering keeps going up silently, so when a new unique value is encountered, the numbering jumps. Value Rank 2 1 3 2 4 3 4 3 4 3 5 6 Every value 4 receives rank 3. Rank 4 and 5 are skipped here, while value 5 receives the next rank, rank 6. Alternatively the duplicates can be assigned the average of their lowest and highest rank. Value Rank 2 1 3 2 4 4 4 4 4 4 5 6 The values 4 would have received rank 3, 4 and 5. The average of 3 and 5 is 4, so all values receive rank 4. #### Calculating the correlation coefficient Now that we can calculate the rank of each element in a list of values, we can calculate the Spearman’s rank correlation coefficient as follows $\frac{cov(x',y')}{std(x')std(y')}$ Where $$x'$$ and $$y'$$ are the list of values obtained by replacing each value in $$x$$ and $$y'$$ by its rank. #### Significance of the correlation coefficient Just having the coefficient isn’t enough to make a conclusion. To know how significant the result is, which depends on how much samples we have, we need to calculate the probability. To do this, we first calculate the tscore as $t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$ Then we can calculate the probability using TDIST(t, n-2, 2) in Excel or pt(-abs(x), n-2) 2 in R. The lower the probability, the more unusual our result is. Usually a cutoff value of 0.05 is used. If our probability is lower than 0.05, we say that there is a correlation between the parameters, as there is a very small chance that our samples come from a distribution where the parameters are unrelated. ### Cramer’s V This method can be used when both our data series are using categories rather than numbers. To calculate Cramer’s V, we first calculate the frequency of each pair. If series x has categories from 1 to k, and series y has categories from 1 to r, then $$n_{ij}$$ holds the count of all data pairs where x is i and y is j. $$n_i$$ is the amount of data pairs where x is i, and $$n_j$$ is the amount of data pairs where y is j. $$n$$ itself is the amount of samples. Then we calculate $$\chi^{2}$$ or Chi squared from these frequencies $\chi^2=\sum_{i,j}{\frac{(n_{ij}-\frac{n_{i}n_{j}}{n})^2}{\frac{n_{i}n_{j}}{n}}}$ And we can calculate V from $$\chi^{2}$$ as follows $V=\sqrt{\frac{\chi^2/n}{\min(k-1,r-1)}}$ Where 0 means that the parameters are independent, and 1 that the parameters are dependent. So the value of V tells us how much the two parameters are related. What exactly are we calculating though? Let’s take a step back, and use an example. Let’s say we have a list of records with people’s gender and hair color. We can create a table listing the frequency of each combination Black Brown Red Blond Male 56 143 34 46 Female 52 143 37 81 So 56 of the people in our set have black hair and are male. Our table now contains the $$n_{ij}$$ values, from which we can calculate $$n_i$$ (108, 286, 71, 127), $$n_j$$ (279, 313) and $$n$$ (592) Black Brown Red Blond Sum Male 56 143 34 46 279 Female 52 143 37 81 313 Sum 108 286 71 127 592 Now we create a table which contains $$\frac{n_{i}n_{j}}{n}$$ for each i and j Black Brown Red Blond Male 50.89864865 134.7871622 33.46114865 59.85304054 Female 57.10135135 151.2128378 37.53885135 67.14695946 This table contains what we call expected values. It is the table you would get if the parameters gender and hair color were completely unrelated. For example males with black hair. we know there are 108 people with black hair, and 279 people of 592 are male. This means that in a perfect case of unrelated parameters 108279/592 or 50.89864865 would be males with black hair. If we replace $$\frac{n_{i}n_{j}}{n}$$ with $$E_{ij}$$ or the expected value at ij, we can rewrite the formula for $$\chi^2$$ as $\chi^2=\sum_{i,j}{\frac{(n_{ij}-E_{ij})^2}{E_{ij}}}$ So we are looking at the scaled squared distance between the observed and expected values, and summing these scaled distances. So the $$\chi^2$$ value tells us how much the two sample distributions are apart. In our case $$\chi^2$$ is 7.994244189. This is actually what CHITEST(observed, expected) uses in Excel. It calculates $$\chi^2$$ like we did and calculates the degrees of freedom df $df=max(1, r-1)max(1, c-1)$ Then it calculates the probability. We can do this using CHIDIST($$\chi^2$$, df) in Excel or pchisq($$\chi^2$$, df, lower.tail=FALSE) in R. In our case p is 0.04613081084. If the probability is beneath a threshold, 0.05 for example, it means that the probability that the samples are from the same population is lower than 5%, so our observed values are probably not from a population where the categories are unrelated, thus there might be a relation between the two parameters. Since 0.04613081084 is slightly less than 0.05, we might say that there is more than 0 correlation between gender and hair color. How much though, is hard to tell. Cramer’s V however, gives an easier to interpret value between 0 and 1, without the need to choose a threshold. In our case V is 0.1162058125, which means there is a slight correlation between gender and hair color. In R, cramersV(x) gives V from the table x. ### The Uncertainty Coefficient Cramer’s V gives us the same value irrespective of the order in we pass our parameters. So we can’t see the direction of a possible relation. However it might be that it is easier to guess whether a person is female when we know the hair color is blond than knowing the hair color when we know the person is female, V won’t provide this information. The uncertainty coefficient will tell us how certain it is that we can guess x given y. To calculate this, we need a few other concepts first. The uncertainty coefficient is based on information theory and the concept of entropy. The entropy of a list of data can be calculated as $H(X)=-\sum_{i=1}^{n}{P(x_{i})\log_{e}P(x_{i})}$ With $$P(x_{i})$$ the probability of $$x_{i}$$ occurring. The higher the entropy, the more uncertainty or surprise there is. This happens when some probabilities are small. This is because the smaller the probability, the larger the logarithm of the probability is. Since we use $$log_e$$, our probability of 0 to 1 maps onto -∞ to 0. So the - sign makes sure the entropy is positive. When we have events involving the probability that one parameter has a certain value and another parameter has a certain value, we can calculate the conditional entropy. The entropy of the data pairs with parameters X and Y can be written as $H(X,Y)=-\sum_{i=1,j=1}^{n}{P(x_{i},y_{j})\log_{e}{P(x_{i},y_{j})}}$ With $$P(x_{i},y_{j})$$ the probability of $$x_{i}$$ and $$y_{i}$$ occurring simultaneously. Now we can calculate conditional entropy. If we subtract the entropy of Y from H(X, Y), because we know Y, we get $H(X\|Y)=H(X,Y)-H(Y)$ or $H(X\|Y)=-\sum_{i=1,j=1}^{n}{P(x_{i},y_{j})\log_{e}{P(x_{i},y_{j})}}+\sum_{i=1}^{n}{P(y_{i})\log_{e}{P(y_{i})}}$ Which gives us $H(X\|Y)=\sum_{i,j}P(x_{i},y_{j})\log_{e}{\frac{P(y_{j})}{P(x_{i},y_{j})}}$ This tells us how much uncertainty remains in X when we know Y. Given these definitions, we can now calculate the uncertainty coefficient as $U(X\|Y)=\frac{H(X)-H(X\|Y)}{H(X)}$ This gives us a number between 0 and 1 which is a measure of how good we can predict X from Y. To use this on our example, we first calculate the frequency lists like we did with Cramer’s V. For X we get Black Brown Red Blond 108 286 71 127 And for Y Male Female 279 313 We can calculate probabilities from these by dividing by the amount of samples, 592. Black Brown Red Blond 108 286 71 127 0.1824324324 0.4831081081 0.1199324324 0.214527027 Male Female 279 313 0.4712837838 0.5287162162 Now we can calculate the entropy H(X) = 1.2464359225967288, as well as H(Y) = 0.6914970305474704, which give the entropy of X and Y respectively. Next we calculate the conditional entropy of H(X\|Y). For this we need our pair data Black Brown Red Blond Male 56 143 34 46 Female 52 143 37 81 We calculate probabilities for these by dividing by our sample count, 592. Black Brown Red Blond Male 0.09459459459 0.2415540541 0.05743243243 0.0777027027 Female 0.08783783784 0.2415540541 0.0625 0.1368243243 And calculate H(X\|Y) which gives 1.239600755. Finally we calculate the uncertainty coefficient U(X\|Y) which is 0.00548376956. This measures how good we can guess the hair color given that we know the gender. Not null, but not very high, so the correlation is there, but very low. We can also calculate U(Y\|X) to see if the reverse is really different. U(Y\|X) is 0.009884593959. This measures how good we can guess the gender given that we know the hair color. Also not very high, but double of U(X\|Y), which means that it is slightly easier to guess the gender than the hair color. In R, given a table x, the uncertainty coefficient can be found using nom.uncertainty(x). There is also a symmetric form of the uncertainty coefficient $U(X,Y)=2\frac{H(X)+H(Y)-H(X,Y)}{H(X)+H(Y)}$ Which, like Cramer’s V is not dependent on the order of the parameters, so $U(X,Y)=U(Y,X)$ While this is not necessarily true for U(X\|Y) and U(Y\|X). In our case H(X,Y) is 0.003527040169, which again points to a very weak correlation. ### Correlation Ratio When we have one qualitative parameter and one quantitative parameter, we can use the correlation ratio to look for correlation. We start by sorting all the quantitative values into buckets, one for each possible value of the qualitative parameter. So if our qualitative parameter is gender, and our quantitative parameter is the salary, we make two lists, one for each gender, with all the salaries for that particular gender. Then we calculate the average for each list, which is the average salary by gender, $$\overline{y_i}$$. We also calculate the average of all salaries, $$\overline{y}$$. Now we can calculate the correlation ratio as $\frac{\sum_i{n_i(\overline{y_i}-\overline{y})^2}}{\sum_i{(y-\overline{y})^2}}$ We can see that the numerator gives us the deviation of each bucket’s average to the total average, scaled by the amount of elements in the bucket. While the denominator gives us the deviation of all values to the total average. What does this mean? If one or more buckets have a disproportionate amount of larger numbers, the average of those buckets will be greater, and thus the deviation to the total average will be greater. This happens when there is a correlation, as in the case of a correlation, the bucket a value is in will influence its magnitude. Updated:
Total 9 Posts [Makeover] Central Park & These Tragic “Write An Expression” Problems tl;dr. I made another digital math lesson in collaboration with Christopher Danielson and our friends at Desmos. It’s called Central Park and you should check out the Walkthrough. Here are two large problems with the transition from arithmetic to algebra: Variables don’t make sense to students. We give students variable expressions like the exponential one above, which they had no hand in developing, and ask them to evaluate the expression with a number. The student says, “Ohhh-kay,” and might do it but she doesn’t know what pianos have to do with exponential equations nor does she know where any of those parameters came from. She may regard the whole experience as one of those nonsensical rites of school math which she’ll forget about as soon as she’s legally allowed. Variables don’t seem powerful to students. In school, using variables is harder than using arithmetic. But what does that difficulty buy us, except a grade and our teacher’s approval? Meanwhile, in the world, variables are responsible for anything powerful you have ever done with a computer. Students should experience some of that power. One solution. Our attempt at solving both of those problems is Central Park. It proceeds in three phases. Guesses We ask the students to drag parking lines into a lot to make four even spaces. Students have no trouble stepping over this bar. We are making sure the main task makes sense. Numbers We transition to calculation by asking the students “What measurements would you need to figure out the exact space between the dividers?” This question prepares them to use the numbers we give them next. Now they use arithmetic to calculate the space width for a given lot. They do that three times, which means they get a sense of the parts of their arithmetic that change (the width of the lot, the width of the parking lines) and those that don’t (dividing by the four lots). This will be very helpful as we take the next big leap. Variables We give students numbers and variables. They can calculate the space width arithmetically again but it’ll only work for one lot. When they make the leap to variable equations, it works for all of them. It works for sixteen lots at once. Variables should make sense and make students powerful. That’s our motto for Central Park. 2014 Jul 28. Here is Christopher Danielson’s post about Central Park on the Desmos blog. Featured Comment In thinking further about your complaint about â€œWrite an expressionâ€ I think what is also going on in this app is a NEEDED slowing down of the learning process. The text (and too many teachers) are quick to jump to algorithms before the students understands their nature and value. Look how long it takes to get to the concept of an appropriate expression in the app: you build to it slowly and carefully. I think this is at the heart of the kind of induction needed for genuine understanding, where the learner is helped, by scaffolding, to draw thoughtful and evidence-based conclusions; test them in a transfer setting; and learn from the feedback â€“ i.e. the essence of what we argue understanding is in UbD. One reason I like this activity so much is that it hits the sweet spot where â€œWhat can you do with it?â€ and â€œWhat does it mean?â€ overlap. “I would eat the extra meatball.” Simon Terrell recaps his lesson study trip to Japan with Akihiko Takahashi, who was the subject of Elizabeth Green’s American math article last week: In one case, a teacher was teaching a lesson about division with remainders and the example was packaging meatballs in pack of 4. When faced with the problem of having 13 meatballs and needing 4 per pack, one student’s solution was “I would eat the extra meatball and then they would all fit.” It was so funny and joyful to see that all thinking was welcomed and the teacher artfully led them to the general thinking that she wanted by the end of the lesson. I can trace my development as a teacher through the different reactions I would have had to “I would eat the extra meatball,” from panic through irritation to some kind of bemusement. BTW. The comments here have been on another level lately, team, including Simon’s, so thanks for that. I’ve lifted a bunch of them into the main posts of Rand Paul Fixes Calculus and These Tragic â€œWrite An Expressionâ€ Problems. Two quick meta-items about blogging from the last week: • I attended Twitter Math Camp 2014 in Jenks, OK, in which 150 math teachers who generally only interact online get together in person. I gave a keynote that could probably best be described as “data-rich,” in which I downloaded and analyzed details on 12,000 blogging and tweeting math teachers. Here are links to my slides and speech as well as the CSVs if you want to analyze some data yourself. (Who doesn’t!) • A doctoral student in Canada is interested in blogging as “unmediated professional growth” and sent me a survey about my blogging. Here is a link to my responses. How would you have answered? Rand Paul Fixes Calculus If you have one person in the country who is, like, the best at explaining calculus, that person maybe should teach every calculus class in the country. It’d be helpful if we could work through the idea that good teaching is just good explaining and vice versa. Someone here at Twitter Math Camp mentioned she conducts a math night for parents at the start of school. “I wish I had learned math like this as a kid,” they tell her. That realization, that there is and should be a difference between how math was taught then and now, is a giant first step. This shows the idea that childrenâ€™s minds are empty vessels that need to be filled with knowledge and teachers are the keepers of that knowledge, whose sole job is to effectively pour said knowledge into the vessel. And if their minds didnâ€™t get filled with our knowledge the fault must lie with our explanations. This flies in the face of what we know about teaching and learning. None of these reforms about math education can happen in a vacuum. Thereâ€™s always a political side to what happens to peopleâ€™s children, and if the way you help children learn math is important then the way you communicate with parents is also important. [Makeover] These Tragic “Write An Expression” Problems We need to do something about these problems, which recur all throughout school Algebra. The original title of this post called them “horrible,” but they’re truly “tragic” â€“ the math education equal to Julius Caesar, Othello, and Hamlet â€“ full of potential but overwhelmed by their nature. Here’s the thing about variable expressions: they’re used by programmers and students both, but those two groups hold variables in very different regard. Ask programmers what their work life would be like without variables and they’ll likely respond that their work life would be impossible. Variables enable every single function of whatever device you’re reading this post on. But ask students what their school life would be like without variables and they’ll likely respond that their school life would be great. What can we do? The moral of this story isn’t “teach Algebra 1 through programming” or “teach computational thinking.” At least I don’t think so. I’ve been down that road and it’s winding. But in some way, however small, we should draw closer together the wildly diverging opinions students and programmers have about variables. Ideas? I’ll offer one on Monday. 2014 Jul 25. I appreciate how Evan Weinberg has thought through this makeover (now and earlier). Featured Comment Dylan Kane restates our task here in a useful way: In terms of making these problems a little better, students should feel a need for the expression. I think this question stinks in part because the expression itâ€™s asking for is so trivial â€” itâ€™s extra work, compared to just multiplying by 3/4 or doing some simple proportional thinking. Jennifer offers an example of that kind of need: I like to introduce the idea of expressions by having the students playing the game of 31 with a deck of cards!their goal- play until they can predict how they can win every time! This will take less than 15 minutes, and a whole class summary of verbal descriptions on â€˜how to winâ€™ are shared. Verbal descriptions become cumbersome to write on the board, so â€˜shorthandâ€™ in the form of clearly named and defined symbols are used to make the summarising more efficient. the beauty of this is that the idea of equivalent expressions presents itself. i think the ability to generalize and write a rule with variables is really important, but you can come to that through lots of nice activities and investigations as well. for example, i did danâ€™s â€œtaco cartâ€ with my students with a few notable changes. instead of telling the students how fast dan and ben walked, i had each group decide on the speed of the two men themselves and list that along with other assumptions they made in the problem. when we did the whole class summary, i told them that i had written down a formula on my paper that would allow me to check if their answers were correct and that i needed that since everyone used their own speed. i shouldâ€™ve asked them all to take a minute to try to come up with the formula i used, but instead i elicited it at that moment and one student gave me the correct formula. the need for two variables (speed on sand and speed on pavement) was obvious. in part two of taco cart, when the students were trying to figure out where the taco cart should be so that the two men reached it at the same time, one group did seemingly endless guess and checks. i suggested to them that this wasnâ€™t a good method and asked if maybe it wasnâ€™t better to write an equation with a variable. again, they could see the need. once they started with a variable, the rest of the problem started to fall into place. hereâ€™s the thing: these â€œwrite an expressionâ€ problem want to train students to learn to generalize and write a rule. they want them to be able to see a situation that would best be tackled with a formula of some sort, write said formula with various variables, and use their formula to solve complex problems. but the issue is that these problems donâ€™t actually train students in that way because theyâ€™re so artificial and one-dimensional. what they do is teach students to â€œtranslateâ€ from english to math (an important step along the way, i do believe), but not to recognize a situation in which a formula would be helpful or necessary or how powerful it can be. Two thoughts from a computer science teacherâ€™s point of view: 1) When introducing programming to 15 yrs olds for the first time, we use Python interactive prompt as a calculator first. And the first point is to show the advantages. Variables and funstions (one-liner formulas) simply save work. That is it, that is one of the goals of the whole programming topic anyway. When they do quadratic equations in maths at that time, we are headed that way. And students realize pretty fast: aha, I have to understand how to solve them, but once I do and I describe it properly, I never have to do it by hand anymore. Their understanding of q.e. deepened, they interest in programming increased, and I could naturally introduce a load of important CS concepts on the way. In younger age we do simpler formulas, also like BMI calculation (not only the â€œarea of the circleâ€ kind of stuff, which they, um, do not prefer), but the point is the same: the kids need to see at least some hypothetical benefit for themselves. Having to introduce variables in maths, I would in principle search for a similar approach. (Note: for even younger kids, fun and creativity achieved with Scratch, turtle etc. overweigh the â€œpractical benefitsâ€; but when practicality leads to more fun â€“ win-win!). 2) A good and often forgotten tool between calculation on paper and programming is a spreadsheet. It can store lots of numbers in a structured way and perform basic calculations, what is well understood by kids. And when we want to do anything more complex without getting beards grown, we absolutely need formulas and â€œvariablesâ€. Their advantages are imminent. And the whole time, everything is in plain sight, the level of abstraction is way lower than with programming, making it very accessible for kids. I am of course interested in it â€œfrom the other sideâ€ â€“ after some decent work in spreadsheets, many more advanced concepts are a step away (for-loop, data type, input-output, function, incremental work on more complicated calculations, debugging etc. etc.). But I believe that thoughtful use of spreadsheets can improve understanding in appropriate topics in maths.
Courses Courses for Kids Free study material Offline Centres More Store # Conversion of Improper Fractions Into Mixed Fractions Reviewed by: Last updated date: 14th Jul 2024 Total views: 201.3k Views today: 4.01k ## An Introduction to Fractions and Mixed Fractions In Mathematics, fraction is used to show or represent a part of the entire thing. It can represent things as equal parts of the whole. In the representation of fractions, there are two parts namely numerator and denominator. The number written on the top is known as the numerator. The number written on the bottom part is known as the denominator. The number of equal parts of a thing written on top is called the numerator. The denominator is the whole number in a fraction. If \$\dfrac{5}{10}\$ is a fraction. The number \$5\$ is a numerator, and \$10\$ is a denominator. Visual Representation of a Fraction ## Improper Fraction The fractions in which the numerator value is larger than the denominator are known as improper fractions. E.g. \$\dfrac{7}{3}, \dfrac{9}{5}, \dfrac{7}{2}\$ ## Mixed Fraction It can be defined as the fraction made up of a combination of whole numbers and fractions. e.g. \$3 \dfrac{1}{2}, 5 \dfrac{1}{2}\$ ## How to Convert an Improper Fraction into a Mixed Fraction? There are certain steps to convert an improper fraction into a mixed fraction. These are discussed below. First Step: We need to identify the improper fraction. Second Step: Then, we need to divide the numerator by the denominator and thereby obtain the quotient and remainder. Third Step: Now, the mixed fraction can be written as: Quotient (Remainder/Denominator). ## Simple Way to Convert Improper Fraction to Mixed Fraction To convert an improper fraction to a mixed fraction, divide the numerator (upper portion) by the denominator in order to write an improper fraction as a mixed fraction (bottom part). The numerator is the remainder, and the quotient is the entire number. ## Solved Examples Example 1. Let us convert \$\dfrac{7}{5}\$ into a mixed fraction. Ans: If in a fraction the numerator and denominator have the same number, it will make a whole. In the case of \$\dfrac{7}{5}, \dfrac{5}{5}\$ can be extracted and that can be made a whole. Therefore, the remaining fraction is \$2 / 5\$. So, now we can write \$\dfrac{7}{5}\$ as the mixed fraction, and that is \$1 \dfrac{2}{5}\$ The fraction \$\dfrac{7}{5}\$ means that \$7 \div 5\$. Now we divide \$7\$ by \$5\$ and get \$1\$ as the quotient making \$2\$ as the remainder. In order to convert an improper fraction into a mixed fraction, we need to use the quotient \$1\$ as the whole number and the remainder \$2\$ as the numerator whereas the divisor \$5\$ is used as the denominator of the proper fraction. Mixed and Improper Fraction. Example 2. Convert the improper fractions into mixed fractions: (i) \$\dfrac{17}{4}\$ Ans: According to the question, the numerator is \$17\$, and the denominator is \$4\$ In order to convert the improper fraction into a mixed fraction, we divide the numerator with the denominator. When we divide \$17\$ by \$4\$ Quotient = \$4\$, Remainder = \$1\$, Denominator = \$4\$. Hence, \$\dfrac{17}{4}=4 \dfrac{1}{4}\$ (ii) \$\dfrac{13}{5}\$ Ans: According to the question, the numerator is \$13\$, and the denominator is \$5\$ In order to convert the improper fraction into a mixed fraction. We divide the numerator with the denominator. When we divide \$13\$ by \$5\$ Quotient = \$2\$, Remainder = \$3\$, Denominator = \$5\$. Hence, \$\dfrac{13}{5}=2 \dfrac{3}{5}\$ (iii) \$\dfrac{28}{5}\$ Ans: According to the question, the numerator is \$28\$, and the denominator is \$5\$ In order to convert the improper fraction into a mixed fraction. We divide the numerator with the denominator. When we divide \$28\$ by \$5\$ Therefore, Quotient = \$5\$, Remainder = \$3\$, Denominator = \$5\$ Hence, \$\dfrac{28}{5}=5 \dfrac{3}{5}\$ (iv) \$\dfrac{28}{9}\$ Ans: According to the question, the numerator is \$28\$ and the denominator is \$9\$ In order to convert the improper fraction into a mixed fraction. We divide the numerator with the denominator. When we divide \$28\$ by \$9\$ The Quotient = \$3\$, Remainder = \$1\$, Denominator = \$9\$. Hence, \$\dfrac{28}{9}=3 \dfrac{1}{9}\$ ## Practice Questions Convert the following improper fractions into mixed fractions. Q 1. Convert into mixed fractions: (a) \$\dfrac{7}{3}\$ (b) \$\dfrac{11}{7}\$ (c) \$\dfrac{13}{6}\$ Ans: (a) \$2^\dfrac{1}{3}\$ (b) \$\dfrac{14}{7}\$ (c) \$2^\dfrac{1}{6}\$ Q 2. Write improper fraction of \$\dfrac{3}{9}\$ Ans: \$\dfrac{1}{3}\$ ## Summary Fractions are representations of a single part out of many parts, and they can be in different forms either mixed or improper. Fractions can be converted into either form following a few steps. The steps include identifying an improper fraction, dividing the numerator by the denominator to get a quotient and remainder, and finally, the mixed fraction is written as Quotient (Remainder/Denominator). Fractions are very useful in expressing a lot of things such as time, i.e. every minute is a part of an hour and every hour is a part of a day. ## FAQs on Conversion of Improper Fractions Into Mixed Fractions 1. What is the need to convert an improper fraction into a mixed fraction? Improper fractions need to be converted to mixed fractions because improper fractions are those in which the numerator value exceeds the denominator. A mixed fraction is one that combines both whole numbers and fractions. The primary purpose of this conversion procedure is to streamline calculations that call for incorrect fractions. But since mixed fractions are simpler to understand, we can use them to describe a result. 2. Convert an improper fraction into mixed fraction \$\dfrac{35}{8}\$ According to the question, the numerator is \$35\$ and the denominator is \$8\$. In order to convert the improper fraction into a mixed fraction. We divide the numerator with the denominator. When we divide \$35\$ by \$8\$ , the Quotient = \$4\$, Remainder = \$3\$, Denominator = \$8\$. Hence, the final result is \$\dfrac{35}{8}=4 \dfrac{3}{8}\$ 3. Can mixed fractions be used in arithmetic operations? Mixed fractions are those fractions in which a whole number is combined with a fraction. The mixed fractions can also be added, subtracted, multiplied or divided just like the integers. They need to be converted into a suitable improper fraction first and that makes the process easier to obtain. Thereafter, the operation processes are very similar to normal integer operation.
# 2021 USAMO Problems/Problem 1 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that$$\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.$$Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent. ## Solution Let $D$ be the second point of intersection of the circles $AB_1B$ and $AA_1C.$ Then: \begin{align} \angle ADB &= 180^\circ – \angle AB_1B,&\angle ADC &= 180^\circ – \angle AA_1C\\ \angle BDC &= 360^\circ – \angle ADB – \angle ADC\\ &= 360^\circ – (180^\circ – \angle AB_1B) – (180^\circ – \angle AA_1C)\\ &= \angle AB_1B + \angle AA_1C\\ \angle BDC + \angle BC_1C &= 180^\circ \end{align} Therefore, $BDCC_1B_2$ is cyclic with diameters $BC_1$ and $CB_2$, and thus $\angle CDB_2 = 90^\circ.$ Similarly, $\angle CDA_1 = 90^\circ$, meaning points $A_1$, $D$, and $B_2$ are collinear. Similarly, the points $A_2, D, C_1$ and $C_2, D, B_1$ are collinear. (After USAMO 2021 Solution Notes – Evan Chen)
A Quadratic Equation in Standard Form (a, b, and c can have any value, except that a can't be 0.) Here is an example: ## Graphing You can graph a Quadratic Equation using the Function Grapher, but to really understand what is going on, you can make the graph yourself. Read On! f(x) = x2 And its graph is simple too: This is the curve f(x) = x2 It is a parabola. Now let us see what happens when we introduce the "a" value: f(x) = ax2 • Larger values of a squash the curve inwards • Smaller values of a expand it outwards • And negative values of a flip it upside down ## Play With It Now is a good time to play with the "Quadratic Equation Explorer" so you can see what different values of a, b and c do. Before graphing we rearrange the equation, from this: f(x) = ax2 + bx + c To this: f(x) = a(x-h)2 + k Where: • h = −b/2a • k = f( h ) In other words, calculate h (= −b/2a), then find k by calculating the whole equation for x=h ## But Why? The wonderful thing about this new form is that h and k show us the very lowest (or very highest) point, called the vertex: And also the curve is symmetrical (mirror image) about the axis that passes through x=h, making it easy to graph ### So ... • h shows us how far left (or right) the curve has been shifted from x=0 • k shows us how far up (or down) the curve has been shifted from y=0 Lets see an example of how to do this: ### Example: Plot f(x) = 2x2 - 12x + 16 First, let's note down: • a = 2, • b = −12, and • c = 16 Now, what do we know? • a is positive, so it is an "upwards" graph ("U" shaped) • a is 2, so it is a little "squashed" compared to the x2 graph Next, let's calculate h: h = −b/2a = −(−12)/(2x2) = 3 And next we can calculate k (using h=3): k = f(3) = 2(3)2 - 12·3 + 16 = 18−36+16 = −2 So now we can plot the graph (with real understanding!): We also know: the vertex is (3,−2), and the axis is x=3 ## From A Graph to The Equation What if we have a graph, and want to find an equation? ### Example: you have just plotted some interesting data, and it looks Quadratic: Just knowing those two points we can come up with an equation. Firstly, we know h and k (at the vertex): (h, k) = (1,1) So let's put that into this form of the equation: f(x) = a(x-h)2 + k f(x) = a(x−1)2 + 1 Then we calculate "a": We know the point (0, 1.5) so:f(0) = 1.5 And a(x−1)2 + 1 at x=0 is:f(0) = a(0−1)2 + 1 They are both f(0) so make them equal: a(0−1)2 + 1 = 1.5 Simplify:a + 1 = 1.5 a = 0.5 And so here is the resulting Quadratic Equation: f(x) = 0.5(x−1)2 + 1 Note: This may not be the correct equation for the data, but itâ€™s a good model and the best we can come up with.
# Collinear Definition: What is Collinearity? Statistics Definitions > Collinear Definition ## Definition A set of points is collinear if you can draw one line through them all. The word literally means “together on a line.” Two points are always collinear: no matter where you draw the two points, you can always draw a straight line between them. A general way to write this is: “Points P1, P2 and P3 are collinear”, which can also be written as “point P1 is collinear with points P2 and P3“. It goes without saying that points are non-collinear if they do not fall on the same line. ## How to Show Points are Collinear It seems reasonable that if you can draw a line through a set of points, then those points are collinear. The trouble is, those points may not be exactly on the same line. One way to work around this is with the knowledge that the points must satisfy the same linear equation. For example, if you are given the linear equation y = 4x + 16, you know that the points (-4,0) and (-1,12) meet the definition because (plugging the x and y values into the equation) we get: 0 = 4(-4) + 16 12 = 4(-1) + 16. ## An Alternate Method A second way is to find the slope between the points (i.e. the slopes of the line segments between points P1 and P2, and P2 and P3); if the slopes are the same then the points are collinear. For example, the set of points in the image below fit the definition if the slope of line segment A equals the slope of line segment B. Sample question: Do the points P1=(−4,0), P2=(−1,12) and P3=(4,32) show collinearity? Step 1: Find the slope for the line segment between the first two points using rise-over-run =(y2−y1)/(x2−x1)=(12−0)/(−1−(−4))= 12/3 =4 Step 2: Find the slope for the line segment between the next two points =(y3−y2)/(x3−x2)=(32−12)/(4-(-1))= 20/5 = 4. Step 3: Compare the slopes you calculated in Steps 1 and 2. The two slopes equal 4, so the points do show collinearity. Check out our YouTube channel for hundreds of videos on elementary probability and statistics, from basic stats to advanced techniques using Excel, SPPS and Minitab. ------------------------------------------------------------------------------ Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. If you rather get 1:1 study help, Chegg Tutors offers 30 minutes of free tutoring to new users, so you can try them out before committing to a subscription. If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.
In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP! # Ex.15.1 Q5 Introduction to Graphs Solutions - NCERT Maths Class 8 Go back to 'Ex.15.1' ## Question Use the tables below to draw linear graphs. (a) The number of days a hill side city received snow in different years. Year $$2003$$ $$2004$$ $$2005$$ $$2006$$ Days $$8$$ $$10$$ $$5$$ $$12$$ (b) Population (in thousands) of men and women in a village in different years. Year $$2003$$ $$2004$$ $$2005$$ $$2006$$ $$2007$$ Number of men $$12$$ $$12.5$$ $$13$$ $$13.2$$ $$13.5$$ Number of Women $$11.3$$ $$11.9$$ $$13$$ $$13.6$$ $$12.8$$ Video Solution Introduction To Graphs Ex 15.1 \| Question 5 ## Text Solution Reasoning: (a) Considering ‘Years’ on horizontal axis and ‘Days’ on vertical axis starting from origin i.e. ($$0,0$$). (b) Considering ‘Years’ on horizontal axis and ‘Number of persons (i.e. ‘Men’ or ‘Women’) on vertical axis starting from origin i.e. ($$0,0$$). Represent ‘Number of men’ as a straight continuous line where as ‘Number of Women’ as a discontinuous straight line ( dashed line). Steps: (a) By taking the years on $$x$$-axis and the number of days on $$y$$-axis and taking scale as $$1$$ unit $$= 2$$ days on $$y$$-axis and $$2$$ unit $$= 1$$ year on $$x$$-axis, the linear graph of the given information can be drawn as follows. (b) By taking the years on $$x$$-axis and population on $$y$$-axis and scale as $$1$$ unit $$= 0.5$$ thousand on $$y$$-axis and $$2$$ unit $$=$$$$1$$ year on $$x$$-axis, the linear graph of the given information can be drawn as follows. Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school
14 Times Table In 14 times table we will learn how to read and write multiplication table of 14. We read fourteen times table as: One time fourteen is 14 Two times fourteen are 28 Three times fourteen are 42 Four times fourteen are 56 Five times fourteen are 70 Six times fourteen are 84 Seven times fourteen are 98 Eight times fourteen are 112 Nine times fourteen are 126 Ten times fourteen are 140 Eleven times fourteen are 154 Twelve times fourteen are 168 We write 14 times table as: 1 × 14 = 14 2 × 14 = 28 3 × 14 = 42 4 × 14 = 56 5 × 14 = 70 6 × 14 = 84 7 × 14 = 98 8 × 14 = 112 9 × 14 = 126 10 × 14 = 140 11 × 14 = 154 12 × 14 = 168 Let us follow the how to read and write multiplication of 14 times table chart: 14 Times Table This is the easiest way to follow 14 times table in the chart. 1. What is 14 × 14 × 14 × 14 × 14 × 14? Answer: 14 × 6 = 84 2. Fill in the blanks: (i) 2 more than 26 = 2 × ___ (ii) 4 times ___ = 56 (iii) 8 times ___ = 112 (iv) 1 less than 99 = 7 × ___ (v) 14 times 3 = ___ 2. (i) 7 more than 77 = ___ × ___ (ii) 2 more than 26 = 2 × 14 (iii) 4 times 14 = 56 (iv) 8 times 14 = 112 (v) 1 less than 99 = 7 × 14 (vi) 14 times 3 = 42 (vii) 7 more than 77 = 6 × 14 3. Using the multiplication time’s table of 14, calculate 14 times 12. Solution: According to the 14 times tables, 14 times 12 = 14 × 12 = 168 [using 14 times table] Therefore, 14 times 12 = 168 4. Each group of students has 14 students. How many students will there be in 22 such groups? Solution: No of students in each group = 14 No of groups = 22 Therefore, total no. of students in 22 groups = 14 × 22 = 308 Answer: There will be 308 students in 22 groups. 5. In a café there are 20 tables. Each table has 14 seats. How many seats are there in the café? Solution: No of seats in 1 table = 14 No of tables = 20 Therefore, total no. of seats in 20 tables = 14 × 20 = 280 Answer: There are 280 seats in the cafe. 6. A box has 14 colour pencils. How many colour pencils are there in 14 such boxes? Solution: No. of colour pencils in 1 box = 14 No. of boxes = 14 Therefore, total no. of colour pencils in 14 boxes = 14 × 14 Answer: There will be 196 colour pencils in 14 boxes. 7. In a jar there are 90 lollipops. How many lollipops are there in 14 such jars? Solution: No. of lollipops in 1 jar = 90 No. of jars = 14 Therefore, total no. of lollipops in 14 jars = 90 × 14 Answer: There will be 1260 lollipops in 14 jars. You might like these • 3 Times Table \| Read and Write Multiplication Table of 3 \| Times Table In 3 times table we will learn how to read and write multiplication table of 3. We read three times table as: One time three is 3 Two times three is 6 Three times three is 9 Four times three is 12 • 7 Times Table \| Read and Write Multiplication Table of 7 \| Seven Times In 7 times table we will learn how to read and write multiplication table of 7. We read seven times table as: One time seven is 7 Two times seven is 14 Three times seven is 21 Four times seven • 2 Times Table \| Read and Write Multiplication Table of 2 \| Times Table In 2 times table we will learn how to read and write multiplication table of 2. We read two times table as: One time two is 2 Two times two is 4 Three times two is 6 Four times two is 8 Five times • 11 Times Table \| Read and Write Multiplication Table of 11\|Times Table In 11 times table we will learn how to read and write multiplication table of 11. We read eleven times table as: One time eleven is 11 Two times eleven are 22 Three times eleven are 33 • 12 Times Table \| Read and Write Multiplication Table of 12\|Times Table In 12 times table we will learn how to read and write multiplication table of 12. We read twelve times table as: One time twelve is 12 Two times twelve are 24 Three times twelve are 36 • 16 Times Table \| Read and Write Multiplication Table of 16\|Times Table In 16 times table we will learn how to read and write multiplication table of 16. We read sixteen times table as: One time sixteen is 16 Two times sixteen are 32 Three times sixteen are 48 • 17 Times Table \|Read and Write Multiplication Table of 17 \|Times Table In 17 times table we will learn how to read and write multiplication table of 17. We read seventeen times table as: One time seventeen is 17 Two times seventeen are 34 Three times seventeen • 8 Times Table \| Read and Write Multiplication Table of 8 \| Times Table In 8 times table we will learn how to read and write multiplication table of 8. We read eight times table as: One time eight is 8 Two times eight is 16 Three times eight is 24 Four times • 5 Times Table \| Read and Write Multiplication Table of 5 \| Times Table In 5 times table we will learn how to read and write multiplication table of 5. We read five times table as: One time five is 5 Two times five is 10 Three times five is 15 Four times five is 20 • 6 Times Table \| Read and Write Multiplication Table of 6 \| Six Table In 6 times table we will learn how to read and write multiplication table of 6. We read six times table as: One time six is 6 Two times six is 12 Three times six is 18 Four times six is 24 • 9 Times Table \| Read and Write Multiplication Table of 9 \| Times Table In 9 times table we will learn how to read and write multiplication table of 9. We read nine times table as: One time nine is 9 Two times nine is 18 Three times nine is 27 Four times nine is 36 • 10 Times Table \| Read and Write Multiplication Table of 10\|Times Table In 10 times table we will learn how to read and write multiplication table of 10. We read ten times table as: One time ten is 10 Two times ten is 20 Three times ten is 30 Four times ten is 40 • 0 Times Table \| Multiplication of 0 Times Table Chart \| Table of 0 In 0 times table we will learn how to read and write multiplication table of 0. One time zero is 0 Two times zero is 0 Three times zero is 0 Four times zero is 0 Five times zero is 0 • 1 Times Table \| Read One Times Table \| Write Multiplication of 1 In 1 times table we will learn how to read and write multiplication table of 1. We read one times table as: One time one is 1 Two times one is 2 Three times one is 3 Four times one is 4 • 4 Times Table \| Read and Write Multiplication Table of 4 \| Four Table In 4 times table we will learn how to read and write multiplication table of 4. We read four times table as: One time four is 4 Two times four is 8 Three times four is 12 Four times four is 16 Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. Recent Articles 1. Constructing a Line Segment \|Construction of Line Segment\|Constructing Aug 14, 24 09:52 AM We will discuss here about constructing a line segment. We know how to draw a line segment of a certain length. Suppose we want to draw a line segment of 4.5 cm length. 2. Construction of Perpendicular Lines by Using a Protractor, Set-square Aug 14, 24 02:39 AM Construction of perpendicular lines by using a protractor is discussed here. To construct a perpendicular to a given line l at a given point A on it, we need to follow the given procedure 3. Construction of a Circle \| Working Rules \| Step-by-step Explanation \| Aug 13, 24 01:27 AM Construction of a Circle when the length of its Radius is given. Working Rules \| Step I: Open the compass such that its pointer be put on initial point (i.e. O) of ruler / scale and the pencil-end be… 4. Practical Geometry \| Ruler \| Set-Squares \| Protractor \|Compass\|Divider Aug 12, 24 03:20 PM In practical geometry, we study geometrical constructions. The word 'construction' in geometry is used for drawing a correct and accurate figure from the given measurements. In this chapter, we shall…
# Divisibility Rules In divisibility rules(test) we find whether a given number is divisible by another number, we perform actual division and see whether the remainder is zero or not. But divisibility tests of a given number by any of the number 2, 3, 4, 5, 6, 7, 8, 9, 10 can be perform simply by examining the digits of the given number. These tests are as under: Divisibility by 2 A number is divisible by 2 if its units place is either 0 or multiple of 2. For example: 346, 3818, 14626, 100, 1994, 1252 All these number is divisible by 2 because their units place in multiple of 2. Divisibility by 3 A number is divisible by 3 if the sum of digits is a multiple of 3. For example: 79851 is divisible by 3 as the sum of its digits, i.e., 7 + 9 + 8 + 5 + 1 = 30 is divisible by 3. Divisibility by 4 A number is divisible by 4 if the number formed by its digits in tens and units place is divisible by 4. For example: 88312 is divisible by 4 because the number formed by its last two digits i.e., 12 is divisible by 4. Divisibility by 5 A number is divisible by 5 if its units place is 0 or 5. For example: 75325 is divisible by 5 as its last digit is 5. Divisibility by 6 A number is divisible by 6 if it is divisible by 2 and 3 both. For example: 85806 is divisible by 6 because it is an even number so divisible by 2 and sum of its digits, i.e., 8 + 5 + 8 + 0 + 6 = 27 27 is divisible by 3. Divisibility by 7 A number of 2 digits is divisible by 7 because 3 × 6 + 3 = 21. 21 is divisible by 7. A number of three or more digits is divisible by 7 if the sum of the numbers formed by the last two digits and twice the number formed by the remaining digits is divisible by 7. For example: (i) 574 is divisible by 7 because 74 + 5 × 2 = 74 + 10 = 84 is divisible by 7. (ii) 2268 is divisible by 7 because 68 + 22 × 2 = 68 + 44 = 112 is divisible by 7. Divisibility by 8. A number is divisible by 8 if the numbers formed by the last three digits is divisible by 8. For example: 54136 is divisible by 8 because if the numbers formed by the last three digits i.e., 136 is exactly divisible by 8. 136 ÷ 8 = 17, Remainder = 0 Divisibility by 9 A number is divisible by 9 if the sum of its digits is divisible by 9. For example: 3888 is divisible by 9 because 3 + 8 + 8 + 8 = 27 is divisible by 9. Divisibility by 10. A number is divisible by 10 if it has zero (0) in its units place. A number is divisible by 11 if the sum of the digits in the oddplaces andthe sum of the digits in the evenplaces difference is a multiple of 11 or zero. For example: Sum of the digits in the even places = 5 + 9 + 8 = 22 Sum of the digits in the odd places = 5 + 1 + 3 + 2 =11 Difference between the two sums = 22 – 11= 11 11 is divisible by 11. Hence, 2839155 is divisible by 11. Notes: 1. A number is divisible by another number if it is also by its co-prime factors. The co-prime factors of 15 are 3 and 5. 2. A number is divisible by : 12, if it is divisible by co-prime 12 i.e., 3 and 4. 15, if it is divisible by co-prime 15 i.e., 3 and 5 18, if it is divisible by co-prime 18 i.e., 2 and 9. 45, if it is divisible by co-prime 45 i.e., 9 and 5. Properties of Divisibility. Divisible by 2. Divisible by 3. Divisible by 4. Divisible by 5. Divisible by 6. Divisible by 7. Divisible by 8. Divisible by 9. Divisible by 10. Divisible by 11. Problems on Divisibility Rules Worksheet on Divisibility Rules
# Question Video: Solving Word Problem Involving Percentages Mathematics • 7th Grade A vase contains some flowers. 20% of the flowers are pink, 48% of the flowers are violet, and the rest are red. Given that the vase contains 10 pink flowers, calculate how many violet flowers and how many red flowers it contains. 02:21 ### Video Transcript A vase contains some flowers. 20 percent of the flowers are pink, 48 percent of the flowers are violet, and the rest are red. Given that the vase contains 10 pink flowers, calculate how many violet flowers and how many red flowers it contains. So we know that 20 percent of the total number of flowers are pink. 48 percent are violet. And the rest are red. All together, the flowers should represent 100 percent. So if we take away the percentage of pink flowers and the percentage of violet flowers, we will find the percentage of red flowers. 100 minus 20 minus 48 will be 32. So, 32 percent of the total number of flowers will be red. And we’re actually given that the vase contains 10 pink flowers. So we can plug that in here. In mathematics, the word “of” means to multiply. So we can replace the word “of” with multiplication. So if we could find the total number of flowers, we could find the number of violet flowers and the number of red flowers. Taking this first equation, we can solve for the total. We would need to divide both sides of the equation by 20 percent. 20 percent can be rewritten by dropping the percent sign and moving the decimal place two places to the left. Or think of moving the numbers two places to the right. So taking 20 and dividing by 100, that’s what would give us the .20. And 10 divided by .20 is 50. So there are total of 50 flowers. So we can replace the word “total” with 50 and then solve for violet and red. 48 percent would be 0.48 dropping the percentage sign and dividing by 100. So .48 times 50 would give us 24. So there will be 24 violet flowers. 32 percent can be written as 0.32. And then multiplying by 50 gives us 16. There are 16 red flowers. So all of these flowers together, the 10 pink, 24 violet, and 16 red, should add together to be 50. In the ones place, four plus six is 10. So we write zero, carry the one, and then one plus one plus two plus one is five. So there’ll be 50 total. So the number of violet flowers will be 24 and the number of red flowers will be 16.
Name: ___________________Date:___________________ Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! ### Grade 4 - Mathematics8.21 More Fraction Problems The fractional part of a collection is determined by dividing the collection into subgroups equal to the number representing the denominator of the fraction. We then take the number of subgroups equal to the number representing the numerator of the fraction. A fraction can be expressed as a division and vice-versa. A fraction written as a combination of a whole number and a proper fraction is called a mixed number. To generate equivalent fractions of any given fraction, we proceed as follows: Multiply the numerator and denominator by the same number (other than 0) or Divide the numerator and denominator by their common factor (other than 1), if any. The two fractions are equivalent if the product of numerator of the first and denominator of the second is equal to the product of the denominator of the first and numerator of the second. A fraction is in its lowest term, if the numerator and denominator have no common factor other than 1. Of the two fractions having the same denominator, the one with greater numerator is greater. Of the two fractions having the same numerator, the one with greater denominator is smaller. Adding and Subtracting Fractions: Read the problems carefully. If the denominators are same, add or subtract the numerators and write the answer. If the denominators are different, they are made common by finding LCM. Example: 9/4 - 3/4 Since the denominators are same, the numerators are subtracted. 9/4 - 3/4 = 6/4 Answer: 6/4 Example: 1/5 + 1/3 Since the denominators are different, find LCM of 5 and 3. LCM of 5 and 3 is 15. 1/5 x 15/15 = 3/15 1/3 x 15/15 = 5/15 Answer: 8/15 Directions: Answer the following questions. Also write at least 5 examples of your own for fraction operations: addition, subtraction, multiplication, and division of fractions. Q 1: Change 2 5/8 into improper fraction.18/2120/821/8 Q 2: There are 360 students in a school. 3/4 of them are boys and rest are girls. How many boys are there in the school?36090270 Q 3: Which of the following fractions is equivalent to 3/5?15/2015/2530/40 Q 4: 3 2/5 = 3 + ___3/52/42/5 Q 5: Reduce the fraction 15/18 to its simplest form.6/55/65/8 Q 6: In a proper fraction, numerator is less than the denominator. In an improper fraction the numerator is ___ than the denominator.greaterlesser Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
# Quick Answer: How Do You Construct A 30 60 90 Triangle? ## How do you solve a 30 60 90 Triangle? A Quick Guide to the 30-60-90 Degree TriangleType 1: You know the short leg (the side across from the 30-degree angle). Double its length to find the hypotenuse. Type 2: You know the hypotenuse. Divide the hypotenuse by 2 to find the short side. Type 3: You know the long leg (the side across from the 60-degree angle).. ## What are the side lengths of a 45 45 90 Triangle? In its simplest form, the ratios of the length of sides in a 45 45 90 special right angle triangles should be 1 : 1 : 2 1:1:\sqrt{2} 1:1:2 . Recall that the 45 45 90 special right angle triangle is an isosceles triangles with two equal sides and the one larger side (i.e. hypotenuse definition). ## What is a 90 degree angle when giving a shot? Subcutaneous injections can be given straight in at a 90 degree angle or at a 45 degree angle. ## How do you find the area of a 45 45 90 Triangle? Correct answer: To find the area of a triangle, multiply the base by the height, then divide by 2. Since the short legs of an isosceles triangle are the same length, we need to know only one to know the other. Since, a short side serves as the base of the triangle, the other short side tells us the height. ## Which is true statement about a 45 45 90 Triangle? 45-45-90 Length of the Hypotenuse The isosceles triangle theorem states that the lengths of the legs in a 45-45-90 triangle are equal because there are two congruent angles. The Pythagorean theorem is used to show the relationship between the length of the legs and the hypotenuse. ## How do you construct a 90 degree triangle? We can construct a 90º angle either by bisecting a straight angle or using the following steps.Step 1: Draw the arm PA.Step 2: Place the point of the compass at P and draw an arc that cuts the arm at Q.Step 3: Place the point of the compass at Q and draw an arc of radius PQ that cuts the arc drawn in Step 2 at R.More items… ## What is the relationship of a 30 60 90 Triangle? Tips for Remembering the 30-60-90 Rules Remembering the 30-60-90 triangle rules is a matter of remembering the ratio of 1: √3 : 2, and knowing that the shortest side length is always opposite the shortest angle (30°) and the longest side length is always opposite the largest angle (90°). ## What is a 30 degree angle called? Acute Angle: An angle whose measure is more than 0° but less than 90° is called an acute angle. Angles having magnitudes 30°, 40°, 60° are all acute angles. ## What is a 30 degree slope? 30 degrees is equivalent to a 58% grade which is another way to describe the magnitude of a slope. ## What is a 240 degree angle called? In geometry, there are three types of angles: acute angle-an angle between 0 and 90 degrees. right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees. straight angle-a 180 degree angle. ## What is a 90 angle called? Acute angles measure less than 90 degrees. Right angles measure 90 degrees. Obtuse angles measure more than 90 degrees. ## How do you prove a 45 45 90 Triangle? In a 45-45-90 triangle, both of the legs have the same length and the ratio of one leg to the hypotenuse is 1:sqrt(2).
# 7.10: Proportions with Angle Bisectors Difficulty Level: At Grade Created by: CK-12 Estimated5 minsto complete % Progress Practice Proportions with Angle Bisectors MEMORY METER This indicates how strong in your memory this concept is Progress Estimated5 minsto complete % Estimated5 minsto complete % MEMORY METER This indicates how strong in your memory this concept is What if you were told that a ray was an angle bisector of a triangle? How would you use this fact to find unknown values regarding the triangle's side lengths? After completing this Concept, you'll be able to solve such problems. ### Guidance When an angle within a triangle is bisected, the bisector divides the triangle proportionally By definition, \begin{align}\overrightarrow{AC}\end{align} divides \begin{align}\angle BAD\end{align} equally, so \begin{align}\angle BAC \cong \angle CAD\end{align}. The proportional relationship is \begin{align}\frac{BC}{CD}=\frac{AB}{AD}\end{align}. Theorem: If a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the lengths of the other two sides. #### Example A Find \begin{align}x\end{align}. Because the ray is the angle bisector it splits the opposite side in the same ratio as the sides. So, the proportion is: \begin{align}\frac{9}{x} &= \frac{21}{14}\\ 21x &= 126\\ x &= 6\end{align} #### Example B Determine the value of \begin{align}x\end{align} that would make the proportion true. You can set up this proportion just like the previous example. \begin{align}\frac{5}{3} &= \frac{4x+1}{15}\\ 75 &= 3(4x+1)\\ 75 &= 12x+3\\ 72 &= 12x\\ 6 &= x\end{align} #### Example C Find the missing variable: Set up a proportion and solve like in the previous examples. \begin{align}\frac{12}{4}&=\frac{x}{3}\\ 36&=4x\\ x&=9\end{align} Watch this video for help with the Examples above. ### Vocabulary Pairs of numbers are proportional if they are in the same ratio. An angle bisector is a ray that divides an angle into two congruent angles. ### Guided Practice Find the missing variables: 1. 2. 3. 1. Set up a proportion and solve. \begin{align} \frac{20}{8}&=\frac{25}{y}\\ 20y&=200 \\ y&=10 \end{align} 2. Set up a proportion and solve. \begin{align} \frac{20}{y}&=\frac{15}{28-y}\\ 15y&=20(28-y)\\ 15y&=560-20y\\ 35y&=560\\ y&=16\end{align} 3. Set up a proportion and solve. \begin{align} \frac{12}{z}&=\frac{15}{9-z}\\ 15z&=12(9-z)\\ 15z&=108=12z\\ 27z&=108\\ z&=4\end{align} ### Practice Find the value of the missing variable(s). Find the value of each variable in the pictures below. Find the unknown lengths. 1. Error Analysis Casey attempts to solve for a in the diagram using the proportion \begin{align}\frac{5}{a}=\frac{6}{5}\end{align}. What did Casey do wrong? Write the correct proportion and solve for \begin{align}a\end{align}. Solve for the unknown variable. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Angle Bisector Theorem The angle bisector theorem states that if a point is on the bisector of an angle, then the point is equidistant from the sides of the angle. Proportion A proportion is an equation that shows two equivalent ratios. Ratio A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects: ## Concept Nodes: Date Created: Jul 17, 2012 Aug 02, 2016 Image Detail Sizes: Medium \| Original MAT.GEO.546.L.2
# High School Math : Finding Domain and Range ## Example Questions ### Example Question #1 : Pre Calculus What is the domain of the function below: Explanation: The domain is defined as the set of possible values for the x variable. In order to find the impossible values of x, we should: a) Set the equation under the radical equal to zero and look for probable x values that make the expression inside the radical negative: There is no real value for x that will fit this equation, because any real value square is a positive number i.e. cannot be a negative number. b) Set the denominator of the fractional function equal to zero and look for probable x values: Now we can solve the equation for x: There is no real value for x that will fit this equation. The radical is always positive and denominator is never equal to zero, so the f(x) is defined for all real values of x. That means the set of all real numbers is the domain of the f(x) and the correct answer is . Alternative solution for the second part of the solution: After figuring out that the expression under the radical is always positive (part a), we can solve the radical and therefore denominator for the least possible value (minimum value). Setting the x value equal to zero will give the minimum possible value for the denominator. That means the denominator will always be a positive value greater than 1/2; thus it cannot be equal to zero by setting any real value for x. Therefore the set of all real numbers is the domain of the f(x). ### Example Question #1 : Pre Calculus What is the domain of the function below? Explanation: The domain is defined as the set of all values of x for which the function is defined i.e. has a real result. The square root of a negative number isn't defined, so we should find the intervals where that occurs: The square of any number is positive, so we can't eliminate any x-values yet. If the denominator is zero, the expression will also be undefined. Find the x-values which would make the denominator 0: Therefore, the domain is . ### Example Question #1 : Functions What is the domain of the function below:
# Discovering Statistics A den for Learning ## JAM 2022 [ 51-60 ] Let A = \left( \begin{matrix} 1 & -1 & 2 \\ -1 & 0 & 1 \\ 2 & 1 & 1 \end{matrix} \right) and let \left( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) be an eigenvector corresponding to the smallest eigenvalue of A, satisfying x_1^2+x_2^2+x_3^2 =1 . Then the value of \|x_1\| +\|x_2\| +\|x_3\| equals ____________________. (round off to 2 decimal points) If \lambda is an Eigen value of A then, \left\|A- \lambda I_3 \right\| =0 \\ \implies (1-\lambda)(-\lambda(1-\lambda)-1) +1 (-1(1-\lambda) -2) \\ + 2(-1+2\lambda) =0 \\ \implies-(1-\lambda)(\lambda^2-\lambda-1) +\lambda-3+4\lambda-2=0 \\ \implies (1-\lambda)(\lambda-3)(\lambda+2)=0 \\ \implies \lambda=1,3,-2 Thus the smallest Eigen value is -2. Now we have, A \underline{x} = -2 \underline x \\ \implies x_1-x_2+2x_3 +2x_1=0 \\ \hskip{1cm} -x_1+x_3 +2x_2 =0 \\ \hskip{1cm} 2x_1+x_2+x_3 + 2x_3 =0 \\ \implies 3x_1-x_2+2x_3 =0 \\ \hskip{1cm} -x_1+ 2x_2 +x_3 =0 \\ \hskip{1cm} 2x_1+x_2+3x_3 =0 \\ \implies 3x_1-x_2+2x_3 =0 \\ \hskip{1cm} -x_1+ 2x_2 +x_3 =0 \\ \hskip{1cm} 2x_1+x_2+3x_3 =0 \\ (3x_1-x_2+2x_3)+3(-x_1+ 2x_2 +x_3) = 0 \\ \implies 0+5x_2+5x_3 = 0 \\ \implies x_2=-x_3 \\ (-x_1+ 2x_2 +x_3)-2(2x_1+x_2+3x_3)=0 \\ \implies -5x_1-5x_3=0 \\ \implies x_1=-x_3 Thus, x_1^2+x_2^2+x_3^2=1 \\ \implies x_1^2=x_2^2=x_3^2=\frac{1}{3} \\ \|x_1\|+\|x_2\|+\|x_3\|= 3\|\frac{1}{\sqrt{3}}\|=\sqrt{3}=1.732
Assignment XII: Fibonacci Sequence by Rui Kang Generate a Fibonacci sequence in the first column using f(0)=1, f(1)=1, and f(n)=f(n-1)+f(n-2) a. Construct the ratio of each pair of adjacent terms in the Fibonacci sequence. What happens as n increases? What about the ratio of every second term? etc. The story behind Fibonnaci Sequence: Medieval mathematician and businessman Fibonacci (Leonardo Pisano) posed the following problem in 1202: How many pairs of rabbits will be produced in a year, beginning with a single pair, if in every month each pair bears a new pair which becomes productive from the second month on? It is easy to see that 1 pair will be produced the first month, and 1 pair also in the second month (since the new pair produced in the first month is not yet mature), and in the third month 2 pairs will be produced, one by the original pair and one by the pair which was produced in the first month. In the fourth month 3 pairs will be produced, and in the fifth month 5 pairs. After these things expand rapidly, and we get the following sequence of numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,... This is an example of a recursive sequence, obeying the simple rule that to calculate the next term one simply sums the preceding two: f(1)=1 f(2)=1 f(n)=f(n-1) + f(n-2) Thus 1 and 1 are 2, 1 and 2 are 3, 2 and 3 are 5 and so on. The above materials is retrieved July 12, 2009 from http://www.mathacademy.com/pr/prime/articles/fibonac/index.asp Now let's look at an Excel-generated Fibonacci Sequence following the above rule: Now let's construct the ratio of each pair of adjacent terms in the Fibonacci sequence as the following: a(0) = 1/1, a(1) = 2/1, a(n) = This gives us a new sequence: As we observe this sequence, we find that although the first several values oscillate up and down, very quickly the sequence approaches to a value close to 1.618034. In fact, this value is called the Golden Ratio. We may find the exact value of the golden ratio using two methods: Method 1: We know by the definition of Fibonacci Sequence that: Method 2: Next let's try to understand and express the exact Golden Ratio using a geometric method. Observe the following triangle: So far, we have considered the ratio of each pair of adjacent terms in the Fibonnaci sequence, but how about sequence , i.e., the ratios between every second terms? i.e., b(0) = 2/1, b(1) = 3/1, b(2)=5/2, : Here we observe that the sequence b(n) also converges to a number, in fact, close to 2.618034 or 1 plus the Golden Ratio! In fact, this observation can be easily proved mathematically: Now let's continue our investigation to , hopefully we will find some relations that we eventually may generalize to all such sequences: This time we observe that c(n) also approaches a number, close to 4.23608, and the behavior of this sequence is highly similar to the previous sequences. But what is the relationship between this number 4.23608 and the Golden Ratio? Let's investigate one more sequence and then we will see whether we get generalize our results to all such upcoming sequences: We notice that d(n) also converges to a number pretty much like our earlier sequences. This time the number is 6.854102. Based on our experience so far, we may guess that this number can also be expressed by the Golden Ratio like: Conclusion 1: b. Explore sequences where f(0) and f(1) are some arbitrary integers other than 1. If f(0)=1 and f(1)=3, then your sequence is a Lucas Sequence. All such sequences, however, have the same limit of the ratio of successive terms. In the second part of this investigation, we want to see whether the two starting values affect the limit of the ratio of successive terms. First, let's look at what happens if we change f(0)=1 and f(1)=3? This sequence in fact is not a Fibonaaci Sequence anymore. It is called a Lucas Sequence: We find that the change from f(1) = 1 to f(1) = 3 did not affect the limit of the ratio of successive terms. Let's try some other numbers, you may ask what happens if the first number is larger than the second number, will that matter? Our observation is that it doesn't matter! The limit of the ratio of successive terms still approaches to the same number as in the Fibonacci sequence! How about we start from 0? Let's look at the next example: Again it doesn't matter! Even if we start our sequence with 0, the limit of the ratio of successive terms still approaches to the same number as in the Fibonacci sequence! Conclusion 2: In fact, except for the case when both of our starting values are zeros, the initial values of all such sequences does not affect the limit of the ratio of successive terms. If you are really a skeptical person and say what happens if we start with negative values, with fractions? Check the following two examples and see if you are convinced! Big conclusion: I believe it is fairly safe to say that the initial values of all such sequences do not affect the limit of the ratio of successive terms! In other words, all such sequences will eventually converge to the same set of values, all related to the Golden Ratio. If you want to explore this investigation yourself, you may want to check out this excel file first. Good Luck!
# What is identity and inverse element? What is identity and inverse element? ## What is identity and inverse element? Evidently 1 is the identity of multiplication for Z (set of integers), Q (set of rational numbers), and R (set of real numbers). Inverse: An element a∈G is said to have its inverse with respect to certain operation ∗ if there exists b∈G such that. a∗b=e=b∗a. e being the identity in G with respect to a. What is the identity of a matrix? In linear algebra, an identity matrix is a matrix of order nxn such that each main diagonal element is equal to 1, and the remaining elements of the matrix are equal to 0. ### What is the relationship between identity and inverse? The reciprocal of a number is its multiplicative inverse. A number and its reciprocal multiply to 1 , which is the multiplicative identity. How do you write an identity matrix? Identity Matrix is denoted with the letter “In×n”, where n×n represents the order of the matrix. One of the important properties of identity matrix is: A×In×n = A, where A is any square matrix of order n×n. #### What is identity inverse? One more important point: the identity element is always its own inverse. For example, if e is the identity element, then e#e=e. So by definition, when e acts on itself on the left or the right, it leaves itself unchanged and gives the identity element, itself, as the result! What is difference between identity and inverse properties? The opposite of a number is its additive inverse. A number and its opposite add to 0, which is the additive identity. ## What is the simplest way to find an inverse matrix? Find the determinant • Find the matrix of minors • Find the matrix of co-factors • Transpose • Divide by the determinant • How do you solve an inverse matrix? – Estimate the determinant of the given matrix – Find the transpose of the given matrix – Calculate the determinant of 2 x 2 matrix. – Prepare the matrix of cofactors – At the last, divide each term of the adjugate matrix by the determinant ### Why do we need to find inverse of matrix? First of all,choose the size of the Square Matrix (for instance,4,5,etc.) • The matrix with input boxes of entered size will be displayed in front of you. Enter the elements of the matrix in their respective places. • Press on the “Calculate” button to find the resultant inverse matrix. • Do you really need to compute that matrix inverse? To calculate the inverse of a matrix, we have to follow these steps: First, we need to find the matrix of minors Now change that matrix into a matrix of cofactors Now find the adjoint of the matrix
# What are the six trigonometry functions? Trigonometry can be defined as the branch of mathematics that determines and studies the relationships between the sides of a triangle and angles subtended by them. Trigonometry is basically used in the case of right-angled triangles. Trigonometric functions define the relationships between the 3 sides and the angles of a triangle. There are 6 trigonometric functions mainly. Before going into the study of the trigonometric functions we will learn about the 3 sides of a right-angled triangle. The three sides of a right-angled triangle are as follows, • Base The side on which the angle Î¸ lies is known as the base. • Perpendicular It is the side opposite to the angle Î¸ in consideration. • Hypotenuse It is the longest side in a right-angled triangle and opposite to the 90Â° angle. ### Trigonometric Functions Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows, • sine It is represented as sin Î¸ and is defined as the ratio of perpendicular and hypotenuse. • cosine It is represented as cos Î¸ and is defined as the ratio of base and hypotenuse. • tangent It is represented as tan Î¸ and is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base. • cosecant It is the reciprocal of sin Î¸ and is represented as cosec Î¸. • secant It is the reciprocal of cos Î¸ and is represented as sec Î¸. • cotangent It is the reciprocal of tan Î¸ and is represented as cot Î¸. ### What are the six trigonometry functions? The six trigonometric functions have formulae for the right-angled triangles, the formulae help in identifying the lengths of the sides of a right-angled triangle, lets take a look at all those formulae, The below table shows the values of these functions at some standard angles, Note: It is advised to remember the first 3 trigonometric functions and their values at these standard angles for ease of calculations. ### Sample Problems Question 1: Evaluate sine, cosine, and tangent in the following figure. Solution: Given Using the trigonometric formulas for sine, cosine and tangent, Question 2: In the same triangle evaluate secant, cosecant, and cotangent. Solution: As it is known the values of sine, cosine and tangent, we can easily calculate the required ratios. Question 3: Given , evaluate sin Î¸.cos Î¸. Solution: Thus P=6, B=8 Using Pythagoras theorem, H2=P2+B2 H2=36+64=100 Therefore, H =10 Now, Question 4: If , evaluate tan2Î¸. Solution: Given Thus Question 5: In the given triangle, verify sin2Î¸+cos2Î¸ = 1 Solution: Given P=12, B=5, H=13 Thus Hence verified. Previous Next
# Difference between revisions of "2010 AIME I Problems/Problem 14" ## Problem For each positive integer n, let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$. Find the largest value of n for which $f(n) \le 300$. Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. ## Solution ### Solution 1 Observe that $f$ is strictly increasing in $n$. We realize that we need $100$ terms to add up to around $300$, so we need some sequence of $2$s, $3$s, and then $4$s. It follows that $n \approx 100$. Manually checking shows that $f(109) = 300$ and $f(110) > 300$. Thus, our answer is $\boxed{109}$. ### Solution 2 Because we want the value for which $f(n)=300$, the average value of the 100 terms of the sequence should be around $3$. For the value of $\lfloor \log_{10} (kn) \rfloor$ to be $3$, $1000 \le kn < 10000$. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$, so $50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500$, and $n = 110$. $f(110) = 301$, so we want to lower $n$. Testing $109$ yields $300$, so our answer is still $\boxed{109}$. ### Solution 3 For any $n$ where the sum is close to $300$, all the terms in the sum must be equal to $2$, $3$ or $4$. Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$). With this definition of $M$ and $N$ the total will be $400 - M - N \le 300$, from which $M + N \ge 100$. Now $M+1$ is the smallest integer $k$ for which $\log_{10}(kn) \ge 4$ or $kn \ge 10000$, thus $$M = \left\lfloor\frac{9999}{n}\right\rfloor.$$ Similarly, $$N = \left\lfloor\frac{999}{n}\right\rfloor = \left\lfloor\frac{M}{10}\right\rfloor.$$ Therefore, $$M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.$$ Since we want the largest possible $n$, the answer is $\boxed{109}$.
Search IntMath Close 450+ Math Lessons written by Math Professors and Teachers 5 Million+ Students Helped Each Year 1200+ Articles Written by Math Educators and Enthusiasts Simplifying and Teaching Math for Over 23 Years # 4. Graphs of tan, cot, sec and csc by M. Bourne The graphs of tan x, cot x, sec x and csc x are not as common as the sine and cosine curves that we met earlier in this chapter. However, they do occur in engineering and science problems. They are interesting curves because they have discontinuities. For certain values of x, the tangent, cotangent, secant and cosecant curves are not defined, and so there is a gap in the curve. [For more on this topic, go to Continuous and Discontinuous Functions in an earlier chapter.] Recall from Trigonometric Functions, that tan x is defined as: tan x=(sin x)/(cos x) Consider the denominator (bottom) of this fraction. For some values of x, the function cos x has value 0. For example, when x=pi/2, the value of cos {:π/2:} is 0, and when x=(3pi)/2, we have cos{:(3π)/2:}=0. When this happens, we have 0 in the denominator of the fraction and this means the fraction is undefined. So there will be a "gap" in the function at that point. This gap is called a discontinuity. The same thing happens with cot x, sec x and csc x for different values of x. For each one, the denominator will have value 0 for certain values of x. ## The Graph of y = tan x Sketch y = tan x. #### Solution As we saw above, tan x=(sin x)/(cos x) This means the function will have a discontinuity where cos x = 0. That is, when x takes any of the values: x = ..., -(5π)/2, -(3π)/2, -π/2, π/2, (3π)/2, (5π)/2, ... It is very important to keep these values in mind when sketching this graph. When setting up a table of values, make sure you include x-values either side of the discontinuities. Recall that -(3pi)/2=-4.7124 and -pi/2= -1.5708. So we take values quite near these discontinuities. x tan x -4.7 -4.5 -4 -3.5 -3.14 -1.58 -1.56 -1 0 0.5 1 1.5 1.57 -80.7 -4.6 -1.2 -0.37 0 108 -92 -1.6 0 0.55 1.6 14.1 1,256 Notice that either side of -pi/2, (our values of -1.58 and -1.56 in the table above), we jump from a large positive number (108), to a small negative number (-92). If we continue our table, we will get similar values (because this is a periodic graph). So we are ready to sketch our curve. Graph of y = tan x: Note that there are vertical asymptotes (the gray dotted lines) where the denominator of tan x has value zero. (An asymptote is a straight line that the curve gets closer and closer to, without actually touching it. You can see more examples of asymptotes in a later chapter, Curve Sketching Using Differentiation.) Note also that the graph of y = tan x is periodic with period π. This means it repeats itself after each π as we go left to right on the graph. ## Interactive Tangent Animation You can see an animation of the tangent function in this interactive. ### Things to do Using the sliders below the graph, you can change: 1. The amount of energy in the wave by changing the amplitude, a 2. The frequency the wave by changing b 3. The phase shift of the wave by changing c 4. The vertical displacement of the wave by changing d The units on the horizontal x-axis are radians (in decimal form). Recall that: So the (initial) graph shown is from -pi/2 to (7pi)/2. The vertical dashed lines are the asymptotes. The pink triangle that appears when you start the animation has base length = 1. The height of that triangle is the tan ratio of the current angle. You may notice the hypotenuse of the triangle is almost vertical when the graph goes off to ±∞. Graph: y = a tan(bx + c) + d = tan(x) (x, y) = a b c d (For more on periodic functions and to see y = tan x using degrees, rather than radians, see Trigonometric Functions of Any Angle.) ## The Graph of y = cot x Recall from Trigonometric Functions that: cot x=1/tanx = (cos x)/(sin x) We now have to consider when sin x has value zero, because this will determine where our asymptotes should go. The function will have a discontinuity where sin x = 0, that is, when x = ..., -3π, -2π, -π, 0, π, 2π, 3π, 4π, 5π, ... Considering the values of cos x and sin x for different values of x (or more simply, finding the values of 1/tanx), we can set up a table of values. We can then sketch the graph of y = cot x as follows. ## The Graph of y = sec x We could laboriously draw up a table with millions of values, or we could work smart and recall that sec x=1/(cos x) We know the sketch for y = cos x and we can easily derive the sketch for y = sec x, by finding the reciprocal of each y-value. (That is, finding 1/y for each value of y on the curve y = cos x.) For example (angles are in radians): x y = cos x 1/y = sec x 0 1 1 1 0.54 1.85 1.55 0.02 48.09 2 −0.42 −2.4 3 −0.99 −1.01 4 −0.65 −1.53 I included a value just less than π/2=1.57 so that we could get an idea of what goes on there. When cos x is very small, sec x will be very large. After applying this concept throughout the range of x-values, we can proceed to sketch the graph of y = sec x. First, we graph y = cos x and then y = sec x immediately below it. Compare the y-values in each of the 2 graphs and assure yourself they are the reciprocal of each other. y = cos x y = sec x We draw vertical asymptotes (the dashed lines) at the values where y = sec x is not defined. That is, when x = ..., -(5π)/2, -(3π)/2, -π/2, π/2, (3π)/2, (5π)/2, ... You will notice that these are the same asymptotes that we drew for y = tan x, which is not surprising, because they both have cos x on the bottom of the fraction. ### Exercise Sketch y = csc x We recall that csc x=1/(sin x) So we will have asymptotes where sin x has value zero, that is: x = ..., -3π, -2π, -π, 0, π, 2π, 3π, 4π, ... We draw the graph of y = sin x first and indicate with dashed lines where the graph has value 0: Next, we consider the reciprocals of all the y-values in the above graph (similar to what we did with the y = sec x table we created above). x y = sin x csc x = 1/(sin x) 0.01 0.01 100 0.5 0.48 2.09 pi/2 1 1 2 0.91 1.10 3 0.14 7.09 3.1 0.04 24.05 I chose values close to 0 and pi, and some values in between. The pattern will be similar for the region from pi to 2pi except it will be on the negative side of the axis. We continue on both sides and realise the pattern will repeat. Now for the graph of y = csc x: You may also be interested in: The next section in this chapter shows some Applications of Trigonometric Graphs.
# Interquartile Mean (IQM) / Midmean Descriptive Statistics > Interquartile Mean ## What is the Interquartile Mean? The interquartile mean (IQM) is the mean of the middle 50 percent of data in a data set. Unlike the “regular” arithmetic mean, it is resistant to outliers. The IQM is the mean of the interquartile range (IQR). ## How to Find the Interquartile Mean The calculation is different depending on if your data is divisible by 4 or not. ## Data is Divisible by Four Example question: Find the IQM for the following data set: 5 6 17 30 44 55 56 8 9 11 13 15 1 3 16 65 Step 1: Sort the data from smallest to largest: 1 3 5 6 8 9 11 13 15 16 17 30 44 55 56 65 Step 2: Discard the bottom 25% and top 25% of numbers. In other words, split the data set into quarters and remove the top and bottom quarters: 1 3 5 6 \| 8 9 11 13 \| 15 16 17 30 \| 44 55 56 65 Step 3: Find the mean of the remaining numbers: (8 + 9 + 11 + 13 + 15 + 16 + 17 + 30 ) / 8 = 14.875 That’s it! ## Data is NOT Divisible by Four Example question: Find the IQM for the following data set: 6 17 30 44 55 56 8 9 11 13 15 1 3 16 65 Step 1: Sort the data from smallest to largest: 1 3 6 8 9 11 13 15 16 17 30 44 55 56 65 Step 2: Divide the number of items in the set by four. The set has 15 items, so 15/4 = 3.75. Step 3: Remove the whole number (Step 2) from the bottom and the top of the set. For this example, the whole number is 3 (from 3.75): 1 3 6 8 9 11 13 15 16 17 30 44 55 56 65 which leaves: 8 9 11 13 15 16 17 30 44 Step 4: Figure out how many items are in the interquartile range. The IQR is the middle two quarters, so there would be 3.75 * 2 = 7.5 numbers. Step 5: Place parentheses around the middle set of numbers using the whole number from Step 4. In this example, the whole number is 7: 8 (9 11 13 15 16 17 30) 44 Step 6: Take the fractional part from Step 4 (.5 in this case) and divide it by two (because there are two numbers on the outside of the parentheses): .5/2 = .25 This means that the numbers 8 and 44 will each contribute 25% to the IQM. Step 7: Multiply the two “outside” numbers (8 and 44 in this case) by the fraction in Step 6: 8 * .25 = 2 44 * .25 = 11 Step 8: Replace the two outside numbers by the fractional numbers (Step 7) and find the mean. When dividing by “n”, use the number of items in the IQR from Step 4 (7.5 in this case), not the actual number count (9 in this example): 82 (9 11 13 15 16 17 30) 4411 = (2 + 9 + 11 + 13 + 15 + 16 + 17 + 30 + 11)/7.5 = 16.53. That’s it! ------------------------------------------------------------------------------ Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. If you'd rather get 1:1 study help, Chegg Tutors offers 30 minutes of free tutoring to new users, so you can try them out before committing to a subscription. If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\text{Set Builder Notation: } \{ x\|x\gt4 \} \\\text{Interval Notation: } (4,\infty)$ $\bf{\text{Solution Outline:}}$ Use the properties of inequality to solve the given inequality, $-16x\lt-64 .$ Write the answer in both set-builder notation and interval notation. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} -16x\lt-64 \\\\ x\gt\dfrac{-64}{-16} \\\\ x\gt4 .\end{array} Hence, the solution set is \begin{array}{l}\require{cancel} \text{Set Builder Notation: } \{ x\|x\gt4 \} \\\text{Interval Notation: } (4,\infty) .\end{array}
# College Algebra : Finding Zeros of a Polynomial ## Example Questions ### Example Question #1 : Finding Roots Find the roots of the function: Explanation: Factor: Double check by factoring: Therefore: ### Example Question #1 : Finding Roots Solve for x. x = 5 x = 4, 3 x = 5, 2 x = –4, –3 x = –5, –2 x = 5, 2 Explanation: 1) Split up the middle term so that factoring by grouping is possible. Factors of 10 include: 1 * 10= 10 1 + 10 = 11 2 * 5 =10 2 + 5 = 7 –2 * –5 = 10 –2 + –5 = –7 Good! 2) Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second. 3) Now pull out the common factor, the "(x-2)," from both terms. 4) Set both terms equal to zero to find the possible roots and solve using inverse operations. x – 5 = 0, x = 5 x – 2 = 0, x = 2 ### Example Question #1 : Finding Roots Solve for : Explanation: To solve for , you need to isolate it to one side of the equation. You can subtract the from the right to the left. Then you can add the 6 from the right to the left: Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6: Finally, you set each binomial equal to 0 and solve for : ### Example Question #1 : Finding Zeros Of A Polynomial Find the roots of the following quadratic expression: Explanation: First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor. This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b. So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping. Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out. Now we factor out the (3x + 4). Setting each factor = 0 we can find the solutions. So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}. ### Example Question #2 : Finding Zeros Of A Polynomial Find the roots of . Explanation: If we recognize this as an expression with form , with and , we can solve this equation by factoring: and and ### Example Question #1 : Finding Zeros Of A Polynomial If the following is a zero of a polynomial, find another zero. Explanation: When finding zeros of a polynomial, you must remember your rules. Without a function this may seem tricky, but remember that non-real solutions come in conjugate pairs. Conjugate pairs differ in the middle sign. Thus, our answer is: Explanation: Explanation:
Question # Evaluate the following integral:$\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}$ Hint: As we have no direct formula for the given integral. Therefore, substitute ${{e}^{x}}=u$ and express integral in terms of u. Then again substitute $u=\sec \theta$ and then finally solve the integral to get the required value. Here, we have to solve the integral $\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}$. Let us consider the integral given in the question, $I=\int{\sqrt{{{e}^{2x}}-1}\text{ }dx}$ Let us take ${{e}^{x}}=u$ We know that $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$ Hence, by differentiating both sides, we get, ${{e}^{x}}dx=du$ Or, $u\text{ }dx=du$ $\Rightarrow dx=\dfrac{du}{u}$ By substituting the value of ${{e}^{x}}=u$ and $dx=\dfrac{du}{u}$ in the given integral, we get, $I=\int{\sqrt{{{\left( u \right)}^{2}}-1}\dfrac{du}{u}}$ Or, $I=\int{\dfrac{\sqrt{{{u}^{2}}-1}}{u}du.....\left( i \right)}$ Now, let us take $u=\sec \theta$ We know that $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$ Therefore by differentiating both sides, we get $du=\sec \theta \tan \theta d\theta$ Now, by substituting $u=\sec \theta$ and $du=\sec \theta \tan \theta d\theta$ in equation (i), we get, $I=\int{\dfrac{\sqrt{{{\sec }^{2}}\theta -1}}{\sec \theta }.\sec \theta .\tan \theta \text{ }d\theta }$ Now, by canceling like terms and substituting ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$ in the above expression, we get $I=\int{\sqrt{{{\tan }^{2}}\theta }.\tan \theta d\theta }$ Or, $I=\int{{{\tan }^{2}}\theta \text{ }d\theta }$ Again, we know that ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$. Therefore we get, $I=\int{\left( {{\sec }^{2}}\theta -1 \right)d\theta }$ We know that $\int{{{\sec }^{2}}x\text{ }dx=\tan x}$ and $\int{kdx=kx}$ By using these, we get, $I=\tan \theta -\theta +C....\left( ii \right)$ We know that $u=\sec \theta$, so we get, $\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}$ $\tan \theta =\sqrt{{{u}^{2}}-1}$ Also, $\theta ={{\sec }^{-1}}\left( u \right)$ By substituting the value of $\tan \theta$ and $\theta$ in equation (ii), we get, $I=\sqrt{{{u}^{2}}-1}-{{\sec }^{-1}}\left( u \right)+C$ Now by substituting $u={{e}^{x}}$ in the above equation, we get, $I=\sqrt{{{\left( {{e}^{x}} \right)}^{2}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C$ Or, $I=\sqrt{{{e}^{2x}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C$ Therefore, we get $\int{\sqrt{{{e}^{2x}}-1}dx=\sqrt{{{e}^{2x}}-1}-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C}$ Hence, option (b) is the right answer. Note: Here, students can cross-check their answers by differentiating the answer and checking if it is giving the original integral or not. Also, students can remember these substitutions in these cases to easily solve the questions of this type. Also, students should always convert the answer of the given integral in its original variable like here we have converted $\theta$ back to x at the end.
# NCERT Solutions for Class 9 Maths Exercise 13.4 Chapter 13 Surface Area And Volume NCERT Solutions for Class 9 Maths Chapter 13 â€“ Surface Areas and Volumes Exercise 13.4, includes step-wise solved problems from the NCERT textbook. The NCERT solutions are created by Maths subject experts along with proper geometric figures and explanations in a step-by-step procedure for good understanding. All the NCERT solutions for Science and Maths subjects are made available in PDF format, hence students can download them easily. The NCERT solutions for class 9 mathsÂ are prepared as per the latest NCERT guidelines and syllabus. It is intended to help the students to score good marks in first and second term and competitive examinations. ### Access Other Exercise Solutions of Class 9 Maths Chapter 13 â€“ Surface Areas and Volumes Exercise 13.1 solution (9 questions) Exercise 13.2 solution (8 questions) Exercise 13.3 solution (9 questions) Exercise 13.5 solution (5 questions) Exercise 13.6 solution (8 questions) Exercise 13.7 solution (9 questions) Exercise 13.8 solution (10 questions) Exercise 13.9 solution (3 questions) ### Access Answers to NCERT Class 9 Maths Chapter 13 â€“ Surface Areas and Volumes Exercise 13.4 1. Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm (Assume Ï€=22/7) Solution: Formula: Surface area of sphere (SA) = 4Ï€r2 (i) Radius of sphere, r = 10.5 cm SA = 4Ã—(22/7)Ã—10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4Ã—(22/ 7)Ã—5.62 = 394.24 Surface area of sphere is 394.24 cm2 (iii) Radius of sphere, r = 14cm SA = 4Ï€r2 = 4Ã—(22/7)Ã—(14)2 = 2464 Surface area of sphere is 2464 cm2 2. Find the surface area of a sphere of diameter: (i) 14cm (ii) 21cm (iii) 3.5cm (Assume Ï€ = 22/7) Solution: (i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm Formula for Surface area of sphere = 4Ï€r2 = 4Ã—(22/7)Ã—72 = 616 Surface area of a sphere is 616 cm2 (ii) Radius (r) of sphere = 21/2 = 10.5 cm Surface area of sphere = 4Ï€r2 = 4Ã—(22/7)Ã—10.52 = 1386 Surface area of a sphere is 1386 cm2 Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2 (iii) Radius(r) of sphere = 3.5/2 = 1.75 cm Surface area of sphere = 4Ï€r2 = 4Ã—(22/7)Ã—1.752 = 38.5 Surface area of a sphere is 38.5 cm2 3. Find the total surface area of a hemisphere of radius 10 cm. [Use Ï€=3.14] Solution: Radius of hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3Ï€r2 = 3Ã—3.14Ã—102 = 942 The total surface area of given hemisphere is 942 cm2. 4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. Solution: Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon) = 4Ï€r12/4Ï€r22 = (r1/r2)2 = (7/14)2 = (1/2)2 = Â¼ Therefore, the ratio between the surface areas is 1:4. 5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume Ï€ = 22/7) Solution: Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2Ï€r2 = 2Ã—(22/7)Ã—(5.25)2 = 173.25 Surface area of hemispherical bowl is 173.25 cm2 Cost of tin-plating 100 cm2 area = Rs 16 Cost of tin-plating 1 cm2 area = Rs 16 /100 Cost of tin-plating 173.25 cm2 area = Rs. (16Ã—173.25)/100 = Rs 27.72 Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72. 6. Find the radius of a sphere whose surface area is 154 cm2. (Assume Ï€ = 22/7) Solution: Let the radius of the sphere be r. Surface area of sphere = 154 (given) Now, 4Ï€r2 = 154 r2 = (154Ã—7)/(4Ã—22) = (49/4) r = (7/2) = 3.5 The radius of the sphere is 3.5 cm. 7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. Solution: If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of moon = Â½Ã—d/4 = d/8 Surface area of moon = 4Ï€(d/8)2 Surface area of earth = 4Ï€(d/2)2 The ratio between their surface areas is 1:16. 8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume Ï€ =22/7) Solution: Given: Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2Ï€r2, where r is radius of hemisphere = 2Ã—(22/7)Ã—(5.25)2 = 173.25Â cm2 Therefore, the outer curved surface area of the bowl is 173.25 cm2. 9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in(i) and (ii). Solution: (i) Surface area of sphere = 4Ï€r2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r CSA of cylinder formula = 2Ï€rh = 2Ï€r(2r) (using value of h) = 4Ï€r2 (iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder) = 4Ï€r2/4Ï€r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1. Exercise 13.4 of Class 9 Maths involves application level real-time problems that help students to think and apply the relevant formula. It helps to apply the total surface area of a sphere and hemisphere. Learn the NCERT solutions of class 9 maths chapter 13 along with other learning materials and notes provided by BYJUâ€™S. The problems are solved in a detailed way with relevant formulas and figures, to score well in first and second term exams. ### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 â€“ Surface Areas and Volume Exercise 13.4 • These NCERT SolutionsÂ help you solve and revise all questions of Exercise 13.4. • Helps to find the radius of a sphere and the total surface area of hemisphere. • It follows NCERT guidelines which help in preparing the students accordingly. • Stepwise solutions given by our subject expert teachers will help you to secure more marks.
# 1st Class Mathematics Subtraction Subtraction of Images ## Subtraction of Images Category : 1st Class ### Subtraction of Images It means taking out picture or pictures from group of pictures and finding the remaining as the result of the subtraction. Take another example: Here, we count the flowers given before the sign of minus and after the sign of minus. We subtract it through the normal method of subtraction. (a) (b) (c) (d) Explanation Subtraction in Column Subtraction in column is done by subtracting the numbers in column. See the given below examples: 27 19 39 -17 24 -12 19 -8 08 22 12 11 49 -36 42 -27 37 -17 13 15 20 31 -15 65 -56 19 -7 16 09 12 38 -17 39 -15 28 -7 21 24 11 Word Problem Based on Subtraction In this section on we will learn about some real life problems based on subtraction. James purchased 18 balloons on his birthday and gave 1 balloon to each of his friends. He has 7 friends. How many balloons left with him? (a) 11 balloons (b) 10 balloons (c) 12 balloons (d) None of these Explanation balloons Smith is 4 years younger than Lina. If the age of Lina is 9 years then find the age of Smith? (a) 4 years (b) 5 years (c) 6 years (d) None of these Explanation years A number isless than. Find the number. (a) 15 (b) 18 (c) 16 (d) 12 Explanation In a magic box if you put a number, three times. 3 is subtracted from it. Serena putsin this magic box. Which one of the following numbers she will get? (a) 14 (b) 15 (c) 16 (d) 18 Explanation A student brought 1 dozen toffees on his birthday in the school. He distributed half dozen toffees among his friends. How many toffees left with him? (a) 4 toffees (b) 6 toffees (c) 8 toffees (d) None of these Explanation dozen toffees =toffees dozen toffees =toffees toffees Smith says that he has a number. If you subtract 5 from it, the result will be 7. Find the number. (a) 11 (b) 13 (c) 15 (d) 12 Explanation In a school there arestudents inand students in. Which section has more students and by how much? (a) Section A by 9 (b) Section C by 9 (c) Section A by 10 (d) Section C by 10 Explanation Sectionhasmore students than section. There are mangoes in the basket. William eats mangoes daily. How many mangoes are left after days? (a) 10 mangoes (b) 8 mangoes (c) 12 mangoes (d) 15 mangoes Explanation Mangoes eaten by William indays Mangoes leftmangoes. A number when subtracted from gives the result. Find the number. (a) 11 (b) 12 (c) 9 (d) None of these Explanation The sum of two numbers is. If bigger number isthen find the smaller number. (a) 8 (b) 7 (c) 6 (d) None of these Explanation Mixed Problems In some questions you have to perform both addition and subtraction. See the given below examples: Solve To solve this sum, first add the numbers with + sign. That is Now subtractfrom. That is. So the answer is. Solve To solve this sum first writethen add all the numbers with negative sign. That is Now subtractfrom. That is. Solve To solve this sum first write Then add all the numbers with negative sign. That is . Now subtractfrom. That is. Solve To solve this sum first add the numbers with + sign. That is . Then add the numbers with - sign. That is Now subtract from.that is . Steve takes a numberand adds 9 to it and then subtractfrom it. Find the number. To solve this sum first addand. That is. Now subtractfrom. That is. Steve has + + and Smith has + , find how much more money Steve has than Smith? To solve this sum first add the money Steve has. That is = 17. Now add the money Smith has. That is7. Now subtract from . That is - = . So Steve has more than Smith. • Subtraction means taking away number of objects from a given collection. • If you subtract 0 from a number the number remains unchanged. • Subtraction is the process of taking away smaller unit from bigger one. • If zero is subtracted from any number the result is number itself. . • Result of subtraction is called subtraction fact.That is • Sign of subtraction is '_' and it is read as 'minus'. • If any number is subtracted from itself, the result is 0. . • If 0 is subtracted from the number then the result is number itself. For example; #### Other Topics You need to login to perform this action. You will be redirected in 3 sec
Question A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two of the balls are 1.5 cm and 2 cm respectively. Find the radius of the third ball. Hint: We are going to solve the given problem by using the formula of finding the volume of a spherical object. It is known that the volume of a sphere with radius $r$ is given by $\frac{4}{3}\pi {r^3}$ Let the radius of the original ball be $R$. So the volume of the original spherical ball with $R = 3$ is equal to $\frac{4}{3}\pi {(3)^3}$ Now the original ball is melted to form three balls of different sizes. Let the radii of three resulting spherical balls be ${r_1},{r_2},{r_3}$ respectively. Given ${r_1} = 1.5cm$ & ${r_2} = 2cm$. $\Rightarrow {V_{total}} = {V_1} + {V_2} + {V_3}$ $\Rightarrow \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi ({r_1}^3 + {r_2}^3 + {r_3}^3)$ $\Rightarrow {r_3}^3 = {3^3} - {1.5^3} - {2^3} = 15.625$ Hence ${r_3} = 2.5cm$ $\therefore$The radius of the third ball = 2.5cm Note: In the given problem it is said that the spherical ball of radius 3cm is melted to form three spherical balls. So the volume of the original spherical ball is equal to the sum of the volumes of the three balls.
<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. # Addition of Fractions ## Add fractions and mixed numbers with like and unlike denominators. Estimated10 minsto complete % Progress Practice Addition of Fractions Progress Estimated10 minsto complete % Lily and Howard ordered a pizza that was cut into 8 slices. Lily ate 3 slices and Howard ate 4 slices. What fraction of the pizza did each person eat? What fraction of the pizza did they eat all together? ### Guidance The problem above can be represented using fraction strips. To add fractions, the fractions must have the same bottom numbers (denominators). In this case, both fractions have a denominator of 5. The answer is the result of adding the top numbers (numerators). The numbers in the numerator are 1 and 2. The sum of 1 and 2 is 3. This sum is written in the numerator over the denominator of 5. Therefore . A number line can also be used to show the addition of fractions, as you will explore in Example C. The sum of two fractions will sometimes result in an answer that is an improper fraction. An improper fraction is a fraction that has a larger numerator than denominator. This answer can be written as a mixed number. A mixed number is a number made up of a whole number and a fraction. In order to add fractions that have different denominators, the fractions must be expressed as equivalent fractions with a least common denominator (LCD). The sum of the numerators can be written over the common denominator. Solution: #### Example B Louise is taking inventory of the amount of water in the water coolers located in the school. She estimates that one cooler is full and the other is full. What single fraction could Louise use to represent the amount of water of the two coolers together? Solution: Use fraction strips to represent each fraction. and are equivalent fractions. . and are equivalent fractions. . The two green pieces will be replaced with eight purple pieces and the one blue piece will be replaced with three purple pieces. The amount of water in the two coolers can be represented by the single fraction . The denominator of 12 is the LCD (least common denominator) of and because it is the LCM (least common multiple) of the numbers 3 and 4. #### Example C What is ? Solution: The number line is labeled in intervals of 4 which indicates that each interval represents . From zero, move to the number 2 plus 3 more intervals to the right. Mark the location. This represents . From here, move to the right or of 4, which is 2 intervals. The sum of and is . #### Concept Problem Revisited Lily ate of the pizza because she ate 3 out of the 8 slices. Howard ate (or ) of the pizza. Together they ate 7 slices which is of the pizza. ### Vocabulary Denominator The denominator of a fraction is the number on the bottom that indicates the total number of equal parts in the whole or the group. has denominator 8. Fraction A fraction is any rational number that is not an integer. Improper Fraction An improper fraction is a fraction in which the numerator is larger than the denominator. is an improper fraction. LCD The least common denominator is the lowest common multiple of the denominators of two or more fractions. The least common denominator of and is 20. LCM The least common multiple is the lowest common multiple that two or more numbers share. The least common multiple of 6 and 4 is 12. Mixed Number A mixed number is a number made up of a whole number and a fraction such as . Numerator The numerator of a fraction is the number on top that is the number of equal parts being considered in the whole or the group. has numerator 5. ### Guided Practice 1. 2. 3. 1. 2. 3. is an improper fraction. An improper fraction is one with a larger numerator than denominator. plus there is left over. This can be written as a whole number and a fraction . This representation is called a mixed number. ### Practice Complete the following addition problems using any method. For each of the following questions, write an addition statement and find the result. Express all answers as either proper fraction or mixed numbers. 1. Karen used of flour to make cookies. Jenny used of flour to make a cake. How much flour did they use altogether? 2. Lauren used of milk, of flour and of oil to make pancakes. How many cups of ingredients did she use in total? 3. Write two fractions with different denominators whose sum is . 4. Allan’s cat ate of food in one week and the next week. How many cans of food did the cat eat in two weeks? 5. Amanda and Justin each solved the same problem. Amanda’s Solution: Justin’s Solution: Who is correct? What would you tell the person who has the wrong answer? ### Vocabulary Language: English Denominator Denominator The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$. Equivalent Fractions Equivalent Fractions Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number. improper fraction improper fraction An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator. inequality inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$. LCD LCD The least common denominator or lowest common denominator (LCD) of two fractions is the smallest number that is a multiple of both of the original denominators. LCM LCM The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both of the original numbers. Least Common Denominator Least Common Denominator The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators. Least Common Multiple Least Common Multiple The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers. Mixed Number Mixed Number A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
## Derivative and the Maximum Area Problem Note: This is the third part of the Derivative Concept Series. The first part is The Algebraic and Geometric Meaning of Derivative and the second part is Derivative in Real Life Context. Introduction The computation of derivative is often seen in maximum and minimum problems. In this article, we will discuss why do we get the derivative of a function and equate it to 0 when we want to get its maximum or minimum. To give you a concrete example, let us consider the problem below. Find the maximum area a rectangle with perimeter 10 units. Without using calculus, we can substitute values for the rectangle’s length, compute for its width and its corresponding area. If we set the interval to 0.5, then we can come up with the table shown in Figure 1. Figure 1 - Table showing the length, width, and area of a rectangle with perimeter 10. Looking at the table above, we can observe that a rectangle of length of 2.5, a square, has the maximum area. If we have prior calculus knowledge, however, we know that whatever the value of our perimeter, a square having the given perimeter will always have the maximum area. Using elementary algebra, if we let $x$ be the width of our rectangle, it follows that the length is $5-x$. Let $f(x)$ be the area of the rectangle. In effect, the area of the rectangle is described by the equation $f(x) = 5x - x^2$. We want to maximize the area, which implies that we want to find the maximum value of $f(x)$. Figure 2 – A rectangle with Perimeter 10 and width x units. In elementary calculus, to compute for the maximum value of $f(x)$, we get its derivative, which is equal to $5 - 2x$, which we will denote $f'(x)$. We then equate the $f'(x)$ to $0$ resulting to the equation $5-2x=0 \Rightarrow x = 2.5$ which is exactly the maximum value in the table above. Derivative and Equation to 0 In the article the Algebraic and Geometric Meaning of Derivative, we have learned that the derivative of a function is the slope of the line tangent to that function at a particular point. From elementary algebra, we also have learned the properties of slopes. If a line is rising to the right, the slope is greater than 0; if the line is rising to the to the left, then the slope is less than 0. We have also learned that a horizontal line has slope 0 and the vertical line has an undefined slope. Figure 3 – Properties of slope of a straight line. In the problem above, we calculated by getting the derivative (the slope of the line tangent to a function at a particular point) and equate it to $0$. But a line with slope $0$ is a horizontal line. In effect, we are looking for a horizontal tangent of $f(x) = 5x-x^2$. To give a clearer picture let us look at the graph of $f(x) = 5x - x^2$. Figure 4 – Tangent lines of 5x – x2. From the graph it is clear that the maximum point of the function is where the tangent line (red line) horizontal. In fact, there are only three possible cases that tangent line could be horizontal as shown in Figure 5: first, the minimum of a function (blue graph); second, the inflection point (red graph); and the third is the maximum of the function (green graph). It should also be noteworthy to say that all the ordered pairs (length, area) or(width, area) in Figure 1 will be on the blue curve in Figure 4. Figure 5 – Cases of a graph where the tangent is horizontal. The derivative has many applications and it is seen in many topics in calculus. In the next Derivative Tutorial, we are going to discuss how the derivative is used in other context. Summary • The derivative of a function is the slope of the line tangent to a function at a particular point. • The horizontal line has slope zero. • In solving maxima and minima problems, we get the derivative of a function and equate to zero to get the minimum or maximum. We do this because geometrically, we want to get the line tangent to a function at a particular point that is horizontal. ## Derivative in Real Life Context Note: This article is the second part of the derivative concept series. The first part is The Algebraic and Geometric Meaning of Derivative and the third part is Derivative and the Maximum Area Problem. If we change the labels of the Figure 2 in the The Algebraic and Geometric Meaning of Derivative article – its x-axis to time and the y-axis to distance – the graph of the secant line is the difference in distance $y_2 - y_1$ = 100 km – 50 km over the difference in time$x_2 - x_1$ = 10:00 – 8:00. This is shown in the figure below. This is equivalent to 50 km/2 hrs or 25 km/hr. From the computation above, it is clear that the interpretation of the slope of the secant line is the total distance over the total time or the average speed. On the other hand, the slope of the tangent line is the speed of the bicycle at exactly 4 o’clock. At exactly 4:00 o’clock the bicycle was traveling 50 km per hour, a lot faster than its average speed. Now, this is reasonable because in real life, the speed of travel is not always constant. The slope of the tangent line or the speed of travel at a particular point is called the instantaneous speed. We have also observed from the previous article that as Q approaches P, the secant line’s approximation of the tangent line becomes better and better. This means that as Q approaches P, the average speed becomes closer and closer to the instantaneous speed at P. ## The Algebraic and Geometric Meaning of Derivative Note: This is the first part of the Derivative Concept Series. The second part is Derivative in Real Life Context and the third part is Derivative and the Maximum Area Problem. *** If we want to get the slope of a line, we need two points. Suppose the points have coordinates $(x_1,y_1)$ and $(x_2,y_2)$, we have learned that the slope is described by the formula $\displaystyle\frac{y_2-y_1}{x_2-x_1}$. In Figure 1, we have line $l$ tangent to the function $f$ at point $P$ where the coordinates of $P$ are $(x,f(x))$. The problem that gave birth to calculus is getting the slope of this tangent line. There is, however, a problem. We need two points to compute for the slope but we have only one point. Note that the word tangent in this problem is different from the definition of tangent on a circle because it is clear that line $l$ will intersect the graph in more than one point. Figure 1 - Line l tangent to the function f at point P. Using the concept of limits we can remedy this problem. First, we create point $Q$ with x coordinate $h$ units to the right of the x-coordinate of $P$. We then draw line $PQ$, a secant line to the function $f$. Figure 2 - A secant line is drawn through P. In effect, the coordinates of $Q$ would be $(x+h, f(x+h))$ and it is clear that the slope of the secant line $PQ$ is described by the formula $\displaystyle\frac{f(x+h)-f(x)}{x+h - x} = \frac{f(x+h)-f(x)}{h}$ If we want to approximate the slope of the tangent line, it is reasonable that we move $Q$ towards $P$ with $P$ fixed. Click here to explore the diagram above using GeoGebra. From the GeoGebra exploration above, if we move $Q$ towards $P$, we observe the following: 1.) The value of $h$ approaches $0$. 2.) The inclination of the secant line approaches the inclination of the tangent line. 3.) The slope of the secant line approaches the slope of the tangent line. 4.) If point $Q$ coincides with point $P$, then the slope of the secant line and is equal to the slope of the tangent line. If we let $m$ be the slope of the secant line and$f'(x)$ be the slope of the tangent line, focusing on observations 1 and 4, we can say the following equivalent statements: • The limit of the slope of the secant line $m$ as $Q$ approaches $P$ is equal to $f'(x)$. • The limit of the slope of the secant line $m$ as $h$ approaches $0$ is equal to $f'(x)$. • The limit of $\displaystyle \frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$ is equal to $f'(x)$. Using the limit notation, we can say that $f'(x) = \lim_{h \to 0} \displaystyle\frac{f(x+h)-f(x)}{h}$ From the above discussion, we can see that the derivative of a function at a particular point is the slope of the line tangent to that function at that particular point. In the next post, we will discuss the meaning of derivative in real life situations. 1 4 5 6 7
GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 29 May 2020, 09:12 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # 10! + 9! can be rewritten as: Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 64242 10! + 9! can be rewritten as: [#permalink] ### Show Tags 25 Apr 2017, 01:16 11 00:00 Difficulty: 35% (medium) Question Stats: 63% (01:24) correct 37% (01:24) wrong based on 182 sessions ### HideShow timer Statistics 10! + 9! can be rewritten as: A. 9!(10) B. 11!/10 C. (10!)^29 D. (9!)^2 + 10 E. 19! _________________ GMAT Club Legend Joined: 11 Sep 2015 Posts: 4878 GMAT 1: 770 Q49 V46 Re: 10! + 9! can be rewritten as: [#permalink] ### Show Tags 07 Mar 2018, 09:16 4 Top Contributor 1 Bunuel wrote: 10! + 9! can be rewritten as: A. 9!(10) B. 11!/10 C. (10!)^29 D. (9!)^2 + 10 E. 19! First recognize that 10! = (10)(9!) Here's why: 10! = (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = (10)(9!) So.... 10! + 9! = (10)(9!) + 9! = 9!(10 + 1) [I factored out the 9!] = 9!(11) Check the answer choice.....9!(11) isn't there! So, we'll need to check each answer choice to see which one can be expressed as (9!)(11) A. 9!(10) We can clearly see that this is not equal to (9!)(11) ELIMINATE A B. 11!/10 11!/10 = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)/(10) = (11)(9)(8)(7)(6)(5)(4)(3)(2)(1) = (11)(9!) PERFECT!! = B Cheers, Brent _________________ Test confidently with gmatprepnow.com ##### General Discussion Intern Joined: 13 Sep 2015 Posts: 25 Re: 10! + 9! can be rewritten as: [#permalink] ### Show Tags 25 Apr 2017, 01:27 1 10! + 9! = (109!)+ 9! =9! (10+1) =9!(11) =(9!1110)/10 =11!/10 Hence B Sent from my D6502 using GMAT Club Forum mobile app Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2800 Re: 10! + 9! can be rewritten as: [#permalink] ### Show Tags 12 Mar 2018, 09:55 Bunuel wrote: 10! + 9! can be rewritten as: A. 9!(10) B. 11!/10 C. (10!)^29 D. (9!)^2 + 10 E. 19! 10! + 9! = 10 x 9! + 9! = 9!(10 + 1) = 9!(11) = 11!/10 _________________ # Jeffrey Miller Jeff@TargetTestPrep.com 214 Reviews 5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE NOW WITH GMAT VERBAL (PRE-BETA) See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Non-Human User Joined: 09 Sep 2013 Posts: 15016 Re: 10! + 9! can be rewritten as: [#permalink] ### Show Tags 09 Feb 2020, 06:34 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: 10! + 9! can be rewritten as: [#permalink] 09 Feb 2020, 06:34
# Whole Numbers In this article, we are learning about Whole Numbers as per the Primary $5$ Maths level. The learning objectives are: 2. Multiplying or Dividing by $10, 100$ or $1000$ and their multiples ## 1. Reading And Writing Numbers Whole numbers include zero and counting numbers. They are without fractions. In Primary $5$, we will learn Numbers up to $10$ million. Let’s place the digits in $675 \;840$ in their respective place values. We read $675 \;840$ as six hundred and seventy-five thousand, eight hundred and forty. The place value of the digit $6$ is hundred thousands. The value of the digit $6$ is $600 \;000$. The place value of the digit $7$ is ten thousands. The value of the digit $7$ is $70 \;000$. The place value of the digit $5$ is thousands. The value of the digit $5$ is $5000$. The place value of the digit $8$ is hundreds. The value of the digit $8$ is $800$. The place value of the digit $4$ is tens. The value of the digit $4$ is $40$. Remember the ‘s’ for the place values! Let us try placing the digits in $9 \;675 \;840$ in their respective place values. We read $9 \;675 \;840$ as nine million, six hundred and seventy-five thousand, eight hundred and forty. The place value of the digit $9$ is $\text{millions}$. The value of the digit $9$ is $9\;000\;000$. It is a good habit to include spaces at the appropriate places when writing numbers in numerals. Remember to include a comma and hyphen(s) at the appropriate places when writing numbers in words. Question 1: Write the following in numerals. Solution: Place the digits in a place value chart as shown below. Three million, three hundred thousand, four hundred and thirty-five $= 3 \;300\;435$ $3\;300\;435$ Question 2: Write the following number in numerals. Solution: Place the digits in a place value chart as shown below. $8\;450\;207$ Question 3: Write the following in numerals. Solution: Place the digits in a place value chart as shown below. $4\;015\;068$ Question 4: Write the following in words. Solution: Place the digits in a place value chart as shown below. $3\;051\;043 =$ Three million, fifty-one thousand and forty-three Three million, fifty-one thousand and forty-three Question 5: Write in words the following number. Solution: Place the digits in a place value chart as shown below. $7\;250\;025 =$ Seven million, two hundred and fifty thousand and twenty-five Seven million, two hundred and fifty thousand and twenty-five Question 6: Write the following in words. Solution: Place the digits in a place value chart as shown below. $4\;325\;630 =$ Four million, three hundred and twenty-five thousand, six hundred and thirty Four million, three hundred and twenty-five thousand, six hundred and thirty ## 2. Multiplying or Dividing by 10, 100 or 1000 and their multiples ### Multiplying by 10, 100 or 1000 When we multiply a whole number by $10$, we add one ‘0’ after the number. Example: $5 × \textbf{10} = 50$ When we multiply a whole number by $100$, we add two $'0'$s after the number. Example: $5 × \textbf{100} = 500$ When we multiply a whole number by $1000$, we add three $'0'$s after the number. Example: $5 × \textbf{1000} = 5000$ Question 1: Match. Solution: ### Multiplying by the multiples of 10, 100 or 1000 Now let’s multiply a number by the multiples of $10, 100$ or $1000$. When we multiply a whole number by a multiple of $10$, we express the multiple of $10$ as a product of a number and $10$ Example: \begin{align} 12 \times \textbf{50} &= 12 \times \textbf{5} \times \textbf{10}\\[2ex] &= 60 \times 10\\[2ex] &= 600 \end{align} When we multiply a whole number by a multiple of $100$, we express the multiple of $100$ as a product of a number and $100$ Example: \begin{align} 12 \times \textbf{500} &= 12 \times \textbf{5} \times \textbf{100}\\[2ex] &= 60 \times 100\\[2ex] &= 6000 \end{align} When we multiply a whole number by a multiple of $1000$, we express the multiple of $1000$ as a product of a number and $1000$ Example: \begin{align} 12 \times \textbf{5000} &= 12 \times \textbf{5} \times \textbf{1000}\\[2ex] &= 60 \times 1000\\[2ex] &= 60\,000 \end{align} Question 1: Do the following multiplication. A. $27 × 1000 =$ __________ B. $27 × 5000 =$ __________ Solution: A. $27 × 1000 = 27\,000$ B. \begin{align} 27 \times \textbf{5000} &= 27 \times \textbf{5} \times \textbf{1000}\\[2ex] &= 135 \times 1000\\[2ex] &= 135\,000 \end{align} A. $27\;000$ B. $135\;000$ Question 2: Do the following multiplication. $319 × 6000 =$ _________ Solution: \begin{align} 319 × 6000 &= 319 × 6 × 1000\\[2ex] &= 1\;914\;000 \end{align} $1\;914\;000$ Question 3: Do the following multiplication. $700 × 9000 =$ _________ Solution: \begin{align} 7 × 100 × 9 × 1000 &= 7 × 9 × 100 × 1000 \\[2ex] &= 6\;300\;000 \end{align} $6 \,300 \,000$ Question 4: Fill in the blanks. A. $12 \;\times$ __________ $= 12 \,000$ B. $12 \;\times$ __________ $= 120 \,000$ Solution: A. $\color{#F00}{12} \times \underline{\quad1000\quad} = \color{#F00}{12} \,000$ B. $\color{#F00}{12} \times \underline{\quad10\;000\quad} = \color{#F00}{12}0 \,000$ A. $1000$ B. $10 \;000$ Question 5: Fill in the blanks $60\;\times$ __________ $= 720$ Solution: $6\color{#F00}{0} \;\times$ ________ $= 72\color{#F00}{0}$ $6\color{#F00}{0} \times \underline{\quad12\quad} = 72\color{#F00}{0}$ $12$ ### Dividing by 10, 100 or 1000 When we divide a whole number by $10$, we remove one $''0’'$ after the number. Example: $70 \;00\textbf{0 ÷ 10} = 7000$ When we divide a whole number by $100$, we remove two $''0’'$s after the number. Example: $70 \;0\textbf{00 ÷ 100} = 700$ When we divide a whole number by $1000$, we remove three $''0’'$s after the number. Example: $70 \textbf{ 000 ÷ 1000} = 70$ Question 1: Do the following division. $880 ÷ 10 =$ __________ Solution: $880 ÷ 10 = 88$ $88$ Question 2: Do the following division. $293 \;000 ÷ 100 =$ __________ Solution: $293 \;000 ÷ 100 = 2930$ $2930$ Question 3: Do the following division. $630 \;000 ÷ 1000 =$__________ Solution: $630 \;000 ÷ 1000 = 630$ $630$ ## Dividing by the multiples of 10, 100 or 1000 Now let’s divide a number by the multiples of $10$, $100$ or $1000$. When we divide a whole number by a multiple of $10$, we break the multiple of $10$ into $10$ and a number. We will divide by $10$ first, followed by the number. Example: \begin{align} 30 \;000 \div \textbf{60} &= 30 000 ÷ \textbf{10 ÷ 6}\\[2ex] &= 3000 ÷ 6\\[2ex] &= 500 \end{align} When we divide a whole number by a multiple of $100$, we break the multiple of $100$ into $100$ and a number. We will divide by $100$ first, followed by the number. Example: \begin{align} 30 \;000 \div \textbf{600} &= 30 \;000 ÷ \textbf{100 ÷ 6}\\[2ex] &= 300 ÷ 6\\[2ex] &= 50 \end{align} When we divide a whole number by a multiple of $1000$, we break the multiple of $1000$ into $1000$ and a number. We will divide by $1000$ first, followed by the number. Example: \begin{align} 30 \;000 \div \textbf{6000} &= 30 000 ÷ \textbf{1000 ÷ 6}\\[2ex] &= 30 ÷ 6\\[2ex] &= 5 \end{align} Question 1: Do the following division. $36 600 ÷ 60 =$ __________ Solution: \begin{align} 36 \;600 ÷ 60 &= 36 \;600 ÷ 10 ÷ 6\\[2ex] &= 3660 ÷ 6\\[2ex] &= 610 \end{align} $610$ Question 2: Do the following division. $44 \;800 ÷ 800 =$ __________ Solution: \begin{align} 44\;800 ÷ 800 &= 44 \;800 ÷ 100 ÷ 8\\[2ex] &= 448 ÷ 8\\[2ex] &= 56 \end{align} $56$ Question 3: Do the following division. $630 \;000 ÷ 9000 =$ __________ Solution: \begin{align} 630 \;000 ÷ 9000 &= 630 \;000 ÷ 1000 ÷ 9\\[2ex] &= 630 ÷ 9\\[2ex] &= 70 \end{align} $70$ Question 4: Fill in the blanks with the correct answer. $8400 \;÷\;$ __________ $= 70$ Solution: \begin{align} 8400 ÷ 70 &= 8400 ÷ 10 ÷ 7\\[2ex] &= 840 ÷ 7\\[2ex] &= 120 \end{align} $120$ ## Conclusion In this article, we learnt about the Whole Numbers as per the Primary $5$ Math level. We learnt the following subtopics in Whole Numbers: • Reading and Writing Numbers up to $10$ million • Multiplication and Division by $10$, $100$, or $1000$ and their multiples. Continue Learning Volume Of A Liquid Decimals - Operations & Conversions Ratio: Introduction Average - Formula Percentage, Fractions And Decimals Whole Numbers Strategy - Equal Stage Angle Properties Table Rates Whole Number Strategy: Gap & Difference Fractions - Addition & Subtraction Ratio Strategy: Repeated Identity Primary Secondary Book a free product demo Suitable for primary & secondary Our Education Consultants will get in touch with you to offer your child a complimentary Strength Analysis. Book a free product demo Suitable for primary & secondary our educational content Start practising and learning. No Error No Error *By submitting your phone number, we have your permission to contact you regarding Let’s get learning! resources now. Error Oops! Something went wrong. Let’s refresh the page! 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### Section MM Matrix Multiplication From A First Course in Linear Algebra Version 2.23 http://linear.ups.edu/ We know how to add vectors and how to multiply them by scalars. Together, these operations give us the possibility of making linear combinations. Similarly, we know how to add matrices and how to multiply matrices by scalars. In this section we mix all these ideas together and produce an operation known as “matrix multiplication.” This will lead to some results that are both surprising and central. We begin with a definition of how to multiply a vector by a matrix. #### Subsection MVP: Matrix-Vector Product We have repeatedly seen the importance of forming linear combinations of the columns of a matrix. As one example of this, the oft-used Theorem SLSLC, said that every solution to a system of linear equations gives rise to a linear combination of the column vectors of the coefficient matrix that equals the vector of constants. This theorem, and others, motivate the following central definition. Definition MVP Matrix-Vector Product Suppose A is an m × n matrix with columns {A}_{1},\kern 1.95872pt {A}_{2},\kern 1.95872pt {A}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {A}_{n} and u is a vector of size n. Then the matrix-vector product of A with u is the linear combination Au ={ \left [u\right ]}_{1}{A}_{1} +{ \left [u\right ]}_{2}{A}_{2} +{ \left [u\right ]}_{3}{A}_{3} + \mathrel{⋯} +{ \left [u\right ]}_{n}{A}_{n} (This definition contains Notation MVP.) So, the matrix-vector product is yet another version of “multiplication,” at least in the sense that we have yet again overloaded juxtaposition of two symbols as our notation. Remember your objects, an m × n matrix times a vector of size n will create a vector of size m. So if A is rectangular, then the size of the vector changes. With all the linear combinations we have performed so far, this computation should now seem second nature. Example MTV A matrix times a vector Consider \eqalignno{ A = \left [\array{ 1 &4& 2 & 3 & 4\cr −3 &2 & 0 & 1 &−2 \cr 1 &6&−3&−1& 5 } \right ] & &u = \left [\array{ 2\cr 1 \cr −2\cr 3 \cr −1 } \right ] & & & & } Then Au = 2\left [\array{ 1\cr −3 \cr 1 } \right ]+1\left [\array{ 4\cr 2 \cr 6 } \right ]+(−2)\left [\array{ 2\cr 0 \cr −3 } \right ]+3\left [\array{ 3\cr 1 \cr −1 } \right ]+(−1)\left [\array{ 4\cr −2 \cr 5 } \right ] = \left [\array{ 7\cr 1 \cr 6 } \right ]. We can now represent systems of linear equations compactly with a matrix-vector product (Definition MVP) and column vector equality (Definition CVE). This finally yields a very popular alternative to our unconventional ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) notation. Theorem SLEMM Systems of Linear Equations as Matrix Multiplication The set of solutions to the linear system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) equals the set of solutions for x in the vector equation Ax = b. Proof This theorem says that two sets (of solutions) are equal. So we need to show that one set of solutions is a subset of the other, and vice versa (Definition SE). Let {A}_{1},\kern 1.95872pt {A}_{2},\kern 1.95872pt {A}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {A}_{n} be the columns of A. Both of these set inclusions then follow from the following chain of equivalences (Technique E), \eqalignno{ &x\text{ is a solution to }ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) & & & & \cr &\kern 3.26288pt \mathrel{⇔}\kern 3.26288pt {\left [x\right ]}_{1}{A}_{1} +{ \left [x\right ]}_{2}{A}_{2} +{ \left [x\right ]}_{3}{A}_{3} + \mathrel{⋯} +{ \left [x\right ]}_{n}{A}_{n} = b & &\text{@(a href="fcla-jsmath-2.23li24.html#theorem.SLSLC")Theorem SLSLC@(/a)} & & & & \cr &\kern 3.26288pt \mathrel{⇔}\kern 3.26288pt x\text{ is a solution to }Ax = b & &\text{@(a href="#definition.MVP")Definition MVP@(/a)} & & & & } Example MNSLE Matrix notation for systems of linear equations Consider the system of linear equations from Example NSLE. \eqalignno{ 2{x}_{1} + 4{x}_{2} − 3{x}_{3} + 5{x}_{4} + {x}_{5} & = 9 & & \cr 3{x}_{1} + {x}_{2} + {x}_{4} − 3{x}_{5} & = 0 & & \cr − 2{x}_{1} + 7{x}_{2} − 5{x}_{3} + 2{x}_{4} + 2{x}_{5} & = −3 & & } has coefficient matrix A = \left [\array{ 2 &4&−3&5& 1\cr 3 &1 & 0 &1 &−3 \cr −2&7&−5&2& 2 } \right ] and vector of constants b = \left [\array{ 9\cr 0 \cr −3 } \right ] and so will be described compactly by the vector equation Ax = b. The matrix-vector product is a very natural computation. We have motivated it by its connections with systems of equations, but here is a another example. Example MBC Money’s best cities Every year Money magazine selects several cities in the United States as the “best” cities to live in, based on a wide array of statistics about each city. This is an example of how the editors of Money might arrive at a single number that consolidates the statistics about a city. We will analyze Los Angeles, Chicago and New York City, based on four criteria: average high temperature in July (Farenheit), number of colleges and universities in a 30-mile radius, number of toxic waste sites in the Superfund environmental clean-up program and a personal crime index based on FBI statistics (average = 100, smaller is safer). It should be apparent how to generalize the example to a greater number of cities and a greater number of statistics. We begin by building a table of statistics. The rows will be labeled with the cities, and the columns with statistical categories. These values are from Money’s website in early 2005. City Temp Colleges Superfund Crime Los Angeles 77 28 93 254 Chicago 84 38 85 363 New York 84 99 1 193 Conceivably these data might reside in a spreadsheet. Now we must combine the statistics for each city. We could accomplish this by weighting each category, scaling the values and summing them. The sizes of the weights would depend upon the numerical size of each statistic generally, but more importantly, they would reflect the editors opinions or beliefs about which statistics were most important to their readers. Is the crime index more important than the number of colleges and universities? Of course, there is no right answer to this question. Suppose the editors finally decide on the following weights to employ: temperature, 0.23; colleges, 0.46; Superfund, − 0.05; crime, − 0.20. Notice how negative weights are used for undesirable statistics. Then, for example, the editors would compute for Los Angeles, (0.23)(77) + (0.46)(28) + (−0.05)(93) + (−0.20)(254) = −24.86 This computation might remind you of an inner product, but we will produce the computations for all of the cities as a matrix-vector product. Write the table of raw statistics as a matrix T = \left [\array{ 77&28&93&254\cr 84 &38 &85 &363 \cr 84&99& 1 &193 } \right ] and the weights as a vector w = \left [\array{ 0.23\cr 0.46 \cr −0.05\cr −0.20 } \right ] then the matrix-vector product (Definition MVP) yields Tw = (0.23)\left [\array{ 77\cr 84 \cr 84 } \right ]+(0.46)\left [\array{ 28\cr 38 \cr 99 } \right ]+(−0.05)\left [\array{ 93\cr 85 \cr 1 } \right ]+(−0.20)\left [\array{ 254\cr 363 \cr 193} \right ] = \left [\array{ −24.86\cr −40.05 \cr 26.21 } \right ] This vector contains a single number for each of the cities being studied, so the editors would rank New York best (26.21), Los Angeles next ( − 24.86), and Chicago third ( − 40.05). Of course, the mayor’s offices in Chicago and Los Angeles are free to counter with a different set of weights that cause their city to be ranked best. These alternative weights would be chosen to play to each cities’ strengths, and minimize their problem areas. If a speadsheet were used to make these computations, a row of weights would be entered somewhere near the table of data and the formulas in the spreadsheet would effect a matrix-vector product. This example is meant to illustrate how “linear” computations (addition, multiplication) can be organized as a matrix-vector product. Another example would be the matrix of numerical scores on examinations and exercises for students in a class. The rows would correspond to students and the columns to exams and assignments. The instructor could then assign weights to the different exams and assignments, and via a matrix-vector product, compute a single score for each student. Later (much later) we will need the following theorem, which is really a technical lemma (see Technique LC). Since we are in a position to prove it now, we will. But you can safely skip it for the moment, if you promise to come back later to study the proof when the theorem is employed. At that point you will also be able to understand the comments in the paragraph following the proof. Theorem EMMVP Equal Matrices and Matrix-Vector Products Suppose that A and B are m × n matrices such that Ax = Bx for every x ∈ {ℂ}^{n}. Then A = B. Proof We are assuming Ax = Bx for all x ∈ {ℂ}^{n}, so we can employ this equality for any choice of the vector x. However, we’ll limit our use of this equality to the standard unit vectors, {e}_{j}, 1 ≤ j ≤ n (Definition SUV). For all 1 ≤ j ≤ n, 1 ≤ i ≤ m, \eqalignno{ {\left [A\right ]}_{ij}& = 0{\left [A\right ]}_{i1} + \mathrel{⋯} + 0{\left [A\right ]}_{i,j−1} + 1{\left [A\right ]}_{ij} + 0{\left [A\right ]}_{i,j+1} + \mathrel{⋯} + 0{\left [A\right ]}_{in} && && \cr & ={ \left [A\right ]}_{i1}{\left [{e}_{j}\right ]}_{1} +{ \left [A\right ]}_{i2}{\left [{e}_{j}\right ]}_{2} +{ \left [A\right ]}_{i3}{\left [{e}_{j}\right ]}_{3} + \mathrel{⋯} +{ \left [A\right ]}_{in}{\left [{e}_{j}\right ]}_{n} &&\text{@(a href="fcla-jsmath-2.23li28.html#definition.SUV")Definition SUV@(/a)} &&&& \cr & ={ \left [A{e}_{j}\right ]}_{i} &&\text{@(a href="#definition.MVP")Definition MVP@(/a)}&&&& \cr & ={ \left [B{e}_{j}\right ]}_{i} &&\text{@(a href="fcla-jsmath-2.23li23.html#definition.CVE")Definition CVE@(/a)} &&&& \cr & ={ \left [B\right ]}_{i1}{\left [{e}_{j}\right ]}_{1} +{ \left [B\right ]}_{i2}{\left [{e}_{j}\right ]}_{2} +{ \left [B\right ]}_{i3}{\left [{e}_{j}\right ]}_{3} + \mathrel{⋯} +{ \left [B\right ]}_{in}{\left [{e}_{j}\right ]}_{n} &&\text{@(a href="#definition.MVP")Definition MVP@(/a)}&&&& \cr & = 0{\left [B\right ]}_{i1} + \mathrel{⋯} + 0{\left [B\right ]}_{i,j−1} + 1{\left [B\right ]}_{ij} + 0{\left [B\right ]}_{i,j+1} + \mathrel{⋯} + 0{\left [B\right ]}_{in}&&\text{@(a href="fcla-jsmath-2.23li28.html#definition.SUV")Definition SUV@(/a)} &&&& \cr & ={ \left [B\right ]}_{ij} && && } So by Definition ME the matrices A and B are equal, as desired. You might notice from studying the proof that the hypotheses of this theorem could be “weakened” (i.e. made less restrictive). We need only suppose the equality of the matrix-vector products for just the standard unit vectors (Definition SUV) or any other spanning set (Definition TSVS) of {ℂ}^{n} (Exercise LISS.T40). However, in practice, when we apply this theorem the stronger hypothesis will be in effect so this version of the theorem will suffice for our purposes. (If we changed the statement of the theorem to have the less restrictive hypothesis, then we would call the theorem “stronger.”) #### Subsection MM: Matrix Multiplication We now define how to multiply two matrices together. Stop for a minute and think about how you might define this new operation. Many books would present this definition much earlier in the course. However, we have taken great care to delay it as long as possible and to present as many ideas as practical based mostly on the notion of linear combinations. Towards the conclusion of the course, or when you perhaps take a second course in linear algebra, you may be in a position to appreciate the reasons for this. For now, understand that matrix multiplication is a central definition and perhaps you will appreciate its importance more by having saved it for later. Definition MM Matrix Multiplication Suppose A is an m × n matrix and B is an n × p matrix with columns {B}_{1},\kern 1.95872pt {B}_{2},\kern 1.95872pt {B}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {B}_{p}. Then the matrix product of A with B is the m × p matrix where column i is the matrix-vector product A{B}_{i}. Symbolically, AB = A\left [{B}_{1}\|{B}_{2}\|{B}_{3}\|\mathop{\mathop{…}}\|{B}_{p}\right ] = \left [A{B}_{1}\|A{B}_{2}\|A{B}_{3}\|\mathop{\mathop{…}}\|A{B}_{p}\right ]. Example PTM Product of two matrices Set \eqalignno{ A = \left [\array{ 1 & 2 &−1& 4 &6\cr 0 &−4 & 1 & 2 &3 \cr −5& 1 & 2 &−3&4 } \right ] & &B = \left [\array{ 1 & 6 & 2 &1\cr −1 & 4 & 3 &2 \cr 1 & 1 & 2 &3\cr 6 & 4 &−1 &2 \cr 1 &−2& 3 &0 } \right ] & & & & & } Then AB = \left [A\left [\array{ 1\cr −1 \cr 1\cr 6 \cr 1 } \right ]\left \vert A\left [\array{ 6\cr 4 \cr 1\cr 4 \cr −2 } \right ]\right .\left \vert A\left [\array{ 2\cr 3 \cr 2\cr −1 \cr 3 } \right ]\right .\left \vert A\left [\array{ 1\cr 2 \cr 3\cr 2 \cr 0 } \right ]\right .\right ] = \left [\array{ 28 & 17 &20&10\cr 20 &−13 &−3 &−1 \cr −18&−44&12&−3 } \right ]. Is this the definition of matrix multiplication you expected? Perhaps our previous operations for matrices caused you to think that we might multiply two matrices of the same size, entry-by-entry? Notice that our current definition uses matrices of different sizes (though the number of columns in the first must equal the number of rows in the second), and the result is of a third size. Notice too in the previous example that we cannot even consider the product BA, since the sizes of the two matrices in this order aren’t right. But it gets weirder than that. Many of your old ideas about “multiplication” won’t apply to matrix multiplication, but some still will. So make no assumptions, and don’t do anything until you have a theorem that says you can. Even if the sizes are right, matrix multiplication is not commutative — order matters. Example MMNC Matrix multiplication is not commutative Set \eqalignno{ A = \left [\array{ 1 &3\cr −1 &2 } \right ] & &B = \left [\array{ 4&0\cr 5&1 } \right ]. & & & & } Then we have two square, 2 × 2 matrices, so Definition MM allows us to multiply them in either order. We find \eqalignno{ AB = \left [\array{ 19&3\cr 6 &2} \right ] & &BA = \left [\array{ 4&12\cr 4&17} \right ] & & & & } and AB\mathrel{≠}BA. Not even close. It should not be hard for you to construct other pairs of matrices that do not commute (try a couple of 3 × 3’s). Can you find a pair of non-identical matrices that do commute? Matrix multiplication is fundamental, so it is a natural procedure for any computational device. See: Computation MM.MMA #### Subsection MMEE: Matrix Multiplication, Entry-by-Entry While certain “natural” properties of multiplication don’t hold, many more do. In the next subsection, we’ll state and prove the relevant theorems. But first, we need a theorem that provides an alternate means of multiplying two matrices. In many texts, this would be given as the definition of matrix multiplication. We prefer to turn it around and have the following formula as a consequence of our definition. It will prove useful for proofs of matrix equality, where we need to examine products of matrices, entry-by-entry. Theorem EMP Entries of Matrix Products Suppose A is an m × n matrix and B is an n × p matrix. Then for 1 ≤ i ≤ m, 1 ≤ j ≤ p, the individual entries of AB are given by \eqalignno{ {\left [AB\right ]}_{ij} & ={ \left [A\right ]}_{i1}{\left [B\right ]}_{1j} +{ \left [A\right ]}_{i2}{\left [B\right ]}_{2j} +{ \left [A\right ]}_{i3}{\left [B\right ]}_{3j} + \mathrel{⋯} +{ \left [A\right ]}_{in}{\left [B\right ]}_{nj} & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & & } Proof Denote the columns of A as the vectors {A}_{1},\kern 1.95872pt {A}_{2},\kern 1.95872pt {A}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {A}_{n} and the columns of B as the vectors {B}_{1},\kern 1.95872pt {B}_{2},\kern 1.95872pt {B}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {B}_{p}. Then for 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [AB\right ]}_{ij}& ={ \left [A{B}_{j}\right ]}_{i} &&\text{@(a href="#definition.MM")Definition MM@(/a)} &&&& \cr & ={ \left [{\left [{B}_{j}\right ]}_{1}{A}_{1} +{ \left [{B}_{j}\right ]}_{2}{A}_{2} +{ \left [{B}_{j}\right ]}_{3}{A}_{3} + \mathrel{⋯} +{ \left [{B}_{j}\right ]}_{n}{A}_{n}\right ]}_{i} &&\text{@(a href="#definition.MVP")Definition MVP@(/a)} &&&& \cr & ={ \left [{\left [{B}_{j}\right ]}_{1}{A}_{1}\right ]}_{i} +{ \left [{\left [{B}_{j}\right ]}_{2}{A}_{2}\right ]}_{i} +{ \left [{\left [{B}_{j}\right ]}_{3}{A}_{3}\right ]}_{i} + \mathrel{⋯} +{ \left [{\left [{B}_{j}\right ]}_{n}{A}_{n}\right ]}_{i}&&\text{@(a href="fcla-jsmath-2.23li23.html#definition.CVA")Definition CVA@(/a)} &&&& \cr & ={ \left [{B}_{j}\right ]}_{1}{\left [{A}_{1}\right ]}_{i} +{ \left [{B}_{j}\right ]}_{2}{\left [{A}_{2}\right ]}_{i} +{ \left [{B}_{j}\right ]}_{3}{\left [{A}_{3}\right ]}_{i} + \mathrel{⋯} +{ \left [{B}_{j}\right ]}_{n}{\left [{A}_{n}\right ]}_{i}&&\text{@(a href="fcla-jsmath-2.23li23.html#definition.CVSM")Definition CVSM@(/a)}&&&& \cr & ={ \left [B\right ]}_{1j}{\left [A\right ]}_{i1} +{ \left [B\right ]}_{2j}{\left [A\right ]}_{i2} +{ \left [B\right ]}_{3j}{\left [A\right ]}_{i3} + \mathrel{⋯} +{ \left [B\right ]}_{nj}{\left [A\right ]}_{in} &&\text{@(a href="fcla-jsmath-2.23li30.html#notation.ME")Notation ME@(/a)} &&&& \cr & ={ \left [A\right ]}_{i1}{\left [B\right ]}_{1j} +{ \left [A\right ]}_{i2}{\left [B\right ]}_{2j} +{ \left [A\right ]}_{i3}{\left [B\right ]}_{3j} + \mathrel{⋯} +{ \left [A\right ]}_{in}{\left [B\right ]}_{nj} &&\text{@(a href="fcla-jsmath-2.23li69.html#property.CMCN")Property CMCN@(/a)} &&&& \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} && && } Example PTMEE Product of two matrices, entry-by-entry Consider again the two matrices from Example PTM \eqalignno{ A = \left [\array{ 1 & 2 &−1& 4 &6\cr 0 &−4 & 1 & 2 &3 \cr −5& 1 & 2 &−3&4 } \right ] & &B = \left [\array{ 1 & 6 & 2 &1\cr −1 & 4 & 3 &2 \cr 1 & 1 & 2 &3\cr 6 & 4 &−1 &2 \cr 1 &−2& 3 &0 } \right ] & & & & & } Then suppose we just wanted the entry of AB in the second row, third column: \eqalignno{ {\left [AB\right ]}_{23} = &{\left [A\right ]}_{21}{\left [B\right ]}_{13} +{ \left [A\right ]}_{22}{\left [B\right ]}_{23} +{ \left [A\right ]}_{23}{\left [B\right ]}_{33} +{ \left [A\right ]}_{24}{\left [B\right ]}_{43} +{ \left [A\right ]}_{25}{\left [B\right ]}_{53} & & \cr = &(0)(2) + (−4)(3) + (1)(2) + (2)(−1) + (3)(3) = −3 & & } Notice how there are 5 terms in the sum, since 5 is the common dimension of the two matrices (column count for A, row count for B). In the conclusion of Theorem EMP, it would be the index k that would run from 1 to 5 in this computation. Here’s a bit more practice. The entry of third row, first column: \eqalignno{ {\left [AB\right ]}_{31} = &{\left [A\right ]}_{31}{\left [B\right ]}_{11} +{ \left [A\right ]}_{32}{\left [B\right ]}_{21} +{ \left [A\right ]}_{33}{\left [B\right ]}_{31} +{ \left [A\right ]}_{34}{\left [B\right ]}_{41} +{ \left [A\right ]}_{35}{\left [B\right ]}_{51} & & \cr = &(−5)(1) + (1)(−1) + (2)(1) + (−3)(6) + (4)(1) = −18 & & } To get some more practice on your own, complete the computation of the other 10 entries of this product. Construct some other pairs of matrices (of compatible sizes) and compute their product two ways. First use Definition MM. Since linear combinations are straightforward for you now, this should be easy to do and to do correctly. Then do it again, using Theorem EMP. Since this process may take some practice, use your first computation to check your work. Theorem EMP is the way many people compute matrix products by hand. It will also be very useful for the theorems we are going to prove shortly. However, the definition (Definition MM) is frequently the most useful for its connections with deeper ideas like the null space and the upcoming column space. #### Subsection PMM: Properties of Matrix Multiplication In this subsection, we collect properties of matrix multiplication and its interaction with the zero matrix (Definition ZM), the identity matrix (Definition IM), matrix addition (Definition MA), scalar matrix multiplication (Definition MSM), the inner product (Definition IP), conjugation (Theorem MMCC), and the transpose (Definition TM). Whew! Here we go. These are great proofs to practice with, so try to concoct the proofs before reading them, they’ll get progressively more complicated as we go. Theorem MMZM Matrix Multiplication and the Zero Matrix Suppose A is an m × n matrix. Then 1. A{O}_{n×p} = {O}_{m×p} 2. {O}_{p×m}A = {O}_{p×n} Proof We’ll prove (1) and leave (2) to you. Entry-by-entry, for 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [A{O}_{n×p}\right ]}_{ij} & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [{O}_{n×p}\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}0 & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.ZM")Definition ZM@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}0 & & & & \cr & = 0 & &\text{@(a href="fcla-jsmath-2.23li69.html#property.ZCN")Property ZCN@(/a)} & & & & \cr & ={ \left [{O}_{m×p}\right ]}_{ij} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.ZM")Definition ZM@(/a)} & & & & } So by the definition of matrix equality (Definition ME), the matrices A{O}_{n×p} and {O}_{m×p} are equal. Theorem MMIM Matrix Multiplication and Identity Matrix Suppose A is an m × n matrix. Then 1. A{I}_{n} = A 2. {I}_{m}A = A Proof Again, we’ll prove (1) and leave (2) to you. Entry-by-entry, For 1 ≤ i ≤ m, 1 ≤ j ≤ n, \eqalignno{ {\left [A{I}_{n}\right ]}_{ij} = &{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [{I}_{n}\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [A\right ]}_{ij}{\left [{I}_{n}\right ]}_{jj} +{ \mathop{∑ }}_{\begin{array}{c}k=1 \\ k\mathrel{≠}j \end{array}}^{n}{\left [A\right ]}_{ ik}{\left [{I}_{n}\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.CACN")Property CACN@(/a)} & & & & \cr & ={ \left [A\right ]}_{ij}(1) +{ \mathop{∑ }}_{k=1,k\mathrel{≠}j}^{n}{\left [A\right ]}_{ ik}(0) & &\text{@(a href="fcla-jsmath-2.23li21.html#definition.IM")Definition IM@(/a)} & & & & \cr & ={ \left [A\right ]}_{ij} +{ \mathop{∑ }}_{k=1,k\mathrel{≠}j}^{n}0 & & & & \cr & ={ \left [A\right ]}_{ij} & & & & } So the matrices A and A{I}_{n} are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. It is this theorem that gives the identity matrix its name. It is a matrix that behaves with matrix multiplication like the scalar 1 does with scalar multiplication. To multiply by the identity matrix is to have no effect on the other matrix. Theorem MMDAA Suppose A is an m × n matrix and B and C are n × p matrices and D is a p × s matrix. Then 1. A(B + C) = AB + AC 2. (B + C)D = BD + CD Proof We’ll do (1), you do (2). Entry-by-entry, for 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [A(B + C)\right ]}_{ij} & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B + C\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}({\left [B\right ]}_{kj} +{ \left [C\right ]}_{kj}) & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MA")Definition MA@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} +{ \left [A\right ]}_{ik}{\left [C\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.DCN")Property DCN@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} +{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [C\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.CACN")Property CACN@(/a)} & & & & \cr & ={ \left [AB\right ]}_{ij} +{ \left [AC\right ]}_{ij} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \left [AB + AC\right ]}_{ij} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MA")Definition MA@(/a)} & & & & } So the matrices A(B + C) and AB + AC are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. Theorem MMSMM Matrix Multiplication and Scalar Matrix Multiplication Suppose A is an m × n matrix and B is an n × p matrix. Let α be a scalar. Then α(AB) = (αA)B = A(αB). Proof These are equalities of matrices. We’ll do the first one, the second is similar and will be good practice for you. For 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [α(AB)\right ]}_{ij} & = α{\left [AB\right ]}_{ij} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MSM")Definition MSM@(/a)} & & & & \cr & = α{\mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}α{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.DCN")Property DCN@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [αA\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MSM")Definition MSM@(/a)} & & & & \cr & ={ \left [(αA)B\right ]}_{ij} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & } So the matrices α(AB) and (αA)B are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. Theorem MMA Matrix Multiplication is Associative Suppose A is an m × n matrix, B is an n × p matrix and D is a p × s matrix. Then A(BD) = (AB)D. Proof A matrix equality, so we’ll go entry-by-entry, no surprise there. For 1 ≤ i ≤ m, 1 ≤ j ≤ s, \eqalignno{ {\left [A(BD)\right ]}_{ij} & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [BD\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}\left ({\mathop{∑ }}_{ℓ=1}^{p}{\left [B\right ]}_{ kℓ}{\left [D\right ]}_{ℓj}\right ) & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{ \mathop{∑ }}_{ℓ=1}^{p}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kℓ}{\left [D\right ]}_{ℓj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.DCN")Property DCN@(/a)} & & & & \text{We can switch the order of the summation since these are finite sums,} \cr & ={ \mathop{∑ }}_{ℓ=1}^{p}{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kℓ}{\left [D\right ]}_{ℓj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.CACN")Property CACN@(/a)} & & & & \text{As ${\left [D\right ]}_{ℓj}$ does not depend on the index $k$, we can use distributivity to move it outside of the inner sum,} \cr & ={ \mathop{∑ }}_{ℓ=1}^{p}{\left [D\right ]}_{ ℓj}\left ({\mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kℓ}\right ) & &\text{@(a href="fcla-jsmath-2.23li69.html#property.DCN")Property DCN@(/a)} & & & & \cr & ={ \mathop{∑ }}_{ℓ=1}^{p}{\left [D\right ]}_{ ℓj}{\left [AB\right ]}_{iℓ} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{ℓ=1}^{p}{\left [AB\right ]}_{ iℓ}{\left [D\right ]}_{ℓj} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.CMCN")Property CMCN@(/a)} & & & & \cr & ={ \left [(AB)D\right ]}_{ij} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & } So the matrices (AB)D and A(BD) are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. The statement of our next theorem is technically inaccurate. If we upgrade the vectors u,\kern 1.95872pt v to matrices with a single column, then the expression {u}^{t}\overline{v} is a 1 × 1 matrix, though we will treat this small matrix as if it was simply the scalar quantity in its lone entry. When we apply Theorem MMIP there should not be any confusion. Theorem MMIP Matrix Multiplication and Inner Products If we consider the vectors u,\kern 1.95872pt v ∈ {ℂ}^{m} as m × 1 matrices then \left \langle u,\kern 1.95872pt v\right \rangle = {u}^{t}\overline{v} Proof \eqalignno{ \left \langle u,\kern 1.95872pt v\right \rangle & ={ \mathop{∑ }}_{k=1}^{m}{\left [u\right ]}_{ k}\overline{{\left [v\right ]}_{k}} & &\text{@(a href="fcla-jsmath-2.23li28.html#definition.IP")Definition IP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{m}{\left [u\right ]}_{ k1}\overline{{\left [v\right ]}_{k1}} & &\text{Column vectors as matrices} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{m}{\left [{u}^{t}\right ]}_{ 1k}\overline{{\left [v\right ]}_{k1}} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.TM")Definition TM@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{m}{\left [{u}^{t}\right ]}_{ 1k}{\left [\overline{v}\right ]}_{k1} & &\text{@(a href="fcla-jsmath-2.23li28.html#definition.CCCV")Definition CCCV@(/a)} & & & & \cr & ={ \left [{u}^{t}\overline{v}\right ]}_{ 11} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & & & & } To finish we just blur the distinction between a 1 × 1 matrix ({u}^{t}\overline{v}) and its lone entry. Theorem MMCC Matrix Multiplication and Complex Conjugation Suppose A is an m × n matrix and B is an n × p matrix. Then \overline{AB} = \overline{A}\kern 1.95872pt \overline{B}. Proof To obtain this matrix equality, we will work entry-by-entry. For 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [\overline{AB}\right ]}_{ij} & = \overline{{\left [AB\right ]}_{ij}} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.CCM")Definition CCM@(/a)} & & & & \cr & = \overline{{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj}} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}\overline{{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj}} & &\text{@(a href="fcla-jsmath-2.23li69.html#theorem.CCRA")Theorem CCRA@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}\overline{{\left [A\right ]}_{ ik}}\kern 1.95872pt \kern 1.95872pt \overline{{\left [B\right ]}_{kj}} & &\text{@(a href="fcla-jsmath-2.23li69.html#theorem.CCRM")Theorem CCRM@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [\overline{A}\right ]}_{ ik}{\left [\overline{B}\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.CCM")Definition CCM@(/a)} & & & & \cr & ={ \left [\overline{A}\kern 1.95872pt \overline{B}\right ]}_{ij} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & } So the matrices \overline{AB} and \overline{A}\kern 1.95872pt \overline{B} are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. Another theorem in this style, and its a good one. If you’ve been practicing with the previous proofs you should be able to do this one yourself. Theorem MMT Matrix Multiplication and Transposes Suppose A is an m × n matrix and B is an n × p matrix. Then {(AB)}^{t} = {B}^{t}{A}^{t}. Proof This theorem may be surprising but if we check the sizes of the matrices involved, then maybe it will not seem so far-fetched. First, AB has size m × p, so its transpose has size p × m. The product of {B}^{t} with {A}^{t} is a p × n matrix times an n × m matrix, also resulting in a p × m matrix. So at least our objects are compatible for equality (and would not be, in general, if we didn’t reverse the order of the matrix multiplication). Here we go again, entry-by-entry. For 1 ≤ i ≤ m, 1 ≤ j ≤ p, \eqalignno{ {\left [{(AB)}^{t}\right ]}_{ ji} = &{\left [AB\right ]}_{ij} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.TM")Definition TM@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [B\right ]}_{ kj}{\left [A\right ]}_{ik} & &\text{@(a href="fcla-jsmath-2.23li69.html#property.CMCN")Property CMCN@(/a)} & & & & \cr & ={ \mathop{∑ }}_{k=1}^{n}{\left [{B}^{t}\right ]}_{ jk}{\left [{A}^{t}\right ]}_{ ki} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.TM")Definition TM@(/a)} & & & & \cr & ={ \left [{B}^{t}{A}^{t}\right ]}_{ ji} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & } So the matrices {(AB)}^{t} and {B}^{t}{A}^{t} are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. This theorem seems odd at first glance, since we have to switch the order of A and B. But if we simply consider the sizes of the matrices involved, we can see that the switch is necessary for this reason alone. That the individual entries of the products then come along to be equal is a bonus. As the adjoint of a matrix is a composition of a conjugate and a transpose, its interaction with matrix multiplication is similar to that of a transpose. Here’s the last of our long list of basic properties of matrix multiplication. Suppose A is an m × n matrix and B is an n × p matrix. Then {(AB)}^{∗} = {B}^{∗}{A}^{∗}. Proof \eqalignno{ {(AB)}^{∗} & ={ \left (\overline{AB}\right )}^{t} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.A")Definition A@(/a)} & & & & \cr & ={ \left (\overline{A}\kern 1.95872pt \overline{B}\right )}^{t} & &\text{@(a href="#theorem.MMCC")Theorem MMCC@(/a)} & & & & \cr & ={ \left (\overline{B}\right )}^{t}{\left (\overline{A}\right )}^{t} & &\text{@(a href="#theorem.MMT")Theorem MMT@(/a)} & & & & \cr & = {B}^{∗}{A}^{∗} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.A")Definition A@(/a)} & & & & } Notice how none of these proofs above relied on writing out huge general matrices with lots of ellipses (“…”) and trying to formulate the equalities a whole matrix at a time. This messy business is a “proof technique” to be avoided at all costs. Notice too how the proof of Theorem MMAD does not use an entry-by-entry approach, but simply builds on previous results about matrix multiplication’s interaction with conjugation and transposes. These theorems, along with Theorem VSPM and the other results in Section MO, give you the “rules” for how matrices interact with the various operations we have defined on matrices (addition, scalar multiplication, matrix multiplication, conjugation, transposes and adjoints). Use them and use them often. But don’t try to do anything with a matrix that you don’t have a rule for. Together, we would informally call all these operations, and the attendant theorems, “the algebra of matrices.” Notice, too, that every column vector is just a n × 1 matrix, so these theorems apply to column vectors also. Finally, these results, taken as a whole, may make us feel that the definition of matrix multiplication is not so unnatural. #### Subsection HM: Hermitian Matrices The adjoint of a matrix has a basic property when employed in a matrix-vector product as part of an inner product. At this point, you could even use the following result as a motivation for the definition of an adjoint. Theorem AIP Suppose that A is an m × n matrix and x ∈ {ℂ}^{n}, y ∈ {ℂ}^{m}. Then \left \langle Ax,\kern 1.95872pt y\right \rangle = \left \langle x,\kern 1.95872pt {A}^{∗}y\right \rangle . Proof \eqalignno{ \left \langle Ax,\kern 1.95872pt y\right \rangle & ={ \left (Ax\right )}^{t}\overline{y} & &\text{@(a href="#theorem.MMIP")Theorem MMIP@(/a)} & & & & \cr & = {x}^{t}{A}^{t}\overline{y} & &\text{@(a href="#theorem.MMT")Theorem MMT@(/a)} & & & & \cr & = {x}^{t}{\left (\overline{\overline{A}}\right )}^{t}\overline{y} & &\text{@(a href="fcla-jsmath-2.23li30.html#theorem.CCM")Theorem CCM@(/a)} & & & & \cr & = {x}^{t}\overline{\left ({\left (\overline{A}\right )}^{t}\right )}\overline{y} & &\text{@(a href="fcla-jsmath-2.23li30.html#theorem.MCT")Theorem MCT@(/a)} & & & & \cr & = {x}^{t}\overline{\left ({A}^{∗}\right )}\overline{y} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.A")Definition A@(/a)} & & & & \cr & = {x}^{t}\overline{\left ({A}^{∗}y\right )} & &\text{@(a href="#theorem.MMCC")Theorem MMCC@(/a)} & & & & \cr & = \left \langle x,\kern 1.95872pt {A}^{∗}y\right \rangle & &\text{@(a href="#theorem.MMIP")Theorem MMIP@(/a)} & & & & } Sometimes a matrix is equal to its adjoint (Definition A), and these matrices have interesting properties. One of the most common situations where this occurs is when a matrix has only real number entries. Then we are simply talking about symmetric matrices (Definition SYM), so you can view this as a generalization of a symmetric matrix. Definition HM Hermitian Matrix The square matrix A is Hermitian (or self-adjoint) if A = {A}^{∗}. Again, the set of real matrices that are Hermitian is exactly the set of symmetric matrices. In Section PEE we will uncover some amazing properties of Hermitian matrices, so when you get there, run back here to remind yourself of this definition. Further properties will also appear in various sections of the Topics (Part T). Right now we prove a fundamental result about Hermitian matrices, matrix vector products and inner products. As a characterization, this could be employed as a definition of a Hermitian matrix and some authors take this approach. Theorem HMIP Hermitian Matrices and Inner Products Suppose that A is a square matrix of size n. Then A is Hermitian if and only if \left \langle Ax,\kern 1.95872pt y\right \rangle = \left \langle x,\kern 1.95872pt Ay\right \rangle for all x,\kern 1.95872pt y ∈ {ℂ}^{n}. Proof () This is the “easy half” of the proof, and makes the rationale for a definition of Hermitian matrices most obvious. Assume A is Hermitian, \eqalignno{ \left \langle Ax,\kern 1.95872pt y\right \rangle & = \left \langle x,\kern 1.95872pt {A}^{∗}y\right \rangle & &\text{@(a href="#theorem.AIP")Theorem AIP@(/a)} & & & & \cr & = \left \langle x,\kern 1.95872pt Ay\right \rangle & &\text{@(a href="#definition.HM")Definition HM@(/a)} & & & & \cr & & & & } () This “half” will take a bit more work. Assume that \left \langle Ax,\kern 1.95872pt y\right \rangle = \left \langle x,\kern 1.95872pt Ay\right \rangle for all x,\kern 1.95872pt y ∈ {ℂ}^{n}. Choose any x ∈ {ℂ}^{n}. We want to show that A = {A}^{∗} by establishing that Ax = {A}^{∗}x. With only this much motivation, consider the inner product, \eqalignno{ \left \langle Ax − {A}^{∗}x,\kern 1.95872pt Ax − {A}^{∗}x\right \rangle & = \left \langle Ax − {A}^{∗}x,\kern 1.95872pt Ax\right \rangle −\left \langle Ax − {A}^{∗}x,\kern 1.95872pt {A}^{∗}x\right \rangle &&\text{@(a href="fcla-jsmath-2.23li28.html#theorem.IPVA")Theorem IPVA@(/a)}&&&& \cr & = \left \langle Ax − {A}^{∗}x,\kern 1.95872pt Ax\right \rangle −\left \langle A\left (Ax − {A}^{∗}x\right ),\kern 1.95872pt x\right \rangle &&\text{@(a href="#theorem.AIP")Theorem AIP@(/a)} &&&& \cr & = \left \langle Ax − {A}^{∗}x,\kern 1.95872pt Ax\right \rangle −\left \langle Ax − {A}^{∗}x,\kern 1.95872pt Ax\right \rangle &&\text{Hypothesis} &&&& \cr & = 0 &&\text{@(a href="fcla-jsmath-2.23li69.html#property.AICN")Property AICN@(/a)}&&&& } Because this first inner product equals zero, and has the same vector in each argument (Ax − {A}^{∗}x), Theorem PIP gives the conclusion that Ax − {A}^{∗}x = 0. With Ax = {A}^{∗}x for all x ∈ {ℂ}^{n}, Theorem EMMVP says A = {A}^{∗}, which is the defining property of a Hermitian matrix (Definition HM). So, informally, Hermitian matrices are those that can be tossed around from one side of an inner product to the other with reckless abandon. We’ll see later what this buys us. 1. Form the matrix vector product of \eqalignno{ \left [\array{ 2& 3 &−1&0\cr 1&−2 & 7 &3 \cr 1& 5 & 3 &2\cr } \right ] & &\text{with} & &\left [\array{ 2\cr −3 \cr 0\cr 5 } \right ] & & & & & & } 2. Multiply together the two matrices below (in the order given). \eqalignno{ \left [\array{ 2& 3 &−1&0\cr 1&−2 & 7 &3 \cr 1& 5 & 3 &2\cr } \right ] & &\left [\array{ 2 & 6\cr −3 &−4 \cr 0 & 2\cr 3 &−1\cr } \right ] & & & & } 3. Rewrite the system of linear equations below as a vector equality and using a matrix-vector product. (This question does not ask for a solution to the system. But it does ask you to express the system of equations in a new form using tools from this section.) \eqalignno{ 2{x}_{1} + 3{x}_{2} − {x}_{3} & = 0 & & \cr {x}_{1} + 2{x}_{2} + {x}_{3} & = 3 & & \cr {x}_{1} + 3{x}_{2} + 3{x}_{3} & = 7 & & } #### Subsection EXC: Exercises C20 Compute the product of the two matrices below, AB. Do this using the definitions of the matrix-vector product (Definition MVP) and the definition of matrix multiplication (Definition MM). \eqalignno{ A = \left [\array{ 2 & 5\cr −1 & 3 \cr 2 &−2 } \right ] & &B = \left [\array{ 1&5&−3& 4\cr 2&0 & 2 &−3 } \right ] & & & & } Contributed by Robert Beezer Solution [637] C21 Compute the product AB of the two matrices below using both the definition of the matrix-vector product (Definition MVP) and the definition of matrix multiplication (Definition MM). \eqalignno{ A & = \left [\array{ 1 &3&2\cr −1 &2 &1 \cr 0 &1&0 } \right ] &B & = \left [\array{ 4&1&2\cr 1&0 &1 \cr 3&1&5} \right ] & & & & } Contributed by Chris Black Solution [637] C22 Compute the product AB of the two matrices below using both the definition of the matrix-vector product (Definition MVP) and the definition of matrix multiplication (Definition MM). \eqalignno{ A & = \left [\array{ 1 &0\cr −2 &1 } \right ] &B & = \left [\array{ 2&3\cr 4&6 } \right ] & & & & } Contributed by Chris Black Solution [637] C23 Compute the product AB of the two matrices below using both the definition of the matrix-vector product (Definition MVP) and the definition of matrix multiplication (Definition MM). \eqalignno{ A & = \left [\array{ 3&1\cr 2&4 \cr 6&5\cr 1&2 } \right ] &B & = \left [\array{ −3&1\cr 4 &2 } \right ] & & & & } Contributed by Chris Black Solution [637] C24 Compute the product AB of the two matrices below. \eqalignno{ A & = \left [\array{ 1&2& 3 &−2\cr 0&1 &−2 &−1 \cr 1&1& 3 & 1 } \right ] &B & = \left [\array{ 3\cr 4 \cr 0\cr 2 } \right ] & & & & } Contributed by Chris Black Solution [638] C25 Compute the product AB of the two matrices below. \eqalignno{ A & = \left [\array{ 1&2& 3 &−2\cr 0&1 &−2 &−1 \cr 1&1& 3 & 1 } \right ] &B & = \left [\array{ −7\cr 3 \cr 1\cr 1 } \right ] & & & & } Contributed by Chris Black Solution [638] C26 Compute the product AB of the two matrices below using both the definition of the matrix-vector product (Definition MVP) and the definition of matrix multiplication (Definition MM). \eqalignno{ A & = \left [\array{ 1&3&1\cr 0&1 &0 \cr 1&1&2} \right ] &B & = \left [\array{ 2 &−5&−1\cr 0 & 1 & 0 \cr −1& 2 & 1 } \right ] & & & & } Contributed by Chris Black Solution [638] C30 For the matrix A = \left [\array{ 1&2\cr 0&1 } \right ], find {A}^{2}, {A}^{3}, {A}^{4}. Find a general formula for {A}^{n} for any positive integer n. Contributed by Chris Black Solution [638] C31 For the matrix A = \left [\array{ 1&−1\cr 0& 1 } \right ], find {A}^{2}, {A}^{3}, {A}^{4}. Find a general formula for {A}^{n} for any positive integer n. Contributed by Chris Black Solution [638] C32 For the matrix A = \left [\array{ 1&0&0\cr 0&2 &0 \cr 0&0&3} \right ], find {A}^{2}, {A}^{3}, {A}^{4}. Find a general formula for {A}^{n} for any positive integer n. Contributed by Chris Black Solution [638] C33 For the matrix A = \left [\array{ 0&1&2\cr 0&0 &1 \cr 0&0&0} \right ], find {A}^{2}, {A}^{3}, {A}^{4}. Find a general formula for {A}^{n} for any positive integer n. Contributed by Chris Black Solution [638] T10 Suppose that A is a square matrix and there is a vector, b, such that ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution. Prove that A is nonsingular. Give a direct proof (perhaps appealing to Theorem PSPHS) rather than just negating a sentence from the text discussing a similar situation. Contributed by Robert Beezer Solution [639] T20 Prove the second part of Theorem MMZM. Contributed by Robert Beezer T21 Prove the second part of Theorem MMIM. Contributed by Robert Beezer T22 Prove the second part of Theorem MMDAA. Contributed by Robert Beezer T23 Prove the second part of Theorem MMSMM. Contributed by Robert Beezer Solution [639] T31 Suppose that A is an m × n matrix and x,\kern 1.95872pt y ∈N\kern -1.95872pt \left (A\right ). Prove that x + y ∈N\kern -1.95872pt \left (A\right ). Contributed by Robert Beezer T32 Suppose that A is an m × n matrix, α ∈ {ℂ}^{}, and x ∈N\kern -1.95872pt \left (A\right ). Prove that αx ∈N\kern -1.95872pt \left (A\right ). Contributed by Robert Beezer T40 Suppose that A is an m × n matrix and B is an n × p matrix. Prove that the null space of B is a subset of the null space of AB, that is N\kern -1.95872pt \left (B\right ) ⊆N\kern -1.95872pt \left (AB\right ). Provide an example where the opposite is false, in other words give an example where N\kern -1.95872pt \left (AB\right )⊈N\kern -1.95872pt \left (B\right ). Contributed by Robert Beezer Solution [640] T41 Suppose that A is an n × n nonsingular matrix and B is an n × p matrix. Prove that the null space of B is equal to the null space of AB, that is N\kern -1.95872pt \left (B\right ) = N\kern -1.95872pt \left (AB\right ). (Compare with Exercise MM.T40.) Contributed by Robert Beezer Solution [641] T50 Suppose u and v are any two solutions of the linear system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ). Prove that u − v is an element of the null space of A, that is, u − v ∈N\kern -1.95872pt \left (A\right ). Contributed by Robert Beezer T51 Give a new proof of Theorem PSPHS replacing applications of Theorem SLSLC with matrix-vector products (Theorem SLEMM). Contributed by Robert Beezer Solution [641] T52 Suppose that x,\kern 1.95872pt y ∈ {ℂ}^{n}, b ∈ {ℂ}^{m} and A is an m × n matrix. If x, y and x + y are each a solution to the linear system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ), what interesting can you say about b? Form an implication with the existence of the three solutions as the hypothesis and an interesting statement about ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) as the conclusion, and then give a proof. Contributed by Robert Beezer Solution [643] #### Subsection SOL: Solutions C20 Contributed by Robert Beezer Statement [629] By Definition MM, \eqalignno{ AB & = \left [\left .\left [\array{ 2 & 5\cr −1 & 3 \cr 2 &−2 } \right ]\left [\array{ 1\cr 2 } \right ]\right \vert \ \left .\left [\array{ 2 & 5\cr −1 & 3 \cr 2 &−2 } \right ]\left [\array{ 5\cr 0 } \right ]\right \vert \ \left .\left [\array{ 2 & 5\cr −1 & 3 \cr 2 &−2 } \right ]\left [\array{ −3\cr 2 } \right ]\right \vert \ \left .\left [\array{ 2 & 5\cr −1 & 3 \cr 2 &−2 } \right ]\left [\array{ 4\cr −2 } \right ]\right .\right ] & & \text{Repeated applications of @(a href="#definition.MVP")Definition MVP@(/a) give} \cr & = \left [\left .1\left [\array{ 2\cr −1 \cr 2 } \right ] + 2\left [\array{ 5\cr 3 \cr −2 } \right ]\right \vert \ \left .5\left [\array{ 2\cr −1 \cr 2 } \right ] + 0\left [\array{ 5\cr 3 \cr −2 } \right ]\right \vert \ \left .−3\left [\array{ 2\cr −1 \cr 2 } \right ] + 2\left [\array{ 5\cr 3 \cr −2 } \right ]\right \vert \ \left .4\left [\array{ 2\cr −1 \cr 2 } \right ] + (−3)\left [\array{ 5\cr 3 \cr −2 } \right ]\right .\right ] & & \cr & = \left [\array{ 12&10& 4 & −7\cr 5 &−5 & 9 &−13 \cr −2&10&−10& 14 } \right ] & & } C21 Contributed by Chris Black Statement [629] AB = \left [\array{ 13&3&15\cr 1 &0 & 5 \cr 1 &0& 1 } \right ]. C22 Contributed by Chris Black Statement [630] AB = \left [\array{ 2&3\cr 0&0 } \right ]. C23 Contributed by Chris Black Statement [630] AB = \left [\array{ −5& 5\cr 10 &10 \cr 2 &16\cr 5 & 5 } \right ]. C24 Contributed by Chris Black Statement [631] AB = \left [\array{ 7\cr 2 \cr 9 } \right ]. C25 Contributed by Chris Black Statement [631] AB = \left [\array{ 0\cr 0 \cr 0 } \right ]. C26 Contributed by Chris Black Statement [632] AB = \left [\array{ 1&0&0\cr 0&1 &0 \cr 0&0&1} \right ]. C30 Contributed by Chris Black Statement [633] {A}^{2} = \left [\array{ 1&4 \cr 0&1 } \right ], {A}^{3} = \left [\array{ 1&6 \cr 0&1 } \right ], {A}^{4} = \left [\array{ 1&8 \cr 0&1 } \right ]. From this pattern, we see that {A}^{n} = \left [\array{ 1&2n \cr 0& 1 } \right ]. C31 Contributed by Chris Black Statement [633] {A}^{2} = \left [\array{ 1&−2 \cr 0& 1} \right ], {A}^{3} = \left [\array{ 1&−3 \cr 0& 1} \right ], {A}^{4} = \left [\array{ 1&−4 \cr 0& 1} \right ]. From this pattern, we see that {A}^{n} = \left [\array{ 1&−n \cr 0& 1} \right ]. C32 Contributed by Chris Black Statement [633] {A}^{2} = \left [\array{ 1&0&0\cr 0&4 &0 \cr 0&0&9} \right ], {A}^{3} = \left [\array{ 1&0& 0\cr 0&8 & 0 \cr 0&0&27 } \right ], and {A}^{4} = \left [\array{ 1& 0 & 0\cr 0&16 & 0 \cr 0& 0 &81 } \right ]. The pattern emerges, and we see that {A}^{n} = \left [\array{ 1& 0 & 0\cr 0&{2}^{n } & 0 \cr 0& 0 &{3}^{n} } \right ]. C33 Contributed by Chris Black Statement [633] We quickly compute {A}^{2} = \left [\array{ 0&0&1\cr 0&0 &0 \cr 0&0&0} \right ], and we then see that {A}^{3} and all subsequent powers of A are the 3 × 3 zero matrix; that is, {A}^{n} = {O}_{ 3,3} for n ≥ 3. T10 Contributed by Robert Beezer Statement [633] Since ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has at least one solution, we can apply Theorem PSPHS. Because the solution is assumed to be unique, the null space of A must be trivial. Then Theorem NMTNS implies that A is nonsingular. The converse of this statement is a trivial application of Theorem NMUS. That said, we could extend our NSMxx series of theorems with an added equivalence for nonsingularity, “Given a single vector of constants, b, the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution.” T23 Contributed by Robert Beezer Statement [634] We’ll run the proof entry-by-entry. \eqalignno{ {\left [α(AB)\right ]}_{ij} = &α{\left [AB\right ]}_{ij} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MSM")Definition MSM@(/a)} & & & & \cr = &α{\mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & \cr = &{ \mathop{∑ }}_{k=1}^{n}α{\left [A\right ]}_{ ik}{\left [B\right ]}_{kj} & &\text{Distributivity in ${ℂ}^{}$} & & & & \cr = &{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}α{\left [B\right ]}_{kj} & &\text{Commutativity in ${ℂ}^{}$} & & & & \cr = &{ \mathop{∑ }}_{k=1}^{n}{\left [A\right ]}_{ ik}{\left [αB\right ]}_{kj} & &\text{@(a href="fcla-jsmath-2.23li30.html#definition.MSM")Definition MSM@(/a)} & & & & \cr = &{\left [A(αB)\right ]}_{ij} & &\text{@(a href="#theorem.EMP")Theorem EMP@(/a)} & & & & } So the matrices α(AB) and A(αB) are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices. T40 Contributed by Robert Beezer Statement [634] To prove that one set is a subset of another, we start with an element of the smaller set and see if we can determine that it is a member of the larger set (Definition SSET). Suppose x ∈N\kern -1.95872pt \left (B\right ). Then we know that Bx = 0 by Definition NSM. Consider \eqalignno{ (AB)x & = A(Bx) & &\text{@(a href="#theorem.MMA")Theorem MMA@(/a)} & & & & \cr & = A0 & &\text{Hypothesis} & & & & \cr & = 0 & &\text{@(a href="#theorem.MMZM")Theorem MMZM@(/a)} & & & & \cr & & & & } This establishes that x ∈N\kern -1.95872pt \left (AB\right ), so N\kern -1.95872pt \left (B\right ) ⊆N\kern -1.95872pt \left (AB\right ). To show that the inclusion does not hold in the opposite direction, choose B to be any nonsingular matrix of size n. Then N\kern -1.95872pt \left (B\right ) = \left \{0\right \} by Theorem NMTNS. Let A be the square zero matrix, O, of the same size. Then AB = OB = O by Theorem MMZM and therefore N\kern -1.95872pt \left (AB\right ) = {ℂ}^{n}, and is not a subset of N\kern -1.95872pt \left (B\right ) = \left \{0\right \}. T41 Contributed by David Braithwaite Statement [635] From the solution to Exercise MM.T40 we know that N\kern -1.95872pt \left (B\right ) ⊆N\kern -1.95872pt \left (AB\right ). So to establish the set equality (Definition SE) we need to show that N\kern -1.95872pt \left (AB\right ) ⊆N\kern -1.95872pt \left (B\right ). Suppose x ∈N\kern -1.95872pt \left (AB\right ). Then we know that ABx = 0 by Definition NSM. Consider \eqalignno{ 0 & = \left (AB\right )x & &\text{@(a href="fcla-jsmath-2.23li20.html#definition.NSM")Definition NSM@(/a)} & & & & \cr & = A\left (Bx\right ) & &\text{@(a href="#theorem.MMA")Theorem MMA@(/a)} & & & & } So, Bx ∈N\kern -1.95872pt \left (A\right ). Because A is nonsingular, it has a trivial null space (Theorem NMTNS) and we conclude that Bx = 0. This establishes that x ∈N\kern -1.95872pt \left (B\right ), so N\kern -1.95872pt \left (AB\right ) ⊆N\kern -1.95872pt \left (B\right ) and combined with the solution to Exercise MM.T40 we have N\kern -1.95872pt \left (B\right ) = N\kern -1.95872pt \left (AB\right ) when A is nonsingular. T51 Contributed by Robert Beezer Statement [635] We will work with the vector equality representations of the relevant systems of equations, as described by Theorem SLEMM. () Suppose y = w + z and z ∈N\kern -1.95872pt \left (A\right ). Then \eqalignno{ Ay & = A(w + z) & &\text{Substitution} & & & & \cr & = Aw + Az & &\text{@(a href="#theorem.MMDAA")Theorem MMDAA@(/a)} & & & & \cr & = b + 0 & &z ∈N\kern -1.95872pt \left (A\right ) & & & & \cr & = b & &\text{@(a href="fcla-jsmath-2.23li23.html#property.ZC")Property ZC@(/a)} & & & & \cr & & & & } demonstrating that y is a solution. () Suppose y is a solution to ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ). Then \eqalignno{ A(y − w) & = Ay − Aw & &\text{@(a href="#theorem.MMDAA")Theorem MMDAA@(/a)} & & & & \cr & = b − b & &\text{$y,\kern 1.95872pt w$ solutions to $Ax = b$} & & & & \cr & = 0 & &\text{@(a href="fcla-jsmath-2.23li23.html#property.AIC")Property AIC@(/a)} & & & & \cr & & & & } which says that y − w ∈N\kern -1.95872pt \left (A\right ). In other words, y − w = z for some vector z ∈N\kern -1.95872pt \left (A\right ). Rewritten, this is y = w + z, as desired. T52 Contributed by Robert Beezer Statement [635] ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) must be homogeneous. To see this consider that \eqalignno{ b & = Ax & &\text{@(a href="#theorem.SLEMM")Theorem SLEMM@(/a)} & & & & \cr & = Ax + 0 & &\text{@(a href="fcla-jsmath-2.23li23.html#property.ZC")Property ZC@(/a)} & & & & \cr & = Ax + Ay − Ay & &\text{@(a href="fcla-jsmath-2.23li23.html#property.AIC")Property AIC@(/a)} & & & & \cr & = A\left (x + y\right ) − Ay & &\text{@(a href="#theorem.MMDAA")Theorem MMDAA@(/a)} & & & & \cr & = b − b & &\text{@(a href="#theorem.SLEMM")Theorem SLEMM@(/a)} & & & & \cr & = 0 & &\text{@(a href="fcla-jsmath-2.23li23.html#property.AIC")Property AIC@(/a)} & & & & } By Definition HS we see that ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) is homogeneous.
# 2004 AIME II Problems/Problem 9 ## Problem A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $n+a_n.$ ## Solution 1 Let $x = a_2$; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$, and in general, $a_{2n} = f(n-1)f(n)$, $a_{2n+1} = f(n)^2$, where $f(n) = nx - (n-1)$.[1] From $$a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19$$, we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$. Thus $f(n) = 4n+1$, and we need to find the largest $n$ such that either $f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000$. This happens with $f(7)f(8) = 29 \cdot 33 = 957$, and this is the $2(8) = 16$th term of the sequence. The answer is $957 + 16 = \boxed{973}$. ^ We can show this by simultaneous induction: since \begin{align}a_{2n} &= 2a_{2n-1} - a_{2n-2} = 2a_{2(n-1)+1} - a_{2(n-1)} \\ &= 2f(n-1)^2 - f(n-2)f(n-1) \\ &= f(n-1)[2f(n-1) - f(n-2)] \\ &= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ &= f(n-1)f(n) \end{align} and \begin{align}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\ \end{align} ## Solution 2 Let $x = a_2$. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$, hence our answer is $957 + 16 = \boxed{973}$. ## Solution 3 We can find the value of $a_{9}$ by its bounds using three conditions: 1. $0 2. $a_{10} < a_{11}$ (note that the sequence must be increasing on all terms, not monotonically increasing) $a_{10} < \frac{a_{10}^2}{a_{9}} \rightarrow a_{9} < a_{10}$ 3. $a_{11} = \frac{a_{10}^2}{a_{9}} = \frac{(646-a_{9})^2}{a_{9}}$, so necessarily $a_{9}$ is a factor of $646^2$, which factorizes to $2^2\cdot 17^2 \cdot 19^2$ Rearranging conditions 1 and 2, we get: $$\frac{646}{3} < a_{9} < \frac{646}{2}$$ trying all the terms from the third condition, it is clear that $a_9 = 289$ is the only solution. Then we can calculate the next few terms from there since we have $a_{10}$ as well, to find that $a_{16} = 957$ and $a_{17} = 1089$, thus we have our answer of $957 + 16 = \boxed{973}$. ~KafkaTamura ## See also 2004 AIME II (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
How to solve radical equations. Video Tutorial and Step by step lesson # How to Solve Radical Equations ## Video Tutorial and practice problems ### How To solve Radical Equations • 1) Isolate radical on one side of the equation • 2) Square both sides of the equation to eliminate radical • 3) Simplify and solve as you would any equations • 4) Substitute answers back into original equation to make sure that your solutions are valid (there could be some extraneous roots that do not satisfy the original equation and that you must throw out) The video below and our examples explain these steps and you can then try our practice problems below. ### Practice Problems ##### Problem 1 Step 1 Step 2 Square both sides. Step 3 Solve expression. x = 10 Step 4 ##### Problem 2 Step 1 Step 2 Square both sides. Step 3 Solve expression. 3x = 23 Step 4 ##### Problem 3 Step 1 Step 2 Square both sides. Step 3 Solve expression. 0 =(x - 4)(x - 5) x = 4, x = 5 Step 4 $$\sqrt{3x -11} = 3x -x \\ \sqrt{3 (\color{Red}{4}) -11} = 3 \cdot (\color{Red}{4}) -\color{Red}{4} \\ \sqrt{1} = 8 \\ 1 \color{red}{ \ne } 8$$ $$\sqrt{3x -11} = 3x -x \\ \sqrt{3 (\color{Red}{5}) -11} = 3(\color{Red}{5}) - \color{Red}{5} \\ \sqrt{15 -11} = 15 - 5 \\ \sqrt{15 -11} = 15 - 5 \\ \sqrt{4} = 10 \\ 2 = 10 \\ \color{red}{ \ne } 10$$
## 10.201 Binomial coefficients and identities • Binomial Theorem • An expression consisting of 2 terms, connected by a + or - sign. • Examples: • $x + a$ • $2x - y$ • $x^2 - y^2$ • $x2 - 3y$ • As we increase power of binomials, expanding them becomes more complicated: • $(x + y)^1 = x + y$ • $(x + y)^2 = x^2 + 2xy + y^2$ • $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ • At very high orders, like $(x + y)^{30}$ binomial theorem becomes very useful. • Theorem: • Let x and y be variables, and n a non-negative integer. • The expansion of $(x + y)^n$ can be formalised as: • $(x + y)^n = \sum\limits_{k=0}^{n} \binom{n}{k} x^ky^{n-k}$ • Example: • What is the coefficient of $x^8y^7$ in the expansion of $(3x - y)^{15}$? • Solution: • We can view the expression as $(3x + (-y))^{15}$ • By the binomial theorem: • $(3x + (-y))^{15} = \sum\limits_{k=0}^{15} \binom{15}{k} (3x)^k(-y)^{15-k}$ • Consequently, the coeeficient of $x^8y^7$ in the expansion is obtained when k = 8: • $\binom{15}{8} (3)^8 (-1)^7 = -3^8 \frac{15!}{8!7!}$ • Application of the binomial theorem • Prove the identity: $2^n = \sum\limits_{k=0}^{n} \binom {n}{k}$ • Using binomial theorem: • With x = 1 and y = 1, from the binomial theorem we see that the identity is verified. • Using sets: • Consider the subsets of a set with n elements. • There are subsets with zero elements, with one element, with two elements and so on...with n elements. • Therefore the total number of subsets is $\sum\limits_{k=0}^{n} \binom{n}{k}$ • Also, since we know that a set with n elements has $2^n$ subsets, we can conclude that: $2^n = \sum\limits_{k=0}^{n} \binom{n}{k}$ • Pascal's identity • If n and k are integers with $n \geq k \geq 1$, then: • $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$ • Proof: • Let T be a set where $\|T\| = n + 1$, $a \in T$, and $S = T - {a}$ • There are $\binom{n+1}{k}$ subsets of T containing k elements. Each of these subsets either: • contains a with k - 1 other elements, or • contains k elements of S and not a • There are • $\binom{n}{k-1}$ subsets of k elements that contain a • $\binom{n}{k}$ subsets of k elements of T that don't contain a • Hence, $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$ • It's the basis for geometic arrangements of the binomial coefficients in a triangle. This is called Pascal's Triangle. • Pascal's triangle is a number triangle with numbers arranged in staggered rows such that $a_{n,r}$ is the binomial coefficients of $\binom{n}{r}$ Using Pascal's identity, we can show the result of adding 2 adjacent coefficients in the triangle is equal to the binomial coefficient in the next row between these 2 coefficients. ## Video: 10.204 Generalised permutations and combinations • Permutations with repetition • The number of r-permutations for a set of n objects when repitition allowed is $n^r$ • Proof: • Since we always have n choices each time: • n possibilities for 1st choice. • n possibilities for 2nd choice. • ... • n possibilities when choosing the last number. • By product rule, multiplying each time: • (do this r times) $n \times n \times n \times ... \times n = n^r$ • Example: • How many strings of length r can be formed if we are using only uppercase letters in the English alphabet? • Solution: • The number of such strings is $26^r$, which is the number of r-permutations with repetition of a set with 26 elements. • Permutations without repetition • When we have permutations without repetition, we reduce the number of choices available each time by 1. • So, the number of r-permutations of a set with n objects without reps is: • (do this r times) $n \times n-1 \times n-2 \times ... \times n-r+1$ • $P(n, r) = n(n -1)(n -2) ... (n - r + 1) = \frac{n!}{(n - r)!}$ • Example • 10 runners take part in running competition. How many different ways can the 1st and 2nd place be awarded? • Solution: • $P(10, 2) = P^2_{10} = \frac{10!}{(10 -2)!} = \frac{10!}{8!} = 90$ • Combination with repetition • The number of ways in which k objects can be selected from n categories of objects, with repetition permitted, can be calculated as: $\binom{k+n-1}{k} = \frac{(k+n-1)!}{k!(n-1)!}$ • It is also the total numebr of ways to put k identical balls into n distinct boxes. • It is also the total number of functions from a set of k identical elements to a set of n distinct elements. • Example • Find all multisets of size 3 from the set {1, 2, 3, 4} • Solution: • Using bars and crossed, think of the values 1, 2, 3, 4 as four categories. • Denote each multiset of size 3 by placing 3 crosses in the various categories. • For example, the multiset {1, 1, 3} is represented by x x \| \| x \| • This countining problem can be modelled as distributing the 3 crosses among the 3+4-1 positions, with the remaining positions occupied by bars • Therefore, the number of multisets of size 3 is: $C(6, 3) = \frac{6!}{3!3!} = 20$ • Combination without repetition • The number of ways in which k objects can be selected from n categories of objects with repetition not permitted can be calculated as: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ • Counting problem is the same as the number of ways to put k identitcal balls into n distinct boxes, where each box receives at most one ball. • It is also the number of one-to-one functions from a set of k identical elements into a set of n distinct elements. • It is also the number of k-element subsets of an n-element set. • Choice of formulas • We have discussed four ways of selecting k objects from a set with n elements: • the order in which choices are made may or may not matter. • repetition may or may not be alloed. • This table summarises the formula in each case: Order matters Order does not matter Repetition is not permitted $\frac{n!}{n - k!}$ $\frac{n!}{k!(n-k)!}$ Repetition is permitted $n^{k}$ $\frac{(k + n - 1)!}{k!(n-1)!}$ • Example • John is a chair of a committee. In how many ways can a committee of 3 be chosen from 10 people, given that John must be one of the people selected? • Solution: • Since John is already chosen, we need to choose another 2 out of 9 people • In choosing a committee, the order doesn't matter, so we need to apply the combination without repetition formula: $C(9, 2) = \frac{9!}{2!(9-2)!} = 36$ ways. ## Video: 10.206 Distinguishable objects and boxes • Many counting problems can be reduced into finding the number of ways objects can be placed into boxes. • Distributing objects into boxes. • We can phrase counting problems in terms of distributing k objects into n boxes, under various conditions: • The objects can be distinguishable or indistinguishable. • The boxes can be either distinguishable or indistinguishable. • The distribution can be done either with exclusion or without exclusion. • Distinguishable means they are marked in someway that allows you to distinguish from each other. • Indistinguishable means objects or boxes are identical so you can't tell them apart. • When placing indistinguishable objects into distinguishable boxes, it makes no difference which object is placed into which box. • The term "with exclusion" means no box can contain more than one object. • "without exclusion" means box can contain more than one object. • Distinguishable objects into distinguishable boxes with exclusion • We want to distribute k balls, numbered from 1 to k, into n boxes, numbered from 1 to n, so that no box gets more than one ball. • Equivalent to making an ordered selection of k boxes from n boxes, where the balls do the selecting for us: • the ball labelled 1 chooses the first box. • ball labelled 2 gets 2nd box. • etc • Theorem: • Distributing $k$ distinguishable balls into n distinguishable boxes, with exclusion, is equivalent to forming a permutation of size $k$ from a set of size $n$. • Therefore, the number of ways of placing k distinguishable balls into n distinguishable boxes is as follows: • $P(n, k) = n(n - 1)(n - 2) ... (n - k + 1) = \frac{n!}{(n-k)!}$ • Distinguishable objects into distinguishable boxes without exclusion • In this case, we want to distribute k balls, numbered from 1 to k, into n boxes, numbered from 1 through n, without restrictions on the number of balls in each box • This is equivalent to making an ordered selection of k boxes from n, with repetition, where the balls do the selecting for us: • the ball labelled 1 chooses the first box. • the ball labelled 2 chooses the 2nd box. • so on... • Theorem: • Distributing k distinguishable balls into n distinguishable boxes, without exclusion, is equivalent to forming a permutation of size k from a set of size n, with repetition. • Therefore, there are: • $n^k$ different ways • Indistinguishable objects into distinguishable boxes with exclusion. • In this case, we want to distribute k balls, into n boxes, numbered from 1 through n, in such a way that no box receives more than one ball. • Theorem: • Distributing k indistinguishable balls into n distinguishable boxes, with exclusion, is equivalent to forming a combination of size k from a set of size n. • Therefore, there are: • $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ different ways. • Indistinguishable objects into distinguishable boxes without exclusion. • Want to distribute k balls, into n boxes, numbered from 1 through n, without restrictions on the number of balls in each box. • Theorem: • Distributing k indistinguishable balls into n distinguishable boxes, without exclusion, is equivalent to forming a combination of size k from a set of size n, with repetition. • Therefore, there are: • $\binom{n+k-1}{k} = \frac{n+k-1}{k!(n-1)!}$ different ways. • Example • How many ways are there of placing 8 indistinguishable balls into 6 distinguishable boxes? • $\binom{8+6-1}{8} = \binom{13}{8} = \frac{13!}{8!5!} = 1287$ # Week 20 - Peer-graded Assignment: 10.208 Combinatorics ## Question Find the number of positive integers not exceeding 2310. Find the number of positive integers not exceeding 2310 that are either the square or the cube of an integer. Hint: use the subtraction rule, also known as the principle of inclusion-exclusion. Let $A$ be the set of all positive integers not exceeding 2310. And $A_1$ the set of all positive integers, perfect square and not exceeding 2310. And $A_2$, the set of all positive integers, perfect cube and not exceeding 2310 Let $A$ be set of all positive integers not exceeding 2310. $A = \{1, 2, 3 ... 2310\}$ $\|A\| = 2310$ Let $A_1$ be the set of all positive integers that are a perfect square and not exceeding 2310. $A_1 = \{1, 4, 9 ... 2304\}$ $\|A_1\| = 48$ Let $A_2$ be the set of all positiev integers that are a perfect cube and not exceeding 2310. $A_2 = \{1, 8, 27...2197\}$ $\|A_2\| = 13$ To find the number of positive integers not exceeding 2310 that are either the square or the cube of an integer, we want to find: $\|A_1 \cup A_2\|$ Using the inclusion-exclusion rule, we know that: $\|A_1 \cup A_2\| = \|A_1\| + \|A_2\| - \|A_1 \cap A_2\|$ We know that: $A_1 \cap A_2 = \{1, 64, 729\}$ $\|A_1 \cap A_2\| = 3$ Therefore: $\|A_1 \cup A_2\| = 48 + 13 - 3$ $\|A_1 \cup A_2\| = 58$
# Parts of a Circle: Definition, Formula, Examples Home » Math Vocabulary » Parts of a Circle: Definition, Formula, Examples ## What Are the Parts of a Circle? There are many parts of a circle, which include center, radius, diameter, circumference, etc. In this article, we will learn about the circle, the different parts of a circle and their definitions in detail. ### Definition of a Circle In our daily lives, we come across shapes such as clocks, plates, round tables, wheels, coins, etc. These are all round, i.e., circular in shape. A circle is a round two-dimensional plane figure, which is formed by the set of all those points that are at a fixed distance from a fixed point in the same plane. The circle looks similar to the letter “O.” Based on the shape and position, there are different parts of a circle in geometry that can be listed as center, radius, circumference, diameter, chord, arc, segment, tangent, secant, and sector. It will be easy to identify and label parts of a circle using the diagram shown below. ## Circle and Its Parts Let’s discuss different parts of circles in detail. Identifying parts of a circle becomes easy once you understand the definitions. ### Center of a Circle The fixed point in the plane is called the center of the circle. The fixed distance from the center to the boundary of the circle is called the radius of the circle. Generally, the radius of a circle is denoted by “r.” ### Circumference of a Circle The total distance measured once around the circle is known as the circumference of the circle. Circumference of a circle $= 2\pi r$ Where, $r =$ radius of the circle. $\pi \simeq 3.14$ or $\frac{22}{7}$ (approx value) ### Diameter of a Circle A line that crosses from the center of the circle starting from one point to the other point of the circumference is called a diameter. ### Chord of a Circle The chord of a circle is a line segment that joins two different points on the circumference of the circle. The diameter is the largest chord of a circle. ### Arc of a Circle A part or segment of the circumference of a circle is called the arc. ### Segment of a Circle A segment of a circle is a part of the circle bounded by a chord and an arc. ### Tangent of a Circle The tangent of a circle is a straight line that touches the circle at a single point. ### Sector of a Circle A sector is a part of a circle that is created by two radii. There are two types of sectors. #### Major sector The sector of the circle that occupies the larger portion is called the major sector. #### Minor sector The sector of the circle that occupies the smaller portion is called the minor sector. ### Secant of a Circle The secant of a circle is the extension of the chord of a circle to the outside of the circle. Let’s summarize the parts of a circle in one picture! ## Regions of Circle Any closed geometric figure has three regions. Since a circle is also a closed figure, it, too, has three regions. A circle has an inside, an outside, and an “on” because we could be right on the circle. A point can lie in any of these three regions of the circle, namely the Interior region, the exterior region, or on the circle. ### Interior Region of the Circle The region inside the circle is called its interior region and the point that lies in this region is called the interior point. ### The Exterior Region of the Circle The region outside the circle is called the exterior region of the circle and any point lying outside the circle is said to be in its exterior region. ### On the Circle Any point on the boundary of the circle is said to be on the circle. In this image, Point A is outside of the circle, Point B is inside of the circle, and Point C is on the circle. ## Facts • The word “circle” is derived from the Greek word κρίκος (krikos), meaning “hoop” or “ring.” • A circle is a curved shape with no edges. • The diameter is the longest distance between two points on a circle. Alternatively, we can also say that the diameter is the longest chord of the circle. • The measurement of the full arc of a circle is 360 degrees. • When we divide the length of the circumference by the length of the diameter, we get 3.141592654 . . ., which is the number $\pi$ (Pi). ## Conclusion In this article, we saw some of the real-life examples of a circle, found out the definition of a circle, and learned about the different parts and regions of a circle. Now, let’s look at some examples and do a few practice questions to better understand the subject. ## Solved Examples 1. Find the radius of a circle with a diameter of 12.5 inches. Solution: Given: the radius of the circle $= 12.5$ inches We know that the diameter of a circle $=$ Radius of circle$\div 2$ $= 12.5$ inches$\div 2 = 6.25$ inches. $\therefore$ Diameter of a circle $= 6.25$ inches. 1. Find the circumference of a coin if its radius is 77 inches. Solution: Given: the radius of the circle $= 77$ inches We know that circumference $= 2\pi r$ $\Rightarrow$Circumference $= 2\pi r = 2 \times \frac{22}{7} \times 77 = 2 \times 22 \times 11 = 484$ inches. $\therefore$ Circumference of the circle $= 484$ inches. 1. Find the diameter of a circle if its circumference is 110 feet. Solution: Given: the circumference $= 12.5$ feet We know that the circumference $= 2\pi r$ $\Rightarrow r =$ Circumference/$2\pi$ $\Rightarrow 2r =$ Circumference/$\pi = \frac{110}{3.14} = 35.03$ $\therefore$Diameter of circle $= 35.03$ feet. 1. Identify the radius, diameter, and chord In the figure given below. Solution: OA, OB, and OC are the radii. AB is the diameter of a circle. Since the diameter is the longest chord, AB is also a chord. Hence, PQ and AB are the chords of the circle. 1. Identify the points and write whether each is the interior or exterior region of the circle or on the circle. Solution: The interior points of the circle are C and D. The exterior points of the circle are A and F. The points on the circle are B and E. ## Practice Problems Parts of a Circle 1 ### Which of the following is a part of a circle? Tangent Secant All of the above. CorrectIncorrect Correct answer is: All of the above. Radius, tangent, and secant are all a part of a circle. 2 ### Which of the following is not a circle? The wheel of a bicycle A car tire Notebook Coin CorrectIncorrect Notebooks are usually rectangular or square. 3 ### The total number of chords of a circle is 1 2 3 Infinite CorrectIncorrect By joining any two points on the circle, we can draw an infinite (uncountable) number of chords of the circle. 4 ### If a circle has a diameter of 11 inches, what is the length of its longest chord? 5.5 inches 11 inches 22 inches 33 inches CorrectIncorrect A diameter is the longest chord of a circle that passes through the center. Thus, the length of its longest chord $=$ diameter of circle $= 11$ inches 5 ### A circle with center O has a radius of 8 inches and OP $= 6$ inches. Where does point P lie? On the circle In the interior of the circle In the exterior of the circle At the center CorrectIncorrect Correct answer is: In the interior of the circle The length of OP is less than the radius of the circle. So point P lies in the interior of the circle. ## Frequently Asked Questions Parts of a Circle The diameter of a circle divides the circular region into two equal parts, each part is called a semicircular region. From two given points, an infinite number of circles can be drawn, as shown in the figure below. Circumference of circle $4= 2\pi r$ Area of circle $= \pi r^{2}$, where r is the radius. Concentric circles are circles having the same center. An incircle is an inscribed circle of a polygon. It is the largest circle that fits inside the polygon and touches each side of the polygon only at one point. For a triangle, the center of an incircle is the point of intersection of three angle bisectors. A circumcircle or circumscribed circle is a circle that passes through each of the three vertices of a triangle. For a triangle, the center of the circumcenter is the point of intersection of the perpendicular bisectors of three sides.
# Do Now 12/7/09 Take out HW from last night. Copy HW in your planner. ## Presentation on theme: "Do Now 12/7/09 Take out HW from last night. Copy HW in your planner."— Presentation transcript: Do Now 12/7/09 Take out HW from last night. Copy HW in your planner. Text page 190, #16-36 evens Copy HW in your planner. Text page 190, #40-68 evens Be ready to copy POTW #4 Homework Text page 190, #16-36 evens 16) 36 18) 16 20) 42 22) 165 24) 12 26) 420 28) 80a³ 30) 51b³ 32) 120s 34) 200a² 36) 24 figures 4 Objective SWBAT find the least common multiple of two numbers. Section 4.4 “Least Common Multiple” A multiple of a number is the product of the number and any other number greater than zero. What are the multiples of 5? Think of counting by fives… 5: ,5 10, 15, 20, 25, 30, 35, 40, 45, 50, 55 … Least Common Multiple Of all the common multiples of two numbers, the smallest is the least common multiple. Least Common Multiple Find the LCM of 180 and 378. When the numbers are too large to list the multiples of the two numbers, find the prime factorization of each number. The LCM is the product of the common prime factors and the factors that are not common. Common Factors Not Common Factors 180: 2 · 2 · 3 · 3 · 5 2 · 3 · · 5 3 · 7 378: 2 · 3 · 3 · 3 · 7 From the list you can see that 2, 3, and 3 are common prime Factors, and the factors that are not common are 2, 3, 5, and 7. The LCM is then 2 · 3 · 3 · 2 · 5 · 3 · 7 which is 3780. Least Common Multiple Find the LCM of 15, 30, and 50. 15: 3 · 5 5 When you have three or more numbers, the common prime factors may only be shared by two of the numbers. The LCM is the product of the common prime factors and the factors that are not common. Common Factors Not Common Factors 15: · 5 5 2 · 3 · 5 30: 2 · 3 · 5 50: 2 · 5 · 5 From the list you can see that 2, 3, and 5 are common prime factors, and the factor that is not common is 5. The LCM is then 2 · 3 · 5 · 5 which is 150. Least Common Multiple Find the LCM of 10ab and 6b. 10ab: 2 · 5 · a · b The LCM is the product of the common prime factors and the factors that are not common. Common Factors Not Common Factors 10ab: 2 · 5 · a · b 2 · b · a 3 6b: 2 · 3 · b From the list you can see that 2 and b are common prime factors, and the factors that are not common are 3, 5, and a. The LCM is then 2 · b · 5 · a · 3 which is 30ab. Least Common Multiple Find the LCM of 2a³b and 3ab. 5 The LCM is the product of the common prime factors and the factors that are not common. Common Factors Not Common Factors 2a³b : 2 · a · a · a · b a · b · a · a 3 · b · b · b · b 5 3ab : 3 · a · b · b · b · b · b From the list you can see that a and b are common prime factors, and the factors that are not common are 2, 3, a, a, b, b, b, and b. The LCM is then a · b · 2 · a · a · 3 · b · b · b · b which is 6a³b. 5 Least Common Denominator The LCD of two or more fractions is the least common multiple of the denominators. Use the LCD to compare and order fractions. “Using the LCD” = = = = = Compare the fractions and . List the equivalent fractions and find two with the same denominator. Then compare. = = same denominator, now compare Therefore, 3/4 is greater than 2/3. = = = Compare the Following Fractions < > < Order the fractions from least to greatest. Find the LCD for 4, 9, & 15. LCD = 180 Rewrite equivalent fractions using the LCD. Order the fractions from least to greatest. Find the LCD for 6, 9, & 3. LCD = 18 Rewrite equivalent fractions using the LCD. Rewrite the variable expressions with a common denominator. Find the LCD for 5b and 4ab². LCD = 20ab² Rewrite equivalent fractions using the LCD. Rewrite the variable expressions with a common denominator. Find the LCD for 5xy and 4y². LCD = 20xy² Rewrite equivalent fractions using the LCD. Rewrite the variable expressions with a common denominator. CHALLENGE Rewrite the variable expressions with a common denominator. Homework Text page 190, #40-68 evens
# By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms. Size: px Start display at page: Download "By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms." Transcription 1 SECTION 5.4 Special Factoring Techniques Special Factoring Techniques OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication of binomials from Section 4.6, we get rules for factoring polynomials in certain forms. OBJECTIVE 1 Factor a difference of squares. The formula for the product of the sum and difference of the same two terms is Factoring a Difference of Squares 1x + y21x - y2 = x 2 - y 2. Reversing this rule leads to the following special factoring rule. x 2 y 2 1x y21x y2 For example, The following conditions must be true for a binomial to be a difference of squares. 1. Both terms of the binomial must be squares, such as x 2, 9y 2 =13y2 2, m 2-16 = m =1m + 421m = 5 2, 1 = 1 2, m 4 =1m The terms of the binomial must have different signs (one positive and one negative). EXERCISE 1 Factor each binomial if possible. x x EXAMPLE 1 Factoring Differences of Squares Factor each binomial if possible. x 2 - y 2 y21x a 2-49 = a =1a + 721a - 72 = 1x + x 2-8 Because 8 is not the square of an integer, this binomial does not satisfy the conditions above. It is a prime polynomial. (d) p Since p is a sum of squares, it is not equal to 1 p p Also, we use FOIL and try the following. 1 p p p p Thus, p is a prime polynomial. y2 = p 2-8p + 16, not p = p 2 + 8p + 16, not p y 2 - m 2 =1 y + m21 y - m2 ANSWERS 1. 1x x prime CAUTION As Example 1(d) suggests, after any common factor is removed, a sum of squares cannot be factored. 2 318 CHAPTER 5 Factoring and Applications EXERCISE 2 Factor each difference of squares. 9t a 2-49b 2 EXAMPLE 2 Factoring Differences of Squares Factor each difference of squares. 25m 2-16 =15m =15m m z 2-64t 2 x 2 =17z2 2-18t2 2 - =17z + 8t217z - 8t2 y 2 = 1x + y21x y2 Write each term as a square. Factor the difference of squares. - NOTE Always check a factored form by multiplying. EXERCISE 3 Factor completely. 16k 2-64 m v EXAMPLE 3 Factor completely. 81y 2-36 = 919y 2-42 = 9313y Factoring More Complex Differences of Squares = 913y y - 22 Neither binomial can be factored further. Don t stop here. m 4-16 p 4-36 =1m =1 p =1 p p 2-62 =1m m 2-42 =1m m + 221m - 22 Factor out the GCF, 9. Write each term as a square. Factor the difference of squares. Write each term as a square. Factor the difference of squares. Factor the difference of squares. Factor the difference of squares again. CAUTION Factor again when any of the factors is a difference of squares, as in Example 3. Check by multiplying. ANSWERS 2. 13t t a + 7b216a - 7b k + 221k m m v v + 521v - 52 OBJECTIVE 2 Factor a perfect square trinomial. The expressions 144, 4x 2, and 81m 6 are called perfect squares because 144 = 12 2, 4x 2 =12x2 2, and 81m 6 =19m A perfect square trinomial is a trinomial that is the square of a binomial. For example, x 2 + 8x + 16 is a perfect square trinomial because it is the square of the binomial x + 4. x 2 + 8x + 16 =1x + 421x + 42 =1x 3 SECTION 5.4 Special Factoring Techniques 319 On the one hand, a necessary condition for a trinomial to be a perfect square is that two of its terms be perfect squares. For this reason, 16x 2 + 4x + 15 is not a perfect square trinomial, because only the term 16x 2 is a perfect square. On the other hand, even if two of the terms are perfect squares, the trinomial may not be a perfect square trinomial. For example, x 2 + 6x + 36 has two perfect square terms, x 2 and 36, but it is not a perfect square trinomial. Factoring Perfect Square Trinomials x 2 2xy y 2 1x y2 2 x 2 2xy y 2 1x y2 2 The middle term of a perfect square trinomial is always twice the product of the two terms in the squared binomial (as shown in Section 4.6). Use this rule to check any attempt to factor a trinomial that appears to be a perfect square. EXERCISE 4 Factor y y EXAMPLE 4 Factoring a Perfect Square Trinomial Factor x x The x 2 -term is a perfect square, and so is 25. Try to factor x x + 25 as 1x To check, take twice the product of the two terms in the squared binomial. 2 # x # 5 = 10x Middle term of x x + 25 Twice First term Last term of binomial of binomial Since 10x is the middle term of the trinomial, the trinomial is a perfect square. x x + 25 factors as 1x EXAMPLE 5 Factoring Perfect Square Trinomials Factor each trinomial. x 2-22x The first and last terms are perfect squares 1121 = 11 2 or Check to see whether the middle term of x 2-22x is twice the product of the first and last terms of the binomial x # x #1-112 = -22x Middle term of x 2-22x Twice First Last term term Thus, x 2-22x is a perfect square trinomial. x 2-22x factors as 1x Same sign ANSWER 4. 1 y Notice that the sign of the second term in the squared binomial is the same as the sign of the middle term in the trinomial. 4 320 CHAPTER 5 Factoring and Applications EXERCISE 5 Factor each trinomial. t 2-18t p 2-28p x 2 + 6x + 4 (d) 80x x x 9m 2-24m + 16 =13m m =13m Twice First Last term term 25y y + 16 The first and last terms are perfect squares. and Twice the product of the first and last terms of the binomial 5y + 4 is which is not the middle term of This trinomial is not a perfect square. In fact, the trinomial cannot be factored even with the methods of the previous sections. It is a prime polynomial. (d) 12z z z = 3z14z z y 2 =15y2 2 2 # 5y # 4 = 40y, 25y y = 4 2 Factor out the common factor, 3z. = 3z312z z z2 + 20z + 25 is a perfect square trinomial. = 3z12z Factor. NOTE 1. The sign of the second term in the squared binomial is always the same as the sign of the middle term in the trinomial. 2. The first and last terms of a perfect square trinomial must be positive, because they are squares. For example, the polynomial x 2-2x - 1 cannot be a perfect square, because the last term is negative. 3. Perfect square trinomials can also be factored by using grouping or the FOIL method, although using the method of this section is often easier. OBJECTIVE 3 Factor a difference of cubes. We can factor a difference of cubes by using the following pattern. Factoring a Difference of Cubes x 3 y 3 1x y21x 2 xy y 2 2 ANSWERS 5. 1t p prime (d) 5x14x This pattern for factoring a difference of cubes should be memorized. To see that the pattern is correct, multiply 1x - y21x 2 + xy + y 2 2. x 2 + xy + y 2 x - y Multiply vertically. (Section 4.5) -y1x 2 + xy + y x 2 y - xy 2 - y 3 x 3 + x 2 y + xy 2 x1x2 + xy + y 2 2 x 3 - y 3 Add. 5 SECTION 5.4 Special Factoring Techniques 321 Notice the pattern of the terms in the factored form of x 3 - y 3. x 3 - y 3 = (a binomial factor)(a trinomial factor) The binomial factor has the difference of the cube roots of the given terms. The terms in the trinomial factor are all positive. The terms in the binomial factor help to determine the trinomial factor. x 3 - y 3 =1x - y21 positive First term product of second term squared + the terms + squared x 2 + xy + y 2 2 CAUTION The polynomial x 3 - y 3 is not equivalent to 1x - y2 3. x 3 - y 3 1x - y2 3 =1x - y21x 2 + xy + y 2 2 =1x - y21x - y21x - y2 =1x - y21x 2-2xy + y 2 2 EXERCISE 6 Factor each polynomial. a t k (d) 125x 3-343y 6 ANSWERS 6. 1a - 321a 2 + 3a t t t k - 421k 2 + 4k (d) 15x - 7y 2 2 # 125x xy y 4 2 EXAMPLE 6 Factor each polynomial. (d) Factoring Differences of Cubes m Let x = m and y = 5 in the pattern for the difference of cubes. m = m =1m - 521m 2 + 5m p 3-27 =12p x 3 y 3 =1m - 521m 2 + 5m =12p p p =12p p 2 + 6p m 3-32 = 41m 3-82 = 41m t 3-216s 6 =15t2 3-16s = 1x - y21x p2 2 = 2 2 p 2 = 4p 2, NOT 2p 2. = 41m - 221m 2 + 2m + 42 =15t - 6s t t16s s =15t - 6s t ts s 4 2 xy + y 2 2 8p 3 =12p2 3 and 27 = 3 3. Let x = 2p, y = 3. Let x = m, y = = 25 Apply the exponents. Multiply. Factor out the common factor, 4. 8 = 2 3 Factor the difference of cubes. Write each term as a cube. Factor the difference of cubes. Apply the exponents. Multiply. 6 322 CHAPTER 5 Factoring and Applications CAUTION A common error in factoring a difference of cubes, such as x 3 - y 3 =1x - y21x 2 + xy + y 2 2, is to try to factor x 2 + xy + y 2. This is usually not possible. OBJECTIVE 4 Factor a sum of cubes. A sum of squares, such as cannot be factored by using real numbers, but a sum of cubes can. m , Factoring a Sum of Cubes x 3 y 3 1x y21x 2 xy y 2 2 Compare the pattern for the sum of cubes with that for the difference of cubes. Positive x 3 - y 3 =1x - y21x 2 + xy + y 2 2 Difference of cubes Same sign Opposite sign Positive The only difference between the patterns is the positive and negative signs. x 3 + y 3 =1x + y21x 2 - xy + y 2 2 Sum of cubes Same sign Opposite sign EXERCISE 7 Factor each polynomial. x a 3 + 8b 3 ANSWERS 7. 1x + 521x 2-5x a + 2b219a 2-6ab + 4b 2 2 EXAMPLE 7 Factoring Sums of Cubes Factor each polynomial. k = k = 3 3 =1k + 321k 2-3k Factor the sum of cubes. =1k + 321k 2-3k + 92 Apply the exponent. 8m n 3 =12m n2 3 8m 3 =12m2 3 and 125n 3 =15n2 3. =12m + 5n2312m2 2-2m15n2 +15n2 2 4 Factor the sum of cubes. =12m + 5n214m 2-10mn + 25n a b 3 =110a b2 3 Be careful: 12m2 2 = 2 2 m 2 and 15n2 2 = 5 2 n 2. =110a 2 + 3b23110a a 2 213b2 +13b2 2 4 Factor the sum of cubes. =110a 2 + 3b21100a 4-30a 2 b + 9b a = a = 100a 4 7 SECTION 5.4 Special Factoring Techniques 323 The methods of factoring discussed in this section are summarized here. Special Factorizations Difference of squares x 2 y 2 1x y21x y2 Perfect square trinomials x 2 2xy y 2 1x y2 2 x 2 2xy y 2 1x y2 2 Difference of cubes x 3 y 3 1x y21x 2 xy y 2 2 Sum of cubes x 3 y 3 1x y21x 2 xy y 2 2 The sum of squares can be factored only if the terms have a common factor. 5.4 EXERCISES Complete solution available on the Video Resources on DVD 1. Concept Check To help you factor the difference of squares, complete the following list of squares. 1 2 = 6 2 = 11 2 = 16 2 = 17 2 = 18 2 = 19 2 = 20 2 = 2. Concept Check The following powers of x are all perfect squares: x 2, x 4, x 6, x 8, x 10. On the basis of this observation, we may make a conjecture (an educated guess) that if the power of a variable is divisible by (with 0 remainder), then we have a perfect square. 3. Concept Check To help you factor the sum or difference of cubes, complete the following list of cubes. 1 3 = 2 2 = 7 2 = 12 2 = 2 3 = 3 2 = 8 2 = 13 2 = 3 3 = 4 2 = 9 2 = 14 2 = 4 3 = 5 2 = 10 2 = 15 2 = 5 3 = 6 3 = 7 3 = 8 3 = 9 3 = 10 3 = 4. Concept Check The following powers of x are all perfect cubes: x 3, x 6, x 9, x 12, x 15. On the basis of this observation, we may make a conjecture that if the power of a variable is divisible by (with 0 remainder), then we have a perfect cube. 5. Concept Check Identify each monomial as a perfect square, a perfect cube, both of these, or neither of these. 64x 6 y t 6 49x 12 (d) 81r Concept Check What must be true for x n to be both a perfect square and a perfect cube? Factor each binomial completely. If the binomial is prime, say so. Use your answers from Exercises 1 and 2 as necessary. See Examples y t x x m k m x r x x a 2-8 8 324 CHAPTER 5 Factoring and Applications p q r 2-25a m 2-100p x w p r x y 4-10, p k Concept Check When a student was directed to factor k 4-81 from Exercise 30 completely, his teacher did not give him full credit for the answer 1k k The student argued that since his answer does indeed give k 4-81 when multiplied out, he should be given full credit. WHAT WENT WRONG? Give the correct factored form. 32. Concept Check The binomial 4x is a sum of squares that can be factored. How is this binomial factored? When can the sum of squares be factored? Concept Check Find the value of the indicated variable. 33. Find b so that x 2 + bx + 25 factors as 1x Find c so that 4m 2-12m + c factors as 12m Find a so that ay 2-12y + 4 factors as 13y Find b so that 100a 2 + ba + 9 factors as 110a Factor each trinomial completely. See Examples 4 and w 2 + 2w p 2 + 4p x 2-8x x 2-10x x x y y x 2-40x y 2-60y x 2-28xy + 4y z 2-12zw + 9w x xy + 9y t tr + 16r h 2-40hy + 8y x 2-48xy + 32y k 3-4k 2 + 9k 52. 9r 3-6r r z 4 + 5z 3 + z x 4 + 2x 3 + x 2 Factor each binomial completely. Use your answers from Exercises 3 and 4 as necessary. See Examples 6 and a m m b k p x y p y w 3-216z x x y 3-8x x 3-16y w 3-216z p q x y a b m 3 + 8p t 3 + 8s r s x 3-125y t 3-64s m 6 + 8n r s x 9 + y x 9 - y 9 9 Summary Exercises on Factoring 325 Although we usually factor polynomials using integers, we can apply the same concepts to factoring using fractions and decimals. z = z 2 - a 3 4 b = A3 4B 2 = az baz b Factor the difference of squares. Apply the special factoring rules of this section to factor each binomial or trinomial. 83. p 84. q b x y t 90. m x 2-1.0x m t y 93. x y Brain Busters Factor each polynomial completely. x m + n2 2-1m - n a - b2 3-1a + b m 2 - p 2 + 2m + 2p 98. 3r - 3k + 3r 2-3k 2 36m PREVIEW EXERCISES Solve each equation. See Sections 2.1 and m - 4 = t + 2 = t + 10 = x = 0 SUMMARY EXERCISES on Factoring As you factor a polynomial, ask yourself these questions to decide on a suitable factoring technique. Factoring a Polynomial 1. Is there a common factor? If so, factor it out. 2. How many terms are in the polynomial? Two terms: Check to see whether it is a difference of squares or a sum or difference of cubes. If so, factor as in Section 5.4. Three terms: Is it a perfect square trinomial? If the trinomial is not a perfect square, check to see whether the coefficient of the seconddegree term is 1. If so, use the method of Section 5.2. If the coefficient of the second-degree term of the trinomial is not 1, use the general factoring methods of Section 5.3. Four terms: Try to factor the polynomial by grouping, as in Section Can any factors be factored further? If so, factor them. (continued) 10 326 CHAPTER 5 Factoring and Applications Match each polynomial in Column I with the best choice for factoring it in Column II. The choices in Column II may be used once, more than once, or not at all x x x 2-17x + 72 I 3. 16m 2 n + 24mn - 40mn a 2-121b p 2-60pq + 25q 2 6. z 2-4z r x 6 + 4x 4-3x w z 2-24z Factor each polynomial completely. II A. Factor out the GCF. No further factoring is possible. B. Factor a difference of squares. C. Factor a difference of cubes. D. Factor a sum of cubes. E. Factor a perfect square trinomial. F. Factor by grouping. G. Factor out the GCF. Then factor a trinomial by grouping or trial and error. H. Factor into two binomials by finding two integers whose product is the constant in the trinomial and whose sum is the coefficient of the middle term. I. The polynomial is prime. 11. a 2-4a a a y 2-6y y y 5-168y a + 12b + 18c 16. m 2-3mn - 4n p 2-17p z 2-6z + 7z z 2-7z m 2-10m x 3 y xy y a 5-8a 4-48a k 2-10k z 2-3za - 10a z x 2-4x - 5x n 2 r nr 3-50n 2 r 29. 6n 2-19n y y x m 2 + 2m y 2-5y m z z x x k 2-12k p p m 2-24z m 2-2m k 2 + 4k a 3 b 5-60a 4 b a 6 b k 3 + 7k 2-70k r - 5s - rs 45. y y 5-30y m - 16m k z y 2 - y k p 10-45p 9-252p m m m m r rm + 9m z 2-12z h hg - 14g z 3-45z z 59. k 2-11k p 2-100m 2 11 Summary Exercises on Factoring k 3-12k 2-15k 62. y 2-4yk - 12k p r m + 2p + mp 66. 2m 2 + 7mn - 15n z 2-8z m 4-400m 3 n + 195m 2 n m 2-36m a 2-81y x 2 - xy + y y z z 2-16z m m m + 12n + 3mn q - 6p + 3pq 77. 6a a y 6-42y 5-120y a 3 - b 3 + 2a - 2b k 2-48k m 2-80mn + 25n y 3 z y 2-24y 4 z k 2-2kh - 3h a 2-7a x a y 2-7yz - 6z m 2-4m a ab - 3b a RELATING CONCEPTS EXERCISES FOR INDIVIDUAL OR GROUP WORK A binomial may be both a difference of squares and a difference of cubes. One example of such a binomial is x 6-1. With the techniques of Section 5.4, one factoring method will give the completely factored form, while the other will not. Work Exercises in order to determine the method to use if you have to make such a decision. 91. Factor x 6-1 as the difference of squares. 92. The factored form obtained in Exercise 91 consists of a difference of cubes multiplied by a sum of cubes. Factor each binomial further. 93. Now start over and factor x 6-1 as the difference of cubes. 94. The factored form obtained in Exercise 93 consists of a binomial that is a difference of squares and a trinomial. Factor the binomial further. 95. Compare your results in Exercises 92 and 94. Which one of these is factored completely? 96. Verify that the trinomial in the factored form in Exercise 94 is the product of the two trinomials in the factored form in Exercise Use the results of Exercises to complete the following statement: In general, if I must choose between factoring first with the method for the difference of squares or the method for the difference of cubes, I should choose the method to eventually obtain the completely factored form. 98. Find the completely factored form of x by using the knowledge you gained in Exercises ### 6.4 Special Factoring Rules 6.4 Special Factoring Rules OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication ### 1.3 Polynomials and Factoring 1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. 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In this section, we will be factoring ### In algebra, factor by rewriting a polynomial as a product of lower-degree polynomials Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Name: Date: Block: Factoring Review Factor: rewrite a number or expression as a product of primes; e.g. 6 = 2 3 In algebra, factor by rewriting ### Greatest Common Factor (GCF) Factoring Section 4 4: Greatest Common Factor (GCF) Factoring The last chapter introduced the distributive process. The distributive process takes a product of a monomial and a polynomial and changes the multiplication ### FACTORING ax 2 bx c. Factoring Trinomials with Leading Coefficient 1 5.7 Factoring ax 2 bx c (5-49) 305 5.7 FACTORING ax 2 bx c In this section In Section 5.5 you learned to factor certain special polynomials. In this section you will learn to factor general quadratic polynomials. ### Using the ac Method to Factor 4.6 Using the ac Method to Factor 4.6 OBJECTIVES 1. Use the ac test to determine factorability 2. Use the results of the ac test 3. Completely factor a trinomial In Sections 4.2 and 4.3 we used the trial-and-error ### 6.1 The Greatest Common Factor; Factoring by Grouping 386 CHAPTER 6 Factoring and Applications 6.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor. ### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method. A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are ### ( ) FACTORING. x In this polynomial the only variable in common to all is x. FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5. The multiples are 5 and. 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Greatest Common Factor Two Terms Three Terms Four Terms. 2008 Shirley Radai Factoring Guidelines Greatest Common Factor Two Terms Three Terms Four Terms 008 Shirley Radai Greatest Common Factor 008 Shirley Radai Factoring by Finding the Greatest Common Factor Always check for ### Tool 1. Greatest Common Factor (GCF) Chapter 4: Factoring Review Tool 1 Greatest Common Factor (GCF) This is a very important tool. You must try to factor out the GCF first in every problem. Some problems do not have a GCF but many do. When ### Operations with Algebraic Expressions: Multiplication of Polynomials Operations with Algebraic Expressions: Multiplication of Polynomials The product of a monomial x monomial To multiply a monomial times a monomial, multiply the coefficients and add the on powers with the ### Factoring Flow Chart Factoring Flow Chart greatest common factor? YES NO factor out GCF leaving GCF(quotient) how many terms? 4+ factor by grouping 2 3 difference of squares? perfect square trinomial? YES YES NO NO a 2 -b ### Factoring Algebra- Chapter 8B Assignment Sheet Name: Factoring Algebra- Chapter 8B Assignment Sheet Date Section Learning Targets Assignment Tues 2/17 Find the prime factorization of an integer Find the greatest common factor (GCF) for a set of monomials. ### Factoring (pp. 1 of 4) Factoring (pp. 1 of 4) Algebra Review Try these items from middle school math. A) What numbers are the factors of 4? B) Write down the prime factorization of 7. C) 6 Simplify 48 using the greatest common ### Factoring Trinomials of the Form x 2 bx c 4.2 Factoring Trinomials of the Form x 2 bx c 4.2 OBJECTIVES 1. Factor a trinomial of the form x 2 bx c 2. Factor a trinomial containing a common factor NOTE The process used to factor here is frequently ### FACTORING OUT COMMON FACTORS 278 (6 2) Chapter 6 Factoring 6.1 FACTORING OUT COMMON FACTORS In this section Prime Factorization of Integers Greatest Common Factor Finding the Greatest Common Factor for Monomials Factoring Out the ### Factoring Trinomials: The ac Method 6.7 Factoring Trinomials: The ac Method 6.7 OBJECTIVES 1. Use the ac test to determine whether a trinomial is factorable over the integers 2. Use the results of the ac test to factor a trinomial 3. For ### Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). Math 50, Chapter 8 (Page 1 of 20) 8.1 Common Factors Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). 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Some examples include 2x+3 ### Determinants can be used to solve a linear system of equations using Cramer s Rule. 2.6.2 Cramer s Rule Determinants can be used to solve a linear system of equations using Cramer s Rule. Cramer s Rule for Two Equations in Two Variables Given the system This system has the unique solution ### EAP/GWL Rev. 1/2011 Page 1 of 5. Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. EAP/GWL Rev. 1/2011 Page 1 of 5 Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. Example: Set the factors of a polynomial equation (as opposed to an ### How To Solve Factoring Problems 05-W4801-AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring ### Veterans Upward Bound Algebra I Concepts - Honors Veterans Upward Bound Algebra I Concepts - Honors Brenda Meery Kaitlyn Spong Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org Chapter 6. Factoring CHAPTER ### Welcome to Math 19500 Video Lessons. Stanley Ocken. Department of Mathematics The City College of New York Fall 2013 Welcome to Math 19500 Video Lessons Prof. Department of Mathematics The City College of New York Fall 2013 An important feature of the following Beamer slide presentations is that you, the reader, move ### Factor Polynomials Completely 9.8 Factor Polynomials Completely Before You factored polynomials. Now You will factor polynomials completely. Why? So you can model the height of a projectile, as in Ex. 71. Key Vocabulary factor by grouping ### x 4-1 = (x²)² - (1)² = (x² + 1) (x² - 1) = (x² + 1) (x - 1) (x + 1) Factoring Polynomials EXAMPLES STEP 1 : Greatest Common Factor GCF Factor out the greatest common factor. 6x³ + 12x²y = 6x² (x + 2y) 5x - 5 = 5 (x - 1) 7x² + 2y² = 1 (7x² + 2y²) 2x (x - 3) - (x - 3) = ### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic ### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers. Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used ### PERFECT SQUARES AND FACTORING EXAMPLES PERFECT SQUARES AND FACTORING EXAMPLES 1. Ask the students what is meant by identical. Get their responses and then explain that when we have two factors that are identical, we call them perfect squares. ### Factors and Products CHAPTER 3 Factors and Products What You ll Learn use different strategies to find factors and multiples of whole numbers identify prime factors and write the prime factorization of a number find square ### Algebra Cheat Sheets Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the concepts ### Lesson 9: Radicals and Conjugates Student Outcomes Students understand that the sum of two square roots (or two cube roots) is not equal to the square root (or cube root) of their sum. 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In ### CHAPTER 7: FACTORING POLYNOMIALS CHAPTER 7: FACTORING POLYNOMIALS FACTOR (noun) An of two or more quantities which form a product when multiplied together. 1 can be rewritten as 3, where 3 and are FACTORS of 1. FACTOR (verb) - To factor ### Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Quarter I: Special Products and Factors and Quadratic Equations Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Time Frame: 20 days Time Frame: 3 days Content Standard: ### Factoring - Greatest Common Factor 6.1 Factoring - Greatest Common Factor Objective: Find the greatest common factor of a polynomial and factor it out of the expression. The opposite of multiplying polynomials together is factoring polynomials. ### Pre-Calculus II Factoring and Operations on Polynomials Factoring... 1 Polynomials...1 Addition of Polynomials... 1 Subtraction of Polynomials...1 Multiplication of Polynomials... Multiplying a monomial by a monomial... Multiplying a monomial by a polynomial... ### Factoring - Grouping 6.2 Factoring - Grouping Objective: Factor polynomials with four terms using grouping. The first thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial like in ### Algebra I Vocabulary Cards Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression ### HIBBING COMMUNITY COLLEGE COURSE OUTLINE HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam, ### Factoring Polynomials Factoring Polynomials 4-1-2014 The opposite of multiplying polynomials is factoring. Why would you want to factor a polynomial? Let p(x) be a polynomial. p(c) = 0 is equivalent to x c dividing p(x). Recall Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the ### #6 Opener Solutions. Move one more spot to your right. Introduce yourself if needed. 1. Sit anywhere in the concentric circles. Do not move the desks. 2. Take out chapter 6, HW/notes #1-#5, a pencil, a red pen, and your calculator. 3. Work on opener #6 with the person sitting across from ### SPECIAL PRODUCTS AND FACTORS SPECIAL PRODUCTS AND FACTORS I. INTRODUCTION AND FOCUS QUESTIONS http://dmciresidences.com/home/20/0/ cedar-crest-condominiums/ http://frontiernerds.com/metal-box http://mazharalticonstruction.blogspot.
# MULTIPLYING POLYNOMIALS PRACTICE AND PROBLEM SOLVING A/B LESSON 14-4 Algebra 1 Linear inequalitites Overview Solving linear inequalities Solving compound inequalities Solving absolute value equations and inequalities Linear inequalities in two variables. When you multiply polynomials where both polynomials have more than one term you just multiply each of terms in the first polynomial with all of the terms in the second polynomial. We just add the like terms to combine the two polynomials into one. The first term of a polynomial is called the leading coefficient. Search Pre-Algebra All courses. Algebra 1 Discovering expressions, equations and functions Overview Expressions and variables Operations in the right order Composing expressions Composing equations and inequalities Representing functions as rules and graphs. We can add polynomials. We just add the like terms to combine the two polynomials into one. Search Pre-Algebra All courses. The degree of the monomial is the sum of the exponents of all included variables. The degree of the polynomial is the greatest degree of its terms. We can add polynomials. If we have a polynomial consisting of only two terms we could instead call it a binomial and a polynomial consisting of three terms can also be called a trinomial. Algebra 1 Discovering expressions, equations and functions Overview Expressions and variables Operations in the right order Composing expressions Composing equations and inequalities Representing functions as rules and graphs. If we have a polynomial consisting of only two mulitplying we could instead call it a binomial and a polynomial consisting of three terms can also be called a trinomial. Just subtract the like terms Or in other words add its opposites. The same goes for subtracting two polynomials. Algebra 1 Linear inequalitites Overview Solving linear inequalities Solving compound inequalities Solving absolute value equations and inequalities Linear inequalities in two variables. Algebra 1 Systems of linear equations and inequalities Overview Graphing linear systems The substitution method for solving linear systems The elimination method for solving linear systems Systems of linear inequalities. MANTEL ESSAY LRB # Multiply binomials by polynomials (practice) \| Khan Academy Algebra 1 How to solve linear equations Overview Properties of equalities Fundamentals in solving equations in one or more steps Ratios and proportions and how to solve them Similar figures Calculating with percents. Algebra 1 Formulating linear equations Overview Writing linear pklynomials using the slope-intercept form Writing linear equations using the point-slope form and the standard form Parallel and perpendicular lines Scatter plots and linear models. Algebra 1 Exploring real numbers Overview Integers and rational numbers Calculating with real numbers The Distributive property Square roots. Algebra 1 Discovering expressions, equations and functions Overview Expressions and variables Operations in the right order Composing expressions Composing equations and inequalities Representing functions as rules and graphs. We can add polynomials. Proble, 1 Exponents and exponential functions Overview Properties of exponents Scientific notation Exponential growth functions. A monomial is a number, a variable or a product of a number and a variable where all exponents are whole numbers. # Monomials and polynomials (Algebra 1, Factoring and polynomials) – Mathplanet Algebra 1 Rational expressions Overview Simplify rational expression Multiply rational expressions Division of polynomials Add and subtract rational expressions Solving rational expressions. A polynomial as oppose to the monomial is a sum of monomials where each monomial is called a term. The degree of the monomial is the sum of the exponents of all included variables. Make the two polynomials into one big polynomial by taking away the parenthesis. When you polgnomials polynomials where both polynomials have more than one term you just multiply each of terms in the first polynomial with all of the terms in the second polynomial. HOLIDAY HOMEWORK RYAN INTERNATIONAL SCHOOL VASANT KUNJ Algebra 1 Visualizing linear functions Overview The coordinate plane Linear equations in the coordinate plane The slope of a linear function The slope-intercept form of a linear equation. Don’t forget to reverse the signs within the second parenthesis since your multiplying all terms with We just add the like terms to combine the two polynomials into one. The degree of the polynomial is the greatest degree of its terms. When multiplying two binomial you can use the leson FOIL to remember how to multiply the binomials. ## Monomials and polynomials A polynomial is usually written with the term with the highest exponent of the variable first and then decreasing from left to right. Algebra 1 Radical expressions Overview The graph of a radical function Simplify radical expressions Radical equations The Pythagorean Theorem The distance and midpoint formulas. Polynomial just means that we’ve got a sum of many monomials. The first term prob,em a polynomial is called the leading coefficient. Constants have the monomial degree of 0. Search Pre-Algebra All courses.
Courses Courses for Kids Free study material Offline Centres More Store # Reenu types $540$ words during half an hour. How many words would she type in $8$ minutes? Last updated date: 20th Jun 2024 Total views: 394.8k Views today: 7.94k Verified 394.8k+ views Hint: In the problem we can clearly notice that the time given is in hours and the target time is in minutes, so first of all we need to convert the given hours into minutes. We know that for every hour there is $60$ minutes i.e. $1$hour $=60$ minutes. From this relation we will convert the given hours into the minutes. Now they have mentioned that Reenu types $540$ words during half an hour. From this statement we will calculate how many words she can type in one minute/unit time by dividing the words typed in half an hour with the minutes in a half an hour. From this value we can calculate how many words she can type in $8$ minutes by multiplying the words she can type in one minute with $8$. Given that, Reenu types $540$ words during half an hour. We know that $1\text{ hour}=60\text{ minutes}$. From this we can calculate how many minutes for half an hour by multiplying the above equation with $\dfrac{1}{2}$ on both sides. \begin{align} & \Rightarrow \dfrac{1}{2}\times 1\text{ hour}=\dfrac{1}{2}\times 60\text{ minutes} \\ & \Rightarrow \dfrac{1}{2}\text{hour}=30\text{ minutes} \\ \end{align} So, we have $30$ minutes for half an hour. Now the words typed by Reenu in one minute can be calculated by dividing the words typed in half an hour with the minutes in half an hour. Mathematically Words typed by Reenu in one minute $=\dfrac{540}{30}=18\text{ wpm}$ Now the words typed by Reenu in $8$ minutes is given by \begin{align} & {{W}_{8}}=18\text{ wpm}\times 8\text{ m} \\ & \Rightarrow {{W}_{8}}=144 \\ \end{align} $\therefore$ Reenu can type $144$ words in $8$ minutes. Note: We can find the words typed by Reenu for any given time period by using the value of Words typed per Minute (wpm). The value of wpm is equal to the words typed in Minute. IF you want to calculate the words typed in two minutes, then ${{W}_{2}}=18\times 2=36\text{ words}$. Again, if you want to calculate the words typed by Reenu in $3$ minutes, then ${{W}_{3}}=18\times 3=54\text{ words}$ likewise we can calculate words typed for any interval of time.
## Elementary Algebra $-47$ We start with the given expression: $3x-2y-4x-x+7y$ We plug in the given values for $x$ and $y$: $3(6)-2(-7)-4(6)-6+7(-7)$ Order of operations states that first, we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. Then, we add and subtract from left to right. We follow order of operations to simplify: First, we multiply from left to right: $18−2(−7)−4(6)−6+7(−7) =18-(-14)−4(6)−6+7(−7)=18-(-14)-24-6+7(-7)=18−(−14)−24−6+(-49)$ We add and subtract from left to right: $32-24-6+(-49)=8-6+(-49)=2+(-49)=-47$
# Class IX Chapter 8 Quadrilaterals Maths Size: px Start display at page: Transcription 1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively. As the sum of all interior angles of a quadrilateral is 360º, 3x + 5x + 9x + 13x = 360º 30x = 360º x = 12º Hence, the angles are 3x = 3 12 = 36º 5x = 5 12 = 60º 9x = 9 12 = 108º 13x = 2 13 12 = 156º Question 2: If the diagonals of a parallelogram are equal, then show that it is a rectangle. Answer: Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º. In ABC and DCB, AB = DC (Opposite sides of a parallelogram are equal) BC = BC (Common) AC = DB (Given) ABC DCB (By SSS Congruence rule) ABC = DCB It is known that the sum of the measures of angles on the same side of transversal is 180º. ABC + DCB = 180º (AB CD) ABC + ABC = 180º 2 ABC = 180º ABC = 90º 3 Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle. Question 3: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Answer: Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and AOB = BOC = COD = AOD = 90º. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal. In AOD and COD, OA = OC (Diagonals bisect each other) AOD = COD (Given) OD = OD (Common) AOD COD (By SAS congruence rule) AD = CD (1) Similarly, it can be proved that AD = AB and CD = BC (2) From equations (1) and (2), 4 AB = BC = CD = AD Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus. Question 4: Show that the diagonals of a square are equal and bisect each other at right angles. Answer: Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º. In ABC and DCB, AB = DC (Sides of a square are equal to each other) ABC = DCB (All interior angles are of 90 ) BC = CB (Common side) ABC DCB (By SAS congruency) AC = DB (By CPCT) Hence, the diagonals of a square are equal in length. In AOB and COD, 5 AOB = COD (Vertically opposite angles) ABO = CDO (Alternate interior angles) AB = CD (Sides of a square are always equal) AOB COD (By AAS congruence rule) AO = CO and OB = OD (By CPCT) Hence, the diagonals of a square bisect each other. In AOB and COB, As we had proved that diagonals bisect each other, therefore, AO = CO AB = CB (Sides of a square are equal) BO = BO (Common) AOB COB (By SSS congruency) AOB = COB (By CPCT) However, AOB + COB = 180º (Linear pair) AOB = 180º 2 AOB = 90º Hence, the diagonals of a square bisect each other at right angles. Question 5: Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Answer: 6 Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and AOB = BOC = COD AOD = = 90º. To prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º. In AOB and COD, AO = CO (Diagonals bisect each other) OB = OD (Diagonals bisect each other) AOB = COD (Vertically opposite angles) AOB COD (SAS congruence rule) AB = CD (By CPCT)... (1) And, OAB = OCD (By CPCT) However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel. AB CD... (2) From equations (1) and (2), we obtain ABCD is a parallelogram. In AOD and COD, 8 (ii) ABCD is a rhombus. Answer: (i) ABCD is a parallelogram. DAC = BCA (Alternate interior angles)... (1) And, BAC = DCA (Alternate interior angles)... (2) However, it is given that AC bisects A. DAC = BAC... (3) From equations (1), (2), and (3), we obtain DAC = BCA = BAC = DCA... (4) DCA = BCA Hence, AC bisects C. (ii)from equation (4), we obtain DAC = DCA DA = DC (Side opposite to equal angles are equal) However, DA = BC and AB = CD (Opposite sides of a parallelogram) AB = BC = CD = DA Hence, ABCD is a rhombus. Question 7: ABCD is a rhombus. Show that diagonal AC bisects A as well as C and diagonal 9 BD bisects B as well as D. Answer: Let us join AC. In ABC, BC = AB (Sides of a rhombus are equal to each other) 1 = 2 (Angles opposite to equal sides of a triangle are equal) However, 1 = 3 (Alternate interior angles for parallel lines AB and CD) 2 = 3 Therefore, AC bisects C. Also, 2 = 4 (Alternate interior angles for lines BC and DA) 1 = 4 Therefore, AC bisects A. Similarly, it can be proved that BD bisects B and D as well. Question 8: ABCD is a rectangle in which diagonal AC bisects A as well as C. Show that: i) ABCD is a square (ii) diagonal BD bisects B as( well as D. 10 Answer: ( i) It is given that ABCD is a rectangle. A = C CD = DA (Sides opposite to equal angles are also equal) However, DA = BC and AB = CD (Opposite sides of a rectangle are equal) AB = BC = CD = DA ABCD is a rectangle and all of its sides are equal. Hence, ABCD is a square. (ii) Let us join BD. In BCD, BC = CD (Sides of a square are equal to each other) CDB = CBD (Angles opposite to equal sides are equal) However, CDB = ABD (Alternate interior angles for AB CD) CBD = ABD BD bisects B. Also, CBD = ADB (Alternate interior angles for BC AD) 11 1 CDB = ABD BD bisects D. Question 9: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that: 12 i) APD CQB ( (ii) AP = CQ iii) AQB CPD ( (iv) AQ = CP (v) APCQ is a parallelogram Answer: (i) In APD and CQB, ADP = CBQ (Alternate interior angles for BC AD) AD = CB (Opposite sides of parallelogram ABCD) DP = BQ (Given) APD CQB (Using SAS congruence rule) ii) As we had observed that APD CQB, ( AP = CQ (CPCT) (iii) In AQB and CPD, ABQ = CDP (Alternate interior angles for AB CD) AB = CD (Opposite sides of parallelogram ABCD) BQ = DP (Given) AQB CPD (Using SAS congruence rule) iv) As we had observed that AQB CPD, ( AQ = CP (CPCT) (v) From the result obtained in (ii) and (iv), AQ = CP and AP = CQ 13 Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram. Question 10: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that i) APB CQD ( (ii) AP = CQ Answer: (i) In APB and CQD, APB = CQD (Each 90 ) AB = CD (Opposite sides of parallelogram ABCD) ABP = CDQ (Alternate interior angles for AB CD) APB CQD (By AAS congruency) (ii) By using the above result APB CQD, we obtain 15 (Opposite sides of a parallelogram are equal and parallel) AD = CF and AD CF (iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram. (v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other. AC DF and AC = DF (vi) ABC and DEF, AB = DE (Given) BC = EF (Given) AC = DF (ACFD is a parallelogram) ABC DEF (By SSS congruence rule) Question 12: ABCD is a trapezium in which AB CD and AD = BC (see the given figure). Show that i) A = B ( ii) C = D ( iii) ABC BAD ( (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] 16 Answer: Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) AD = CE (Opposite sides of parallelogram AECD) However, AD = BC (Given) Therefore, BC = CE CEB = CBE (Angle opposite to equal sides are also equal) Consider parallel lines AD and CE. AE is the transversal line for them. A + CEB = 180º (Angles on the same side of transversal) A + CBE = 180º (Using the relation CEB = CBE)... (1) However, B + CBE = 180º (Linear pair angles)... (2) From equations (1) and (2), we obtain A = B (ii) AB CD A + D = 180º (Angles on the same side of the transversal) Also, C + B = 180 (Angles on the same side of the transversal) A + D = C + B However, A = B [Using the result obtained in (i)] C = D (iii) In ABC and BAD, AB = BA (Common side) BC = AD (Given) B = A (Proved before) ABC BAD (SAS congruence rule) 17 (iv) We had observed that, ABC BAD AC = BD (By CPCT) Exercise 8.2 Question 1: 18 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that: ( i) SR AC and SR = AC ( ii) PQ = SR ( iii) PQRS is a parallelogram. Answer: (i) In ADC, S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. SR AC and SR = AC... (1) (ii) In ABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem, PQ AC and PQ = AC... (2) Using equations (1) and (2), we obtain PQ SR and PQ = SR... (3) PQ = SR 19 (iii) From equation (3), we obtained PQ SR and PQ = SR Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram. Question 2: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. Answer: In ABC, P and Q are the mid-points of sides AB and BC respectively. AC (Using mid-point theorem)... (1) R and S are the mid-points of CD and AD respectively. PQ AC and PQ = In ADC, (2) RS AC and RS = AC (Using mid-point theorem)... From equations (1) and (2), we obtain PQ RS and PQ = RS Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. 20 In quadrilateral OMQN, MQ ON ( PQ AC) QN OM ( QR BD) Therefore, OMQN is a parallelogram. MQN = NOM PQR = NOM However, NOM = 90 (Diagonals of a rhombus are perpendicular to each other) PQR = 90 Clearly, PQRS is a parallelogram having one of its interior angles as 90º. Hence, PQRS is a rectangle. Question 3: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. Answer: Let us join AC and BD. In ABC, P and Q are the mid-points of AB and BC respectively. 21 PQ AC and PQ = AC (Mid-point theorem)... (1) Similarly ADC, in SR AC and SR = AC (Mid-point theorem)... (2) Clearly, PQ SR and PQ = SR Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. PS QR and PS = QR (Opposite sides of parallelogram)... (3) In BCD, Q and R are the mid-points of side BC and CD respectively. QR BD and QR = BD (Mid-point theorem)... (4) However, the diagonals of a rectangle are equal. AC = BD (5) By using equation (1), (2), (3), (4), and (5), we obtain PQ = QR = SR = PS Therefore, PQRS is a rhombus. Question 4: ABCD is a trapezium in which AB DC, BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC. 22 Answer: Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. 23 1 In ABD, EF AB and E is the mid-point of AD. Therefore, G will be the mid-point of DB. As EF AB and AB CD, EF CD (Two lines parallel to the same line are parallel to each other) In BCD, GF CD and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC. Question 5: 24 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD. Answer: ABCD is a parallelogram. AB CD And hence, AE FC Again, AB = CD (Opposite sides of parallelogram ABCD) AB = CD AE = FC (E and F are mid-points of side AB and CD) In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. AF EC (Opposite sides of a parallelogram) In DQC, F is the mid-point of side DC and FP CQ (as AF EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ. DP = PQ... (1) 25 Similarly, in APB, E is the mid-point of side AB and EQ AP (as AF EC). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB. PQ = QB... (2) From equations (1) and (2), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Question 6: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Answer: Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that 26 SP BD and SP = BD... (1) Similarly BCD, in QR BD and QR = BD... (2) From equations (1) and (2), we obtain SP QR and SP = QR In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore, SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other. Question 7: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC 27 It is given that M is the mid-point of AB and MD BC. Therefore, D is the mid-point of AC. (Converse of mid-point theorem) (ii) As DM CB and AC is a transversal line for them, therefore, MDC + DCB = 180º (Co-interior angles) MDC + 90º = 180º MDC = 90º MD AC (iii) Join MC. In AMD and CMD, 28 AD = CD (D is the mid-point of side AC) ADM = CDM (Each 90º) DM = DM (Common) AMD CMD (By SAS congruence rule) Therefore, AM = CM (By CPCT) However, AM = AB (M is the mid-point of AB) Therefore, it can be said that CM = AM = AB ### Class IX Chapter 8 Quadrilaterals Maths 1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles ### 3. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is: Solved Paper 2 Class 9 th, Mathematics, SA 2 Time: 3hours Max. Marks 90 General Instructions 1. All questions are compulsory. 2. Draw neat labeled diagram wherever necessary to explain your answer. 3. ### Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? Class IX - NCERT Maths Exercise (7.1) Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? Solution 1: In ABC and ABD, ### Class IX Chapter 7 Triangles Maths Class IX Chapter 7 Triangles Maths 1: Exercise 7.1 Question In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, ### Class IX Chapter 7 Triangles Maths. Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, AC = AD (Given) CAB = DAB (AB bisects ### TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii) Fill in the blanks Chapter 10 Circles Exercise 10.1 Question 1: (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 Geometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS ### SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c) 1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x ### Mathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. () Exercise 6.4 Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. If EF = 15.4 cm, find BC. Answer 1: 1 () Question 2: Diagonals of a trapezium ABCD with AB DC intersect each other ### Class IX - NCERT Maths Exercise (10.1) Class IX - NCERT Maths 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 ### EXERCISE 10.1 EXERCISE 10.2 NCERT Class 9 Solved Questions for Chapter: Circle 10 NCERT 10 Class CIRCLES 9 Solved Questions for Chapter: Circle EXERCISE 10.1 Q.1. Fill in the blanks : (i) The centre of a circle lies in of the circle. ### RD Sharma Solutions for Class 10 th RD Sharma Solutions for Class 10 th Contents (Click on the Chapter Name to download the solutions for the desired Chapter) Chapter 1 : Real Numbers Chapter 2 : Polynomials Chapter 3 : Pair of Linear Equations ### Triangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z. Triangles 1.Two sides of a triangle are 7 cm and 10 cm. Which of the following length can be the length of the third side? (A) 19 cm. (B) 17 cm. (C) 23 cm. of these. 2.Can 80, 75 and 20 form a triangle? ### Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths Exercise 1.1 1. Find the area of a triangle whose sides are respectively 150 cm, 10 cm and 00 cm. The triangle whose sides are a = 150 cm b = 10 cm c = 00 cm The area of a triangle = s(s a)(s b)(s c) Here ### 8. Quadrilaterals. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ. 8. Quadrilaterals Q 1 Name a quadrilateral whose each pair of opposite sides is equal. Mark (1) Q 2 What is the sum of two consecutive angles in a parallelogram? Mark (1) Q 3 The angles of quadrilateral ### CHAPTER 7 TRIANGLES. 7.1 Introduction. 7.2 Congruence of Triangles CHAPTER 7 TRIANGLES 7.1 Introduction You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a ### Page 1 of 15. Website: Mobile: Exercise 10.2 Question 1: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 ### Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior) 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 ### PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES 1. Find the value of k, if x =, y = 1 is a solution of the equation x + 3y = k.. Find the points where the graph of the equation ### 6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle. 6 CHAPTER We are Starting from a Point but want to Make it a Circle of Infinite Radius A plane figure bounded by three line segments is called a triangle We denote a triangle by the symbol In fig ABC has ### AREAS OF PARALLELOGRAMS AND TRIANGLES AREAS OF PARALLELOGRAMS AND TRIANGLES Main Concepts and Results: The area of a closed plane figure is the measure of the region inside the figure: Fig.1 The shaded parts (Fig.1) represent the regions whose ### 21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 22. Prove that If two sides of a cyclic quadrilateral are parallel, then ### KENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION KENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION SAMPLE PAPER 01 FOR HALF YEARLY EXAM (017-18) SUBJECT: MATHEMATICS(041) BLUE PRINT FOR HALF YEARLY EXAM: CLASS IX Chapter VSA (1 mark) SA I ( marks) SA II ### Rhombi, Rectangles and Squares Rhombi, Rectangles and Squares Math Practice Return to the Table of Contents 1 Three Special Parallelograms All the same properties of a parallelogram apply to the rhombus, rectangle, and square. Rhombus ### Similarity of Triangle Similarity of Triangle 95 17 Similarity of Triangle 17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree 2D VECTORS Question 1 (*) Relative to a fixed origin O, the point A has coordinates ( 2, 3). The point B is such so that AB = 3i 7j, where i and j are mutually perpendicular unit vectors lying on the ### 9 th CBSE Mega Test - II 9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A ### Triangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR. Triangles Two geometric figures having the same shape and size are said to be congruent figures. Two geometric figures having the same shape, but not necessarily the same size, are called similar figures. ### Properties of the Circle 9 Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference ### 9. Areas of Parallelograms and Triangles 9. Areas of Parallelograms and Triangles Q 1 State true or false : A diagonal of a parallelogram divides it into two parts of equal areas. Mark (1) Q 2 State true or false: Parallelograms on the same base ### Chapter 8 Similar Triangles Chapter 8 Similar Triangles Key Concepts:.A polygon in which all sides and angles are equal is called a regular polygon.. Properties of similar Triangles: a) Corresponding sides are in the same ratio b) ### Class 7 Lines and Angles ID : in-7-lines-and-angles [1] Class 7 Lines and Angles For more such worksheets visit www.edugain.com Answer the questions (1) ABCD is a quadrilateral whose diagonals intersect each other at point O such ### Exercise. and 13x. We know that, sum of angles of a quadrilateral = x = 360 x = (Common in both triangles) and AC = BD 9 Exercise 9.1 Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. 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Marks: 90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of ### CBSE Sample Paper-05 (Solved) SUMMATIVE ASSESSMENT I MATHEMATICS Class IX. Time allowed: 3 hours Maximum Marks: 90. Section A. CBSE Sample Paper-05 (Solved) SUMMATIVE ASSESSMENT I MATHEMATICS Class IX Time allowed: hours Maximum Marks: 90 General Instructions: a) All questions are compulsory. b) The question paper consists of ### Question 1 ( 1.0 marks) places of decimals? Solution: Now, on dividing by 2, we obtain = Question 1 ( 1.0 marks) The decimal expansion of the rational number places of decimals? will terminate after how many The given expression i.e., can be rewritten as Now, on dividing 0.043 by 2, we obtain ### Coordinate Geometry. Exercise 13.1 3 Exercise 3. Question. Find the distance between the following pairs of points (i) ( 3) ( ) (ii) ( 5 7) ( 3) (iii) (a b) ( a b) Solution (i) Let A( 3 ) and B( ) be the given points. 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Geometry Regents Exam Based on the diagram below, which statement is true? 0809ge 1 Based on the diagram below, which statement is true? 3 In the diagram of ABC below, AB AC. The measure of B is 40. 1) a b ) a c 3) b c 4) d e What is the measure of A? 1) 40 ) 50 3) 70 4) 100 2012 GCSE Maths Tutor All Rights Reserved www.gcsemathstutor.com This book is under copyright to GCSE Maths Tutor. However, it may be distributed freely provided it is not sold for profit. Contents angles ### Question Bank Tangent Properties of a Circle Question Bank Tangent Properties of a Circle 1. In quadrilateral ABCD, D = 90, BC = 38 cm and DC = 5 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 7 cm. Find ### 0811ge. Geometry Regents Exam 0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 ) 8 3) 3 4) 6 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation ### AREAS OF PARALLELOGRAMS AND TRIANGLES 15 MATHEMATICS AREAS OF PARALLELOGRAMS AND TRIANGLES CHAPTER 9 9.1 Introduction In Chapter 5, you have seen that the study of Geometry, originated with the measurement of earth (lands) in the process of ### CCE PR Revised & Un-Revised D CCE PR Revised & Un-Revised 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 560 00 08 S.S.L.C. EXAMINATION, JUNE, 08 :. 06. 08 ] MODEL ANSWERS : 8-K Date :. 06. 08 ] CODE ### Maharashtra State Board Class X Mathematics Geometry Board Paper 2015 Solution. Time: 2 hours Total Marks: 40 Maharashtra State Board Class X Mathematics Geometry Board Paper 05 Solution Time: hours Total Marks: 40 Note:- () Solve all questions. Draw diagrams wherever necessary. ()Use of calculator is not allowed. ### Triangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y? Triangle Congruence and Similarity Review Score Name: Date: Show all work for full credit. 1. In a plane, lines that never meet are called. 5. In the drawing, what is the measure of angle y? A. parallel ### Geometry - Semester 1 Final Review Quadrilaterals Geometry - Semester 1 Final Review Quadrilaterals 1. Consider the plane in the diagram. Which are proper names for the plane? Mark all that apply. a. Plane L b. Plane ABC c. Plane DBC d. Plane E e. Plane ### Solution 1: (i) Similar (ii) Similar (iii) Equilateral (iv) (a) Equal (b) Proportional Class X - NCERT Maths EXERCISE NO: 6.1 Question 1: Fill in the blanks using correct word given in the brackets: (i) All circles are. (congruent, similar) (ii) All squares are. (similar, congruent) (iii) ### Mathematics. A basic Course for beginers in G.C.E. (Advanced Level) Mathematics Mathematics A basic Course for beginers in G.C.E. 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Geometric Inequalities 4. Let m S, then 3 2 m R. Since the angles are supplementary: 3 2580 4568 542 Therefore, m S 42 and m R 38. Part IV 5. Statements Reasons. ABC is not scalene.. Assumption. 2. ABC has at least 2. Definition ### SSC CGL Tier 1 and Tier 2 Program Gurudwara Road Model Town, Hisar 9729327755 www.ssccglpinnacle.com SSC CGL Tier 1 and Tier 2 Program ------------------------------------------------------------------------------------------------------------------- ### b) What is the area of the shaded region? Geometry 1 Assignment - Solutions Geometry 1 Assignment - Solutions 1. ABCD is a square formed by joining the mid points of the square PQRS. LKXS and IZXY are the squares formed inside the right angled triangles DSC and KXC respectively. ### KENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION KENDRIYA VIDYALAYA SANGATHAN, HYDERABAD REGION SAMPLE PAPER 02 F PERIODIC TEST III EXAM (2017-18) SUBJECT: MATHEMATICS(041) BLUE PRINT : CLASS IX Unit Chapter VSA (1 mark) SA I (2 marks) SA II (3 marks) ### COMMON UNITS OF PERIMITER ARE METRE MENSURATION BASIC CONCEPTS: 1.1 PERIMETERS AND AREAS OF PLANE FIGURES: PERIMETER AND AREA The perimeter of a plane figure is the total length of its boundary. The area of a plane figure is the amount of ### Class X Delhi Math Set-3 Section A Class X Delhi Math Set-3 Section A 1. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30. The distance of the car from the base of the tower (in m.) is: ### 10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2) 10. Circles Q 1 True or False: It is possible to draw two circles passing through three given non-collinear points. Mark (1) Q 2 State the following statement as true or false. Give reasons also.the perpendicular ### Sample Question Paper Mathematics First Term (SA - I) Class IX. Time: 3 to 3 ½ hours Sample Question Paper Mathematics First Term (SA - I) Class IX Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided ### ( )( ) PR PQ = QR. Mathematics Class X TOPPER SAMPLE PAPER-1 SOLUTIONS. HCF x LCM = Product of the 2 numbers 126 x LCM = 252 x 378 Mathematics Class X TOPPER SAMPLE PAPER- SOLUTIONS Ans HCF x LCM Product of the numbers 6 x LCM 5 x 378 LCM 756 ( Mark) Ans The zeroes are, 4 p( x) x + x 4 x 3x 4 ( Mark) Ans3 For intersecting lines: a ### Class IX Chapter 9 Areas of Parallelograms and Triangles Maths Exercise 9.1 Question 1: Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. (i) (ii) (iii) (iv) (v) (vi) (i) ### LLT Education Services 8. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. (a) 4 cm (b) 3 cm (c) 6 cm (d) 5 cm 9. From a point P, 10 cm away from the ### Postulates and Theorems in Proofs Postulates and Theorems in Proofs A Postulate is a statement whose truth is accepted without proof A Theorem is a statement that is proved by deductive reasoning. The Reflexive Property of Equality: a
# Complex Number #### Short Description Maths Question... #### Description 2 Complex numbers This chapter will show you how to: use complex numbers to solve equations such as 3x2 - 2x + 1 = 0 add, subtract, multiply and divide complex numbers represent complex numbers geometrically solve polynomial equations of up to degree 4, such as x4 + 3x3 - 12x - 16 = 0 Refer to 1.1, 1.2 Before you start Check in: 1 Solve a x2 + x - 12 = 0 b 9x2 - 6x = 0 c 4x2 - 4x + 1 = 0 2x2 e.g. Solve + x - 10 = 0 2x2 + x - 10 = (2x + 5)(x - 2) Since 2x2 + x - 10 = 0, (2x + 5)(x - 2) = 0 hence x = - 5 , 2 See C1 for revision. See C1 for revision. FP1 You should know how to: 2 2 Solve a pair of simultaneous equations. e.g. Solve the simultaneous equations x2 + 2y2 = 6, x - y = 3 x2 + 2y2 = 6 and x = y + 3 so (3 + y)2 + 2y2 - 6 = 0 3y2 + 6y + 3 = 0 3(y + 1)2 = 0 Hence y = -1, x = 2 3 Solve a polynomial equation. e.g. Solve the equation x3 - 2x2 - x + 2 = 0 given that x = 2 is one of its roots. ) x2 2 Solve each pair of simultaneous equations. a 2x2 + y2 = 9 x+y=3 b 2x2 + 2y2 = 1 4xy = 1 c x4 - y4 = 15 x2 - y2 = 3 3 Solve the equation 3x3 - 5x2 - 4x + 4 = 0 given that x = -1 is one of its roots. −1 x − 2 x − 2x − x + 2 3 2 x 3 − 2x 2 −x+2 −x+2 0 x3 - 2x2 - x + 2 = 0 so (x - 2)(x + 1)(x - 1) = 0 Hence x = 2 or x = ±1 15 2.1 The imaginary number i The square root of any number, x . 0, is a real number, but what is the square root of a negative number such as -9? Real numbers lie on the number line. −2 −1 0 1 2 Consider the equation x2 + 9 = 0 then x2 = -9 x = −9 = 9 × −1 = 9 × −1 since ab = a× b = ±3 ´ −1 You can write the square root of any negative number as the product of a real number and −1. FP1 −1 is an imaginary number and is denoted by the letter i. So −9 = ±3i and the solution to the equation x2 + 9 = 0 is x = ±3i. x = ±3i are the roots of the equation x2 + 9 = 0 You can check each answer is correct by substituting it into the given equation. When x = 3i, EXAMPLE 1 Find −25, −12 and −7 . ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •••••••••••••••••••••••••••• −25 = ± 25 i = ±5 i −12 = ± 12 i = ± 4 × 3 i = ± 2 3 i −7 16 x2 + 9 = (3i)2 + 9 = 9i2 + 9 = -9 + 9 = 0 as required = ± 7i (3i)2 = 3i ´ 3i = 9i2 9i2 = 9 ´ -1 = -9 Use this method to check that x = -3i is also a root of this equation. 2 Complex numbers Exercise 2.1 1 Express each square root in terms of the imaginary number i. a −16 b −36 c −100 d −20 2 Solve these equations, giving your answers in terms of the imaginary number i. Check each answer by substitution. a x2 + 4 = 0 b x2 + 49 = 0 c x2 + 225 = 0 d 2x2 + 50 = 0 e 3x2 + 27 = 0 f g 9x2 + 4 = 0 h 27x2 + 6 = 0 4x2 + 1 = 0 3 Solve these equations, giving your answers in simplified surd form. b x2 + 18 = 0 c 3x2 + 36 = 0 d 4x2 + 180 = 0 e 5x2 + 120 = 0 f FP1 a x2 + 8 = 0 1 x2 + 14 = 0 2 h 1 x2 + 1 = 0 3 4 g 12x2 + 1 = 0 4 Solve each equation, giving answers in terms of the positive number a. a x 2 + a2 = 0 b x2 + 9a2 = 0 c x2 + a4 = 0 d (ax)2 + 1 = 0 5 The equation ax2 + b = 0, where a and b are positive integers, has a root 5 2 i. 2 a Show that b = 25 a 2 b Given that b < 50, find the value of a and the value of b. 6 Solve these equations. a x2 + 2 = 0 c 1 +9=0 x2 b x4 = 16 2 d 2x +2 25 = 1 x 17 2.2 Complex numbers You can use the imaginary number i to solve a quadratic equation. EXAMPLE 1 Solve the equation x2 - 6x + 13 = 0. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• If x2 - 6x + 13 = 0, x= = − ( −6 ) ± ( −6 )2 − 4 × 1 × 13 2 ×1 6 ± −16 2 Using the quadratic formula with a = 1, b = -6, c = 13. See C1 for revision. Discriminant = 36 - 52 = -16 = 6 ± 4i −16 = 2 16 × −1 = ±4i = 3 ± 2i Divide by 2: The solutions to the equation x2 - 6x + 13 = 0 are x = 3 ± 2i FP1 3 + 2i is an example of a complex number. An expression of the form a + ib where a and b are real numbers, is called a complex number. You write a, b Î R for ‘a and b are real numbers’. You can use the letter z to denote a complex number. z = a + ib b is the imaginary part of z b = Im(z) z is a complex number a is the real part of z a = Re(z) You can write the equation in Example 1 using the letter z instead of x. The equation z2 - 6z + 13 = 0 has solutions z = 3 ± 2i You need to be able to identify the real and imaginary parts of a complex number. EXAMPLE 2 Find the real and imaginary parts of a z = 5 - 2i b w = −1 + 3i 2 You can use other letters, such as w, to denote a complex number. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a Compare 5 - 2i with a + ib: a = 5, b = -2 so Re(z) = 5, Im(z) = -2 b Express so 18 −1 + 3i as − 1 + 3 i: 2 2 2 Re(w) = − 1 and Im(w) = 3 2 2 You can divide the real and imaginary parts by the denominator. 2 Complex numbers A complex number with real part zero is a purely imaginary number. e.g. The number z = 0 + 3i is purely imaginary and is abbreviated to z = 3i. A complex number with zero imaginary part is a real number. e.g. z = 3 + 0i is treated as the real number 3 and is abbreviated to z = 3. The zero complex number z = 0 + 0i, has real and imaginary parts both zero. This number is abbreviated to z = 0. The quadratic equation ax2 + bx + c = 0 has complex roots if the discriminant b2 - 4ac < 0. See C1 for revision of the discriminant. FP1 EXAMPLE 3 By finding the value of the discriminant state the nature of the roots of these equations. a x2 + 4x - 5 = 0 b x2 - 4x + 5 = 0 c x2 + 2x + 1 = 0 •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a x2 + 4x - 5 = 0 a = 1, b = 4, c = -5 b2 - 4ac = 16 + 20 = 36 2 ∴ x + 4x - 5 = 0 has two real roots. b x2 - 4x + 5 = 0 a = 1, b = -4, c = 5 b2 - 4ac = 16 - 20 = -4 ∴ x2 - 4x + 5 = 0 has two complex roots. c x2 + 2x + 1 = 0 a = 1, b = 2, c = 1 b2 - 4ac = 4 - 4 =0 ∴ x2 + 2x + 1 = 0 has one real root. 19 2 Complex numbers Exercise 2.2 1 Write down the real and imaginary parts of each complex number. a z = 3 + 2i b z = 4 - 5i c z = -1 + 4i d z = 2 - 6i e z =1+ i f z = -5 - 3i g z = -7 h z= 2 2 i 3 −4 z= 2 Find the real and imaginary parts of each complex number. Give your answers in simplified surd form where appropriate. a z = 4 − 6i b z = −8 − 4i c z= i d z = 2 + 4i 2 3 6 e z = 8 − 4i FP1 2 2 f z = 3 + 6i 3 3 a Find all complex numbers of the form z = a + a2i for a Î R, with imaginary part 9. b Find all complex numbers of the form z = 2k + k2i for k Î R such that Re(z) = Im(z). 4 By finding the discriminant state the nature of the roots of these equations. a x2 - 3x + 5 = 0 b x2 + 3x + 1 = 0 c 2x2 + x - 1 = 0 d 4x2 - 4x + 1 = 0 e 4x2 + 7x + 5 = 0 5 Solve these equations. Give your answers in the form a + ib where a, b Î R. 20 a z2 + 4z + 5 = 0 b z2 - 2z + 5 = 0 c z2 - 4z + 13 = 0 d 2z2 + 2z + 5 = 0 e 5z2 + 2 = 2z f 8z2 + 17 = 12z 2 Complex numbers 6 Solve these equations. Give each answer in the form a + ib, where a, b Î R are exact. a z2 + 3z + 3 = 0 b 2z2 - z + 1 = 0 c 1 z2 + 3 + 2z = 0 d 4z2 + 3 = 5z e 3(z2 + 1) = 5z f z(11 - 5z) = 7 2 For exact answers, leave in surd form. 7 Find, in simplified surd form, the solutions to each equation. a z2 - z + 1 = 0 b 3z2 - 2z + 3 = 0 c 2z2 + 4z + 3 = 0 d z2 + 3z + 9 = 0 e 2z2 + 4z + 5 = 0 f 9z2 - 6z + 4 = 0 8 Solve each equation. Express the imaginary part of each solution as an integer multiple of 3. a z2 - 2z + 4 = 0 b z2 + 6z + 12 = 0 c z2 - 6z + 21 = 0 d z2 + 2z + 49 = 0 FP1 9 Solve these equations. Give answers in simplified surd form where appropriate. a z(z - 3) = -5 b 2z(1 - z) = 7 c z(3z - 1) = z - 5 d 4 =3−z e 3z = 1 z2 + 3 z f 2z + 7 = 4 z 10 a Show that (z - 1)(z2 + z + 1) º z3 - 1 b Hence solve the equation z3 = 1 21 2.3 The arithmetic of complex numbers You can add, subtract, multiply and divide complex numbers. You can apply many of the rules for ordinary arithmetic to complex numbers. When calculating with complex numbers always replace i2 with -1. FP1 EXAMPLE 1 If z = 2 + 3i and w = 5 - 2i find a z+w b z-w c zw This example shows you how to add, subtract and multiply two complex numbers, z and w. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a z + w = (2 + 3i) + (5 - 2i) = (2 + 5) + i(3 - 2) =7+i b z - w = (2 + 3i) - (5 - 2i) = 2 + 3i - 5 + 2i = (2 - 5) + i(3 + 2) = -3 + 5i c zw = (2 + 3i)(5 - 2i) = 10 - 4i + 15i - 6i2 = 10 + 11i + 6 = 16 + 11i Collect the real and imaginary parts separately. -(5 - 2i) = -5 - (-2)i You may find it helpful to use brackets when subtracting. Expand the brackets in the usual way. -6i2 = -6 ´ -1 = 6 When multiplying a complex number by a number k, multiply its real part by k and its imaginary part by k. e.g. and 3(4 - i) = (3 ´ 4) + (3 ´ -i) = 12 - 3i 2i(3 + 5i) = 6i + 10i 2 = -10 + 6i The product of z with itself is written as z2. Similarly, z3 means z ´ z ´ z. e.g. If z = 3 + i then 2z2 = 2(3 + i)2 = 2(9 + 6i - 1) = 16 + 12i You can solve simultaneous equations involving complex numbers. 22 2i ´ 5i = 10i2 = -10 z2 = z ´ z, as in normal algebra. i2 = -1 2 Complex numbers EXAMPLE 2 The complex numbers z and w satisfy the simultaneous equations 3z + w = 11 - 10i (1) z + w = 5 - 2i (2) Solve the equations to find z and w, giving each answer in the form a + ib for a, b Î R. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Subtract (2) from (1) to eliminate w: 3z + w - z - w = 11 - 10i - (5 - 2i) 2z = 11 - 10i - 5 + 2i 2z = 6 - 8i z = 6 − 8i 2 Collect the real parts and imaginary parts separately. Simplify the real and imaginary parts as much as possible. z = 3 - 4i Substitute z = 3 - 4i into (2) to find w: (5 - 2i) - (3 - 4i) = 5 - 2i - 3 + 4i = 2 + 2i FP1 z + w = 5 - 2i (3 - 4i) + w = 5 - 2i w = (5 - 2i) - (3 - 4i) = 2 + 2i (2) is easier to work with than (1). Hence z = 3 - 4i and w = 2 + 2i You can divide a complex number, z, by another complex number, w, provided w ¹ 0 + 0i. EXAMPLE 3 Find An expression such as z w is called a quotient. 5 in the form a + bi, where a and b are 2 + 3i real numbers. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• You need to transform the divisor into a real number. Make the denominator real by multiplying both the top and the bottom of the fraction by 2 - 3i: ( 5 = 5 × 2 − 3i 2 + 3i 2 + 3i 2 − 3i ( ) = ( 5 2 )−( 3i ) 2 + 3i 2 − 3i = 10 − 15i 13 Hence 5 = 10 − 15 i 2 + 3i 13 13 ) The divisor is 2 + 3i 2 - 3i is the complex conjugate of 2 + 3i. See Section 2.4. Compare with rationalising See C1 5 2+ 3 for revision. (2 + 3i)(2 - 3i) = 4 - 6i + 6i - 9i2 =4+9 = 13, a real number 23 2 Complex numbers EXAMPLE 4 Find the quotient 1 + 2i in the form a + bi, where a, b Î R. 3 − 4i •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Multiply both the top and the bottom of the fraction by 3 + 4i to eliminate the complex divisor: 1 + 2i = (1 + 2i ) ( 3 + 4i ) 3 − 4i ( 3 − 4i ) ( 3 + 4i ) = 3 + 4i + 6i + 8i 2 9 + 16 = 3 + 10i − 8 25 −5 + 10i = 25 8i2 = 8 ´ (-1) = -8 Hence 1 + 2i = − 1 + 2 i 3 − 4i 5 (3 - 4i)(3 + 4i) = 9 + 12i - 12i - 16i2 = 9 + 16 Divide the real and imaginary parts by 25 and separate them. 5 EXAMPLE 5 FP1 You can use division to solve simultaneous equations involving complex numbers. The complex numbers z and w satisfy the simultaneous equations 3z + w = 6 + i (1) z + iw = 3 + 4i (2) Solve the equations to find z and w, giving each answer in the form a + ib for a, b Î R. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Multiply (2) by 3: 3z + 3iw = 9 + 12i (3) Subtract (1) from (3): 3z + 3iw - 3z - w = 9 + 12i - 6 - i (-1 + 3i)w = 3 + 11i Hence w = 3 + 11i −1 + 3i To express w in the required form, multiply the top and bottom of 3 + 11i by (-1 - 3i): −1 + 3i ( )( ) w = 3 + 11i = ( 3 + 11i ) ( −1 − 3i ) −1 + 3i −1 + 3i −1 − 3i (3 + 11i)(-1 - 3i) = -3 - 9i - 11i - 33i2 = 30 - 20i (-1 + 3i)(-1 - 3i) = 1 + 3i - 3i - 9i2 = 1 + 9 = 10 = 30 − 20i 10 = 3 - 2i Substitute w = 3 - 2i into equation (1) to find z: 3z + w = 6 + i 3z + (3 - 2i) = 6 + i 3z = 3 + 3i z=1+i 24 Hence z = 1 + i and w = 3 - 2i (1) (1) is easier to use to find z than (2). (6 + i) - ( 3 - 2i) = 6 + i - 3 + 2i = 3 + 3i Divide the real and imaginary parts of (3 + 3i) by 3. 2 Complex numbers Exercise 2.3 Unless indicated otherwise, give answers in the form a + ib where a, b Î R. 1 Simplify a (3 + 2i) + (4 + 3i) b (8 - i) - (4 + 2i) c (-4 - 2i) + (-2 - 3i) d (3 - 6i) - (2 - 6i) e 3(2 + 3i) f 3i(1 + 4i) g 4 + (5 + 3i)i h 3(1 - 2i) + i(2 + i) i i(2 + 3(2i + 1)) 2 By making z the subject, or otherwise, solve these equations. a 4z - 3 = 5 + 4i b 4z - 3i = 4 + 3i c 6z + 5 = i(3 - 5i) 3 Solve each pair of simultaneous equations. a 3z + 2w = -3 + 11i z + w = 1 + 4i b 2z + 3w = 2 + 11i 3z - w = 3 c 4z + 3w = 5 + 2i 3z - 2w = 8 - 7i b (-5 + 3i)(2 - 4i) c (-2 + i)2 4 Simplify a (-2 + 3i)(4 + 2i) a 5 + 3i b 2+i 2−i 1 − 3i c FP1 5 Find the following quotients. Give each answer in the form a + bi where a, b Î R are exact. 3 − 4i 2i 6 By making z the subject, or otherwise, solve these equations. a (3 + i) z = 3 + 11i d 2 + 3i = 3 − 2i i(z + i) b 7 − 17i = 2 − 3i z +1 c i =1+ i z −i e z = 3i z+i f 2z − i = 1 iz + 2 7 Solve these pairs of simultaneous equations. b 3z + w = -9 + 8i z + iw = 7i a z + w = 2 + 4i z + iw = 1 - i 8 Solve the equation 1 + 1 z −1 z +1 c 2z + 3w = 20 - 7i iz - 2w = -6 + 6i = 1 z 9 Solve each pair of simultaneous equations. For part c consider the expansion (z + iw)2 a z2 - w2 = 10 z-w=1-i b z2 + w2 = 5 - 2i z + iw = 2 + 5i c z2 - w2 = -2 + 16i z + iw = 4 + 4i 25 2.4 Real parts, imaginary parts and complex conjugates Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. If z = a + bi and w = c + di with a, b c, d Î R then z = w Û a = c and b = d. Complex numbers do not behave like real numbers. 2 + 5 = 3 + 4 but 2 + 5i ¹ 3 + 4i Û means ‘implies and is implied by’. EXAMPLE 1 Given that a - 3i = 7 + bi where a, b Î R, find the value of a and the value of b. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Equate the real parts of each number: Re(a - 3i) = a, Re(7 + bi) = 7 Since a - 3i = 7 + bi, the real parts must be equal. Hence a = 7 Equate the imaginary parts of each number: Im(a - 3i) = -3, FP1 Hence Im(7 + bi) = b b = -3 You can solve an equation in z by replacing z with the expression a + bi, where a, b Î R. EXAMPLE 2 Solve the equation z2 = 3 + 4i •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Let z = a + bi then z2 = (a + ib)2 = (a2 - b2) + 2abi Replace z 2 with (a2 - b2) + 2abi in the original equation: (a2 - b2) + 2abi = 3 + 4i Equate real parts: a2 - b2 = 3 Equate imaginary parts: 2ab = 4, Make b the subject of (2): Substitute b = 2 into (1): a Multiply through by a2: ab = 2 (a + ib)2 = a2 + 2abi + i2b2 = (a2 - b2) + 2abi is a useful expansion: You should learn it. (1) (2) b=2 a a2 − 42 = 3 This is a quadratic equation in a2. a a4 - 3a2 - 4 = 0 (a2 + 1)(a2 - 4) = 0 so a2 = -1 or a2 = 4 a = ±2 When a = 2, b = 1 and so z = 2 + i When a = -2, b = -1 giving z = -2 - i 26 A general complex number z has the form z = a + bi for a, b Î R. Hence the solution is z = 2 + i or z = -2 - i a2 = -1 has no solution since a is a real number. Using b = 2 a You can write the solution as z = ±(2 + i). These numbers are the square roots of 3 + 4i. 2 Complex numbers For any complex number z = a + bi, where a, b Î R, the complex conjugate of z is z* = a - bi The conjugate of z is found by multiplying its imaginary part by (-1). EXAMPLE 3 If z = 1 + 3i and w = -6 -7i find a z* b w* c (z - w)* •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a If z = 1 + 3i, b If w = -6 - 7i, z* = 1 - 3i w* = -6 + 7i The real part is the same in a complex number, z, and its conjugate, z. Re(z) = Re(z) Im(z) = -Im(z) c Since z - w = (1 + 3i) - (-6 - 7i) = 7 + 10i (z - w) = 7 - 10i You can solve an equation involving z* by replacing z with a + bi, where a, b Î R. FP1 EXAMPLE 4 Solve the equation 4z - 2z* = 12 + 3i •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• If z = a + bi then z* = a - bi Substitute z = a + bi and z* = a - bi into the equation: 4z - 2z* = 12 + 3i 4(a + bi) - 2(a - bi) = 12 + 3i Collect real and imaginary parts: (4a - 2a) + (4b + 2b)i = 12 + 3i Simplify: 2a + 6bi = 12 + 3i Equate real and imaginary parts: a = 6, b = 1 If 2a = 6, a = 3 2 Hence the solution, z = a + bi, is given by z = 6 + 1 i 2 For any complex number z = a + bi, where a, b Î R z + z* = 2a z - z* = 2bi zz* = a2 + b2 = 2Re(z) = 2Im(z)i 6b = 3 so b = 1 2 The notation zz* means z ´ (z) not (z ´ z). 27 2 Complex numbers Exercise 2.4 Where appropriate, give answers in the form a + ib where a, b Î R. 1 Find the real numbers a and b for which a a + 2i = 7 + bi b 4 + bi = a - 2i c 3 - bi = a + 8i d a - 5i = 3 - bi e a + bi = 2(1 + 2i) f a + bi = -2 2 Find all real numbers a and b for which a 4a + abi = 12 - 6i b a + abi = 3(2 + i) c a2 + bi = a(4 - i) d a(b + ai) = 3i e a2 + 2abi = 1 - 4i f (a + ib)2 = -12 + 4bi 3 Given that p(p + 5qi) = 4 - q2i, where p, q Î R, q > 0, a show that p = -2, justifying your answer FP1 b find the value of q. 4 Solve these equations. a z2 = 15 + 8i b z2 = -21 + 20i c z2 - 12 + 16i = 0 d z2 = 8i e 2z2 - 4 + 3i = 0 f (2z + 1)(1 - 2z) = 8(2 - i) 5 Write down the conjugate of each of these complex numbers. a 3 + 7i b -2 + 5i c 3 - 9i d z2 = 8i e -(4 - 5i) f 3i 4 6 Given that z = 2 + 4i, evaluate these expressions. a (2z)* b z - z* c 2z + 3z* d iz* e i(z + iz) 7 Solve these equations. 28 a z + 3z = 4 - 6i b 2z + 5z* = 21 + 6i c 3z - iz* = 17 - 11i d iz + 2z* = 3(3 + 2i) e z(2 + i) - z* = 12i f i(3z* + iz) = z + 1 2 Complex numbers 8 Solve these pairs of simultaneous equations. a z + w* = 8 + 7i z* - w = 4 - i b z + iw = 1 + 4i * z + iw* = 7 + 6i 9 For the complex number z = a + ib, where a, b Î R, b ¹ 0 = iz prove that 2bz ∗ z −z 10 Prove these results for any complex number z = a + ib, where a, b Î R. a z + z* = 2a b z - z* = 2bi c z (z) = a2 + b2 d (z2) = (z)2 e iz + (iz)* = 0 f ( 1z ) = 1∗ z 11 If z = a + bi and w = b + ai, where a, b Î R and a + b ¹ 0 b prove that Re FP1 a simplify z∗ w ( z +z w ) = 12 12 Given that z = a + bi, where a, b Î R are non-zero and z2 = c + di for c, d Î R ( ) a prove that a2 = 1 c + c 2 + d 2 and find a similar 2 expression for b2 b hence show that (a2 + b2)2 = c2 + d2 c Apply the result of part b to z = 5 + 12i to express 134 as the sum of two non-zero square numbers. 29 2.5 Representing complex numbers geometrically You can represent a complex number geometrically using an Argand diagram. The Argand diagram is a plane with a real horizontal axis and an imaginary vertical axis, which divide the plane into four quadrants. Im 2nd quadrant z a O EXAMPLE 1 FP1 Re Represent the complex number z = 2 + 3i on an Argand diagram. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Identify the real and imaginary parts of z: Re(z) = 2, Im(z) = 3 Draw the diagram with real horizontal axis and imaginary vertical axis. Plot a point with real coordinate 2 and imaginary coordinate 3. Im 3 P 2 1 −3 −2 −1 O 1 2 3 Re Point P on the Argand diagram represents the complex number z = 2 + 3i 30 You can place a complex number z = a + ib on the Argand diagram by plotting its real and imaginary parts on the relevant axes. In the diagram Re denotes the real axis and lm denotes the imaginary axis. 2 Complex numbers You can refer to a complex number as a point in an Argand diagram. EXAMPLE 2 Write down the complex numbers z and w represented by the points P and Q respectively on the Argand diagram. You may assume integer values for each real and imaginary part. Im Q 3 P 2 1 −3 −2 −1 O 1 2 3 Re •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• P has real coordinate 3 and imaginary coordinate 2 and so z has real part 3 and imaginary part 2. Im 3 P 2 1 Hence z = 3 + 2i −3 −2 −1 O Similarly w = -2 + 3i 1 2 3 Re 1 2 3 Re Im FP1 Q 3 2 1 −3 −2 −1 O The point O on the Argand diagram where the axes cross represents the zero complex number 0 + 0i. The real and imaginary parts of this number are both zero. You can represent a complex number z by drawing the vector from O to z. Im 3 P z = 2 + 3i e.g. The vector OP represents the complex number z = 2 + 3i 2 1 −3 −2 −1 O 1 2 3 Re Multiplying a complex number by a real number has a stretching effect on its geometric representation. 31 2 Complex numbers EXAMPLE 3 The points P, Q and R on the Argand diagram represent the complex numbers u, v and w respectively. On a copy of the diagram, illustrate the complex numbers b 1v a 2u 2 Im Q P R Re O c -2w •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a 2u is represented by P ¢, where OP ¢ is a stretch, scale factor 2, of the vector OP. Im Q Pʹ P b 1 v is represented by Q ¢, 2 where OQ ¢ is a stretch, scale R O factor 1 , of the vector OQ. FP1 2 Re Rʹ c -2w is represented by R ¢, where OR ¢ is a stretch, scale factor 2, of the vector RO (the direction OR has been reversed). You can represent the sum z + w, of the complex numbers z and w, on an Argand diagram by joining the vector representing w on to the end of the vector representing z. The diagram shows vectors representing z = 2 + 3i, w = 3 + i and their sum z + w = 5 + 4i. Multiplying z by a negative number reverses the direction of the vector representing z. Im 4 3 w z z+w 2 The dotted line, which is a translation of the vector representing w, has been joined on to the end of the vector representing z. The result represents the sum z + w. Since z - w can be expressed as z + (-w), you can represent z - w geometrically by adding the vector representing (-w) on to the vector representing z. In the diagram, z and w are any pair of complex numbers. The dotted vector, which represents -w, has been joined on to the end of the vector representing z. The result is a representation of z - w. 1 w O 1 2 3 4 5 6 Re Im −w z−w z w O Re Reversing the direction of the vector representing w shows -w. 32 2 Complex numbers For the complex numbers z and w, the sum z + w is constructed by joining the vector representing one of the numbers on to the vector representing the other. The difference z - w is constructed by joining the vector representing -w on to the vector representing z. Alternatively, to represent z - w, you can draw a line from w to z and then translate this line to the point O. EXAMPLE 4 Points P and Q in the Argand diagram represent the complex numbers z and w respectively, where OP = 6 and OQ = 10. The alternative construction for z - w is useful for solving geometrical problems. Im Q 10 P 6 a Determine the quadrant to which w - z belongs. Re O FP1 b Given that OR = 8, where R represents w - z, prove that triangle OPQ is right-angled. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a For w - z, draw a vector from P to Q and then translate that vector to the point O: Im Using the alternative method of construction. Im Q Q ‘w - z’ means ‘go from z to w’. w−z 10 10 6 O P P 6 Re Re O Hence w - z lies in the second quadrant. b The Argand diagram shows point R, which represents w - z. From the way in which w - z was constructed, you can see that the length of OR equals the length of PQ. Im Q R 8 10 8 6 O Hence PQ = 8 and so the side lengths 6, 8, and 10 of triangle OPQ obey Pythagoras’ Theorem. This proves triangle OPQ is right-angled. P Re 62 + 82 = 36 + 64 = 100 = 102 33 2 Complex numbers Exercise 2.5 1 The Argand diagram shows the complex numbers p, u, v, w and z. Im p 3 2 1 v −3 −2 −1 O −1 1 2 3 −2 z Re u w a Write down each number in the form a + bi for integers a and b. FP1 b Without using a calculator, determine which of the numbers on the diagram is the complex number resulting from rounding, to the nearest integer, the real and imaginary parts of p - i 3. 2 Show these complex numbers on the same Argand diagram. a z = -3 + 2i b w = 1 - 2i c u = -2 - 3i d v = 2.5 + 3i e p = -1 f q = -2.5i 3 The Argand diagram shows the complex numbers z and w. Im w z O Re a On a copy of the diagram, draw a vector from O which represents i z+w ii the complex number v such that z + w + v = 0 b Identify the quadrant in which the complex number w + v lies. 34 You may assume that the real and imaginary parts of each number are integers. 2 Complex numbers 4 The Argand diagram shows points A and B, which represent the complex numbers z and w respectively. Im B a On a copy of the diagram, indicate i the point C representing z + w ii the point D representing 2z iii the point E representing 2z + w. A Re O b Completely describe the quadrilateral OCED. 5 For the complex numbers z and w, the vectors OP and OQ on the Argand diagram represent z + w and z - w respectively. You may assume that P and Q are correctly positioned relative to the value of the constant k marked on each axis. Use an Argand diagram to demonstrate that Im Q k P a z lies in the first quadrant O Re k b w lies in the fourth quadrant. a On a copy of the diagram, draw a vector i OA to represent z + w ii OB to represent w - z. Im P FP1 6 In the Argand diagram, the complex numbers z and w, in the first quadrant, are represented by the points P and Q respectively. P and Q lie on a common circle, centre O, part of which is shown in the diagram. Q Re O b Completely describe the quadrilateral OPAQ. c Given that point B lies on this circle, show that angle POQ = 60°. 7 The Argand diagram shows vectors OP and OQ in the first quadrant, representing the non-zero complex numbers z and w respectively. You may assume that the position of Q is correct relative to P. Im Q P Points A and B, not shown on the diagram, represent z + w and z - w respectively. O Re Given that the lengths OA and OB are equal, show that z is real and that w is purely imaginary. 35 2.6 The modulus and argument of a complex number You can define the modulus and argument of a complex number z = a + bi using the vector that represents z on an Argand diagram. The modulus of a complex number z, written \|z\|, is the length r of the vector which represents z. Im z = a + bi r b q O Re a An angle, q, measured from the positive real axis to the vector representing z is called an argument of z, written as arg z. You can show this using Pythagoras’ Theorem. If z = a + bi then \|z\| = a2 + b2 You can show this using EXAMPLE 1 FP1 If arg z = q then tan q = b a tan q = Find the modulus and an argument of z = 2 + 2i. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Illustrate z on an Argand diagram: Im z = 2 + 2i \|z\| = 2 +2 2 z = 2 + 2i 2 2 = 8 =2 2 q tan q = 2 = 1 O 2 2 Re q = tan-1(1) =p 4 Hence \|z\| = 2, arg z = 1 p 4 An argument of a complex number is negative if it is measured in a clockwise direction. 36 2 Complex numbers The diagram shows a complex number z in the second quadrant. q1 and q2 are two possible arguments of z where q1 > 0 and q2 < 0. Im z q1 You can see from this that a complex number has more than one argument. Re O q2 An argument q of z is principal if -p < q - p. Every complex number has exactly one principal argument. Unless told otherwise you should assume that arg z refers to the principal argument of z. You can evaluate arg z, where z = a + bi, (for a ¹ 0) by calculating tan −1 () b a ( ) q = tan−1 ( ) necessary, according to the quadrant in which z lies. (b) q = tan−1 a Re O b −p a q = tan−1 FP1 q = tan−1 b +p a ( ) b a EXAMPLE 2 Find, in exact form, arg (z) when z = -1 + i 3 . •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Compare z = -1 + 3i with z = a + bi: a = -1 and b = 3 (a)   tan −1 b = tan −1  −3  = − 1 p  1 3 Use the result tan-1(-q) º -tan-1 q Since the complex number -1 + i 3 lies in the second quadrant, (a) arg z = tan-1 b + p = - 1 p + p Im z = −1 + √3i 3 q= = 2p 2 p 3 3 The principal argument of z is arg z = 2p 3 O Re 37 2 Complex numbers If z = a + bi has zero real part, then a = 0 giving z = bi. In this case, arg z = ± p depending on the sign of b. Im 2 z = bi b > 0 The argument of the zero complex number, z = 0 + 0i, is undefined. arg z = 1 p 2 You can solve geometrical problems involving complex numbers by using properties of the modulus and argument. EXAMPLE 3 In the Argand diagram, point P represents z = -3 + 2i and point Q represents w = 2 + 4i P 2 −3 O a Find the exact distance PQ. b Find angle QOP. Give your answer in radians, correct to one decimal place. Re 1 arg z = − p 2 z = bi b < 0 Im 4 O Q 2 Re •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a Draw an Argand diagram to help Refer to Section 2.5. FP1 you visualise the problem: PQ is equal to the modulus of the complex number (w - z). Im w−z \|w - z\| = length of vector joining w and z w z w - z = (2 + 4i) - (-3 + 2i) = 5 + 2i Re O PQ = \|w - z\| 52 + 22 = \|5 + 2i\| = 29 = 29 b Use a diagram to see which angle Im you need to find: Angle QOP = a = arg z - arg w Calculate arg z and arg w: w Q z P (2) In radians, arg w = tan −1 4 a O arg (w) Re = 1.107… ( 3) arg z = tan −1 − 2 + p = 2.553. . . Hence angle QOP = 2.553 - 1.107 = 1.446 = 1.4 to 1 d.p. 38 arg z = tan −1 ( ) + p since − 2 3 z is in the second quadrant. 2 Complex numbers Exercise 2.6 All arguments in these questions are principal and are in radians. 1 Find the modulus and argument of each of these complex numbers. Give your answers to one decimal place. a z = 3 + 7i b z = -4 + 5i c z = 6 - 2i d z = -1 - 3i e z = 3 − 2i f 4 3 z= 2 + 4.2 i 2 Find the modulus and argument of each complex number. Give your answers for arg z correct to one decimal place. a z = 3 + 4i b z = -5 + 12i c z = 10 - 10.5i d z = -4.5 - 6i 1 e z = 9 (2 + 5i ) f z= 3 + 6i a z = 3 + 3i b z = -2 + 2i c z = − 2 −i 6 d z= e z = 5 + i 15 f g z= 2 h z = −i 18 z= FP1 3 Find the modulus and argument of each of these complex numbers. Give z in simplified surd form (where appropriate) and arg z in terms of p. 3 3 − i 2 2 − 1 + 2i 2 2 4 For the complex numbers z = 3 + i and w = 1 + 4i, calculate these quantities. Give your answers correct to two decimal places where appropriate. a \|z + w\| b arg(z - w) \|iz + w\| d arg(z + iw) e \|zw - 1\| f arg(zw - i) c 39 2 Complex numbers 5 The complex number z = a + 3i, where a < 0 is a real number, has modulus 5. a Find the value of a. b Find the argument of z. Give your answer to one decimal place. 6 Given that z = 2 + bi, where b Î R, has argument − 1 p 3 a find the exact value of b b calculate z . 7 Find the modulus and argument of each of these complex numbers. Give your answers in exact form. a z= 1 1−i 3 c z = 1−i FP1 i e z= i 1−i 3 b z= 3 2 + 2i d z= 2 1+i f z= i( 3 3 + i) 8 The complex number z satisfies the equation z + a = 2, where z − ia a Î R is negative. a Show that z = a(1 + 2i) b Find arg z, giving your answer to two decimal places. c Express z in terms of a. 9 A and B on the Argand diagram represent the complex numbers z = 2 + 6i and w = 3 - i 3 respectively. Im A Re B a Show that triangle OAB is right-angled. b Hence show that the area of triangle OAB is 2 6 . 40 2 Complex numbers 10 For each of the following, calculate the distance AB. Give your answers in surd form. a b Im −2√3 + 6i Im A B 3 + 3i 3 2 1 A −3 −2 −1 O Re O Re 1 2 3 B −3i 11 Points P and Q on the Argand diagram represent the complex numbers z = 2(1 + i 3 ) and w = - 3 + i respectively. Im P Q FP1 Re O a Find, in terms of p, arg z and arg w. b Hence show that angle POQ = 1 p 2 c Find the area of triangle OPQ. d Use a geometrical approach to show that, for this particular pair of numbers, \|z + w\|2 = \|z\|2 + \|w\|2 12 On the Argand diagram, points P and Q represent the complex numbers z = 3 + i and w = -3 + 3 i respectively. The obtuse angle q = POQ has been marked on the diagram. Im Q P q Re O a Show that q = 2 p 3 b Find the acute angle OPQ, giving your answer to two decimal places. 41 2.7 The modulus-argument form for complex numbers You can express a complex number in modulus-argument form. In the Argand diagram, the complex number z = a + ib has modulus r and argument q. Im z = a + ib r b q O Refer to Section 2.6. The complex number z = a + ib is in cartesian form since you position z on the Argand diagram in the same way that you plot a point with cartesian coordinates (a, b) on a pair of axes. Re a You can see that cos q = a , so a = rcos q r b sin q = , so b = rsin q r and FP1 Hence Þ z = a + ib = rcos q + i(rsin q) z = r (cos q + isin q) z is in cartesian form. z is in modulus-argument form. If z has modulus r and argument q then in modulus-argument form, z = r(cos q + isin q) You can convert a complex number from cartesian form into modulus-argument form and vice versa. EXAMPLE 1 Express z = 3 - 3i in modulus-argument form. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• In modulus-argument form, z = r(cos q + isin q) where r = z and q = arg z z = 3 - 3i Þ r = \|z\| = 32 + ( −3 ) = 18 = 3 2 2 z = 3 - 3i lies in the fourth quadrant ( 3) Þ arg z = tan-1 − 3 Use tan-1(-a) = -tan-1(a): q = arg z = − 1 p 4 O q 3 Re 3 r z = 3 − 3i Hence in modulus-argument form ( ( 14 p ) + isin ( 14 p )) z = 3 2 cos 42 Leave cos ( p ) and sin( p ) − 1 4 1 4 2 Complex numbers EXAMPLE 2 5 The complex number z has modulus 8 and argument 6 p . Find z in cartesian form. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• r = 8, q = 5 p 6 Refer to Hence z = r(cos q + isin q) ( (6 ) ( ) ( ) sin ( 5 p ) = sin ( p ) = 6 ( 6 )) cos 5 p = −cos 1 p = = 8 cos 5 p + isin 5 p ( 2 ) 6 3 2 1 2 z = -4 3 + 4i Þ 6 1 6 = 8 −1 3 + 1i 2 C2 If z and w are two complex numbers then the modulus of their product is \|zw\| = \|z\|\|w\| and, provided that w ¹ 0, the modulus of their quotient is FP1 z = z w w EXAMPLE 3 If z = 3 + i and w = 2 - 2i calculate b \|z2\| a \|zw\| 2 c w3 z •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• First find the modulus of each given complex number: z = 3 + i Þ \|z\| = 3 + 1 = 4 w = 2 - 2i Þ \|w\| = 4 + 4 = 8 8 =2 2 a \|zw\| = \|z\|\|w\| = 4×2 2 = 8 2 b \|z2\| = \|z ´ z \| = \|z\| ´ \|z\| = 4 ´ 4 = 16 This shows that \|z2\| = \|z\|2 In general, \|zn\| = \|z\|n where n is a positive integer. 2 c w2 w = 3 3 z z 8 1 = = 64 8 \|w2\| = \|w\|2, \|z3\| = \|z\|3 43 2 Complex numbers Exercise 2.7 1 Express these complex numbers in modulus-argument form. In each case give the modulus in simplified surd form and the argument to two decimal places. a z = 1 + 3i b z = 2 - 5i 1 1 c z = −3 + 2i d z = -2 - 4i e z = 2 + 3i f z= 1 − 1 i 2 3 2 Express each number in cartesian form. Give real and imaginary parts of each number to two decimal places. a z, where \|z\| = 3 and arg z = 1.5c The symbol c stands for radians. b w, where \|w\| = 6 and arg w = -2.8c c u, where \|u\| = 1.5 and arg u = 7p FP1 12 d v, where \|v\| = 10 and arg v = - 5p 11 3 Copy and complete the following table. Give exact answers where appropriate. Modulus-argument form Cartesian form z = 4  cos  1 p  + i sin  1 p   3  3  z= w w = -4 + 4i  1  1  p = 6  cos  − p  + i sin  − p    6   6  p= q= q = − 3 −i 4 z = 1 + i and w = 1 - i 3 a Find the exact values of \|z\| and \|w\|. b Hence find the exact value of i \|z3w\| ii z5 w2 c Show that \|z - w\| = 1 + 3 and hence find the exact value of \|z2 - zw\|. 44 Angles which involve p are assumed to be in radians. 2 Complex numbers 5 It is given that z = 3 + 3i and z = 2 6 for w a particular w complex number. a Show that \|w\| = 1 2 2 b Given further that arg w = - 1 p , express w in 4 i modulus-argument form, ii cartesian form. ( 3 )) ( (3 ) 6 z = 3 cos 1 p − i sin 1 p a Briefly explain why this expression is not the modulus-argument form for z. b Express z in exact cartesian form. c Hence, or otherwise, express z in modulus-argument form. ( )) and w = 4 + bi where b < 0. ( ( ) FP1 1 2 2 7 z = 4 cos 3 p + i sin 3 p It is given that \|zw2\| = 16 a Show that b = −4 3 and hence express w in exact modulus-argument form. b Illustrate, on a single Argand diagram, the complex numbers z and w. c Hence, or otherwise, find the value of \|z + w\|. 8 z = 1 + i, w = 2 + i and u = 3 + i a Show that \|zw\| = \|u\| and find the value of wu z b Given that arg(wu) = arg w + arg u, find the exact value of (2) (3) tan −1 1 + tan −1 1 c Illustrate, on a single Argand diagram, the points A and B representing the numbers z and w respectively. d Find the area of triangle OAB and hence show that sin BÔA = 1 . 10 9 For z = 1 + 3i, use the information given to find, in exact cartesian form, a w, given that \|zw\| = 32 and arg (w) = − 1 p 4 2 b v, given that v = 2 and arg v = 5 p z 6 45 2 Complex numbers 10 Given that \|z\| = 3 and w = 1 + 2i, find the exact value of a \|zw - z\| c b \|z + zw\| w −i z2 d \|w2 - 1\| 11 In the Argand diagram, points A and B represent the complex numbers z = 1 + 3i and w = 2 - 2i respectively. Point C represents zw. Im A C Re O B a Find the exact value of \|zw\|. FP1 b Given that arg(zw) = 1 p , find the exact area of 12 quadrilateral OACB. 12 Points A and B on the Argand diagram represent the complex numbers z = 3 + i and w = 3 2 ( −1 + i ) respectively. 2 Point C represents w . z Im B C A O Re a Express z and w in modulus-argument form. (z) b Given that arg w = arg w − arg z , show that angle COB = 1 p c Hence i find the area of triangle OBC ii calculate the distance AC giving your answer correct to two decimal places. 46 6 2 Complex numbers 13 In the Argand diagram, point A represents z = 1 + i and point B represents zw for w, a particular complex number. Im B √8 A Re O The line AB has length 8. a State the value of \|zw - z\| and hence find the value of \|w - 1\|. b Given further that arg(w - 1) = 1 p , find w in exact 6 cartesian form. 14 In the Argand diagram, points A, B and C represent the complex numbers z, w and w respectively. z 4 FP1 OA = 2, OB = 5 and the angle between OB and OC is 1 p radians. Im C B A 1 p 4 2 5 Re O Given that AB2 = 39 find, in modulus-argument form, the complex number w. You may assume the result arg w = arg w - arg z (z) 47 2.8 Roots of polynomial equations A polynomial in the complex number z is an expression involving non-negative integer powers of z. e.g. P(z) = 2z4 - p z2 + 1 is a polynomial in z. EXAMPLE 1 If P(z) = 5z2 - 2z + 1, evaluate P(1 + i). •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• P(z) = 5z2 - 2z + 1 Substitute z = 1 + i: P(1 + i) = 5(1 + i)2 - 2(1 + i) + 1 = 5(2i) - 2(1 + i) + 1 = 10i - 2 - 2i + 1 = -1 + 8i Therefore Given the polynomial P(z), any value a for which P(a) = 0 is called a root of the equation P(z) = 0. FP1 (1 + i)2 = (1 + i) (1 + i) = 1 + 2i + i2 = 2i P(1 + i) = -1 + 8i P(a) = 0 means P(a) = 0 + 0i, the zero complex number. EXAMPLE 2 Verify that 1 + 2i is a root of the equation z2 - 2z + 5 = 0 •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Define P(z) = z2 - 2z + 5 Substitute z = 1 + 2i: P(1 + 2i) = (1 + 2i)2 - 2(1 + 2i) + 5 = (-3 + 4i) - 2(1 + 2i) + 5 = 0 + 0i Since P(1 + 2i) = 0, the complex number 1 + 2i is a root of the equation z2 - 2z + 5 = 0 Exercise 2.8 1 Given that P(z) = z2 + z + 1, evaluate each expression. Give your answers in the form a + bi for a, b Î R. Use exact values for a and b. a P(i) ( 2) c P −1i 48 You have to find the value of 5z2 - 2z + 1 when z = 1 + i b P(2i) d P (i 2 ) If 1 + 2i is a root of the equation z2 - 2z + 5 = 0, then P(1 + 2i) = 0 (1 + 2i)2 = 1 + 4i + 4i2 = 1 + 4i - 4 = -3 + 4i 2 Complex numbers 2 Evaluate a P(3i) for P(z) = 2z2 + 3z - 1 b P(2 + 3i) for P(z) = 3z2 - 2z + 4 c P(2i) for P(z) = z3 + z2 + z + 1 d P(-1 + 2i) for P(z) = z3 + 2z - 1 3 Verify that each complex number, a, is a root of the given equation. a a = 1 + i, z2 - 2z + 2 = 0 b a = 3 + 2i, c a = 2 + i, z3 - 5z2 + 9z - 5 = 0 d a = 1 - i, z2 - 6z + 13 = 0 z4 + 4 = 0 4 a Given that 2 - i is a root of the equation z2 + az + (1 - a) = 0, find the value of the real constant, a. b Given that 3i is a root of the equation P(z) = 0, where P(z) = z3 - z2 + az + b, find the value of the real constants, a and b. 5 Given that 2 + 3i is a root of the equation P(z) = 0, where P(z) = z2 + az + b, for a, b Î R FP1 a show that a = -4 and find the value of b b find the value of the real number k for which z = 2 + ki is another root of the equation P(z) = 0 6 Find the value of real constants, a and b, such that a + i is a root of the equation z2 - 4z + b = 0 7 It is given that 1 - i is a root of the equation P(z) = 0, where P(z) = z3 + kz + h, for real constants h and k. a Show that k = -2 and find the value of h. b Show that P(z + 1) = z3 + 3z2 + z + 3 and hence write down a complex number which is a root of the equation z3 + 3z2 + z + 3 = 0 8 a Verify that 1 + i is a root of the equation z4 + z2 - 2z + 6 = 0 b Hence, by making the substitution w = 1 z , find a complex number 2 which satisfies the equation 8w4 + 3 = 2w(1 - w) 9 It is given that i 5 is a root of the equation P(z) = 0, where P(z) = z4 + kz2 + 5 for k a real constant. a Show that k = 6 b Find all the roots of the equation P(z) = 0 49 2.9 Complex roots and conjugate pairs You can find the roots of the equation z2 - 2z + 5 = 0 by using the quadratic formula. z2 - 2z + 5 = 0 so z = 2± −16 2 = 2 ± 4i = 1 ± 2i 2 The roots of this equation are 1 + 2i and 1 - 2i. If P(z) is a polynomial with real coefficients and P(a) = 0 then P(a) = 0 You may notice that one root is the complex conjugate of the other. (1 + 2i) = 1 - 2i and (1 - 2i)* = 1 + 2i a* is the complex conjugate of a. Refer to Section 2.4. EXAMPLE 1 FP1 It follows from this that the roots of a polynomial equation with real coefficients occur in conjugate pairs. Given that 2 + i is a root of the equation P(z) = 0 where P(z) = z3 - 2z2 - 3z + 10 a write down another root of P(z) b completely factorise P(z). An exam question may use the term ‘solution’ in place of ‘root’. •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a All the coefficients of P(z) are real numbers. Hence 2 - i is also a root of the equation P(z) = 0 This follows since (2 + i)* = 2 - i and roots occur in conjugate pairs. b 2 + i and 2 - i are roots of the equation P(z) = 0 Hence (z - (2 + i)) and (z - (2 - i)) are factors of P(z). \[z - (2 + i)][z - (2 - i)] is also a factor of P(z) Factor theorem. Refer to C2 Expand the expression [z - (2 + i)] [z - (2 - i)] and simplify: [z - (2 + i)][z - (2 - i)] = z2 - z(2 - i) - z(2 + i) + (2 + i)(2 - i) = z2 - 2z + zi - 2z - zi + 4 - 2i + 2i - i2 = z2 - 4z + 5 The terms in zi cancel. 4 - i2 = 4 + 1 = 5 Hence (z2 - 4z + 5) is a factor of P(z). Divide P(z) by z2 - 4z + 5 to find the remaining factor: z+2 ) z 2 − 4z + 5 z 3 − 2z 2 − 3z + 10 z 3 − 4z 2 + 5z 2z 2 − 8z + 10 2z 2 − 8z + 10 Zero remainder confirms that z2 - 4z + 5 is a factor of P(z). 0 The remaining factor of P(z) is (z + 2). 50 Hence P(z) = (z2 - 4z + 5)(z + 2) = (z - (2 + i))(z - (2 - i))(z + 2) Make sure you give the complete factorisation of P(z). 2 Complex numbers You can find remaining factors by inspection rather than long division. EXAMPLE 2 Given that (x2 + 2x + 3) is a factor of P(x) = 2x3 + 5x2 + 8x + 3, solve completely the equation 2x3 + 5x2 + 8x + 3 = 0 •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• To solve P(x) = 0, you need to find the remaining factor of P(x). Since (x2 + 2x + 3) is a quadratic factor of the cubic P(x), the remaining factor must be a linear expression, say (ax + b), for constants a and b to be determined. 2x3 + 5x2 + 8x + 3 º (ax + b)(x2 + 2x + 3) In this question, you only need to consider the first and last terms in the expansion in order to find a and b. º ax3 + . . . + 3b Equate coefficients of x3: The identity symbol º means that corresponding coefficients must be equal. a=2 Equate constant terms: b=1 FP1 3 = 3b so b = 1 Hence 2x3 + 5x2 + 8x + 3 º (2x + 1)(x2 + 2x + 3) If 2x3 + 5x2 + 8x + 3 = 0 then (2x + 1)(x2 + 2x + 3) = 0 so either 2x + 1 = 0 or x2 + 2x + 3 = 0 The equation 2x + 1 = 0 has solution x = − 1 2 2 The equation x + 2x + 3 = 0 has roots given by − x = 2± −8 2 = −2 ±i 8 2 = −1 ± i 2 Discriminant = 22 - 4 × 1 × 3 = -8 −8 = 2i 2 Hence the equation 2x3 + 5x2 + 8x + 3 = 0 has roots x = − 1, 2 x = −1 ± i 2 . Exercise 2.9 1 Use an appropriate method to find the remaining factor of the given polynomial. a P(z) = z3 + z2 + 8z - 10, given that (z2 + 2z + 10) is a factor of P(z). b Q(z) = z3 - 2z2 - 13z - 10, given that (z2 - 4z - 5) is a factor of Q(z). c R(z) = 2z3 - 9z2 + 14z - 5, given that (z2 - 4z + 5) is a factor of R(z). 51 2 Complex numbers 2 1 - 2i is a root of the equation P(z) = 0, where P(z) = z3 - z2 + 3z + 5 a Write down another complex root of this equation. b Factorise P(z) completely. 3 In each case a is a root of the equation P(z) = 0 Use this root to find all the factors of P(z). a a = 3 + i, P(z) = z3 - 7z2 + 16z - 10 b a = 2 - i, P(z) = 2z3 - 9z2 + 14z - 5 c a = 2 + 3i, P(z) = z3 - 3z + 52 4 Solve completely a 2x3 - 15x2 + 26x + 17 = 0 given that 4 - i is a root b x3 - 8x + 32 = 0 given that 2 - 2i is a root. FP1 5 Given that 2i is a root of the equation P(z) = 0, where P(z) = z4 - 3z3 - 12z - 16 a write down another root and hence show that (z2 + 4) is a factor of P(z) b completely factorise P(z) and hence solve the equation P(z) = 0 6 In each case, use the given root, a, to solve the equation. Give your answers in the form a + bi where a, b Î R. a a = 1 - 2i, x4 - 2x3 + x2 + 8x - 20 = 0 b a = 2 + i, x4 - 4x3 + 9x2 - 16x + 20 = 0 c a = 2 + 3i x4 - x3 + 3x2 + 31x + 26 = 0 d a = -3 - i x4 -16x2 + 100 = 0 7 It is given that 1 + 2i is a root of the equation z3 + z + k = 0, where k Î R. a Simplify (1 + 2i)3 and hence show that k = 10 b Solve completely the equation z3 + z + k = 0 8 Given that i 2 is a root of the equation P(z) = 0, where P(z) = 2z3 + 3z2 + hz + k, for real constants h and k, a find the value of h and the value of k 52 b completely factorise P(z). 2 Complex numbers 9 Points A, B and C on the Argand diagram represent the three roots of the equation Im z3 - 7z2 + 15z - 25 = 0 A One of the roots of this equation is 1 + 2i. a Find each of the numbers represented by the points B and C. O C Re B b Prove that triangle OAC is right-angled. c Describe fully the quadrilateral OACB and find its area. 10 2 - i is a root of the equation z3 - 11z + 20 = 0 a Find the other two roots of the equation. b On a single Argand diagram, illustrate the three roots of this equation. c Show that the perimeter of the triangle with vertices defined by these three roots is 2(1 + 37) . x4 - 2x3 + 4x2 - 4x + 4 = 0 FP1 11 It is given that 1 + i is a root of the equation a Write down another root of this equation. b Find all the roots of this equation, giving each answer in the form a + bi where a, b Î R. c Illustrate, on a single Argand diagram, the four points which represent the roots of this equation. d Give a complete description of the quadrilateral with vertices defined by these points and calculate its exact area. 12 Prove that if the polynomial P(z) has real coefficients and if a is 2 a complex root of the equation P(z) = 0 then (z 2 − 2z Re (a ) + a ) is a factor of P(z). 13 a By using the fact that its roots occur in conjugate pairs, or otherwise, prove that any cubic equation with real coefficients must have at least one real root. b Give an example of a cubic equation which has i exactly one real root ii no real roots. 53 Review 2 1 Solve these equations. Give answers in simplified surd form where appropriate. a 2z2 - 5z + 4 = 0 b 5 = 3z 2−z z − 4 = 2z 2z + 1 1 +z =1 z −1 c d 2 It is given that z = 3 + 4i and w = 2 - i a Express these quantities in the form a + bi, for a, b Î R. i 2z - 3w ii 1 w−z FP1 iii (z + w)2 iv 1 + i z b w Show that w2 = z* and hence state the value of arg (z - w 2). 3 By making the substitution z = a + ib, for a and b real numbers, solve the equation 2z + iz* = 3 + 5i 4 The non-zero complex number z is such that z + i = l , where z −i l Î R is real. a Show that z is purely imaginary and write down its imaginary part in terms of l. b Hence write down, in terms of p, the argument of z when i l>1 ii 0 < l < 1 5 Solve the equation 2z + 1 = 3 − i z 54 2 Complex numbers ( 6 In modulus-argument form, the complex number z = 12 cos p + i sin p The complex number w is such that z = 1 + i 3 w 6 6 a Express z in exact cartesian form and hence show that w = 3( 3 − i ) b Mark, on the same Argand diagram, points B and C ) representing the numbers w and z respectively. w c Show that triangle OBC is right-angled and hence find the length BC, giving your answer in simplified surd form. d Hence, or otherwise, show that \|w2 - z\| = 12 10 7 On the Argand diagram, point P represents the complex number z = 2 + 2i The line OQ is the result of rotating the line OP through 7p radians anticlockwise. 12 Im P 7p 12 FP1 Q O Re a Find the exact modulus and argument of z. b Hence find the complex number represented by point Q. Give your answer in exact cartesian form. 8 It is given that z = ki , where k < 0 is a constant. 3+i a Express z in exact cartesian form, giving real and imaginary parts in terms of k. b Show that z = − 1 k and find arg z, giving your answer in terms p. c Show that z3 is a real number and determine its sign. 2 9 Given that 2 + i is a root of the equation P(z) = 0, where P(z) = 2z3 - 5z2 - 2z + 15 a write down another complex root of this equation b solve the equation P(z) = 0. 55 2 Complex numbers 10 It is given that 1 + 2i is a root of the equation z3 + az + 10 = 0, where a is a real constant. a Show that a = 1. b Solve the equation z 2 + 10 + 1 = 0 for z ¹ 0. z 11 It is given that 2 (1 + i 3 ) is a root of the equation z3 + 64 = 0 a Find the other two roots of this equation. b Represent these three roots as points on a single Argand diagram. c Show that these points lie on a common circle and state its radius. d Give a geometrical reason why the sum of the roots of this equation is zero. 12 On the Argand diagram, points A, B and C represent the three roots of the equation x3 + 5x2 + 11x +15 = 0 Point D on the real axis is such that angle CBD = 1 p . FP1 2 Im B O C D Re A It is given that -1 - 2i is a root of this equation. a Find the other roots of the equation. Hence write down each of the complex numbers represented by point A, point B and by point C. b Show that triangle ABC is right-angled. c Show that OD = 1 and hence find a polynomial equation of degree 4 whose roots on the Argand diagram are represented by the points A, B, C and D. Give your answer as a series of descending powers of x. 13 Given that z = 22 + 4i and z = 6 − 8i, find w 56 a w, in the form a + bi, where a and b are real b the argument of z, in radians to two decimal places. [(c) Edexcel Limited 2002] 2 Complex numbers 14 a Find the roots of the equation z2 + 2z + 17 = 0 giving your answers in the form a + ib, where a and b are integers. b Show these roots on an Argand diagram. [(c) Edexcel Limited 2007] 15 The complex numbers z and w satisfy the simultaneous equations 2z + iw = -1 z - w = 3 + 3i a Use algebra to find z, giving your answers in the form a + ib, where a and b are real. b [(c) Edexcel Limited 2006] 16 The complex number z = a + ib, where a and b are real numbers, satisfies the equation z2 +16 - 30i = 0 Show that ab = 15 b Write down a second equation in a and b and hence find the roots of z2 + 16 - 30i = 0 [(c) Edexcel Limited 2004] 17 Given that z = 2 - 2i and w = a FP1 a 3 +i find the modulus of wz2. It is given that arg (wz2) = 1 p 3 b Show on an Argand diagram the points A, B and C which represent z, w and wz2 respectively, and determine the size of angle BOC. 18 Given that 3 -2i is a solution of the equation x4 - 6x3 + 19x2 - 36x + 78 = 0, a solve the equation completely. b Show on a single Argand diagram the four points that represent the roots of the equation. [(c) Edexcel Limited 2006] 57 2 Exit FP1 Summary Refer to Complex numbers have the form z = a + bi, where a and b are real numbers and i is the imaginary number −1 a = Re(z) is the real part of z, and b = Im(z) is the imaginary part of z If z = a + bi and w = c + di then z ± w = (a + c) ± (b + d)i To divide z by w, multiply both top and bottom of z by w* w The conjugate of z = a + bi is z* = a - bi You can represent any complex number by a point, or by a line drawn from O, on an Argand diagram The modulus of z = a + bi, written \|z\| = a2 + b2 , is the length r of the line representing z An argument of z = a + bi, written arg z, is an angle q between a line representing z and the positive real axis. Angles measured clockwise are defined to be negative An argument is principal if -p < arg z - p Every non-zero complex number has a unique principal argument. Any complex number has modulus-argument form z = r(cos q + i sin q) where r = \|z\| and q is an argument of z z = z \|zw\| = \|z\|\|w\| (for w ¹ 0) w w if a is a root of the polynomial equation P(z) = 0, then P(a) = 0 If P(z) is a polynomial with real coefficients and P(a) = 0 then P(a*) = 0 Links Complex numbers are very important in analysing many aspects of the physical world. Equations involving complex numbers can be used to predict the motion of electronic particles allowing engineers to design integrated circuits. These circuits are central to the development of much modern technology such as cars, computers and mobile phones. 58 Many of today’s blockbuster films would not be possible without modern mathematics. Animation techniques rely on complex numbers called quarternions (these are numbers of the form q = a0 + a1i + a2 j + a3k where i2 = j 2 = k2 = ijk = -1 and a0, a1, a2 and a3 are real numbers) which can be used to represent rotations in 3-dimensional space. 2.1, 2.2 2.3 2.3 2.4 2.5 2.6 2.6 2.6 2.7 2.7 2.8 2.9
The Common Core content standards for mathematics are all part of specific domains, or categories, that are often relevant to multiple grade levels. The idea behind this approach is that certain concepts in math progress in complexity over several years. ## Counting and cardinality Counting and cardinality involves getting comfortable with what numbers represent and how they’re used. Students count numbers 1 to 100, work on writing numbers 1 to 20, and solidify their understanding of numbers as representative of the total quantity of objects in a group. You may also see this domain referenced as developing “number sense.” Counting and cardinality is the first step in a conceptual staircase of mathematics that students climb over the course of their school years. ## Operations and algebraic thinking The emphasis in the operations and algebraic thinking domain (Grades K-5) is to develop comfort and fluency when using numbers to add, subtract, multiply, and divide. Numbers emerge as tools that students can use to identify and represent quantities, relationships, and patterns. Starting with understanding how to take apart and put together numbers within 10 and understanding relationships between parts and wholes, students build upon these skills until they’re comfortable with multiplying and dividing up to 100, adding and subtracting decimals fluently, and multiplying and dividing decimals to the hundredths place. This leads to the application of these skills in finding missing parts, solving problems with multiple steps, finding the solution to numerical expressions, and using these concepts to determine patterns and represent data on a coordinate plane. ## Number and operations in base ten This domain (Grades K-5) stresses place value — the value of a digit according to its position in a number (for example, in the number 346, the 4 represents 40). Over the course of these grades, students gain an understanding of the use and function of numbers, place value with whole numbers, and eventually place value with multi-digit numbers and decimals. ## Number and operations — fractions Fractions (Grades 3-5) can be challenging for many students. The standards in this domain are designed to help students see fractions as numbers, or more specifically as parts of a whole, instead of as foreign symbols that are difficult to comprehend. Students explore the relationship of parts to wholes; the notation involved in expressing fractions; the ordering of fractions based on their values; and the use of fractions in addition, in subtraction, and in limited applications of multiplication and division. ## Measurement and data In early grades (Grades K-5), students explore measurement and data by measuring objects and quantities and telling time as a means of collecting data. This undergirds the central understanding that numbers may be used as a means of classification based on quantity. Students then begin to perform other operations, such as addition and subtraction, using gathered data. Geometric concepts and shapes are also introduced as a part of this domain. High school students pursue the use of data more extensively with an emphasis on statistics and probability. ## Geometry Starting with a basic understanding of shapes in kindergarten, students explore properties of shapes and other geometric concepts in later grades. As students proceed through Grades 5-8 and into high school, they begin to focus on the application of geometry to real-world settings. ## Ratios and proportional relationships In Grades 6-7, students strive to gain mastery of ratios and proportions. With a firm understanding of the basic tenets of multiplication and division, students explore relationships between sets of numbers and use mathematical terms to describe any relationships that exist. Significant emphasis is placed on real-world application. ## The number system Students extend their abilities in addition, subtraction, multiplication, and division in Grades 6-8 as they apply these operations to multi-digit whole numbers and to fractions. The concept of rational and irrational numbers is also a component of these grades. Students continue to graph data and closed figures in a coordinate plane and use their results to make determinations for specific problems. ## Expressions and equations In Grades 6-8, students apply their understanding of basic arithmetic and part/whole reasoning to solve algebraic expressions, discovering how to substitute letters for numbers and solve to find unknown values. Students then begin to compare various expressions to find out whether they’re equivalent. Teachers introduce students to the idea of using exponents to denote significantly large or small numbers and show them how to solve linear equations and graph them in a coordinate plane. ## Functions Understanding relationships between numbers and relationships between ratios is an essential aspect of Grade 8. Students must know what a function is: a rule, equation, or expression that produces only one output for every input. For example, height is generally a function of age, because the older you are, the taller you get. If you measure your height on your birthday every year, your data set includes only one height for every birthday. Students must also be able to display functions in various ways — with words, graphs, or numbers, for example. As they progress in their ability to recognize and use functions in Grade 8 and in high school, students are required to use functions to show the relationship between objects in real life. ## Statistics and probability Students dive into statistics and probability in Grades 6-8 and again in high school. They explore basic concepts of variability first so they can get a handle on the fact that data varies from person to person. They also practice displaying sets of data in various forms. After grasping the basics, students apply their understanding to make determinations about larger populations based on data samples. Students eventually use graphs, scatterplots, and tables to represent multiple forms of data that may include more than one variable. ## Domains in high school The high-school content standards are grouped within six domains: number and quantity, algebra, functions, modeling, geometry, and statistics and probability. Some of these domains are extensions of domains that are addressed in lower grades. The rest are new domains that build on previously learned concepts.
?> How to Multiply Trinomials? \| Math@TutorCircle.com Sales Toll Free No: 1-855-666-7446 # How to Multiply Trinomials? Top Polynomials with three terms are known as trinomials. We use the distributive method when we want to multiply two trinomials, when both multiplicand and multiplier are trinomial. Once u know how to multiply two binomials then it will not be hard to multiply two trinomial. You should have to follow these two steps: Now we are going to discuss about how to multiply trinomials. To multiply trinomials we need to follow below mentioned steps: Step 1: First you have to multiply each and every term of the second trinomial with the first term of the first trinomial and then repeat the multiplication by multiplying the second term of the second trinomial with the second term of the first trinomial and until the final term of second trinomial with the last term of first trinomial. We can apply both horizontal and vertical multiplication method. Step 2: After this, group the terms and add them. Let’s take some example to understand the basic behind above method: example1:find the multiplication of (3p2+3p+1) and (3p2+5p+1) Solution: (3p2+3p+1) ---->first trinomial (3p2+5p+1) ----> second trinomial Horizontal method of multiplication------- Step 1: we have to multiply the first term of first trinomial with each second trinomial term. 3p2(3p2+5p+1)=9p4+15p3+3p2 -------(a) Multiply the second term of first trinomial with second trinomial 3p(3p2+5p+1)=9p3+15p2+3p ---------(b) Multiply the last term of first trinomial with the second trinomial. 1*(3p2+5p+1) = 3p2+5p+1 ------------(c) Step 2: Add all the equations (a),(b),(c) 9p4+15p3+3p2+ 9p3+15p2+3p+ 3p2+5p+1 Group the terms having same power 9p4+15p3+ 9p3 +3p2+15p2+3p2+3p+5p+1 Now add the likely terms 9p4+ 24p3 +21p2+8p+1 So the multiplication of (3p2+3p+1) and (3p2+5p+1) is 9p4+ 24p3 +21p2+8p+1 (solved).
Presentation is loading. Please wait. # Algebra 6-5 Solving Open Sentences Involving Absolute Value ## Presentation on theme: "Algebra 6-5 Solving Open Sentences Involving Absolute Value"— Presentation transcript: Algebra 6-5 Solving Open Sentences Involving Absolute Value – 3 – 2 – \| \| \| \| \| \| \| \| \| \| – 5 – 4 – 3 – 2 – \| \| \| \| \| \| \| \| \| \| Harbour Solving Open sentences involving absolute vale Section 6-5 Solving Open sentences involving absolute vale Solving Open Sentences Involving Absolute Value Algebra 6-5 Solving Open Sentences Involving Absolute Value There are three types of open sentences that can involve absolute value. Consider the case \| x \| = n. \| x \| = 5 means the distance between 0 and x is 5 units If \| x \| = 5, then x = – 5 or x = 5. The solution set is {– 5, 5}. Solving Open Sentences Involving Absolute Value Harbour Algebra 6-5 Solving Open Sentences Involving Absolute Value When solving equations that involve absolute value, there are two cases to consider: Case 1 The value inside the absolute value symbols is positive. Case 2 The value inside the absolute value symbols is negative. Equations involving absolute value can be solved by graphing them on a number line or by writing them as a compound sentence and solving it. Harbour Solve an Absolute Value Equation Method 1 Graphing means that the distance between b and –6 is 5 units. To find b on the number line, start at –6 and move 5 units in either direction. The distance from –6 to –11 is 5 units. The distance from –6 to –1 is 5 units. Answer: The solution set is Solve an Absolute Value Equation Method 2 Compound Sentence Write as or Case 1 Case 2 Original inequality Subtract 6 from each side. Simplify. Answer: The solution set is Example 5-1a Solve an Absolute Value Equation Answer: {12, –2} Write an Absolute Value Equation Write an equation involving the absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1. The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. So, an equation is . Write an Absolute Value Equation Answer: Check Substitute –4 and 6 into Write an Absolute Value Equation Write an equation involving the absolute value for the graph. Answer: Algebra 6-5 Solving Open Sentences Involving Absolute Value Consider the case \| x \| < n. \| x \| < 5 means the distance between 0 and x is LESS than 5 units If \| x \| < 5, then x > – 5 and x < 5. The solution set is {x\| – 5 < x < 5}. Harbour Algebra 6-5 Solving Open Sentences Involving Absolute Value When solving equations of the form \| x \| < n, find the intersection of these two cases. Case 1 The value inside the absolute value symbols is less than the positive value of n. Case 2 The value inside the absolute value symbols is greater than negative value of n. Harbour Solve an Absolute Value Inequality (<) Then graph the solution set. Write as and Case 1 Case 2 Original inequality Add 3 to each side. Simplify. Answer: The solution set is Solve an Absolute Value Inequality (<) Then graph the solution set. Answer: Algebra 6-5 Solving Open Sentences Involving Absolute Value Consider the case \| x \| > n. \| x \| > 5 means the distance between 0 and x is GREATER than 5 units If \| x \| > 5, then x < – 5 or x > 5. The solution set is {x\| x < – 5 or x > 5}. Harbour Algebra 6-5 Solving Open Sentences Involving Absolute Value When solving equations of the form \| x \| > n, find the union of these two cases. Case 1 The value inside the absolute value symbols is greater than the positive value of n. Case 2 The value inside the absolute value symbols is less than negative value of n. Harbour Solve an Absolute Value Inequality (>) Then graph the solution set. Write as or Case 1 Case 2 Original inequality Add 3 to each side. Simplify. Divide each side by 3. Simplify. Solve an Absolute Value Inequality (>) Answer: The solution set is Solve an Absolute Value Inequality (>) Then graph the solution set. Answer: Algebra 6-5 Solving Open Sentences Involving Absolute Value In general, there are three rules to remember when solving equations and inequalities involving absolute value: If then or (solution set of two numbers) If then and (intersection of inequalities) If then or (union of inequalities) Harbour Assignment Study Guide 6-5 (In-Class) Pages #’s 14-19, 24-35, 40, 41. (Homework) Download ppt "Algebra 6-5 Solving Open Sentences Involving Absolute Value" Similar presentations Ads by Google
What Strategies Can Be Used to Multiplicate Whole Numbers? When you learn to calculate the square root of a number, you might not be thinking in terms of what strategies can be used to multiply whole numbers. But for those of you who are, here are some things you need to know. For starters, when you get to the root of a number, it means that all the numbers up to the first number are represented by the same number. And then if the next number is smaller than the first, then the two will cancel each other out. What are the best strategies to use to multiply whole numbers and find the square root? The best way is to start with a base number. The numbers you choose to work with depend on your purpose. Here are some possibilities: If you’re working with very small numbers, like say, digits ranging from -2 to 12, the best thing to do would be to find all the roots of the number. These can be found using the method known as cross-product. This involves finding the sum, divided by the prime number (also called a factor). This is a great way of finding the square roots of numbers as they’ll always be either positive or negative. Here’s how you can apply this strategy to finding square roots of the numbers you want to find: If your number is between zero and one, you’ll need to go through all zeros. You’ll find that the first number has three roots. To make your life easier, you can also make sure your are only looking at the first few digits. If not, you’ll have to do the whole thing backwards. This means you’ll have to go though all zeros one through nine again. What strategies can be used to multiply whole numbers if you’re dealing with large numbers, say, 4 digits long? Well, let’s do a little trick. All you have to do is multiply each digit individually, then add them together. For example, if your digit is 7, you can multiply it by seven, then by nine, and so on. This will give you your final result, which will in turn give you the square roots. Another question might be – “How about if I’m not looking for the square roots?” How can you possibly use this when you’re looking for square roots? Well, in this case, you have to multiply each digit independently. Once you have the roots, you can find the actual number. Then you can figure out what the number is and multiply it by itself. So now, we’ve covered the basics of what strategies can be used to multiply whole numbers. Now, there are other things you can do as well, but these two are probably the most useful. If you have a calculator on your desk, make sure you use it. You can easily plug in numbers that you’re calculating, then check the value right away. It’s a great way to double check your calculations! Also, make sure your calculator has a memory function so you can go back to previous tables or the results you were getting previously. These two simple tips should be able to answer any question about what strategies can be used to multiply whole numbers. Now, let’s talk about the second part of the equation. The real answer is not just the product of the quotient and the roots, but also the sum of all the squared roots. This will always be equal to the multiplicative total of all the numbers involved. What strategies can be used to multiply whole numbers? The best answer is none, unless you want to buy some lottery tickets! For the rest of you, keep on using your calculators. For the square root, if your calculator produces a negative answer, check your math skills. There’s probably a hidden problem somewhere. Now, we can go on to using the solutions of the quadratic formula and logarithms. These are a little more involved, but the methods are relatively easy. We can also use the solutions of other areas such as geometric forms and algorithms. If you’re interested in trying more than one method, this is an option as well. What strategies can be used to multiply whole numbers can be confusing at first if you don’t know how the formula works. However, once you understand the basics of it, you can start figuring out different methods of application and multiplication in no time.
# Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m Question: Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m. Solution: (i) Let: $a=85 \mathrm{~m}$ and $b=154 \mathrm{~m}$ Given : Perimeter $=324 \mathrm{~m}$ or, $a+b+c=324$ $\Rightarrow c=324-85-154=85 \mathrm{~m}$ $\therefore s=\frac{324}{2}=162 \mathrm{~m}$ By Heron's formula, we have: Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{162(162-85)(162-154)(162-85)}$ $=\sqrt{162 \times 77 \times 8 \times 77}$ $=\sqrt{1296 \times 77 \times 77}$ $=\sqrt{36 \times 77 \times 77 \times 36}$ $=36 \times 77$ $=2772 \mathrm{~m}^{2}$ (ii) We can find out the height of the triangle corresponding to 154 m in the following manner: We have: Area of triangle $=2772 \mathrm{~m}^{2}$ $\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=2772$ $\Rightarrow$ Height $=\frac{2772 \times 2}{154}=36 \mathrm{~m}$
Courses Courses for Kids Free study material Offline Centres More Store # Prime Factorization By Division Method Last updated date: 09th Apr 2024 Total views: 100.8k Views today: 1.00k ## Introduction Prime factorization is a method to find the prime factors of the given number with the help of different composite numbers. We know that a composite number has more than two factors; so this method is applicable for all the composite numbers. For example: 5 is a prime number which has two factors 5×1, whereas a composite number has more than two factors present in it. For example 15 has three factors such as 1×3×5. ### Prime Factorization by Division Method We know that there are two different methods of prime factorization: • Division Method • Factor Tree Method We can hear discussion about the division method. In this method we will divide the large number to the small prime numbers to find the factors. In other words we can say that the division method is used to find out the prime factors of a large number by dividing the number by different prime numbers. ### Division by Primes In the prime factorization method we have to divide the large number with the small prime numbers. This method is known as the division method. For example 60 is a composite number and we have to find its prime factors. So we have different methods but we have to divide with the prime number, 60 = 2×2×3×5 ### Prime Factorization Steps These are the steps of prime factorization • Divide the given number by the smallest prime numbers ( in this case we have to find the smallest prime number which can divide the number exactly). • Again divide the quotient by the smallest prime number (it can be the same or a different prime number). • Repeat the procedure until the quotient becomes 1. • At the end multiply all the prime factors. ( One thing we have to remember is that the product is the number itself). ### Prime Factors of 16 First of all 16 is a composite number and we have to find its prime factors; • As we consider it to be the smallest prime number. • We get 8 and again by 2 • Then we get 4 again we dividing by 2 • Then we get 2 again dividing by the smallest prime number 2. • We get a quotient as 1. 16=2×2×2×2 Prime Factorization of 16 ### How to Get Prime Factorization We can get prime factorization by using the both methods; 1) factor tree method, 2) division method. These methods help us to get the prime factorization. For example: Factor Tree Method and Continuous Method ### Solved Examples 1. What is the prime factorization of 90. Solution: Step 1: divided by the smallest prime number 2 90÷2 = 45 Step 2: divided by 3 45÷3 = 15 Step 3: divided by 3 15÷3 = 5 Step 4: divided by 5 5÷5 = 1 Ans. 2×3×3×5= 90 ### Solved Questions 1. Find the prime factors of 40. Solution Prime Factorization of 40. Ans. 2×2×2×5 = 40 2. Find prime factorization of 24. Solution: Prime Factorization of 24 Ans. 24=2×2×2×3 ## Summary In this article we learn about the different types of prime factorization, mainly we learn about the division method by prime factorization and its rules or steps. The most important uses of prime factorization are cryptography, HCF and LCM. Prime factorization is really helpful for us. ## FAQs on Prime Factorization By Division Method 1. What is prime factorization? Prime factorization of any number means to define the number as the product of different small prime numbers. For example; the prime factorization of 20 is 2×2×5 Here to add 5 are the prime factors of 20. 2. What are the types of prime factorization? There are two types of prime factorization 1. Factor Tree Method 2. Division Method 3. What are prime numbers? A prime number is a natural number which is greater than 1 and it can be divided by one or itself. 4. What is a composite number? A natural number which is greater than 1 and not a prime number is called a composite number. 5. What are prime factors? Any prime number that can be multiplied with others to give the original number is called prime factors.
# Difference between revisions of "2003 Pan African MO Problems/Problem 6" ## Problem Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that: $$f(x^2)-f(y^2)=(x+y)(f(x)-f(y))$$ for $x,y \in \mathbb{R}$. ## Solution By letting $x = 0, y = n$, we have $f(0) - f(n^2) = n (f(0) - f(n))$, and by letting $x = 0, y = -n$, we have $f(0) - f(n^2) = -n (f(0) - f(-n))$. Substitution and simplification results in \begin{align} n (f(0) - f(n)) &= -n (f(0) - f(-n)) \\ f(0) - f(n) &= -f(0) + f(-n) \\ 2 f(0) &= f(n) + f(-n). \end{align} Therefore, if $f(n) = f(0) + m$, where $m$ is a real number, then $f(-n) = f(0) - m$. Thus, the function $f(x)$ has rotational symmetry about $(0,f(0))$. Additionally, multiplying both sides by $x - y$ results in $(x-y)(f(x^2)-f(y^2))=(x^2 - y^2)(f(x)-f(y))$, and rearranging results in $\frac{f(x^2) - f(y^2)}{x^2 - y^2} = \frac{f(x)-f(y)}{x-y}$. The equation seems to resemble a rearranged slope formula, and the rotational symmetry seems to hint that $f(x)$ is a linear function. To prove that $f(x)$ must be a linear function, we need to prove that the slope is the same from $(0,f(0))$ to all the other points of the function. By letting $x = n, y = 0$, we have $\frac{f(n^2) - f(0)}{n^2} = \frac{f(n) - f(0)}{n}$. Additionally, by letting $x = n, y = 1$, we have $\frac{f(n^2) - f(1)}{n^2-1} = \frac{f(n)-f(1)}{n-1}$. By rearranging the prior equation, we have \begin{align} f(n^2) - f(1) &= (n^2 - 1) \cdot \frac{f(n)-f(1)}{n-1} \\ f(n^2) - f(1) &= (n+1) f(n) - (n+1) f(1) \\ f(n^2) &= (n+1) f(n) - n f(1) \\ f(1) &= \frac{(n+1) f(n) - f(n^2)}{n} \end{align} The slope from points $(0,f(0))$ and $(1,f(1))$ is $\frac{f(1)-f(0)}{1-0} = f(1) - f(0)$. By substitution, \begin{align} f(1) - f(0) &= \frac{(n+1) f(n) - f(n^2)}{n} - \frac{n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(n^2) - n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(n^2) + f(0) - f(0) - n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - \frac{f(n^2) - f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - n \cdot \frac{f(n^2) - f(0)}{n^2} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - n \cdot \frac{f(n) - f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - \frac{n f(n) - n f(0)}{n} \\ &= \frac{f(n) - f(0)}{n}. \end{align} The slope from point $(0,f(0))$ to $(1,f(1))$ is the same as the slope from point $(0,f(0))$ to any other point on the function, so $f(x)$ must be a linear function. Let $f(x) = mx+b$. Using the function on the original equation results in \begin{align} f(x^2)-f(y^2) &= (x+y)(f(x)-f(y)) \\ mx^2 + b - my^2 - b &= (x+y)(mx+b-my-b) \\ mx^2 - my^2 &= (x+y)(mx-my) \\ mx^2 - my^2 &= mx^2 - my^2 \end{align} Thus, $f(x)$ can be any linear function, so $\boxed{f(x) = mx+b}$, where $m, b$ are real numbers.
0 # What is the lowest common multiple of 2 4 and 12? Updated: 9/26/2023 Wiki User 6y ago It is: 12 Wiki User 6y ago Earn +20 pts Q: What is the lowest common multiple of 2 4 and 12? Submit Still have questions? Related questions ### What is the lowest common multiple of 2 3 and 4? The lowest common multiple of 2, 3, and 4, is 12. L.C.M is 12 ### Lowest common multiple of 12? To find the lowest common multiple you must have at least two numbers. ### What is the LCM of 12 and 62? 2 is the lowest common multiple of 12 and 62. ### What are the multiples of 2 8 and 12? The lowest common multiple is 24 ### What is the lowest common multiple of 2 and 7? Lowest common multiple of 2 and 7 is 14. ### What is the least common multiple of 4 2 and 12? The least common multiple of 2, 4, and 12 is 12. ### What is the lowest common multiple of 2 and 6? 6.The LCM of 2 & 6... is 6 ! ### What is the least common multiple of 6 12 and 2? The least common multiple (LCM) is often also called the lowest common multiple or smallest common multiple. Keep in mind that these different terms all refer to the same thing: the smallest positive integer which is a multiple of two or more numbers. The least common multiple of 2, 6, and 12 is 12. ### What is the lowest commmon multiple of 12 and 30? The least common multiple of 12 and 30 is 60. 60/12=5 and 60/30=2. ### What is the lowest multiple of 4 and 6? To work out the lowest common multiple of two or more numbers we need to resolve each number to its prime factors. Then, we can multiply those factors together to give us the lowest common multpile. The factors of 4 are 2, 2 The factors of 4 are 2, 3 To be a multiple of 4, the lowest common multiple must have 2 and 2 in its list to be a multiple of 4, the lowest common multiple must have 2 and 3 in its list So, starting with the 4 we have (2, 2) As there is already a 2 in the mix, when we come to the 6 we only need to add the 3 making our list (2, 2, 3) If we multiple all these numbers together we get 12. 4 x 3 = 12 6 x 2 = 12 ### What is the lowest common multiple of 2 and 4? 4 is the lowest common multiple of 2 and 4. (2x2 and 4x1)
Search by Topic Resources tagged with Divisibility similar to What Numbers Can We Make Now?: Filter by: Content type: Stage: Challenge level: There are 54 results Broad Topics > Numbers and the Number System > Divisibility What Numbers Can We Make Now? Stage: 3 Challenge Level: Imagine we have four bags containing numbers from a sequence. What numbers can we make now? What Numbers Can We Make? Stage: 3 Challenge Level: Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make? Differences Stage: 3 Challenge Level: Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number? Dozens Stage: 2 and 3 Challenge Level: Do you know a quick way to check if a number is a multiple of two? How about three, four or six? Three Times Seven Stage: 3 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? Going Round in Circles Stage: 3 Challenge Level: Mathematicians are always looking for efficient methods for solving problems. How efficient can you be? Elevenses Stage: 3 Challenge Level: How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results? Counting Factors Stage: 3 Challenge Level: Is there an efficient way to work out how many factors a large number has? Stage: 3 Challenge Level: List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? Repeaters Stage: 3 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. Legs Eleven Stage: 3 Challenge Level: Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have? Take Three from Five Stage: 4 Challenge Level: Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? Flow Chart Stage: 3 Challenge Level: The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does. Remainder Stage: 3 Challenge Level: What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2? What an Odd Fact(or) Stage: 3 Challenge Level: Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5? Oh! Hidden Inside? Stage: 3 Challenge Level: Find the number which has 8 divisors, such that the product of the divisors is 331776. Ben's Game Stage: 3 Challenge Level: Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters. Eminit Stage: 3 Challenge Level: The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M? Remainders Stage: 3 Challenge Level: I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number? Divisively So Stage: 3 Challenge Level: How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7? Gaxinta Stage: 3 Challenge Level: A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N? Ewa's Eggs Stage: 3 Challenge Level: I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket? Square Routes Stage: 3 Challenge Level: How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number? Digat Stage: 3 Challenge Level: What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A AB Search Stage: 3 Challenge Level: The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B? Digital Roots Stage: 2 and 3 In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers. Stage: 3 Challenge Level: Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true. Powerful Factorial Stage: 3 Challenge Level: 6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!? Skeleton Stage: 3 Challenge Level: Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum. Just Repeat Stage: 3 Challenge Level: Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence? Expenses Stage: 4 Challenge Level: What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? Mod 3 Stage: 4 Challenge Level: Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3. Factoring Factorials Stage: 3 Challenge Level: Find the highest power of 11 that will divide into 1000! exactly. LCM Sudoku Stage: 4 Challenge Level: Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it. The Remainders Game Stage: 2 and 3 Challenge Level: A game that tests your understanding of remainders. American Billions Stage: 3 Challenge Level: Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3... Big Powers Stage: 3 and 4 Challenge Level: Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas. Peaches Today, Peaches Tomorrow.... Stage: 3 Challenge Level: Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for? Fac-finding Stage: 4 Challenge Level: Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful. Knapsack Stage: 4 Challenge Level: You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value. Obviously? Stage: 4 and 5 Challenge Level: Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6. 396 Stage: 4 Challenge Level: The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396. Squaresearch Stage: 4 Challenge Level: Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares? Multiplication Magic Stage: 4 Challenge Level: Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . . Sixational Stage: 4 and 5 Challenge Level: The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . . Transposition Fix Stage: 4 Challenge Level: Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . . Check Codes Stage: 4 Challenge Level: Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . . The Chinese Remainder Theorem Stage: 4 and 5 In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5." Odd Stones Stage: 4 Challenge Level: On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.
# How do you find the displacement of a vector problem? ## What is displacement vector physics? Displacement Formula Displacement = Final position – initial position = change in position. ## What is the formula for displacement? What does displacement mean? If an object moves relative to a reference frame—for example, if a professor moves to the right relative to a whiteboard, or a passenger moves toward the rear of an airplane—then the object’s position changes. This change in position is known as displacement. ## What are examples of displacement? Answer: Starting position, s_i = 0. His final position s_f is the distance traveled North minus the distance traveled South. Calculating displacement, s = s_f – s_i. ## What is the formula of displacement and distance? To find the direction of the displacement vector, calculate the inverse tangent of the ratio of the displacement components in the y- and x-directions. In this example, the ratio of the displacement components is 15÷5 and calculating the inverse tangent of this number gives 71.6 degrees. ## How do you find the direction of a displacement vector? the formula to determine the magnitude of a vector (in two dimensional space) v = (x, y) is: \|v\| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the formula to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: \|V\| = √(x2 + y2 + z2) ## What is vector formula? The correct option is D 0.6i + 0.8j. ## What is the unit vector of P 3i 4j? Displacement is a vector that is the shortest distance from the initial to the final position, as Wikipedia accurately states. ## Is vector same as displacement? The distance between two vectors v and w is the length of the difference vector v – w. There are many different distance functions that you will encounter in the world. We here use “Euclidean Distance” in which we have the Pythagorean theorem. ## What is the distance between two vectors? The vector sum v → + w → may be imagined as the displacement of an object resulting from first applying and then . It is also the same as applying and then , i.e., vector addition is commutative: v → + w → = w → + v → . ## Is displacement a vector sum? To find the displacement when the velocity is changing, a velocity-time graph is needed. Normally, velocity is plotted on the y-axis (the vertical axis) and time is plotted on the x-axis (the horizontal axis). The area under the line on a velocity-time graph is equal to the displacement of the object. ## Is displacement a vector or scalar? Displacement is a vector quantity that refers to “how far out of place an object is”; it is the object’s overall change in position. ## How do you find displacement in physics graph? Calculator Use The average velocity of the object is multiplied by the time traveled to find the displacement. The equation x = ½( v + u)t can be manipulated, as shown below, to find any one of the four values if the other three are known. ## How do you find displacement with velocity and time? The resultant displacement formula is written as: S = √x²+y². “S” stands for displacement. X is the first direction that the object is traveling and Y is the second direction that the object is traveling. If your object only travels in one direction, then Y = 0. ## How do you find the displacement of two points? Speed is a scalar quantity – it is the rate of change in the distance travelled by an object, while velocity is a vector quantity – it is the speed of an object in a particular direction. ## Is speed a vector? Displacement can be positive, negative or zero. ## Can the displacement be negative? Distance is a scalar quantity as it only depends upon the magnitude and not the direction. Displacement is a vector quantity as it depends upon both magnitude and direction. Distance gives the detailed route information that is followed while travelling from one point to another. ## What are the 5 difference between distance and displacement? Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement has direction as well as magnitude. Distance traveled is the total length of the path traveled between two positions. ## What is total displacement in physics? Displacement is the vector difference between the ending and starting positions of an object. It may be very different from the distance the object has travelled along the way. Velocity is the rate at which displacement changes with time. ## What is displacement in velocity? MAGNITUDE AND DIRECTION OF A VECTOR Given a position vector →v=⟨a,b⟩,the magnitude is found by \|v\|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba), as illustrated in Figure 8.8.
# Write a system of equations that has no solution in math Solution Step 1 Our purpose is to add the two equations and eliminate one of the unknowns so that we can solve the resulting equation in one unknown. Many students forget to multiply the right side of the equation by And if you add 7x to the right hand side, this is going to go away and you're just going to be left with a 2 there. Step 5 Check the solution in both equations. Inconsistent equations The two lines are parallel. These are known as Consistent systems of equations but they are not the only ones. I'll add this 2x and this negative 9x right over there. In the top line x we will place numbers that we have chosen for x. What are the coordinates of the origin? So a linear equation with no solutions is going to be one where I don't care how you manipulate it, the thing on the left can never be equal to the thing on the right. Thus we refer to such systems as being inconsistent because they don't make any mathematical sense. And you say, hey, Sal how did you come up with that? Given an ordered pair, locate that point on the Cartesian coordinate system. As a result, when solving these systems, we end up with equations that make no mathematical sense. You already understand that negative 7 times some number is always going to be negative 7 times that number. Step 2 Adding the equations, we obtain Step 3 Solving for y yields Step 4 Using the first equation in the original system to find the value of the other unknown gives Step 5 Check to see that the ordered pair - 1,3 is a solution of the system. The system is consistent since there are no inconsistent rows. This region is shown in the graph. We indicate this solution set with a screen to the left of the dashed line. A linear inequality graphs as a portion of the plane. All three planes have to parallel Any two of the planes have to be parallel and the third must meet one of the planes at some point and the other at another point. The number of variables is always the number of columns to the left of the augmentation bar. The system is consistent since there are no inconsistent rows. The arrows indicate the number lines extend indefinitely.A system has no solution if the equations are inconsistent, they are contradictory. for example 2x+3y=10, 2x+3y=12 has no solution. is the rref form of the matrix for this system. The rref of the matrix for an inconsistent system has a row with a nonzero number in the last column and 0's in all other columns, for example 0 0 0 0 1. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. Resources / Lessons / Math / Precalculus / Systems of Equations / Consistent and Inconsiste GO. Consistent and Inconsistent Systems of Equations Consistent and Inconsistent Systems of Equations. we say that the system of equations has NO SOLUTION. Thus we refer to such systems as being inconsistent because they don't. Together they are a system of linear equations. Can you discover the values of x and y yourself? (Just have a go, play with them a bit.). formulated in terms of systems of linear equations, and we also develop two methods for solving these equations. In addition, we see how matrices (rectangular arrays of numbers) can be used to write systems of linear equations in compact form. We then go on to consider some real-life Finally, in the third case, the system has no solution. With this direction, you are being asked to write a system of equations. You want to write two equations that pertain to this problem. Solution from fmgm2018.com We need to write two equations. 1. The cost 2. The number of small prints based on large prints. Writing a System of Equations by: Anonymous. Write a system of equations that has no solution in math Rated 5/5 based on 79 review
How many factors does 24 have? # Factors and Common Factors This Math quiz is called 'Factors and Common Factors' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14. It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us A factor is a number that can be evenly divided into other whole number without having a remainder left over. For example, 5 is a common factor of 5, 10, 15, 20, 25, etc. Whole numbers begin at “0” and continue through infinity. Fractions and decimals are not found in common factors making finding common factors rather easy. A common factor is a number that can be evenly divided into a collection or series of whole numbers. For example, the number 1 is a common number that can be evenly divided into any series of whole numbers. Finding the factors of a whole number begins with starting with the number “1” and then dividing it with the given number. For example, let’s say that our given whole number is “6”. 6 ÷ 1 = 6 We see that when we divide “6” by “1” we get a whole number, i.e., “6”. We then continue to divide the number “6” with each ascending number as follows: 6 ÷ 2 = 3 6 ÷ 3 = 2 6 ÷ 4 = 1.5 6 ÷ 5 = 1.2 6 ÷ 6 = 1 Once you divide the whole number into itself, this will give you the last factor. The last factor of 6 is “6”. The factors of 6 are then, 1, 2, 3 and 6 because the answer derived is a whole number. 4 and 5 are not factors because they did not divide evenly and the answer included either a fraction or decimal. 1. For the whole number given below, find the right answer that lists all of the number’s factors. 21 1, 7, 21 1, 3, 7, 21 1, 3, 7, 9, 21 1, 3, 9, 21 21 ÷ 1 = 21; 21 ÷ 2 = 10.5; 21 ÷ 3 = 7; 21 ÷ 4 = 5.25; 21 ÷ 5 = 4.2; 21 ÷ 6 = 3.5; 21 ÷ 7 = 3; 21 ÷ 8 = 2.62; 21 ÷ 9 = 2.33; 21 ÷ 10 = 2.1; 21 ÷ 11 = 1.90; 21 ÷ 12 = 1.75; 21 ÷ 13 = 1.61; 21 ÷ 14 = 1.5; 21 ÷ 15 = 1.4; 21 ÷ 16 = 1.31; 21 ÷ 17 = 1.23; 21 ÷ 18 = 1.16; 21 ÷ 19 = 1.10; 21 ÷ 20 = 1.05; 21 ÷ 21 = 1 The factors of 21 are 1, 3, 7 and 21 2. For the whole number given below, find the right answer that lists all of the number’s factors. 16 1, 2, 3, 4, 8, 16 1, 2, 4, 6, 8, 16 1, 2, 4, 8, 16 1, 2, 3, 6, 8, 16 16 ÷ 1 = 16; 16 ÷ 2 = 8; 16 ÷ 3 = 5.33; 16 ÷ 4 = 4; 16 ÷ 5 = 3.2; 16 ÷ 6 = 2.66; 16 ÷ 7 = 2.28; 16 ÷ 8 = 2; 16 ÷ 9 = 1.77; 16 ÷ 10 = 1.6; 16 ÷ 11 = 1.45; 16 ÷ 12 = 1.33; 16 ÷ 13 = 1.23; 16 ÷ 14 = 1.14; 16 ÷ 15 = 1.06; 16 ÷ 16 = 1 The factors of 16 are 1, 2, 4, 8 and 16 3. For the whole number given below, find the right answer that lists all of the number’s factors. 27 1, 2, 3, 9, 27 1, 3, 9, 27 1, 3, 6, 9, 27 1, 3, 9, 12, 27 27 ÷ 1 = 27; 27 ÷ 2 = 13.5; 27 ÷ 3 = 9; 27 ÷ 4 = 6.75; 27 ÷ 5 = 5.4; 27 ÷ 6 = 4.5; 27 ÷ 7 = 3.85; 27 ÷ 8 = 3.37; 27 ÷ 9 = 3; 27 ÷ 10 = 2.7; 27 ÷ 11 = 2.45; 27 ÷ 12 = 2.25; 27 ÷ 13 = 2.07; 27 ÷ 14 = 1.92; 27 ÷ 15 = 1.8; 27 ÷ 16 = 1.68; 27 ÷ 17 = 1.58; 27 ÷ 18 = 1.5; 27 ÷ 19 = 1.42; 27 ÷ 20 = 1.35; 27 ÷ 21 = 1.28; 27 ÷ 22 = 1.22; 27 ÷ 23 = 1.17; 27 ÷ 24 = 1.12; 27 ÷ 25 = 1.08; 27 ÷ 26 = 1.03; 27 ÷ 27 = 1 The factors of 27 are 1, 3, 9 and 27 4. For the whole number given below, find the right answer that lists all of the number’s factors. 9 3, 9 1, 9 1, 3, 9 1, 3, 6, 9 9 ÷ 1 = 9; 9 ÷ 2 = 4.5; 9 ÷ 3 = 3; 9 ÷ 4 = 2.25; 9 ÷ 5 = 1.8; 9 ÷ 6 = 1.5; 9 ÷ 7 = 1.28; 9 ÷ 8 = 1.12; 9 ÷ 9 = 1 The factors of 9 are 1, 3 and 9 5. For the whole number given below, find the right answer that lists all of the number’s factors. 19 1, 3, 19 1, 3, 4, 19 1, 7, 19 1, 19 19 ÷ 1 = 19; 19 ÷ 2 = 9.5; 19 ÷ 3 = 6.33; 19 ÷ 4 = 4.75; 19 ÷ 5 = 3.8; 19 ÷ 6 = 3.16; 19 ÷ 7 = 2.71; 19 ÷ 8 = 2.37; 19 ÷ 9 = 2.11; 19 ÷ 10 = 1.9; 19 ÷ 11 = 1.72; 19 ÷ 12 = 1.58; 19 ÷ 13 = 1.46; 19 ÷ 14 = 1.35; 19 ÷ 15 = 1.26; 19 ÷ 16 = 1.18; 19 ÷ 17 = 1.11; 19 ÷ 18 = 1.05; 19 ÷ 19 = 1 The factors of 19 are 1 and 19 6. For the whole number given below, find the right answer that lists all of the number’s factors. 12 1, 2, 3, 4, 6, 10, 12 1, 2, 3, 4, 6, 12 1, 2, 3, 4, 10, 12 1, 2, 4, 6, 12 12 ÷ 1 = 12; 12 ÷ 2 = 6; 12 ÷ 3 = 4; 12 ÷ 4 = 3; 12 ÷ 5 = 2.4; 12 ÷ 6 = 2; 12 ÷ 7 = 1.71; 12 ÷ 8 = 1.5; 12 ÷ 9 = 1.33; 12 ÷ 10 = 1.2; 12 ÷ 11 = 1.09; 12 ÷12 = 1 The factors of 12 are 1, 2, 3, 4, 6 and 12 7. For the whole number given below, find the right answer that lists all of the number’s factors. 18 1, 2, 3, 9, 18 1, 2, 3, 6, 10, 18 1, 2, 6, 9, 10, 18 1, 2, 3, 6, 9, 18 18 ÷ 1 = 18; 18 ÷ 2 = 9; 18 ÷ 3 = 6; 18 ÷ 4 = 4.5; 18 ÷ 5 = 3.6; 18 ÷ 6 = 3; 18 ÷ 7 = 2.57; 18 ÷ 8 = 2.25; 18 ÷ 9 = 2; 18 ÷ 10 = 1.8; 18 ÷ 11 = 1.63; 18 ÷ 12 = 1.5; 18 ÷ 13 = 1.38; 18 ÷ 14 = 1.28; 18 ÷ 15 = 1.2; 18 ÷ 16 = 1.12; 18 ÷ 17 = 1.05; 18 ÷ 18 = 1 The factors of 18 are 1, 2, 3, 6, 9 and 18 8. For the whole number given below, find the right answer that lists all of the number’s factors. 13 1, 13 1, 3, 13 1, 3, 6, 13 1, 6, 13 13 ÷ 1 = 13; 13 ÷ 2 = 6.5; 13 ÷ 3 = 4.33; 13 ÷ 4 = 3.25; 13 ÷ 5 = 2.6; 13 ÷ 6 = 2.16; 13 ÷ 7 = 1.85; 13 ÷ 8 = 1.62; 13 ÷ 9 = 1.44; 13 ÷ 10 = 1.3; 13 ÷ 11 = 1.18; 13 ÷ 12 = 1.08; 13 ÷ 13 = 1 The factors of 13 are 1 and 13 9. For the whole number given below, find the right answer that lists all of the number’s factors. 36 1, 2, 3, 6, 9, 12, 36 1, 2, 3, 4, 6, 12, 36 1, 2, 3, 4, 5, 9, 18, 36 1, 2, 3, 4, 6, 9, 12, 18, 36 36 ÷ 1 = 36; 36 ÷ 2 = 18; 36 ÷ 3 = 12; 36 ÷ 4 = 9; 36 ÷ 5 = 7.2; 36 ÷ 6 = 6; 36 ÷ 7 = 5.14; 36 ÷ 8 = 4.5; 36 ÷ 9 = 4; 36 ÷ 10 = 3.6; 36 ÷ 11 = 3.27; 36 ÷ 12 = 3; 36 ÷ 13 = 2.76; 36 ÷ 14 = 2.57; 36 ÷ 15 = 2.4; 36 ÷ 16 = 2.25; 36 ÷ 17 = 2.11; 36 ÷ 18 = 2; 36 ÷ 19 = 1.89; 36 ÷ 20 = 1.8; 36 ÷ 21 = 1.71; 36 ÷ 22 = 1.63; 36 ÷ 23 = 1.56; 36 ÷ 24 = 1.5; 36 ÷ 25 = 1.44; 36 ÷ 26 = 1.38; 36 ÷ 27 = 1.33; 36 ÷ 28 = 1.28; 36 ÷ 29 = 1.24; 36 ÷ 30 = 1.2; 36 ÷ 31 = 1.16; 36 ÷ 32 = 1.12; 36 ÷ 33 = 1.09; 36 ÷ 34 = 1.05; 36 ÷ 35 = 1.02; 36 ÷ 36 = 1 The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36 10. For the whole number given below, find the right answer that lists all of the number’s factors. 24 1, 2, 3, 4, 6, 8, 12, 24 1, 2, 3, 4, 6, 12, 24 1, 2, 4, 6, 8, 12, 24 1, 2, 3, 4, 8, 12, 24 24 ÷ 1 = 24; 24 ÷ 2 = 12; 24 ÷ 3 = 8; 24 ÷ 4 = 6; 24 ÷ 5 = 4.8; 24 ÷ 6 = 4; 24 ÷ 7 = 3.42; 24 ÷ 8 = 3; 24 ÷ 9 = 2.66; 24 ÷ 10 = 2.4; 24 ÷ 11 = 2.18; 24 ÷ 12 = 2; 24 ÷ 13 = 1.84; 24 ÷ 14 = 1.71; 24 ÷ 15 = 1.6; 24 ÷ 16 = 1.5; 24 ÷ 17 = 1.41; 24 ÷ 18 = 1.33; 24 ÷ 19 = 1.26; 24 ÷ 20 = 1.2; 24 ÷ 21 = 1.14; 24 ÷ 22 = 1.09; 24 ÷ 23 = 1.04; 24 ÷ 24 = 1 The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 Author: Christine G. Broome
+0 # fractions +1 137 2 +62 Darius, Ellen and Farman each had some marbles. They traded marbles with one another. In the 1st round, Darius gave 1/5 of his marbles to Ellen. In the 2nd round, Ellen gave 1/4 of his marbles to Farhan. In the 3rd round, Farhan gave 1/2 of his marbles to Darius. Darius had 45 marbles, Ellen had 30 marbles and Farhan had 25 marbles left in the end. How many marbles had each of them at first? Jul 9, 2021 #1 +318 +4 Information given: At the end, Number of marbles Darius has = 45 marbles Number of marbles Ellen has = 30 marbles Number of marbles Farhan has = 25 marbles Calculations: Let’s say Darius has x number of marbles, Ellen has y number of marbles, and Farhan has z number of marbles at the beginning. 1st round Darius gave 1/5 marbles to Ellen. => Number of marbles Darius has = x – x/5 = 4x/5 Number of marbles Ellen has = y + x/5 2nd round Ellen gave 1/4 marbles to Farhan => Number of marbles Ellen has = (y + x/5) – 1/4 (y + x/5) = 3/4 (y + x/5) Number of marbles Farhan has = z + 1/4 (y + x/5) 3rd round Farhan gave 1/2 of his marbles to Darius. => Number of marbles Farhan has = 1/2 [z + 1/4 (y + x/5)] Number of marbles Darius has = 4x/5 + 1/2 [z + 1/4 (y + x/5)] So, it can be written as, 1/2 [z + 1/4 (y + x/5)] = 25 4x/5 + 1/2 [z + 1/4 (y + x/5)] = 45 3/4 (y + x/5) = 30 Simplifications: 4x/5 + 25 = 45 => x = 5(45-25)/4 = 25 3/4 (y + x/5) = 30 => (y + 25/5) = (30)4/3 = 40 => y + 5 = 40 => y = 40 - 5 = 35 1/2 [z + 1/4 (y + x/5)] = 25 => [z + 1/4 (y + x/5)] = 2 (25) = 50 => z + 1/4 (35 + 25/5) = 50 => z + 1/4 (35 + 5) = 50 => z + 1/4 (40) = 50 => z + 10 = 50 => z = 50 - 10 = 40 Conclusions: Darius has 25 number of marbles, Ellen has 35 number of marbles, and Farhan has 40 number of marbles at the beginning. Jul 9, 2021 #2 +26 +3 Let's solve this problem by working backwards. Third Round: Since Farhan gave half of his marbles last round and he now has 25 marbles, he had 25 x 2 = 50 marbles in the third round. Darius has 45 - 25 = 20 marbles in the third round as well. Second Round: Ellen gave a fourth of her marbles during the second round and now has 30 marbles left. This means that 3/4 of her previous marbles = 30, so she has 30 x 4/3 = 40 marbles in the second round. Ellen also gave 10 marbles to Farhan during this round, so Farhan had 50 - 10 = 40 before then. First Round: Darius gave 1/5 of his marbles and ended up with 20 marbles. This means that Darius originally had 20 x 5/4 = 25 marbles. Ellen received 5 marbles and ended up with 40 marbles, so she originally had 40 - 5 = 35 marbles. Let's now write down the number of marbles each person has after working backwards: Ellen has 35 marbles. Darius has 25 marbles. Farhan has 40 marbles. Jul 9, 2021
## Real Mathematics – Geometry #15 Drawing a square I am dealing with geometry and I imagine that I am in ancient Greece again. Aegean sea is in front of me and I am sitting on a marble between two huge white columns while holding an unmarked ruler and a compass. First I draw a circle that has center at A and has radius r: Then I draw the same circle but taking its center at B this time: I connect the points A and B with a straight line. Then I draw two perpendiculars from the endpoints of the line AB: I connect the point E to the point F and end up with the ABEF square which has side lengths r: Biggest circle that can be drawn into ABEF will have diameter r and touch the square at exactly four points: Area In order to find the area of a square one can take the square of one side that gives r2. To find a circle’s area one should multiply the square of the radius with π. In our inscribed circle we calculate the area as πr2/4. Ratio of these areas would give π/4. Weight Now let’s make an experiment. For that all you need is some kind of cardboard cut as a square and a precision scale. Using the scale find the weight of the square-shaped cardboard. Then draw the biggest possible circle inside this square. Cut that circle out and find its weight with the scale. Since we are using the same material ratio of the weights should be equal to the ratio of the areas. From here one can easily find an approximation for the number π: 0,76/0,97 = π/4 3,1340… = π One of the main reasons why we only found an approximation is that the cardboard might not be homogeneous. In other words the cardboard might not have equal amount of material on every point of itself. Another reason for finding an approximation is that I didn’t cut the square and the circle perfectly. One wonders… Draw a circle and then draw the biggest-possible square inside that circle. Find their areas and measure their weights. See if you found an approximation. M. Serkan Kalaycıoğlu ## Real Mathematics – Life vs. Maths #2 Matches For thousands of years people tried find a precise value for the number π (3,1415192…). At first this special number was thought to be seen only when there is a circle around. Within time π started to appear in places where scientist didn’t expect it to be. One of them was an 18th century scientist Georges Buffon. Buffon came up with a probability problem named “Buffon’s needle problem” in 1777 when he came across with the number π. As I didn’t possess that many needles, I modified the problem as “Serkan’s matches problem”. Buffon’s Needle Problem: Take a piece of paper and draw perpendicular lines on it with specific amount of space between them. Buffon wondered if one can calculate the probability of a needle that will land on one of the lines. To start Serkan’s matches problem you need at least 100 matches, a piece of empty paper, a ruler, pen/pencil and a calculator. First of all, draw perpendicular lines with 2 matches-length spaces between them. Then just throw the matches on the paper randomly. Start collecting the matches which land on a line. At last you should use your calculator to divide the total number of matches to the number of matches landed on a line. In my experiment out of 100 matches, 32 of them landed on a line. That gave me 3,125 which is close to the magical number π. In fact, 100 matches are not enough for this experiment. In my second try 34 matches landed on one of the lines which gave 100/34=2,9411… Obviously this is not close to π. More matches we use, closer we will get to π. In an experiment back in 1980 2000 needles were used to analyze Buffon’s needle problem. Result was 3,1430… which is seriously close to the number π. ##### You could go to https://mste.illinois.edu/activity/buffon/ and use this simulator which uses 1000 needle. In my first try I got 3,1496… You should try and see the result yourself. In the future I will be talking about why a needle (or a match) is connected to the number π. One wonders… Try to do your own experiment and repeat Buffon’s needle problem for five times. Take the arithmetic average of your solutions and see how close you are to π? M. Serkan Kalaycıoğlu
# Arc Measure: Definition & Formula Coming up next: Finding Instantaneous Rate of Change of a Function: Formula & Examples ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:04 Arc of a Circle • 0:37 Arc Measure vs Arc Length • 2:46 Arc Measure Formula • 3:46 Lesson Summary Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Laura Pennington Laura has taught collegiate mathematics and holds a master's degree in pure mathematics. Arc measure is a way of identifying an arc. In this lesson, we'll learn how to define and calculate arc measure while discussing the different elements involved in this calculation. ## Arc of a Circle Let's start with a simple definition. An arc is a segment of a circle. For example, you've probably seen a rainbow at some point in your life, so imagine using a giant piece of paper to draw a full circle that includes the rainbow. Do you see how rainbow is a segment of a circle? As we can see in the picture above, rainbows come in all different lengths, as do arcs. An arc can be any length up to a full circle; therefore, it's important for us to know how to identify and measure different arcs. Let's take a few minutes to learn how to do that now. ## Arc Measure vs. Arc Length There are two different ways to identify an arc. We can identify it by its length or by the measure of the angle that the arc creates in the center of a circle. The span of the arc is called the arc length, and the measure of the angle that the arc creates is called the arc measure. Check out the picture below, which illustrates both characteristics of an arc: To understand an arc measure, we need to be familiar with the measure of an angle in both degrees and radians. Let's now talk for a bit about degrees and radians and the relationship between the two. Angles have two different units of measure. A circle measures 360 degrees. The degree of an angle will represent the same fraction of a circle as the angle's corresponding arc. For example, 90 degrees is 1/4 of 360 degrees, so a 90-degree angle has a corresponding arc that is 1/4 of a circle. The other unit of measurement we use to measure an angle is the radian. The relationship between radians and degrees allows us to convert between the two using the rules below. Keep in mind that a full circle equals 360 degrees, or 2pi radians, and that a single radian equals 180 degrees. • The measure of an angle in degrees equals the measure of an angle in radians: To convert degrees to radians, we multiply the degree measure by pi / 180. • The measure of an angle in radians equals the measure of an angle in degrees: To convert radians to degrees, we multiply the radian measure by 180 / pi. For example, let's calculate the radian measure of a 90-degree angle. • 90 multiplied by pi / 180 = 90(pi / 180). • 90pi / 180 = pi / 2 radians. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
# 2012 USAJMO Problems/Problem 1 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\overline{AB}$ and $\overline{AC}$, respectively, such that $AP = AQ$. Let $S$ and $R$ be distinct points on segment $\overline{BC}$ such that $S$ lies between $B$ and $R$, $\angle BPS = \angle PRS$, and $\angle CQR = \angle QSR$. Prove that $P$, $Q$, $R$, $S$ are concyclic (in other words, these four points lie on a circle). ## Solution Since $\angle BPS = \angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$. $[asy] import markers; unitsize(0.5 cm); pair A, B, C, O, P, Q, R, S; A = (2,12); B = (0,0); C = (14,0); P = intersectionpoint(A--B,Circle(A,8)); Q = intersectionpoint(A--C,Circle(A,8)); O = extension(P, P + rotate(90)(A - P), Q, Q + rotate(90)(A - Q)); S = intersectionpoint(B--C,arc(O, abs(O - P), 180, 270)); R = intersectionpoint(B--C,arc(O, abs(O - P), 270, 360)); draw(A--B--C--cycle); draw(Circle(O, abs(O - P))); draw(S--P--R); draw(S--Q--R); label("A", A, N); label("B", B, SW); label("C", C, SE); label("P", P, W); label("Q", Q, NE); label("R", R, SE); label("S", S, SW); markangle(1, B, P, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, P, R, S, radius=4mm, marker(markinterval(stickframe(n=1,2mm),true))); markangle(1, R, Q, C, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); markangle(1, R, S, Q, radius=4mm, marker(markinterval(stickframe(n=2,2mm),true))); [/asy]$ For the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle. Since $AP = AQ$, $A$ lies on the radical axis of both circles. However, both circles pass through $R$ and $S$, so the radical axis of both circles is $RS$. Hence, $A$ lies on $RS$, which is a contradiction. Therefore, the two circumcircles are the same circle. In other words, $P$, $Q$, $R$, and $S$ all lie on the same circle. ## Solution 2 Note that (as in the first solution) the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$. Similarly, since $\angle CQR = \angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$. Now, suppose these circumcircles are not the same circle. They already intersect at $R$ and $S$, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points $M$ and $N$, with $M$ on the circumcircle of triangle $PRS$. By Power of a Point, $AQ^2 = AM \cdot AS$ and $AP^2 = AN \cdot AS$. Hence, because $AP = AQ$, $AM = AN$, a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle.
28 March, 18:22 # Patricia, Hugo and Sun work at a music store. Each week, Patricia works three more than twice the number of hours that Hugo works. Sun works 2 less than Hugo. (a) Let x represent the number of hours that Hugo works each week. The number of hours that Hugo, Patricia, and Sun work can be modeled is shown below. Write an expression that represents each person's number of hours. +1 1. 28 March, 19:57 0 No. of hours Patrica works = 2x+3 No. of hours Sun works = x-2 No. of hours Hugo works = x Step-by-step explanation: Given : Patricia, Hugo and Sun work at a music store. Each week, Patricia works three more than twice the number of hours that Hugo works. Sun works 2 less than Hugo. To Find : Write an expression that represents each person's number of hours. Solution : Let Let x represent the number of hours that Hugo works each week. Since we are given that Patricia works three more than twice the number of hours that Hugo works. So, Patrica works for hours = 2x+3 Since Sun works 2 less than Hugo. = x-2 Hence no. of hours Patrica works = 2x+3 No. of hours Sun works = x-2 No. of hours Hugo works = x
# Question 19822 Dec 30, 2016 $D . 5$ #### Explanation: Let $R$ be the radius of the incircle. Radius of incircle of a right triangle R = (a*b)/((a+b+c), where a and b are the legs of the triangle and c is the hypotenuse. To get the largest possible radius of the incircle, 12 should be the smallest side of the triangle. As this is a multiple-choice (objective) question, let's start with trying Option $D \left(R = 5\right)$ Let $a$ be the smallest side =$12$ $\implies R = \frac{12 b}{12 + b + c} = 5$ $\implies c = \frac{7 b - 60}{5}$ ...............(1) As the triangle is right-angled, $\implies {c}^{2} = {a}^{2} + {b}^{2}$ $\implies {c}^{2} = 144 + {b}^{2}$ ....... (2) Substituting (1) into (2), we get $b = 35 , c = 37$ So the triangle is in the ratio of : $12 : 35 : 37$ Check : ${12}^{2} + {35}^{2} = 1369 = {37}^{2}$, (OK) $R = \frac{a \times b}{a + b + c} = \frac{12 \times 35}{12 + 35 + 37} = \left(\frac{420}{84}\right) = 5$, Hence, option $D$ is the answer. ........................................................................................................... Footnote : the solution below is a much better one. A big "thank you" to $\mathrm{dk}$_$c h$ for kindly providing such an excellent solution. We have c^2−b^2=144, where $c$ is the hypotenuse and $b$ is another side. $c > b$. By the given condition both $c \mathmr{and} b$ are integers. Now (c+b)(c−b)=144=72⋅2...[1] Here both (c+b)and(c−b) are even integers as their product $144$ is even one. To get maximum integer value satisfying the given condition both $c \mathmr{and} b$ will have large integer value and the value of c−b will have minimum possible even integer value. So the minimum integer value of (c−b) should be $2$, i.e., (c−b)=2. So from relation [1] we have (c+b)=72 and (c−b)=2# Thus we get $c = 37 \mathmr{and} b = 35$.
Combine Like Terms Combine Like Terms Key Terms A variable can represent a variety of numbers. A number by itself, not directly affected by a variable, is called a constant . An algebraic expression is a collection of numbers, variables, grouping symbols, and operation symbols. When an algebraic expression consists of several parts, each part that is added is called a term . The numerical part of a term is called its numerical coefficient (or coefficient). In the term 5x, the 5 is the numerical coefficient. 5x means that the variable, x, is multiplied by 5. Whenever a term appears without a numerical coefficient, we assume that the coefficient is 1. For example, x is the same as 1x, and -x is the same as -1x. Like Terms Like terms (or similar terms) are terms that have the same variables with the same exponents. Constants, such as 4 or -3 are like terms because they are both numbers with a definite value. Only the coefficients may differ for terms to be categorized as like terms. Combining Like Terms To combine like terms means to add or subtract the like terms in an expression. 1. Determine which terms are like terms. 2. Add or subtract the coefficients of the like terms. 3. Multiply the number found in step 2 by the common variable(s). Practice 1. Combine like terms in the algebraic expressions below. a) 2x + 5x + 7 b) 2ab + 3a + 9 c) x + 3 + y - 1/2 d) x2 + 3x - 4x2 e) 12 + x - x2 - 7 f) 3ab + 3ab2 + 3a2b2 + 2ab2 g) 5y - y + 6y 2. Combine like terms (simplify the expressions). a) 3/5x - 2/3x b) 6.47b - 8.2b c) 3b + 5a - 4 - 2a d) -2x2 + 4y - 3x2 + 2 - y + 5 1.) a.) 7x + 7 b.) 2ab + 3a + 9 c.) x + 5/2 + d.) -3x2 + 3x e.) 5 + x - x2 f.) 3ab + 5ab2 + 3a2b2 g.) 10y 2.) a.) (-1/15)x b.) -1.73b c.) 3a + 3b -4 d.) -5x2 + 3y + 7
# Measurement Word Problems (Grade 4) Videos and lessons to help Grade 4 students learn to use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale. Common Core: 4.MD.2 ### Suggested Learning Targets • I can use +, -, ×, and ÷ to solve word problems. • I can solve measurement word problems that include whole numbers, fractions, and decimals. • I can convert larger units into equivalent smaller units to solve a problem. Related Topics: The following activities are obtained from the Howard County Public School System. Activity 1: In PE, Zack and his friends had to measure their heights. They each used a different measurement tool, and then recorded their heights in the chart below. Student Zack Duncan Cameron Height 1 yard 3 1/2feet 34 inches Order the three boys by height, writing their names from tallest to shortest. Explain how you figured out which boy was the tallest of the three. A fourth student, Ryan, measured himself using a meter stick, and he found that he was exactly one meter tall. If a meter measures approximately 39 inches, how does Ryan's height compare to the heights of the other three boys? Tell how Ryan compares in height to Zack, Duncan, and Cameron. Then, use what you know about yards, feet, inches, and meters to explain your thinking. Activity 2: Jack and Abby are in swim class together. They challenged each other to see who could hold their breath underwater the longest. Jack and Abby's coach, Coach Foster, timed them with a stopwatch when they went underwater. Jack stayed underwater for 10 1/4 seconds, while Abby was able to stay under for 14 seconds. Show on a number line the amount of time Jack and Abby each stayed underwater. How much longer could Abby stay underwater than Jack? Coach Foster told the kids that he was a champion swimmer, and he used to train himself to hold his breath. The longest he was ever able to hold his breath was three minutes. How much longer would Abby have to hold her breath underwater to match Coach Foster's time? Explain your thinking using words, numbers, and/or symbols. Activity 3: Brandon and Kelly are training to run in a 5-kilometer race next month. Each morning, Brandon runs a route through the neighborhood park while Kelly runs on the racetrack at the high school. On Monday, Brandon ran 3 1/2 kilometers before he needed to take a break. Kelly ran 7 laps on the track, and then she needed to rest. If each lap Kelly ran was 400 meters, who ran a longer distance on Monday: Brandon or Kelly? Explain how you know which person ran a longer distance. On Wednesday, Kelly was able to run 9 laps, while Brandon ran 3 kilometers. How much further did Kelly run than Brandon on Wednesday? On Friday, Kelly ran a certain number of laps, and Brandon ran a certain number of kilometers. They ended up running the same distance as each other. How far could each of them have run? Activity 4: Tyler has been saving up for a new video game that costs \$60. He earned \$28.75 over the past two weeks by mowing lawns, and his grandmother sent him \$25 for his birthday.Â He knew he didn't have enough for the video game, so he decided to take the twenty half-dollar coins he had in his coin collection. Tyler didn't figure out the exact amount of money he had, but he knew that his total was more than enough to purchase the video game.Â Explain how could Tyler have known that he had enough money without figuring out the exact amount he had. When Tyler went to the store to buy the game, he discovered that the game cost \$60, but the sales tax was an additional \$7.20. With the added sales tax, he didn't have enough money after all. How much more money does Tyler need to be able to buy the game? The Iditarod & Math Grade 4 / Math / Tech Integration CCSS: Math.4.MD.A.2 Lesson Objective Calculate elapsed time and distance using real-time data Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Finding the Volume of a T-shaped Rectangular Prism Save Next Video: How to Determine the Volume of a Conical Solid....5 Finding the volume of a T-shaped rectangular prism will be a great way to get experience working with irregular shapes. Find the volume of a t-shaped rectangular prism with help from an experienced mathematics educator in this free video clip. Part of the Video Series: Math Lessons & Tips Promoted By Zergnet ## Video Transcript Hi, my name is Marija, and today I'm gonna show you how to find the volume of a t-shaped rectangular prism. But first it's important to understand that if a prism is t-shaped, then it's not necessarily rectangular, because rectangular means that the length and width are each the same size, right. So, what we're actually gonna have to do is split this t-shaped prism into two different prisms that are rectangular. So now we're looking at this bottom piece and this top piece, right? So it will be 3D, and probably look a lot better drawn where you have it, but we're looking at this prism down here, this rectangular prism, and this separate rectangular prism going this way. So the way that you would find the volume of this whole t-shape prism is by adding the two separate volumes of the rectangular prisms together. And to find the volume of a rectangular prism, all we have to do is length times width times height. So if you're given all the dimensions, let's say we're given that this is ten, going down we have a two, and then the depth we'll give it a three, we're doing ten times two is 20, and 20 times three is 60, so if these were let's say feet, now we have 60 feet cubed. And then if we're working with this other prism, let's say that we have going down another two, going this way four, and five. Five times four is 20 and 20 times two is 40, so this would be 40. Again if it was feet, we'll say feet cubed. Now I take these two separate volumes that I have and because I've put together these shapes in order to come up with this t, I'm gonna put together these volumes by adding. So I'm adding the 60 and the 40 to get 100 feet cubed as my volume of this t-shaped prism. ## Related Searches M Is DIY in your DNA? Become part of our maker community.
## Intermediate Algebra (12th Edition) $(5c-2+d)(5c-2-d)$ $\bf{\text{Solution Outline:}}$ To factor the given expression, $25c^2-20c+4-d^2 ,$ group the first $3$ terms since these form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms above results to \begin{array}{l}\require{cancel} (25c^2-20c+4)-d^2 .\end{array} The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (5c-2)^2-d^2 .\end{array} The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is \begin{array}{l}\require{cancel} [(5c-2)+d][(5c-2)-d] \\\\= (5c-2+d)(5c-2-d) .\end{array}
S k i l l i n A L G E B R A 13 # EXPONENTS Three rules WHEN All THE FACTORS OF A PRODUCT are equal -- 6· 6· 6· 6 -- we call the product a power of that factor (Lesson 1). Do the problem yourself first! Problem 1. What number is a) the third power of 2? 2· 2· 2 = 8 b) the fourth power of 3? = 81 c) the fifth power of 10? = 100,000 d) the first power of 8? = 8 Now, rather than write the third power of 2 as 2· 2· 2, we write 2 just once -- and place an exponent: 23. 2 is called the base. The exponent indicates the number of times to repeat the base as a factor. The student must take care not to confuse 3a, which means 3 times a, with a3, which means a times a times a. 3a = a + a + a, a3 = a· a· a. Problem 2. What does each symbol mean? a) x5 = xxxxx b) 53 = 5· 5· 5 c) 5· 3 = 3 + 3 + 3 + 3 + 3. d) (5a)3 = 5a· 5a· 5a e) 5a3 = 5aaa In part d), the parentheses indicate that the base is 5a. In part e), only a is the base. The exponent does not apply to 5. Problem 3. 34 = 81. a) Which number is called the base? 3 b) Which number is the power? 81 is the power of 3. c) Which number is the exponent? 4. It indicates the power, namely the 4th. Problem 4. Write out the meaning of these symbols. a) a2a3 = aa· aaa b) (ab)3 = ab· ab· ab c) (a2)3 = a2· a2· a2 Problem 5. Write out the meaning of these symbols. In each one, what is the base? a) a4aaaa. The base is a. b) −a4 = −aaaa. The base again is a. This is the negative of a4.A minus sign always signifies the negative of the number that follows. −5 is the negative of 5. And −a4 is the negative of a4. c) (−a)4 = (−a)(−a)(−a)(−a). Here, the base is (−a). Problem 6. Evaluate. a) 24 = 16. b) −24 = −16. This is the negative of 24. The base is 2. See Problem 5b) above. a = (−1)a, for any number a. (Lesson 6.) Therefore −24 = (−1)24. And therefore according to the order of operations (Lesson 1), this is (−1)16 = −16. c) (−2)4 = +16, according to the Rule of Signs (Lesson 4). The parentheses indicate that the base is −2. See Problem 5c). Example 1. Negative base. (−2)3 = (−2)(−2)(−2) = −8, again according to the Rule of Signs. Whereas, (−2)4 = +16. When the base is negative, and the exponent is odd, then the product is negative. But when the base is negative, and the exponent is even, then the product is positive. Problem 7. Evaluate. a) (−1)2 = 1 b) (−1)3 = −1 c) (−1)4 = 1 d) (−1)5 = −1 e) (−1)100 = 1 f) (−1)253 = −1 g) (−2)4 = 16 h) (−2)5 = −32 Problem 8. Rewrite using exponents. a) xxxxxx = x6 b) xxyyyy = x2y4 c) xyxxyx = x4y2 d) xyxyxy = x3y3 Problem 9. Rewrite using exponents. a) (x + 1)(x + 1) = (x + 1)2 b) (x − 1)(x − 1)(x − 1) = (x − 1)3 c) (x + 1)(x − 1)(x + 1)(x − 1) = (x + 1)2(x − 1)2 d) (x + y)(x + y)2 = (x + y)3 Three rules Rule 1. Same Base aman = am + n "To multiply powers of the same base, add the exponents." For example, a2a3 = a5. Why do we add the exponents? Because of what the symbols mean. Problem 4a. Example 2. Multiply 3x2· 4x5· 2x Solution. The problem means (Lesson 5): Multiply the numbers, then combine the powers of x : 3x2· 4x5· 2x = 24x8 Two factors of x -- x2 -- times five factors of x -- x5 -- times one factor of x, produce a total of 2 + 5 + 1 = 8 factors of x : x8. Problem 10. Multiply. Apply the rule Same Base. a) 5x2· 6x4 = 30x6 b) 7x3· 8x6 = 56x9 c) x· 5x4 = 5x5 d) 2x· 3x· 4x = 24x3 e) x3· 3x2· 5x = 15x6 f) x5· 6x8y2 = 6x13y2 g) 4x· y· 5x2· y3 = 20x3y4 h) 2xy· 9x3y5 = 18x4y6 i) a2b3a3b4 = a5b7 j) a2bc3b2ac = a3b3c4 k) xmynxpyq = xm + pyn+ q l) apbqab = ap + 1bq + 1 Problem 11. Distinguish the following: x· x and x + x. x· x = x². x + x = 2x. Example 3. Compare the following: a) x· x5 b) 2· 25 Solution. a) x· x5 = x6 b) 2· 25 = 26 Part b) has the same form as part a). It is part a) with x = 2. One factor of 2 multiplies five factors of 2 producing six factors of 2. 2· 2 = 4 is not correct here. Problem 12. Apply the rule Same Base. a) xx7 = x8 b) 3· 37 = 38 c) 2· 24· 25 = 210 d) 10· 105 = 106 e) 3x· 36x6 = 37x7 Problem 13. Apply the rule Same Base. a) xnx2 = xn + 2 b) xnx = xn + 1 c) xnxn = x2n d) xnx1 − n = x e) x· 2xn − 1 = 2xn f) xnxm = xn + m g) x2nx2 − n = xn + 2 Rule 2: Power of a Product of Factors (ab)n = anbn "Raise each factor to that same power." For example, (ab)3 = a3b3. Why may we do that? Again, according to what the symbols mean: (ab)3 = ab· ab· ab = aaabbb = a3b3. The order of the factors does not matter: ab· ab· ab = aaabbb. Problem 14. Apply the rules of exponents. a) (xy)4 = x4y4 b) (pqr)5 = p5q5r5 c) (2abc)3 = 23a3b3c3 d) x3y2z4(xyz)5 = x3y2z4· x5y5z5 Rule 2, = x8y7z9 Rule 1. Rule 3: Power of a Power (am)n = amn "To take a power of a power, multiply the exponents." For example, (a2)3 = a2· 3 = a6. Why do we do that? Again, because of what the symbols mean: (a2)3 = a2a2a2 = a3· 2 = a6 Problem 15. Apply the rules of exponents. a) (x2)5 = x10 b) (a4)8 = a32 c) (107)9 = 1063 Example 4. Apply the rules of exponents: (2x3y4)5 Solution. Within the parentheses there are three factors: 2, x3, and y4. According to Rule 2 we must take the fifth power of each one. But to take a power of a power, we multiply the exponents. Therefore, (2x3y4)5 = 25x15y20 Problem 16. Apply the rules of exponents. a) (10a3)4 = 10,000a12 b) (3x6)2 = 9x12 c) (2a2b3)5 = 32a10b15 d) (xy3z5)2 = x2y6z10 e) (5x2y4)3 = 125x6y12 f) (2a4bc8)6 = 64a24b6c48 Problem 17. Apply the rules of exponents. a) 2x5y4(2x3y6)5 = 2x5y4· 25x15y30 = 26x20y34 b) abc9(a2b3c4)8 = abc9· a16b24c32 = a17b25c41 Problem 18. Use the rules of exponents to calculate the following. a) (2· 10)4 = 24· 104 = 16· 10,000 = 160,000 b) (4· 102)3 = 43· 106 = 64,000,000 c) (9· 104)2 = 81· 108 = 8,100,000,000 Example 5. Square x4. Solution. (x4)2 = x8. Thus to square a power, double the exponent. Problem 19. Square the following. a) x5 = x10 b) 8a3b6 = 64a6b12 c) −6x7 = 36x14 d) xn = x2n Part c) illstrates: The square of a number is always positive. (−6)(−6) = +36. The Rule of Signs. Except 02 = 0. Problem 20. Apply a rule of exponents -- if possible. a) x2x5 = x7, Rule 1. b) (x2)5 = x10, Rule 3. c) x2 + x5 Not possible. The rules of exponents apply onlyto multiplication. In summary: Add the exponents when the same base appears twice: x2x4 = x6. Multiply the exponents when the base appears once -- and in parentheses: (x2)5 = x10. Problem 21. Apply the rules of exponents. a) (xn)n = xn· n = xn² b) (xn)2 = x2n Problem 22. Apply a rule of exponents or add like terms -- if possible. a) 2x2 + 3x4 Not possible. These are not like terms (Lesson 1). b) 2x2· 3x4 = 6x6. Rule 1. c) 2x3 + 3x3 = 5x3. Like terms. The exponent does not change. d) x2 + y2 Not possible. These are not like terms. e) x2 + x2 = 2x2. Like terms. f) x2· x2 = x4. Rule 1. g) x2· y3 Not possible. Different bases. h) 2· 26 = 27. Rule 1. i) 35 + 35 + 35 = 3· 35 (On adding those like terms) = 36. We will continue the rules of exponents in Lesson 21. Next Lesson: Multiplying out. The distributive rule. Please make a donation to keep TheMathPage online. Even \$1 will help.
<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. Division of Polynomials Using long division to divide polynomials Estimated20 minsto complete % Progress Practice Division of Polynomials Progress Estimated20 minsto complete % Polynomial Long Division and Synthetic Division While you may be experienced in factoring, there will always be polynomials that do not readily factor using basic or advanced techniques. How can you identify the roots of these polynomials? Watch This http://www.youtube.com/watch?v=brpNxPAkv1c James Sousa: Dividing Polynomials-Long Division Guidance There are numerous theorems that point out relationships between polynomials and their factors. For example there is a theorem that a polynomial of degree must have exactly solutions/factors that may or may not be real numbers. The Rational Root Theorem and the Remainder Theorem are two theorems that are particularly useful starting places when manipulating polynomials. • The Rational Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction , where is an integer factor of the constant term and is an integer factor of the leading coefficient. Example A shows how all the possible rational solutions can be listed using the Rational Root Theorem. • The Remainder Theorem states that the remainder of a polynomial divided by a linear divisor is equal to . The Remainder Theorem is only useful after you have performed polynomial long division because you are usually never given the divisor and the remainder to start. The main purpose of the Remainder Theorem in this setting is a means of double checking your application of polynomial long division. Example B shows how the Remainder Theorem is used. Polynomial long division is identical to regular long division. Synthetic division is a condensed version of regular long division where only the coefficients are kept track of. In Example B polynomial long division is used and in Example C synthetic long division is used. Example A Identify all possible rational solutions of the following polynomial using the Rational Root Theorem. Solution: The integer factors of 5 are 1, 5. The integer factors of 12 are 1, 2, 3, 4, 6 and 12. Since pairs of factors could both be negative, remember to include While narrowing the possible solutions down to 24 possible rational answers may not seem like a big improvement, it surely is. This is especially true considering there are only a handful of integer solutions. If this question required you to find a solution, then the Rational Root Theorem would give you a great starting place. Example B Use Polynomial Long Division to divide: Solution: First note that it is clear that 3 is not a root of the polynomial because of the Rational Root Theorem and so there will definitely be a remainder. Start a polynomial long division question by writing the problem like a long division problem with regular numbers: Just like with regular numbers ask yourself “how many times does go into ?” which in this case is . Now multiply the by and copy below. Remember to subtract the entire quantity. Combine the rows, bring down the next number and repeat. The number 37 is the remainder. There are two things to think about at this point. First, interpret in an equation: Second, check your result with the Remainder Theorem which states that the original function evaluated at 3 must be 37. Notice the notation indicating to substitute 3 in for Example C Use Synthetic Division to divide the same rational expression as the previous example. Solution: Synthetic division is mostly used when the leading coefficients of the numerator and denominator are equal to 1 and the divisor is a first degree binomial. Instead of continually writing and rewriting the symbols, synthetic division relies on an ordered spacing. Notice how only the coefficients for the denominator are used and the divisor includes a positive three rather than a negative three. The first coefficient is brought down and then multiplied by the three to produce the value which goes beneath the 2. Next the new column is added. , which goes beneath the column. Now, multiply , which goes underneath the -5 in the column. And the process repeats… The last number, 37, is the remainder. The three other numbers represent the quadratic that is identical to the solution to Example B. Concept Problem Revisited Identifying roots of polynomials by hand can be tricky business. The best way to identify roots is to use the rational root theorem to quickly identify likely candidates for solutions and then use synthetic or polynomial long division to quickly and effectively test them to see if their remainders are truly zero. Vocabulary Polynomial long division is a procedure with rules identical to regular long division. The only difference is the dividend and divisor are polynomials. Synthetic division is an abbreviated version of polynomial long division where only coefficients are used. Guided Practice 1. Divide the following polynomials. 2. Completely factor the following polynomial. 3. Divide the following polynomials. 1. 2. Notice that possible roots are . Of these 14 possibilities, four will yield a remainder of zero. When you find one, repeat the process. 3. Practice Identify all possible rational solutions of the following polynomials using the Rational Root Theorem. 1. 2. 3. 4. 5. Completely factor the following polynomials. 6. 7. 8. 9. 10. Divide the following polynomials. 11. 12. 13. 14. 15. Vocabulary Language: English Denominator Denominator The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$. Dividend Dividend In a division problem, the dividend is the number or expression that is being divided. divisor divisor In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend. Polynomial long division Polynomial long division Polynomial long division is the standard method of long division, applied to the division of polynomials. Rational Expression Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator. Rational Root Theorem Rational Root Theorem The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$. Remainder Theorem Remainder Theorem The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$. Synthetic Division Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used. Explore More Sign in to explore more, including practice questions and solutions for Division of Polynomials.
# How do you solve using the completing the square method 2x^2-8x+3=0? May 6, 2016 $x = 2 \pm \frac{\sqrt{10}}{2}$ #### Explanation: Complete the square then use the difference of squares identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$ with $a = \left(2 x - 4\right)$ and $b = \sqrt{10}$ as follows. Multiply by $2$ first to make the leading term a perfect square: $0 = 2 \left(2 {x}^{2} - 8 x + 3\right)$ $= 4 {x}^{2} - 16 x + 6$ $= {\left(2 x\right)}^{2} - 2 \left(2 x\right) \left(4\right) + 6$ $= {\left(2 x - 4\right)}^{2} - 16 + 6$ $= {\left(2 x - 4\right)}^{2} - 10$ $= {\left(2 x - 4\right)}^{2} - {\left(\sqrt{10}\right)}^{2}$ $= \left(\left(2 x - 4\right) - \sqrt{10}\right) \left(\left(2 x - 4\right) + \sqrt{10}\right)$ $= \left(2 x - 4 - \sqrt{10}\right) \left(2 x - 4 + \sqrt{10}\right)$ $= \left(2 \left(x - 2 - \frac{\sqrt{10}}{2}\right)\right) \left(2 \left(x - 2 + \frac{\sqrt{10}}{2}\right)\right)$ $= 4 \left(x - 2 - \frac{\sqrt{10}}{2}\right) \left(x - 2 + \frac{\sqrt{10}}{2}\right)$ Hence: $x = 2 \pm \frac{\sqrt{10}}{2}$
# CAT Practice : Number System: Factorial You are here: Home CAT Questionbank CAT Quant Number System: Factorial Question 7 We can find the number of trailing zeroes a factorial have. Can you find a number that has a specific number of trailing zeroes? Can you give a certain number of trailing zeroes that are not present in any factorial? ## Factorials - Trailing zeroes Q.7: Find the least number n such that no factorial has n trailing zeroes, or n + 1 trailing zeroes or n + 2 trailing zeroes. 1. 153 2. 126 3. 624 4. 18 Choice A. 153 ## Detailed Solution The previous question includes a detailed discussion on how to find the number of trailing zeroes of n!, for any natural number n. We see that 24! has ${{{{\rm{24}}} \over {\rm{5}}}}$ = 4 zeroes 25! ends with $\left[ {{{{\rm{25}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{25}}} \over {\rm{5}}}} \right]$ = 6 zeroes. There is no natural number m such that m! has exactly 5 zeroes. Similarly, we see that 49! ends with$\left[ {{{{\rm{49}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{49}}} \over {{\rm{25}}}}} \right]$ = 10 zeroes, whereas 50! ends with $\left[ {{{{\rm{50}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{50}}} \over {{\rm{25}}}}} \right]$ = 12 zeroes. No factorial ends with 11 zeroes. So, any time we have a multiple of 25, we 'skip' a zero. This is because a multiple of 25 adds two zeroes to the factorial. Extrapolating this, we can see that 125 might actually 'skip' two zeroes. 124! ends with $\left[ {{{{\rm{124}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{124}}} \over {{\rm{25}}}}} \right]$ = 24 + 4 = 28 zeros, whereas 125! has $\left[ {{{{\rm{125}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{125}}} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{125}}} \over {{\rm{125}}}}} \right]$ = 25 + 5 + 1 = 31 zeros. There is no factorial with 29 or 30 zeros. In order to jump three zeros, think about what we need to look at. Every multiple of 25 gives us one 'skipped' zero. Every multiple of 125 gives us two 'skipped' zeroes. In order to have three skipped zeroes, we need to look at 624! and 625! 624! has $\left[ {{{624} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{624} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{624}}} \over {{\rm{125}}}}} \right]$ = 124 + 24 + 4 = 152 zeros 625! has $\left[ {{{625} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{625} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{625}}} \over {{\rm{125}}}}} \right]{\rm{ + }}\left[ {{{{\rm{625}}} \over {{\rm{625}}}}} \right]$ = 125 + 25 + 5 + 1 = 156 zeros There is no factorial with 153, 154 or 155 zeros. Or the least value of n such that no factorial ends with n, (n + 1) or (n + 2) zeroes is 153. ## Our Online Course, Now on Google Playstore! ### Fully Functional Course on Mobile All features of the online course, including the classes, discussion board, quizes and more, on a mobile platform. ### Cache Content for Offline Viewing Download videos onto your mobile so you can learn on the fly, even when the network gets choppy! ## More questions from Number System - Factorial This idea is so good that it comes with an exclamation mark. N! holds marvels that you might not have noticed before. Enter here to see those.
# Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.3 Last Updated : 30 Apr, 2021 ### (i) f (x) = 1/x Solution: We are given, f (x) = 1/x. Here, f (x) is defined for all real values of x, except for the case when x = 0. Therefore, domain of f = R â€“ {0} ### (ii) f (x) = 1/(xâˆ’7) Solution: We are given, f (x) = 1/(xâˆ’7). Here, f (x) is defined for all real values of x, except for the case when x â€“ 7 = 0 or x = 7. Therefore, domain of f = R â€“ {7} ### (iii) f (x) = (3xâˆ’2)/(x+1) Solution: We are given, f (x) = (3xâˆ’2)/(x+1). Here, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = â€“1. Therefore, domain of f = R â€“ {â€“1} ### (iv) f (x) = (2x+1)/(x2âˆ’9) Solution: We are given, f (x) = (2x+1)/(x2âˆ’9). Here, f (x) is defined for all real values of x, except for the case when x2 â€“ 9 = 0. => x2 â€“ 9 = 0 => (x + 3)(x â€“ 3) = 0 => x + 3 = 0 or x â€“ 3 = 0 => x = Â± 3 Therefore, domain of f = R â€“ {â€“3, 3} ### (v) f (x) = (x2+2x+1)/(x2â€“8x+12) Solution: We are given, f (x) = (x2+2x+1)/(x2â€“8x+12). Here, f(x) is defined for all real values of x, except for the case when x2 â€“ 8x + 12 = 0. => x2 â€“ 8x + 12 = 0 => x2 â€“ 2x â€“ 6x + 12 = 0 => x(x â€“ 2) â€“ 6(x â€“ 2) = 0 => (x â€“ 2)(x â€“ 6) = 0 => x â€“ 2 = 0 or x â€“ 6 = 0 => x = 2 or 6 Therefore, domain of f = R â€“ {2, 6} ### Question 2. Find the domain of each of the following real-valued functions of a real variable: (i) f (x) = âˆš(xâ€“2) Solution: We are given, f (x) = âˆš(xâ€“2). Here, f (x) takes real values only when x â€“ 2 â‰¥ 0 as the square of a real number cannot be negative. => x â€“ 2 â‰¥ 0 => x â‰¥ 2 => x âˆˆ [2, âˆž) Therefore, domain of f = [2, âˆž) ### (ii) f (x) = 1/(âˆš(x2â€“1)) Solution: We are given, f (x) = 1/(âˆš(x2â€“1)). Here, f (x) takes real values only when x2 â€“ 1 > 0 as the square of a real number cannot be negative and the denominator x2 â€“ 1 cannot be zero. => x2 â€“ 1 > 0 => (x + 1) (x â€“ 1) > 0 => x < â€“1 or x > 1 => x âˆˆ (â€“âˆž, â€“1) âˆª (1, âˆž) Therefore, domain of f = (â€“âˆž, â€“1) âˆª (1, âˆž) ### (iii) f (x) = âˆš(9â€“x2) Solution: We are given, f (x) = âˆš(9â€“x2). Here, f (x) takes real values only when 9 â€“ x2 â‰¥ 0 as the square of a real number cannot be negative. => 9 â€“ x2 â‰¥ 0 => 9 â‰¥ x2 => x2 â‰¤ 9 => x2 â€“ 9 â‰¤ 0 => (x + 3)(x â€“ 3) â‰¤ 0 => x â‰¥ â€“3 and x â‰¤ 3 => x âˆˆ [â€“3, 3] Therefore domain of f = [â€“3, 3] ### (iv) f (x) = âˆš[(xâ€“2)/(3â€“x)] Solution: We are given, f (x) = âˆš[(xâ€“2)/(3â€“x)]. Here, f (x) takes real values only when x â€“ 2 and 3 â€“ x are both positive and negative. Case 1. x â€“ 2 â‰¥ 0 and 3 â€“ x â‰¥ 0 => x â‰¥ 2 and x â‰¤ 3 Therefore, x âˆˆ [2, 3] Case 2. x â€“ 2 â‰¤ 0 and 3 â€“ x â‰¤ 0. => x â‰¤ 2 and x â‰¥ 3 This case is not possible as the intersection of these sets is null set. Hence, x âˆˆ [2, 3] â€“ {3} => x âˆˆ [2, 3) Therefore, domain of f = [2, 3) ### (i) f (x) = (ax+b)/(bxâ€“a) Solution: We are given, f (x) = (ax+b)/(bxâ€“a). Here, f(x) is defined for all real values of x, except for the case when bx â€“ a = 0 or x = a/b. So, domain of f = R â€“ (a/b) Let f (x) = y. So, (ax+b)/(bxâ€“a) = y. => ax + b = y(bx â€“ a) => ax + b = bxy â€“ ay => ax â€“ bxy = â€“ay â€“ b => x(a â€“ by) = â€“(ay + b) => x = â€“ (ay+b)/(aâ€“by) When a â€“ by = 0 or y = a/b. Hence, f(x) cannot take the value a/b. Therefore, range of f = R â€“ (a/b) ### (ii) f (x) = (axâ€“b)/(cxâ€“d) Solution: We are given, f (x) = (axâ€“b)/(cxâ€“d). Here, f(x) is defined for all real values of x, except for the case when cx â€“ d = 0 or x = d/c. So, domain of f = R â€“ (d/c) Let f (x) = y. So, (axâ€“b)/(cxâ€“d) = y => ax â€“ b = y(cx â€“ d) => ax â€“ b = cxy â€“ dy => ax â€“ cxy = b â€“ dy => x(a â€“ cy) = b â€“ dy => x = (bâ€“dy)/(aâ€“cy) When a â€“ cy = 0 or y = a/c. Hence, f(x) cannot take the value a/c. Therefore, range of f = R â€“ (a/c) ### (iii) f (x) = âˆš(xâ€“1) Solution: We are given, f (x) = âˆš(xâ€“1). Here, f(x) takes real values only when x â€“ 1 â‰¥ 0. => x â‰¥ 1 => x âˆˆ [1, âˆž) So, domain of f = [1, âˆž) When x â‰¥ 1, we have x â€“ 1 â‰¥ 0. So, âˆš(xâ€“1) â‰¥ 0. => f (x) â‰¥ 0 => f(x) âˆˆ [0, âˆž) Therefore, range of f = [0, âˆž) ### (iv) f (x) = âˆš(xâ€“3) Solution: We are given, f (x) = âˆš(xâ€“3). Here, f (x) takes real values only when x â€“ 3 â‰¥ 0. => x â‰¥ 3 => x âˆˆ [3, âˆž) So, domain of f = [3, âˆž) When x â‰¥ 3, we have x â€“ 3 â‰¥ 0. Hence, âˆš(xâ€“3) â‰¥ 0 => f (x) â‰¥ 0 => f(x) âˆˆ [0, âˆž) Therefore, range of f = [0, âˆž) ### (v) f (x) = (xâ€“2)/(2â€“x) Solution: We are given, f (x) = (xâ€“2)/(2â€“x). Here, f(x) is defined for all real values of x, except for the case when 2 â€“ x = 0 or x = 2. So, domain of f = R â€“ {2} And also, f (x) = â€“(2â€“x)/(2â€“x) = â€“1 Hence, when x â‰ 2, f(x) = â€“1 Therefore, range of f = {â€“1} ### (vi) f (x) = \|xâ€“1\| Solution: We are given, f (x) = \|xâ€“1\|. Clearly, f(x) is defined for all real numbers x. So, domain of f = R Let f (x) = y. So, \|xâ€“1\| = y. Therefore, y can take only the positive values. So, y â‰¥ 0. Therefore, range of f = (0, âˆž] ### (vii) f (x) = â€“\|x\| Solution: We are given, f (x) = â€“\|x\|. Clearly, f(x) is defined for all real numbers x. So, domain of f = R Let f (x) = y. So, y = â€“\|x\|. Therefore, y can take only the negative values. So, y â‰¤ 0. Therefore, range of f = (â€“âˆž, 0] ### (viii) f (x) = âˆš(9â€“x2) Solution: We are given, f (x) = âˆš(9â€“x2) Here, f(x) takes real values only when 9 â€“ x2 â‰¥ 0. => 9 â‰¥ x2 => x2 â‰¤ 9 => x2 â€“ 9 â‰¤ 0 => (x + 3)(x â€“ 3) â‰¤ 0 => x â‰¥ â€“3 and x â‰¤ 3 => x âˆˆ [â€“3, 3] So, domain of f = [â€“3, 3] When, x âˆˆ [â€“3, 3], we have 0 â‰¤ 9 â€“ x2 â‰¤ 9. => 0 â‰¤ âˆš(9â€“x2) â‰¤ 3 => 0 â‰¤ f (x) â‰¤ 3 => f (x) âˆˆ [0, 3] Therefore, range of f = [0, 3] Previous Next
# Solving One-Step Linear Inequalities: A Comprehensive Guide Unlock the power of one-step linear inequalities with our easy-to-follow guide. Learn essential techniques, practice with real-world examples, and boost your algebra skills today. Now Playing:Solve one step linear inequalities– Example 0 Intros 1. i) What is an inequality? ii) How to solve linear inequalities? iii) How to graph inequalities on a number line? Examples 1. Solve. 1. $7x \le 63$ 2. $- \frac{1}{8}x$ > $- 5$ 3. $3 \ge \frac{x}{{1.3}}$ 4. $2.4x$ < $-1.8$ Practice Express linear inequalities graphically and algebraically Notes We learn how to solve one-step linear inequalities in this lesson. Besides typical math questions, we are going to see word problems related to one-step linear inequalities and money. Concept ## Introduction to One-Step Linear Inequalities One-step linear inequalities are fundamental concepts in algebra that extend our understanding beyond equations. The introduction video provides a crucial foundation for grasping this topic. Unlike equations, which have a single solution, inequalities represent a range of values that satisfy a given condition. For example, while x = 5 is an equation with one solution, x > 5 is an inequality with multiple solutions. Inequalities use symbols like <, >, , or to show relationships between expressions. The solution set of an inequality includes all values that make the statement true. Understanding one-step linear inequalities is essential for solving inequalities and more complex problems in mathematics and real-world applications. By mastering this concept, students can develop critical thinking skills and a deeper appreciation for mathematical relationships. The introduction video serves as a gateway to exploring these ideas, setting the stage for more advanced topics in algebra and beyond. Example Solve. $7x \le 63$ #### Step 1: Understand the Inequality When solving one-step linear inequalities, it's important to understand that the process is similar to solving linear equations. The main difference is the inequality sign (, , <, >) instead of an equal sign (=). In this case, we have the inequality $7x \le 63$. Our goal is to isolate the variable $x$ on one side of the inequality. #### Step 2: Treat the Inequality Like an Equation To simplify the process, treat the inequality as if it were an equation. This means you can perform the same operations on both sides of the inequality as you would with an equation. For instance, if you had $7x = 63$, you would solve for $x$ by dividing both sides by 7. The same approach applies here. #### Step 3: Divide Both Sides by 7 To isolate $x$, divide both sides of the inequality by 7. This operation will help you get $x$ by itself on one side of the inequality. Here's how it looks: $7x \le 63$ Divide both sides by 7: $\frac{7x}{7} \le \frac{63}{7}$ When you divide 7 by 7 on the left side, it cancels out, leaving you with $x$. On the right side, 63 divided by 7 equals 9. So, the inequality simplifies to: $x \le 9$ #### Step 4: Interpret the Solution The final inequality $x \le 9$ means that $x$ can be any number less than or equal to 9. This is the solution to the inequality. In other words, any value of $x$ that is 9 or smaller will satisfy the original inequality $7x \le 63$. #### Step 5: Verify the Solution To ensure the solution is correct, you can test a few values of $x$ that are less than or equal to 9. For example, if $x = 9$, substituting it back into the original inequality gives: $7(9) \le 63$ $63 \le 63$ This is true. You can also test a value less than 9, such as $x = 5$: $7(5) \le 63$ $35 \le 63$ This is also true. Therefore, the solution $x \le 9$ is verified. FAQs 1. What is a one-step linear inequality? A one-step linear inequality is a mathematical statement that compares two expressions using inequality symbols (<, >, , ) and can be solved in one step. For example, x + 3 > 7 is a one-step linear inequality. 2. How do you solve a one-step linear inequality? To solve a one-step linear inequality, perform the inverse operation on both sides of the inequality to isolate the variable. For example, to solve x + 3 > 7, subtract 3 from both sides: x > 4. 3. What happens when you multiply or divide an inequality by a negative number? When you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality symbol. For example, if -2x < 6, dividing both sides by -2 gives x > -3. 4. How do you graph the solution of a one-step linear inequality on a number line? To graph the solution, use an open circle () for strict inequalities (< or >) and a closed circle () for inclusive inequalities ( or ) at the boundary point. Then, shade the line in the direction that satisfies the inequality. 5. Can you give an example of a real-life application of one-step linear inequalities? A real-life application could be determining how many hours you need to work to earn at least a certain amount of money. For instance, if you earn $15 per hour and want to make at least$300, you can set up the inequality 15x 300, where x is the number of hours worked. Prerequisites Understanding the foundation of mathematics is crucial when tackling more advanced concepts like solving one-step linear inequalities. To excel in this area, it's essential to grasp several key prerequisite topics that form the building blocks of algebraic problem-solving. One fundamental skill is solving equations, particularly those involving distance and time. This ability helps students develop a strong sense of algebraic reasoning and equation manipulation, which directly translates to working with inequalities. By mastering these types of problems, students can more easily transition to understanding the subtle differences between equations and inequalities. While one-step linear inequalities are relatively straightforward, it's beneficial to have exposure to more complex concepts like graphing inequalities. This broader perspective allows students to visualize how inequalities work on a coordinate plane, reinforcing their understanding of the number line and inequality symbols used in simpler problems. Additionally, familiarity with compound inequalities can provide valuable context. Although one-step inequalities are less complex, understanding how multiple inequalities can be combined helps students appreciate the versatility and importance of inequality concepts in mathematics. A solid grasp of basic arithmetic operations is also crucial. Specifically, proficiency in multiplying by negative numbers and dividing by negative numbers is essential when solving inequalities. These operations often come into play when manipulating inequality expressions, and understanding how they affect the inequality sign is vital for arriving at correct solutions. By mastering these prerequisite topics, students build a strong foundation for tackling one-step linear inequalities. The ability to solve equations provides the basic algebraic skills needed, while exposure to graphing and compound inequalities offers a broader context. Proficiency in working with negative numbers in multiplication and division ensures students can confidently manipulate inequality expressions. As students progress through these prerequisite topics, they develop critical thinking skills and mathematical intuition that are invaluable when solving one-step linear inequalities. Each concept builds upon the last, creating a comprehensive understanding of algebraic relationships and inequalities. This solid foundation not only aids in solving one-step linear inequalities but also prepares students for more advanced mathematical concepts they will encounter in their future studies. In conclusion, taking the time to thoroughly understand these prerequisite topics will greatly enhance a student's ability to solve one-step linear inequalities with confidence and accuracy. It's an investment in mathematical knowledge that pays dividends throughout one's academic journey and beyond.
# NCERT Solutions for Class 7 Maths Exercise 1.2 ## myCBSEguide App CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more. NCERT solutions for Class 7 Integers Maths ## NCERT Solutions for Class 7 Maths Integers ###### Question 1.Write down a pair of integers whose: (a) sum is {tex} – 7{/tex} (b) difference is {tex} – 10{/tex} (c) sum is 0 (a) One such pair whose sum is {tex} – 7{/tex}: {tex} – 5 + \left( { – 2} \right) = – 7{/tex} (b) One such pair whose difference is {tex} – 10{/tex}: {tex} – 2 – 8 = – 10{/tex} (c) One such pair whose sum is 0: {tex} – 5 + 5 = 0{/tex} NCERT Solutions for Class 7 Maths Exercise 1.2 ###### Question 2.(a) Write a pair of negative integers whose difference gives 8. (b) Write a negative integer and a positive integer whose is {tex} – 5.{/tex} (c) Write a negative integer and a positive integer whose difference is {tex} – 3.{/tex} (a) {tex} – 2 – \left( { – 10} \right) – 2 + 10 = 8{/tex} (b) {tex}\left( { – 7} \right) + 2 = – 5{/tex} (c) {tex}\left( { – 2} \right) – 1 = – 2 – 1 = – 3{/tex} NCERT Solutions for Class 7 Maths Exercise 1.2 ###### Question 3.In a quiz, team A scored {tex} – 40,10,0{/tex} and team B scores 10, 0, {tex} – 40{/tex} in three successive rounds. Which team scored more? Can we say that we can add integers in any order? Team A scored {tex} – 40,10,0{/tex} Total score of Team A = {tex} – 40 + 10 + 0 = – 30{/tex} Team B scored {tex}10,0, – 40{/tex} Total score of Team B = {tex}10 + 0 + \left( { – 40} \right) = 10 + 0 – 40 = – 30{/tex} Thus, scores of both teams are same. Yes, we can add integers in any order due to commutative property. NCERT Solutions for Class 7 Maths Exercise 1.2 ###### Question 4.Fill in the blanks to make the following statements true: (i){tex}\left( { – 5} \right) + \left( { – 8} \right) = \left( { – 8} \right) + \left( {…….} \right){/tex} (ii){tex} – 53 + ……. = – 53{/tex} (iii)17 + ……. = 0 (iv){tex}\left[ {13 + \left( { – 12} \right)} \right] + \left( {…….} \right) = 13 + \left[ {\left( { – 12} \right) + \left( { – 7} \right)} \right]{/tex} (v){tex}\left( { – 4} \right) + \left[ {15 + \left( { – 3} \right)} \right] = \left[ { – 4 + 15} \right] + …….{/tex} (i) {tex}\left( { – 5} \right) + \left( { – 8} \right) = \left( { – 8} \right) + \underline {\left( { – 5} \right)} {/tex} [Commutative property] (ii) {tex} – 53 + \underline 0 = – 53{/tex} [Zero additive property] (iii) {tex}17 + \underline {\left( { – 17} \right)} = 0{/tex} (Additive identity] (iv){tex}\left[ {13 + \left( {12} \right)} \right] + \underline {\left( { – 7} \right)} = 13 + \left[ {\left( { – 12} \right) + \left( { – 7} \right)} \right]{/tex} [Associative property] {tex}\left( { – 4} \right) + \left[ {15 + \left( { – 3} \right)} \right] = \left[ { – 4 + 15} \right] + \underline {\left( { – 3} \right)} {/tex} [Associative property] ## NCERT Solutions for Class 7 Maths Exercise 1.2 NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide. ## CBSE app for Students To download NCERT Solutions for Class 7 Maths, Social Science Computer Science, Home Science, Hindi English, Maths Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website. ### 4 thoughts on “NCERT Solutions for Class 7 Maths Exercise 1.2” 1. one mistake it is class not lass 2. Dude ur app has help us a lot . 3. Very nice 4. Ncert math 1.2 ka ans book me nahi mil raha hai
### 20110630 Chapter 9: Concluding Activity - Viva voce (a Practice) Each member in the class has been assigned a number: #1, #2, #3, #4 or #5 You have been assigned to attempt the question your number has been assigned to: Revision Exercise 9 (page 21-22_ Question 2 - attempted by those who have been assigned #1 Question 4 - attempted by those who have been assigned #2 Question 6 - attempted by those who have been assigned #3 Question 8 - attempted by those who have been assigned #4 Question 10 - attempted by those who have been assigned #5 You shall do the following: 1. Attempt the question in your notebook - work out the solution clearly 2. Using Quicktime Player or PhotoBooth, record your explanation of the solution as a video clip. 3. You may upload your video clip into Youtube before embedding it in a Post in the Maths Blog. 4. In the Maths blog • Title: (Viva Voce) Chap 9 Exercise 9 Question {2, 4, 6, 8 or 10} • Embed the video clip • You may include a 'brief' write-up of the steps/ solutions 5. Complete the posting before the next lesson. We will discuss the rubric and give peer feedback in the next lesson. ~~~~~~~~~~~~~~~ Question 2: Car X travels 60 km in 45 minutes. Car Y travels 72 km in 1 h 20 min. Find (a) The average speed of car X (b) The average speed of car Y (c) The ratio of average speeds of cars X and Y Question 4: Alcohol and water are mixed in the ratio 1:4 by volume. If the volume of the solution is 600cm^3 (a) Find the volume of alcohol in the solution (b) Find the volume of water in the solution (c) how much alcohol must be added to the solution so that the ratio of alcohol to water in the solution becomes 1:3 by volume? Question 6: (a) Simplify each of the following ratios (i) a : b = 1 1/2 : 2 2/5 (ii) b : c = 0.105 : 0.350 (b) Find the ratio a : b : c using the data in (a) (c) Alan, Bob and Cathy share \$500 in the ratio a : b : c found in (b). Find Alan's share correct to 2 decimal places. Question 8: A man took 4 1/2 hours to drive 360 km from Singapore to Kuala Lumpur. He used 37.5 litres of petrol for the entire journey. (a) Find his average speed. (b) Find the petrol consumption rate in km/l (c) He drove an average speed of 110 km/h on a highway for 2 hours during his journey. Find his average speed for the remaining part of his journey. Question 10: Towns P and Q are 120 km apart. Mr Tan drove from P to Q and was scheduled to reach Q after 2 hours. His average speed was 54 km/h for the first 40 minutes. (a) What was his average speed for the remaining journey if he managed to arrive just on time? (b) The time taken for his return journey is 2 hours and 10 minutes. Find his average speed for (i) the return journey (ii) the whole trip
# How do you simplify (root3(6)*root4(6))^12? Mar 19, 2017 ${\left(\sqrt[3]{6} \cdot \sqrt[4]{6}\right)}^{12} = {6}^{7} = 279936$ #### Explanation: We need the rules: • $\sqrt[m]{x} = {x}^{\frac{1}{m}} \text{ "" } \textcolor{red}{\star}$ • ${x}^{a} \cdot {x}^{b} = {x}^{a + b} \text{ "" } \textcolor{g r e e n}{\star}$ • ${\left({x}^{c}\right)}^{d} = {x}^{c d} \text{ "" } \textcolor{b l u e}{\star}$ Using $\textcolor{red}{\star}$, we see that: ${\left(\sqrt[3]{6} \cdot \sqrt[4]{6}\right)}^{12} = {\left({6}^{\frac{1}{3}} \cdot {6}^{\frac{1}{4}}\right)}^{12}$ Now using $\textcolor{g r e e n}{\star}$, this becomes: ${\left({6}^{\frac{1}{3}} \cdot {6}^{\frac{1}{4}}\right)}^{12} = {\left({6}^{\frac{1}{3} + \frac{1}{4}}\right)}^{12}$ Note that $\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$. ${\left({6}^{\frac{1}{3} + \frac{1}{4}}\right)}^{12} = {\left({6}^{\frac{7}{12}}\right)}^{12}$ Now using $\textcolor{b l u e}{\star}$, we multiply the exponents: ${\left({6}^{\frac{7}{12}}\right)}^{12} = {6}^{\frac{7}{12} \times 12}$ And we see that $\frac{7}{12} \times 12 = 7$: ${6}^{\frac{7}{12} \times 12} = {6}^{7}$ All of which we did without a calculator! For an expanded value, we could plug in ${6}^{7}$ into a calculator to see that ${6}^{7} = 279936$.
Vector Notes ```Vectors and Scalars AP Physics C Scalar A SCALAR is ANY quantity in physics that has MAGNITUDE, but NOT a direction associated with it. Magnitude – A numerical value with units. Scalar Example Magnitude Speed 20 m/s Distance 10 m Age 15 years Heat 1000 calories Vector A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION.     v , x, a, F Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude. Vector Velocity Magnitude & Direction 20 m/s, N Acceleration 10 m/s/s, E Force 5 N, West Polar Notation Polar notation defines a vector by designating the vector’s magnitude \|A\| and angle θ relative to the +x axis. Using that notation the vector is written: In this picture we have a force vector with magnitude 12 Newtons oriented at 210 degrees with the + x axis. It would be characterized as F = 12 < 210 Polar Notation In this picture we have a force vector of 12 Newtons oriented along the -x axis. However, polar notation is relative to the + x axis. Therefore, it would be characterized by F = 12 N < 180 In this last picture we have 2 vectors. They are characterized by: C = 2 N < 30 D = 4 N < - 50 or D = 4 N< 310 Scalar Multiplication Multiplying a vector by “-1” does not change the magnitude, but it does reverse it's direction or in a sense, it's angle. Multiplying a vector by a scalar will ONLY CHANGE its magnitude. Thus if A = 12 < 105, Then 2A = 24 < 105 -1/2A Thus if A = 12 < 105, then -A = 12 < 285 If A = 12 < 105, then (-1/2)A = 6 < 285 Unit Vector Notation An effective and popular system used in engineering is called unit vector notation. It is used to denote vectors with an x-y Cartesian coordinate system. Unit Vector Notation =3j J = vector of magnitude “1” in the “y” direction i = vector of magnitude “1” in the “x” direction = 4i The hypotenuse in Physics is called the RESULTANT or VECTOR SUM. The LEGS of the triangle are called the COMPONENTS A  4 iˆ  3 ˆj 3j 4i Horizontal Component Vertical Component NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw Unit Vector Notation iˆ - unit vecto r  1 in the  x direction ˆj - unit vecto r  1 in the  y direction kˆ - unit vecto r  1 in the  z direction The proper terminology is to use the “hat” instead of the arrow. So we have i-hat, j-hat, and k-hat which are used to describe any type of motion in 3D space. How would you write vectors J and K in unit vector notation? J  2 iˆ  4 ˆj K  2 iˆ  5 ˆj Applications of Vectors VECTOR ADDITION – If 2 similar vectors point in the SAME  Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started? 54.5 m, E + 84.5 m, E 30 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION. Applications of Vectors VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.  Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started? 54.5 m, E 30 m, W 24.5 m, E - Non-Collinear Vectors When 2 vectors are perpendicular, you must use the Pythagorean theorem. A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT. c  a b  c  2 55 km, N a b 2 c  Resultant c 95 km,E 2  2 95  55 2 12050  109 . 8 km 2 2 BUT…..what about the VALUE of the angle??? Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle. 109.8 km 55 km, N  N of E 95 km,E To find the value of the angle we use a Trig function called TANGENT. Tan   opposite side   Tan 1  side ( 0 . 5789 )  30 55  0 . 5789 95   109 . 8 km @ 30 NofE So the COMPLETE final answer is : 109 . 8 km  30 95 iˆ km  55 ˆj km What if you are missing a component? Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components? H.C. = ? V.C = ? 25 65 m The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions sine and cosine. cosine   hypotenuse sine   opposite side hypotenuse opp  hyp sin  adj  V .C .  65 cos 25  58 . 91 m , N or 58 . 91 ˆj m opp  H .C .  65 sin 25  27 . 47 m , E or 27 . 47 iˆ m Example A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement. - 12 m, W - = 23 m, E = 14 m, N 6 m, S R  20 m, N 14  23 2 Tan   14 2  26 . 93 m  . 6087 23 35 m, E 14 m, N R  23 m, E   Tan  1 ( 0 . 6087 )  31 . 3 26 . 93 m @ 31 . 3 NofE 26 . 93 m  31 . 3 23 iˆ m  14 ˆj m  Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. Rv  8.0 m/s, W 8  15 Tan     Tan  8 2  17 m / s  0 . 5333 15 15 m/s, N Rv 2 1 ( 0 . 5333 )  28 . 1   17 m / s @ 28 . 1 WofN 17 m / s  118 . 1   8 iˆ m / s  15 ˆj m / s Example A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components. cosine   hypotenuse H.C. =? 32 63.5 m/s V.C. = ? sine   opposite side hypotenuse opp  hyp sin  adj  H .C .  63 . 5 cos 32  53 . 85 m / s , E or 53 . 85 iˆ opp  V .C .  63 . 5 sin 32  33 . 64 m / s , S or  33 . 64 ˆj What if the vectors are not at right angles?  A plane is traveling at 150 mph with a heading of 60 N of W. The wind is pushing the plane East at 40 mph. What is the resulting velocity of the plane? The “Dot” Product (Vector Multiplication) Multiplying 2 vectors sometimes gives you a SCALAR quantity which we call the SCALAR DOT PRODUCT. In polar notation consider 2 vectors: A = \|A\| < θ1 & B = \|B\| < θ2 The dot product between A and B produces a SCALAR quantity. The magnitude of the scalar product is defined as: Where  is the NET angle between the two vectors. As shown in the figure. The Scalar Dot Product Let A = \|12\| < 30, Let B = \|5\| < 65 What is A "dot" B? A  B  A B cos   12 5 cos 35 A  B  49 . 15 In unit vector notation, it looks a little different. Consider: The "Dot" product between these is equal to: The Scalar Dot Product What is the SIGNIFICANCE of the dot product? The significance of the dot product In this figure, vector B has been split into 2 components, one PARALLEL to vector A and one PERPENDICULAR to vector A. Notice that the component parallel to vector A has a magnitude of \|B\|Cos θ THEREFORE when you find the DOT PRODUCT, the result is: i) The MAGNITUDE of one vector, in this case \|A\| and, ii) The MAGNITUDE of the 2nd vector's component that runs parallel to the first vector. (That is where the cosine comes from) Dot Products in Physics Consider this situation: A force F is applied to a moving object as it transverses over a frictionless surface for a displacement, d. As F is applied to the object it will increase the object's speed! But which part of F really causes the object to increase in speed? It is \|F\|Cos θ ! Because it is parallel to the displacement d In fact if you apply the dot product, you get (\|F\|Cos θ)d, which happens to be defined as "WORK" (check your equation sheet!) A  B  A B cos  W  F  x  F x cos  Work is a type of energy and energy DOES NOT have a direction, that is why WORK is a scalar or in this case a SCALAR PRODUCT (AKA DOT PRODUCT). Example  A particle moving in the xy plane undergoes a displacement given by: r = (2.0i + 3.0j) m as a constant force F=(5.0i+2.0j) N acts on the particle. Calculate the work done on the particle. (16 J) The “Cross” Product (Vector Multiplication) Multiplying 2 vectors sometimes gives you a VECTOR quantity which we call the VECTOR CROSS PRODUCT. In polar notation consider 2 vectors: A = \|A\| < θ1 & B = \|B\| < θ2 The cross product between A and B produces a VECTOR quantity. The magnitude of the vector product is defined as: Where  is the NET angle between the two vectors. As shown in the figure.  B A The Vector Cross Product  A B A  B  A B sin   12 5 sin 150 A  B  30 kˆ What about the direction???? Positive k-hat??? We can use what is called the RIGHT HAND THUMB RULE. •Fingers are the first vector, A •Palm is the second vector, B •Thumb is the direction of the cross product. •Cross your fingers, A, towards, B so that they CURL. The direction it moves will be either clockwise (NEGATIVE) or counter clockwise (POSITIVE) In our example, the thumb points OUTWARD which is the Z axis and thus our answer would be 30 k-hat since the curl moves counter clockwise. The significance of the cross product In this figure, vector A has been split into 2 components, one PARALLEL to vector B and one PERPENDICULAR to vector B. Notice that the component perpendicular to vector B has a magnitude of \|A\|sin θ THEREFORE when you find the CROSS PRODUCT, the result is: i) The MAGNITUDE of one vector, in this case \|B\| and, ii) The MAGNITUDE of the 2nd vector's component that runs perpendicular to the first vector. ( that is where the sine comes from) Cross Products in Physics There are many cross products in physics. You will see the matrix system when you learn to analyze circuits with multiple batteries. The cross product system will also be used in mechanics (rotation) as well as understanding the behavior of particles in magnetic fields. A force F is applied to a wrench a displacement r from a specific point of rotation (ie. a bolt). Common sense will tell us the larger r is the easier it will be to turn the bolt. But which part of F actually causes the wrench to turn? \|F\| Sin θ A  B  A B sin    F  r  F r sin  Cross Products in Physics A  B  A B sin    F  r  F r sin  Which way will the wrench turn? Counter Clockwise Is the turning direction positive or negative? Positive Which way will the BOLT move? IN or OUT of the page? OUT You have to remember that cross products give you a direction on the OTHER axis from the 2 you are crossing. So if “r” is on the x-axis and “F” is on the y-axis, the cross products direction is on the z-axis. In this case, a POSITIVE k-hat. Example  A force of F=(2.00i+3.00 j)N is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at a point located at r=(4.00i+5.00j) m . Find the torque vector(2.0 k) Nm ```
Mathematics Questions and Answers – Calculus Application – Maxima and Minima – 1 « » This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Maxima and Minima – 1”. 1. For which value of x will (x – 1)(3 – x) have its maximum? a) 0 b) 1 c) 2 d) -2 Explanation: Let, y = (x – 1)(3 – x) = 4x – x2 – 3 Then, dy/dx = 0 Or 4 – 2x = 0 Or 2x = 4 Or x = 2 Now, [d2y/dx2] = -2 which is negative. Therefore, (x – 1)(3 – x) will have its maximum at x = 2. 2. What will be the values of x for which the value of cosx is minimum? a) (2m + 1)π b) (2m)π c) (2m + 1)π/2 d) (2m – 1)π Explanation: Let, f(x) = cosx Then, f’(x) = -sinx and f”(x) = -cosx At an extreme point of f(x), we must have, f’(x) = 0 Or -sinx = 0 Or x = nπ where, n is any integer. If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at, x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1 So, f”(x) is positive at x = (2m + 1)π Hence, f(x) = cosx is minimum at x = (2m + 1)π. 3. What will be the value of x for which the value of cosx is minimum? a) 0 b) -1 c) 1 d) Cannot be determined Explanation: Let, f(x) = cosx Then, f’(x) = -sinx and f”(x) = -cosx At an extreme point of f(x) we must have, f’(x) = 0 Or -sinx = 0 Or x = nπ where, n – any integer. If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at, x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1 So, f”(x) is positive at x = (2m + 1)π Hence, f(x) = cosx is minimum at x = (2m + 1)π. So, the minimum value of cosx is cos(2mπ + π) = cosπ = -1. Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now! 4. What will be the point of maximum of the function 2x3 + 3x2 – 36x + 10? a) -1 b) -2 c) -3 d) -4 Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1) Differentiating both sides of (1) with respect to x we get, dy/dx = 6x2 + 6x – 36 And d2y/dx2 = 12x + 6 For maxima or minima value of y, we have, dy/dx = 0 Or 6x2 + 6x – 36 = 0 Or x2 + x – 6 = 0 Or (x + 3)(x – 2) = 0 Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2 Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30, which is < 0. 5. What will be the point of minimum of the function 2x3 + 3x2 – 36x + 10? a) 1 b) 2 c) 3 d) 4 Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1) Differentiating both sides of (1) with respect to x we get, dy/dx = 6x2 + 6x – 36 And d2y/dx2 = 12x + 6 For maxima or minima value of y, we have, dy/dx = 0 Or 6x2 + 6x – 36 = 0 Or x2 + x – 6 = 0 Or (x + 3)(x – 2) = 0 Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2 Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30, which is > 0. 6. What will be the maximum value of the function 2x3 + 3x2 – 36x + 10? a) 71 b) 81 c) 91 d) 0 Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1) Differentiating both sides of (1) with respect to x we get, dy/dx = 6x2 + 6x – 36 And d2y/dx2 = 12x + 6 For maxima or minima value of y, we have, dy/dx = 0 Or 6x2 + 6x – 36 = 0 Or x2 + x – 6 = 0 Or (x + 3)(x – 2) = 0 Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2 Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30 < 0 Putting x = -3 in (1) we get its maximum value as, 2x3 + 3x2 – 36x + 10 = 2(-3)3 + 3(-3)2 – 36(-3) + 10 = 91 7. What will be the minimum value of the function 2x3 + 3x2 – 36x + 10? a) -31 b) 31 c) -34 d) 34 Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1) Differentiating both sides of (1) with respect to x we get, dy/dx = 6x2 + 6x – 36 And d2y/dx2 = 12x + 6 For maxima or minima value of y, we have, dy/dx = 0 Or 6x2 + 6x – 36 = 0 Or x2 + x – 6 = 0 Or (x + 3)(x – 2) = 0 Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2 Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30 > 0 Putting x = 2 in (1) we get its minimum value as, 2x3 + 3x2 – 36x + 10 = 2(2)3 + 3(2)2 – 36(2) + 10 = -34 8. What will be the maxima for the function f(x) = x4 –8x3 + 22x2 –24x + 8? a) 0 b) 1 c) 2 d) 3 Explanation:We have, x4 –8x3 + 22x2 –24x + 8 ……….(1) Differentiating both sides of (1) with respect to x, we get, f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2) At an extremum of f(x), we have f’(x) = 0 Or 4x3 – 24x2 + 44x – 24 = 0 Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0 Or (x – 1)(x2 – 5x + 6) = 0 Or (x – 1)(x – 2)(x – 3) = 0 So, x = 1, 2, 3 Now, f”(x) = 12x2 – 48x + 44 f”(1) = 8 > 0 f”(2) = -4 < 0 f”(3) = 8 < 0 So, f(x) has maximum at x = 2. 9. What will be the minima for the function f(x) = x4 – 8x3 + 22x2 – 24x + 8? a) -1 b) 0 c) 2 d) 3 Explanation: We have, x4 – 8x3 + 22x2 – 24x + 8 ……….(1) Differentiating both sides of (1) with respect to x, we get, f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2) At an extremum of f(x), we have f’(x) = 0 Or 4x3 – 24x2 + 44x – 24 = 0 Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0 Or (x – 1)(x2 – 5x + 6) = 0 Or (x – 1)(x – 2)(x – 3) = 0 So, x = 1, 2, 3 Now, f”(x) = 12x2 – 48x + 44 f”(1) = 8 > 0 f”(2) = -4 < 0 f”(3) = 8 < 0 So, f(x) has minimum at x = 1 and 3. 10. What is the nature of the function f(x) = 2/3(x3) – 6x2 + 20x – 5? a) Possess only minimum value b) Possess only maximum value c) Does not possess a maximum or minimum value Explanation: We have, f(x) = 2/3(x3) – 6x2 + 20x – 5 ……….(1) Differentiating both side of (1) with respect to x, we get, f’(x) = 2x2 – 12x + 20 Now, for a maximum and minimum value of f(x) we have, f’(x) = 0 Or 2x2 – 12x + 20 = 0 Or x2 – 6x + 10= 0 So, x = [6 ± √(36 – 4*10)]/2 x = (6 ± √-4)/2, which is imaginary. Hence, f’(x) does not vanishes at any point of x. Thus, f(x) does not possess a maximum or minimum value. Sanfoundry Global Education & Learning Series – Mathematics – Class 12. To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. 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# If α and β are the Zeros of the Quadratic Polynomial f(x) = ax2 + bx + c, then Evaluate By Mohit Uniyal\|Updated : May 14th, 2023 If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: (i) α − β (ii) 1/α − 1/β (iii) 1/α + 1/β - 2αβ (iv) α2β − αβ2 (v) α4 + β4 If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then we know the following basic things which can be used to find the desired results: • α + β = -b/a • αβ = c/a • Quadratic polynomial can be factored as f(x) = a(x - α)(x - β) Solution: ## (i) α − β To find α - β, we need to know the values of α and β, which are the zeroes of the quadratic polynomial. Let's assume the quadratic polynomial is f(x) = ax2 + bx + c, and its zeroes are α and β. Then we know that: α + β = -b/a We can rearrange this equation to get: α = -b/a - β Now, we substitute this expression for α into the equation f(α) = 0, which gives: 2 + bα + c = 0 Substituting for α, we get: a(-b/a - β)2 + b(-b/a - β) + c = 0 Expanding and simplifying, we get: 2 + (-2ab/a)β + (a/b)c - b/a = 0 Multiplying both sides by b/a, we get: abβ2 - 2abβ + ac - b2 = 0 Dividing both sides by ab, we get: β2 - (b/a)β + (c/a) = 0 This is a quadratic equation in β, with coefficients -b/a and c/a. So we can use the quadratic formula to solve for β: β = [-(b/a) ± √((b/a)2 - 4(c/a))]/2 Once we have found the values of α and β, we can easily find α - β: α - β = (-b/a - β) - β α - β = -b/a - 2β α - β = -b/a - 2[(-b/a) ± √((b/a)2 - 4(c/a))]/2 Simplifying, we get: α - β = ±√(b2 - 4ac)/a ## (ii) 1/α − 1/β 1/α - 1/β = (β - α)/(αβ) Using the quadratic formula, we have: α,β = [-b ± sqrt(b2 - 4ac)]/(2a) So, substituting the expression for α and β: 1/α - 1/β = [(b + sqrt(b2 - 4ac)) - (b - sqrt(b2 - 4ac))]/[2a(sqrt(b2 - 4ac))] Simplifying the numerator: 1/α - 1/β = 2sqrt(b2 - 4ac)/[2a(sqrt(b2 - 4ac))] 1/α - 1/β = ±sqrt(b2 - 4ac)/c Therefore, the correct answer for (ii) in terms of a, b, and c is: 1/α - 1/β = ±sqrt(b2 - 4ac)/c ## (iii) 1/α + 1/β - 2αβ We can first find a common denominator for the two fractions by multiplying them by (αβ)/(αβ): 1/α + 1/β - 2αβ = (β + α) / (αβ) - 2αβ = (β + α - 2αβ²) / (αβ) using αβ = c/a and α + β = -b/a 1/α + 1/β - 2αβ = (-b/a) / (c/a) - 2c/a On solving we will get, = -(b/c + 2c/a) Therefore, 1/α + 1/β - 2αβ = -b/c - 2c/a. ## (iv) α2β − αβ2 We can rewrite the expression α2β − αβ2 as αβ(α − β) On putting αβ = c/a and α − β = -b/a, we get = c/a(− b/a) = -cb/a2 ## (v) α4 + β4 Simplifying: α4 + β4 = (α2 + β2)2 - 2α2β2 α + β = -b/a αβ = c/a Using these values, we can find: α2 + β2 = (α + β)2 - 2αβ α2 + β2 = (b/a)2 - 2(c/a) α2 + β2 = (b2 - 2ac)/a2 Substituting this back into the original identity, we get: α4 + β4 = ((b2 - 2ac)/a2)2 - 2α2β2 α4 + β4 = (b4 - 4b2ac + 4a2c2)/a4 - 2(c/a)2 α4 + β4 = (b4 - 4b2ac + 4a2c2 - 2c2a2)/a4 Simplifying this expression, we get: α4 + β4 = (b4 - 4b2ac + 4a2c2 - 2c2a2)/a4 α4 + β4 = (b4 - 4b2ac + 4c2(a2 - c))/a4 Therefore, the value of α4 + β4 in terms of a, b, and c is: α4 + β4 = (b4 - 4b2ac + 4c2(a2 - c))/a4 or α4 + β4 = (b2 - 2ac)2- 2a2c2))/a4 ## If α and β are the Zeros of the Quadratic Polynomial f(x) = ax2 + bx + c, then Evaluation Answers (i) α − β = ±sqrt(b2 - 4ac)/a (ii) 1/α − 1/β = ±sqrt(b2 - 4ac)/c (iii) 1/α + 1/β - 2αβ = -b/c - 2c/a (iv) α2β − αβ2 = -cb/a2 (v) α4 + β4 = (b2 - 2ac)2- 2a2c2))/a4 Similar Questions:
A summer program and resource for middle school students showing high promise in mathematics ### Why the Cube-root Algorithm works To show why the cube-root algorithm works we use Propositions 1 and 2 given below. To prove the next result, we need the following. Claim: According to Proposition 2, an integer having number of digits in the range from 3n-2 to 3n, inclusive, has n digits in the integer part of its cube-root. Thus, for instance, for n=1, each of the numbers 1 to 999 has a single digit in the integer part of its cube-root. So why does the cube-root algorithm work? Let us start with a number, say 91125. Since it has five digits and 5 lies between 3(2)-2 and 3(2), 911251/3 has 2 digits, say a1a0, in the integer part, by Proposition 2. That is, (a1a0)3 is the largest cube not exceeding 91125. But, by Proposition 1, (a1a0)3 = 103a13 + (30 a1(a1a0)+ a02)a0. That is, 91,125 ≥ (a1a0)3 = 103a13 + (30 a1(a1a0)+ a02)a0. ------ Step () In particular, 91,125 ≥ 103a13. So, 91,000 ≥ 103a13. That is, a13 ≤ 91. Suppose a13 is not the largest cube not exceeding 91, say 91 ≥ b13 > a13. The first part of this inequality is 91 ≥ b13 which implies (10b1)3 < 91,125. The second part of the inequality is (b1)3 > (a1)3 which gives b1 > a1 and so 10b1 > a1a0 or (10b1)3 > (a1a0)3. This means 91,125 > (10b1)3 > (a1a0)3, which contradicts the assumption that (a1a0)3 is the largest cube not exceeding 91,125. Thus (a1)3 is indeed the largest cube not exceeding 91. That is, a1 = 4. But then, from Step () above, 91,125 - 64,000 ≥ (30 × 4 × (4a0) + a02)a0, where 4a0 denotes the integer whose units digit is a0 and tens digit is 4. Suppose now that there exists a0' such that 91,125 - 64,000 ≥ (30 × 4 × (4a0') + a0'2)a0' and a0' > a0. But then, 91,125 ≥ (a1a0')3 > (a1a0)3, which contradicts the assumption that (a1a0)3 is the largest cube not exceeding 91,125. So, a0 is the largest integer such that (91,125 - 64,000) - (30 × 4 × (4a0) + a02)a0) ≥ 0. That is, a0 = 5. Since (91,125 - 64,000) - (30 × 4 × (4a0) + a02)a0) = 0, the cube-root of 91125 is 45. Observe that, in the "division" process, the subtraction of 64,000 is implemented by subtracting 64 from 91 whose positional value is 91000. By successivley bringing down all groups in the parsing of 91125, we have, in effect, subtracted 64,000 from 91,125. This also illustrates how the powers 103i that appear on the right of the equation in Proposition 1 are taken care of by subracting the quantity Qi whose positional value is 103i × Qi. This implements the decomposition in terms of 103i((30 × an...ai+1 × anan-1...ai) + ai2)ai appearing on the right side of the equation in Proposition 1. Another way of seeing this is that i-1 groups of three digits each are brought down after the subtraction of (30 × an...ai+1 × anan-1...ai) + ai2)ai so that the subtracted quantity has a place value of 103i. What if we are looking for the cube-root of 91126? Then we have a remainder of 1 and so we think of 91126 as 91126.000 and bring down the three zeros. Why? Since two consecutive integers can not be both perfect cubes, 91126 can not be a perfect cube. So its cube-root must be of the form x = anan-1...a1a0.a-1a-2...a-k, say, and, so, equals 10-k(anan-1...a1a0a-1a-2...a-k). Let n+k = m. Rename ai as ak+i, i = n, n-1,...,2,1,0,-1,...,-k. Then, x = 10-k(amam-1...akak-1...a0) and, from Proposition 2, . This means that, in finding the cube-root of a number that is not a perfect cube, if all the digits to the left of the decimal point have been brought down in the division process of this algorithm, we continue the process beginning from the immediate right of the decimal point, successively bringing down groups of three digits and each time repeating the "division" as before. After k groups of three digits each are brought down beginning from the digit to the immediate right of the decimal point, we have obtained the cube root of (10kx)3. Therefore, the value of x which is the cube-root of the given number is obtained up to k decimal digits by dividing the number obtained in the division process by 10k, or by inserting a decimal point just before the last k digits of the "quotient." We have shown the following. Proposition 3 (The cube-root algorithm) Let N be a number in decimal representation. The cube-root of N is obtained by the following procedure. (1) Beginning at the decimal point, parse N in groups Ai, i = 0,..,n, of three digits going left from the decimal point and then in groups A-j, j = 1,...,k, of three digits going right from the decimal point so that N is given by the concatenation AnAn-1...A1A0.A-1A-2...Ak-1A-k, where the left-most group An on the left of the decimal point and the right-most group A-k, on the right of the decimal point, each has one, two or three digits, depending on the number of digits of N. (2) Choose an so that (an)3 is the largest cube not exceeding An. Let Bn-1 be the number obtained by concatenating (An - an3) with An-1. That is, Bn-1 = (An - an3)An-1. (3) For i = n-1 to -k, choose ai, being the largest single-digit number, so that ((30 × an...ai+1 × anan-1...ai) + ai2)ai does not exceed Bi. Let Bi-1 = (Bi - ((30 × an...ai+1 × anan-1...ai) + ai2)ai)Ai. (4) Then the cube-root of N is anan-1...a1a0.a-1a-2...a-k. Observe that when N is an integer, Ai = 000, for i = -1 to -k. And when N is an integer that is a perfect cube, B-1 = 0 as well. Ü BACK MathPath - "BRIGHT AND EARLY" Send suggestions to webmaster@mathpath.org
# Estimating a Product of Whole Numbers The whole numbers are first rounded as specified, i.e., rounded to the nearest ten, hundred and so on. Then the product of the rounded whole numbers is found to estimate the product of whole numbers. Estimate the product 573 × 94 by first rounding each number so that it has only one non-zero digit. ### Solution Step 1: We round each number such that it has only one non-zero digit 573 is a three-digit number. So its first digit is going to be the only non-zero digit and the other two digits would be zeros. It means rounding to nearest hundred. Since the tens digit, 7 is greater than 5, we round up 57󠄀3 to 600. Step 2: 94 is a two-digit number. Its first digit is going to be the only non-zero digit and the other digit would be zero. It means rounding to nearest ten. Since the ones digit, 4 is less than 5, we round down 94 to 90. Step 3: The estimate of the product after rounding = 600 × 90 = 54,000 Estimate the product 2092 × 167 by first rounding each number so that it has only one non-zero digit. ### Solution Step 1: We round each number such that it has only one non-zero digit 2092 is a four-digit number. So its first digit is going to be the only non-zero digit and the other three digits would be zeros. It means rounding to nearest thousand. Since the hundreds digit, 0 is less than 5, we round down 2092 to 2000. Step 2: 167 is a three-digit number. Its first digit is going to be the only non-zero digit and the other two digits would be zero. It means rounding to nearest hundred. Since the tens digit, 6 is greater than 5, we round up 167󠄀 to 200. Step 3: The estimate of the product after rounding = 2000 × 200 = 400,000 ##### Kickstart Your Career Get certified by completing the course
# RS Aggarwal Chapter 11 Class 9 Maths Exercises 11.1 (ex 11a) Solutions RS Aggarwal Chapter 11 Class 9 Maths Exercise 11.1 Solution: Areas of Parallel Quadrilateral and Triangle is an essential chapter from the point of view of next year’s board examination. RS Aggarwal is a dedicated practice in Chapter 11 of Class 9 Maths Solutions, which has 38 questions for your practice. Some of these questions also have 4-5 subtypes for a better review. The main topics covered in this exercise are axioms of the Euclidean region, which figures on the same base and between the same parallel lines, and the parallelogram also has the same base and the same parallels. Calculation of the base and height of a parallelogram is also included in this practice. This chapter of RS Aggarwal Solutions for Class 9 Maths has only one dedicated exercise which covers the full range of questions: The first is to identify the figures along a joint base and between the corresponding parallel lines. It has six subtypes. The next three elements try to find the area of a parallelogram and the length of its sides. The next three questions are based on trapezoidal. They mainly focus on finding the area. The following questions are based on proving various theorems and their applications. These problems mainly focus on the triangle and its properties with variations of quadrilateral and rhombus. Some questions on · 27 are based on finding the area of a parallelogram with intermediate difficulty. About After this, by the end, all the questions are trying different applications on triangles and trapezoids with much higher difficulty levels. Know more about the chapter here. EXERCISE11 ## Important Definition for RS Aggarwal Chapter 11 Class 9 Maths Exercise 11.1 Solution Area: It is the amount of planar surface that is being covered by a closed geometric figure in an even or uneven manner. Every closed figure imparts some areas that can always be calculated. Identifying the Figures on the Common Base and Between the Same Parallels It can only be said true for any two figures if: 1. a) They both have a common side. 2. b) The sides that are parallel to the common base and also the vertices opposite to that side lie on a similar straight line, which is parallel to the base. Triangle: A plane figure that is bounded by three straight lines is called a triangle. It is the simplest form of a polygon. It is a closed figure that is formed by joining three line segments. Area of a Parallelogram Area of a parallelogram = b×h where ‘b′ is known as the base, and ‘h′ is the corresponding altitude or height. Area of a Triangle Area of a triangle = (1/2)×b×h where “b” is known as the base, and “h” is the corresponding altitude. Many theorems need to be learned for comfortable solving of problems. Theorem 1 Statement: Diagonals of a parallelogram divides it into two triangles of equal area. Theorem 2 Statement: Parallelograms that are on a similar base and between the same parallel lines are equal in area. Theorem 3 Statement: Triangles that share a similar base and are formed between the same parallels are equal in area. Theorem 4 Statement: Triangles having equal areas and having one side of one of the triangles similar to one side of the other, always their corresponding altitudes similar. Know more here.
# Important Questions Class 8 Maths Chapter 7: Cubes and Cube Roots In this article, we will provide important questions for class 8 maths chapter 7 – Cubes and Cube Roots. These questions are prepared with reference to NCERT book as per CBSE syllabus (2021-22) The problems are provided with solutions here to make students prepare for exams and score good marks in their final exam. The chapter-cubes and cube roots will comprise of finding the cubes & cube roots of numbers. We will also solve word problems based on this concept here. ## Important Questions With Solutions For Maths Class 8 Chapter 7 (Cubes and Cube Roots) Q.1: Find the cube of 3.5. Solution: 3.53 = 3.5 x 3.5 x 3.5 = 12.25 x 3.5 = 42.875 Q.2: Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product is a perfect cube. Solution: The prime factorisation of 392 gives: 392 = 2 x 2 x 2 x 7 x 7 Since, we can see, number 7 cannot be paired in a group of three. Therefore, 392 is not a perfect cube. To make it a perfect cube, we have to multiply the 7 by the original number. Thus, 2 x 2 x 2 x 7 x 7 x 7 = 2744, which is a perfect cube, such as 23 x 73 or 143. Hence, the smallest natural number which should be multiplied to 392 to make a perfect cube is 7. Q.3: Find the smallest number by which 128 must be divided to obtain a perfect cube. Solution: The prime factorisation of 128 gives: 128 = 2×2×2×2×2×2×2 Now, if we group the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2 Here, 2 cannot be grouped into triples of equal factors. Therefore, we will divide 128 by 2 to get a perfect cube Q.4: Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube? Solution: Given, side of the cube is 5 cm, 2 cm and 5 cm. Therefore, volume of cube = 5×2×5 = 50 The prime factorisation of 50 = 2×5×5 Here, 2, 5 and 5 cannot be grouped into triples of equal factors. Therefore, we will multiply 50 by 2×2×5 = 20 to get perfect square. Hence, 20 cuboid is needed. Q.5: Find the cube root of 13824 by prime factorisation method. Solution: First let us prime factorise 13824: 13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 23 × 23 × 23 × 33 3√13824 = 2 × 2 × 2 × 3 = 24 Q.6: Find the cube root of 17576 through estimation. Solution: The given number is 17576. Step 1: Form groups of three digits starting from the rightmost digit. • Here, one group has three digits i.e., 576 whereas the other group has only two digits i.e.,17. Step 2: Take 576. • The digit 6 is at its one’s place. • We will take here the one’s place of the required cube root as 6. Step 3: Take the other group, i.e., 17. • Cube of number 2 is 8 and cube of number 3 is 27. • 17 lies between 8 and 27. • The smaller number between 2 and 3 is 2. • The one’s place of 2 is 2 itself. • Now, take digit 2 as ten’s place of the required cube root. Hence, the cube root of 17576 is; 3√17576 = 26 Q.7: You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768. Solution: By grouping the digits, we get 1 and 331 Since, the unit digit of cube is 1, the unit digit of cube root is 1. Therefore, we get 1 as the unit digit of the cube root of 1331. The cube of 1 matches with the number of the second groups. Therefore, the ten’s digit of our cube root is taken as the unit place of the smallest number. We know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1. Therefore, ∛1331 = 11 By grouping the digits, we get 4 and 913 We know that, since the unit digit of the cube is 3, the unit digit of the cube root is 7. Therefore, we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8. Thus, 1 is taken as ten-digit of the cube root. Therefore, ∛4913 = 17 By grouping the digits, we get 12 and 167. Since the unit digit of the cube is 7, the unit digit of the cube root is 3. Therefore, 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27, 8 > 12 > 27. Thus, 2 is taken as the tenth digit of the cube root. Therefore, ∛12167= 23 By grouping the digits, we get 32 and 768. Since, the unit digit of the cube is 8, the unit digit of the cube root is 2. Therefore, 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64. Thus, 3 is taken as ten-digit of the cube root. Therefore, ∛32768= 32 ### Class 8 Maths Chapter 7 Extra Questions 1. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. 1. 72 2. 675 2. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube 1. 81 2. 192 3. Find the cube root of 8000. 4. Find the cube root of each of the following numbers by prime factorisation method. 1. 91125 2. 110592 5. State true or false. 1. Cube of any odd number is even. 2. A perfect cube does not end with two zeros. 3. If the square of a number ends with 5, then its cube ends with 25. 4. There is no perfect cube which ends with 8. 5. The cube of a two-digit number may be a three-digit number. 6. The cube of a two-digit number may have seven or more digits. 7. The cube of a single-digit number may be a single-digit number. 6. What number do you get when you multiply your number by three times? (a)Square numbers (b) Perfect numbers (c) Cube numbers 7. Which of these is not a perfect cube? (a) 1000 (b) 1728 (c)100 • I bring to you the BEST for students of Class 9th - 12th. I (Balkishan Agrawal) aim at providing complete preparation for CBSE Board Exams (Maths) along with several other competitive examinations like NTSE, NSO, NSEJS, PRMO, etc. & Maths Reasoning, Vedic Maths at the school level. 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## Thinking Mathematically (6th Edition) Assume that the thief has stolen Aplants. After giving one-half of what he has and 2 more plants to first guard, he has $B$plants with him. And after giving to the second guard one-half of remaining and 2 more, he has C plants. And after giving to the third guard, he has D plants with him. Express it in algebraic format to obtain, \begin{align} & A-\left[ \frac{1}{2}A+2 \right]=B \\ & B-\left[ \frac{1}{2}B+2 \right]=C \\ & C-\left[ \frac{1}{2}C+2 \right]=D \\ \end{align} It is provided that at last he has 1 plant with him so, value of Dis 1, now put the value of D in last equation and solving it further we get all values. \begin{align} & C-\left[ \frac{1}{2}C+2 \right]=1 \\ & \frac{C}{2}-2=1 \\ & \frac{C}{2}=3 \\ & C=6 \end{align} Now substitute back 6 for C in the second equation and solve for B. \begin{align} & B-\left[ \frac{1}{2}B+2 \right]=6 \\ & \frac{B}{2}-2=6 \\ & \frac{B}{2}=8 \\ & B=16 \end{align} Finally, substitute back 16 for B in the second equation and solve for A. \begin{align} & A-\left[ \frac{1}{2}A+2 \right]=16 \\ & \frac{A}{2}-2=16 \\ & \frac{A}{2}=18 \\ & A=36 \end{align} So, here the number of plants the thief stole was 36. The thief stole 36 plants.
# Rotation - Mathematics Form 2 Notes ## Introduction • A transformation in which a plane figure turns around a fixed center point called center of rotation. • A rotation in the anticlockwise direction is taken to be positive whereas a rotation in the clockwise direction is taken to be negative. • For example a rotation of 900 clockwise is taken to be negative. − 900 while a rotation of anticlockwise 900 is taken to be +900. For a rotation to be completely defined the center and the angle of rotation must be stated. Illustration • To rotate triangle A through the origin, angle of rotation +1/4 turn. • Draw a line from each point to the center of rotation ,in this case it’s the origin. • Measure 900 from the object using the protacter and make sure the base line of the proctacter is on the same line as the line from the point of the object to the center. • The 0 mark should start from the object. • Mark 900 and draw a straight line to the center joining the lines at the origin. The distance from the point of the object to the center should be the same distance as the line you drew.This give you the image point • The distance between the object point and the image point under rotation should be the same as the center of rotation in this case 900 Illustration. To find the center of rotation. • Draw a segment connecting point’s and ′ • Using a compass, find the perpendicular bisector of this line. • Draw a segment connecting point’s and ′.Find the perpendicular bisector of this segment. • The point of intersection of the two perpendicular bisectors is the center of rotation. Label this point . Justify your construction by measuring angles ′ and ′. Did you obtain the same measure? The angle between is the angle of rotation. The zero mark of protector should be on the object to give you the direction of rotation. ## Rotational Symmetry of Plane Figures • The number of times the figure fits onto itself in one complete turn is called the order of rotational symmetry. Note; • The order of rotational symmetry of a figure figure = 360/angle between two identical parts of the figure • Rotational symmetry is also called point symmetry. Rotation preserves length, angles and area, and the object and its image are directly congruent. • ✔ To read offline at any time. • ✔ To Print at your convenience • ✔ Share Easily with Friends / Students ### Related items . Subscribe now access all the content at an affordable rate or Buy any individual paper or notes as a pdf via MPESA and get it sent to you via WhatsApp
# Sequence and Series – Types, Difference, & Formulas You have studied various patterns in mathematics such as shape patterns, letter patterns, number patterns, etc. Sequences and series are number patterns that are widely used in different fields. One of the many applications of sequences and series occurs in financial mathematics. Let’s use examples to understand what a sequence and series are and the types and differences between the two. ## What are Sequence and Series? A sequence is a group of numbers arranged in a particular order or following a set of rules. For example, $1$, $3$, $5$, $7$, … is a sequence that starts with a number $1$ and the rule to generate a next term is ‘add $2$ to the previous term’. Therefore, we can say that a sequence is an arrangement of any objects or a set of numbers in a particular order followed by some rule. If $a_1$, $a_2$, $a_3$, $a_4$, … etc. denote the terms of a sequence, then $1$, $2$, $3$, $4$, … denotes the position of the term. A sequence can be i.e. either a finite sequence or an infinite sequence depending on whether there are finite or finite terms in a sequence. A series is formed by adding the terms of a sequence. For example $1 + 3 + 5 + 7 + …$ is a seeries. If $a_1$, $a_2$, $a_3$, $a_4$, … is a sequence, then the corresponding series is given by $\text{S}_n = a_1+ a_2 + a_3 + .. + a_n$. ## Sigma Notation of a Series Series are often represented in compact form, called sigma notation, using the Greek letter $\sum{}$(sigma) as means of indicating the summation involved. Thus, the series $a_1 + a_2 + a_3 + … + a_n$ is abbreviated as $\sum_{k=1}^{n}{{a_k}}$. Note: When the series is used, it refers to the indicated sum, not to the sum itself. For example, $1 + 3 + 5 + 7$ is a finite series with four terms. When we use the phrase “sum of a series,” we will mean the number that results from adding the terms, the sum of the series is $16$. ## General Term of a Sequence The general term for a sequence follows a certain pattern. The successive terms are obtained by performing the mathematical operation present in the general term in indicated order to the previous term. Sometimes each term of the series follows an expression. ### Examples of General Term of a Sequence Let’s consider some examples to understand the general term of a sequence. Example 1: Write the first five terms of a sequence whose general term is $a_n = 3n – 5$. The general term of the sequence is $a_n = 3n – 5$. When $n = 1$, $a_1 = 3 \times 1 – 5 = 3 – 5 = -2$. When $n = 2$, $a_2 = 3 \times 2 – 5 = 6 – 5 = 1$. When $n = 3$, $a_3 = 3 \times 3 – 5 = 9 – 5 = 4$. When $n = 4$, $a_4 = 3 \times 4 – 5 = 12 – 5 = 7$. When $n = 5$, $a_5 = 3 \times 5 – 5 = 15 – 5 = 10$. Therefore, the first five terms of the sequence are $-2$, $1$, $4$, $7$, and $10$. Example 2: Write the first five terms of a sequence whose general term is $a_n = \frac{n + 2}{3}$. The general term of the sequence is $a_n = \frac{n + 2}{3}$. When $n = 1$, $a_1 = \frac{1 + 2}{3} = \frac{3}{3} = 1$. When $n = 2$, $a_2 = \frac{2 + 2}{3} = \frac{4}{3}$. When $n = 3$, $a_3 = \frac{3 + 2}{3} = \frac{5}{3}$. When $n = 4$, $a_4 = \frac{4 + 2}{3} = \frac{6}{3} = 2$. When $n = 5$, $a_2 = \frac{5 + 2}{3} = \frac{7}{3}$. Therefore, the first five terms of the sequence are $1$, $\frac{4}{3}$, $\frac{5}{3}$, $2$, and $\frac{7}{3}$. Example 3: Write the first five terms of a sequence whose general term is $a_n = (n + 1)(n – 2)(n + 3)$. The general term of the sequence is $a_n = a_n = (n + 1)(n – 2)(n + 3)$. When $n = 1$, $a_1 = (1 + 1)(1 – 2)(1 + 3) = 2 \times (-1) \times 4 = -8$. When $n = 2$, $a_2 = (2 + 1)(2 – 2)(2 + 3) = 3 \times 0 \times 5 = 0$. When $n = 3$, $a_3 = (3 + 1)(3 – 2)(3 + 3) = 4 \times 1 \times 6 = 24$. When $n = 4$, $a_4 = (4 + 1)(4 – 2)(4 + 3) = 5 \times 2 \times 7 = 70$. When $n = 5$, $a_5 = (5 + 1)(5 – 2)(5 + 3) = 6 \times 3 \times 8 = 144$. Therefore, the first five terms of the sequence are $-8$, $0$, $24$, $70$, and $144$. Example 4: Write the first five terms of a sequence which is defined as $a_1 = 1$, $a_n = a_{n-1} + 2, n \ge 2$. Here the first term is $a_1 = 1$ and the next term $a_n$ is given in terms of the previous term $a_{n – 1}$, defied by $a_n = a_{n-1} + 2$. When $n = 1$, $a_1 = 1$. When $n = 2$, $a_2 = a_{2-1} + 2 = a_1 + 2 = 1 + 2 = 3$. When $n = 3$, $a_3 = a_{3-1} + 2 = a_2 + 2 = 3 + 2 = 5$. When $n = 4$, $a_4 = a_{4-1} + 2 = a_3 + 2 = 5 + 2 = 7$. When $n = 5$, $a_5 = a_{5-1} + 2 = a_4 + 2 = 7 + 2 = 9$. Therefore, the first five terms of the sequence are $1$, $3$, $5$, $7$, and $9$. Example 5: Write the first six terms of a sequence which is defined as $a_1 = a_2 = 2$, $a_n = a_{n – 1} – 1, n > 2$. Here the first two terms is $a_1 = 2$, and $a_2 = 2$ and the next term $a_n$ is given in terms of the previous term $a_{n – 1}$, defied by $a_n = a_{n – 1} – 1$. When $n = 1$, $a_1 = 1$. When $n = 2$, $a_2 = 1$. When $n = 3$, $a_3 = a_{3 – 1} – 1 = a_2 – 1 = 1 – 1 = 0$. When $n = 4$, $a_4 = a_{4 – 1} – 1 = a_3 – 1 = 0 – 1 = -1$. When $n = 5$, $a_5 = a_{5 – 1} – 1 = a_4 – 1 = -1 – 1 = -2$. When $n = 6$, $a_6 = a_{6 – 1} – 1 = a_5 – 1 = -2 – 1 = -3$. Therefore, the first six terms of the sequence are $1$, $1$, $0$, $-1$, $-2$, and $-3$. Example 6: Write the first six terms of a sequence which is defined as $a_1 = a_2 = 1$ and $a_n = a_{n – 1} + a_{n – 2}, n > 2$. Here the first two terms is $a_1 = 1$, and $a_2 = 1$ and the next term $a_n$ is given in terms of the previous terms $a_{n – 1}$ and $a_{n – 2}$, defied by $a_n = a_{n – 1} + a_{n – 2}$. When $n = 1$, $a_1 = 1$. When $n = 2$, $a_2 = 1$. When $n = 3$, $a_3 = a_{3 – 1} + a_{3 – 2} = a_2 + a_1 = 1 + 1 = 2$. When $n = 4$, $a_4 = a_{4 – 1} + a_{4 – 2} = a_3 + a_2 = 2 + 1 = 3$. When $n = 5$, $a_5 = a_{5 – 1} + a_{5 – 2} = a_4 + a_3 = 3 + 2 = 5$. When $n = 6$, $a_6 = a_{6 – 1} + a_{6 – 2} = a_5 + a_4 = 5 + 3 = 8$. Therefore, the first six terms of the sequence are $1$, $1$, $2$, $3$, $5$, and $8$. Note: This sequence is a very popular sequence and is known as Fibonacci Sequence. ## Difference Between Sequence and Series The following are the differences between sequence and series. ## Types of Sequences and Series There are various types of sequences and series. The most common of these are • Arithmetic Sequences and Series: An arithmetic sequence is a sequence where the successive terms are either the addition or subtraction of the common term known as common difference. For example, $5$, $9$, $13$, $17$, $21$, … is an arithmetic sequence. Here the first term $a = 5$, and the common difference $d = 9 – 5 = 13 – 9 = 17 – 13 = 4$. A series formed by using an arithmetic sequence is known as the arithmetic series for example $5 + 9 + 13 + 17…$ is an arithmetic series. • Geometric Sequences and Series: A geometric sequence is a sequence where the successive terms have a common ratio. For example, $1$, $4$, $16$, $64$, … is a geometric sequence. Here the first term $a = 1$, and the common ratio $r = \frac{4}{1} = \frac{16}{4} = \frac{64}{16} = 4$. A series formed by using a geometric sequence is known as a geometric series for example $1 + 4 + 16 + 64 …$ is a geometric series. • Harmonic Sequences and Series: A harmonic sequence is a sequence where the sequence is formed by taking the reciprocal of each term of an arithmetic sequence. For example, $\frac{1}{5}$, $\frac{1}{9}$, $\frac{1}{13}$,$\frac{1}{17}$, $\frac{1}{21}$, … is a harmonic sequence. A series formed by using harmonic sequence is known as a harmonic series for example $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \frac{1}{17}+ \frac{1}{21} …$ is a harmonic series. ## Key Takeaways • In an arithmetic sequence and series, $a$ is represented as the first term, $d$ is a common difference, $a_n$ is the $n^{th}$ term, and $n$ is the number of terms. • In general, the arithmetic sequence can be represented as $a$, $a+d$, $a+2d$, $a+3d$, .., where $a$ is the first term, $d$ is the common difference. • Each successive term is obtained in a geometric progression by multiplying the common ratio by its preceding term. • The formula for the nth term of a geometric progression whose first term is $a$ and the common ratio is $r$ is $a_n = ar_{n−1}$ • The sum of the infinite GP formula is given as $\text{S}_n = \frac{a}{1−r}$ where $\|r\|<1$. ## Practice Problems 1. What is a sequence? 2. What is a series? 3. What is the difference between sequence and series? 4. Write the first six terms of a sequence which is defined as • $a_n = n(n + 3)$ • $a_n = \frac{2n}{n – 1}$ • $a_n = 2^{n – 1}$ • $a_n = 2.3^n$ • $a_n = \frac{4n – 5}{6}$ • $a_n = (-1)^n.4^{-n}$ • $a_n = (-1)^{n-1}.2^n$ • $a_n = \frac{n \left(n^2 – 4 \right)}{3}$ • $a_1 = -2$, $a_n = 3a_{n-1} + 2$ • $a_1 = 1$, $a_n = \frac{a_{n-1}}{2}$ ## FAQs ### Give an example of sequence and series. An example of a sequence is $1$, $4$, $7$, $10$, $13$, … An example of a series is $1 + 4 + 7 + 10 + 13 + …$ ### What are some of the common types of sequences? The three most common types of sequences are arithmetic sequence, geometric sequence, and harmonic sequence. ### What is the difference between sequence and series? In sequence, elements are placed in a particular order following a particular set of rules, a definite pattern of the numbers is important, and the order of appearance of the numbers is important. In series, the order of the elements is not necessary, the pattern of the numbers is not important, and the order of appearance is not important. ### What is the similarity between sequence and series? The sequence and the series of the same type, both are made up of the same elements, i.e, the elements that follow a pattern. A series is formed by using the elements of the sequence and adding them by the addition symbol. ### What sigma notation of a series? Series are often represented in compact form, called sigma notation, using the Greek letter $\sum{}$(sigma) as means of indicating the summation involved. Thus, the series $a_1 + a_2 + a_3 + … + a_n$ is abbreviated as $\sum_{k=1}^{n}{{a_k}}$. ## Conclusion A sequence is a group of numbers arranged in a particular order or following a set of rules, whereas a series is formed by adding the terms of a sequence. If $a_1$, $a_2$, $a_3$, $a_4$, … etc. denote the terms of a sequence, then the corresponding series is given by $\text{S}_n = a_1+ a_2 + a_3 + .. + a_n$.
Ex 5.4 (Optional) Chapter 5 Class 10 Arithmetic Progressions Serial order wise ### Transcript Ex 5.4, 2 (Optional) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. We know that nth term of an AP is an = a + (n − 1)d Hence, 3rd term of AP = a3 = a + 2d and 7th term of AP = a7 = a + 6d Given Sum of third & seventh terms is 6 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 Also, Product of the third and seventh terms is 8 a3 × a7 = 8 (a + 2d) (a + 6d) = 8 From (1) a + 4d = 3 a = 3 − 4d (3 − 4d + 2d) (3 − 4d + 6d) = 8 (3 − 2d) (3 + 2d) = 8 (3)2 − (2d)2 = 8 9 – 4d2 = 8 4d2 = 1 (2d)2 = (1)2 2d = ± 1 d = ± 𝟏/𝟐 Finding value of a For d = 𝟏/𝟐 a = 3 − 4d a = 3 − 4(1/2) a = 3 − 2 a = 1 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 Therefore, when a = 1, d = 𝟏/𝟐 And when a = 5, d = (−𝟏)/𝟐 Now, we need to find the Sum of first Sixteen Terms Sum of n terms of an AP is Sn = 𝒏/𝟐 [𝟐𝒂+(𝒏 −𝟏)𝒅] Taking a = 1 and d = 𝟏/𝟐 S16 = 16/2 [(2" × 1" )+(16 −1)(1/2)] = 8 [2+15/2] = 8 [(4 + 15)/2] = 8 × 19/2 = 76 Taking a = 5 and d = (−𝟏)/𝟐 S16 = 16/2 [(2" × " 5)+(16 −1)((−1)/2)] = 8 [10−15/2] = 8 [(20 − 15)/2] = 8 × 5/2 = 20 Hence, If a = 1 and d = 𝟏/𝟐 , the sum of first sixteen terms of the AP is 76 and If a = 5 and d = (−𝟏)/𝟐, the sum of first sixteen terms of the AP is 20.
GeeksforGeeks App Open App Browser Continue # Difference between Simpson ‘s 1/3 rule and 3/8 rule In Simpson’s 1/3 rule, we approximate the polynomial based on quadratic approximation. In this, each approximation actually covers two of the subintervals. This is why we require the number of subintervals to be even. Some of the approximations look more like a line than a quadric, but they really are quadratics. Formula of Simpson’s¹/₃ rule ₐ∫ᵇ f (x) dx = h/₃ [(y₀ + yₙ) + 4 (y₁ + y₃ + ..) + 2(y₂ + y₄ + ..)] where, a, b is the interval of integration h = (b – a)/ n y₀ means the first terms and yₙ means last terms. (y₁ + y₃ + ..) means the sum of odd terms. (y₂ + y₄ + …) means sum of even terms. Example: Find Solution using Simpson’s 1/3 rule. Solution: Using Simpson’s 1/3 rule ₐ∫ᵇ f (x) dx = h/₃ [(y₀ + yₙ) + 4 (y₁ + y₃ + …) + 2 (y₂ + y₄ + …)] h = 0.1 ₐ∫ᵇ f (x) dx = 0.1/3 [(1+0.8604)+4×(0.9975+0.9776)+2×(0.99)] ₐ∫ᵇ f (x) dx = 0.1/3 [(1+0.8604)+4×(1.9751)+2×(0.99)] ₐ∫ᵇ f (x) dx = 0.39136 Solution of Simpson’s 1/3 rule = 0.39136 In Simpson’s 3/8 rule, we approximate the polynomial based on quadratic approximation. However, each approximation actually covers three of the subintervals instead of two. Formula of Simpson’s 3/8 rule ₐ∫ᵇ f (x) dx = 3h/8[(y₀ + yₙ) + 3(y₁ + y₂ + y₄ + …) + 2(y₃ + y₆ +…)] where, a, b is the interval of integration h = (b – a )/ n y₀ means the first terms and yₙ means the last terms. ( y₁ + y₂ + y₄ + … ) means the sum of remaining terms. ( y₃ + y₆ +…) means the multiples of 3. Example: Find a Solution using Simpson’s 1/3 rule. Solution: Using Simpson’s 3/8 rule: ₐ∫ᵇ f (x) dx = 3h/8[(y₀ + yₙ) + 3(y₁ + y₂ + y₄ + …) + 2(y₃ + y₆ +…)] h = 0.1 ₐ∫ᵇ f (x) dx = 3h/8 [(y0+y4)+2(y3)+3(y1+y2)] ₐ∫ᵇ f (x) dx = 3* 0.1/8 [(1+0.8604)+2×(0.9776)+3×(0.9975+0.99)] ₐ∫ᵇ f (x) dx = 3* 0.1/8 [(1+0.8604)+2×(0.9776)+3×(1.9875)] ₐ∫ᵇ f (x) dx = 0.36668 Solution of Simpson’s 3/8 rule = 0.36668 Following is a table of differences between the Simpson’s 1/3 rule and Simpson’s 3/8 rule My Personal Notes arrow_drop_up Related Tutorials
Math Train # Number Friends In this lesson, children will learn to count and recognize the number of objects in a set, match numbers to quantities and practice simple addition 30–45 Min. Profis Kita und Kindergarten ## Connect • Gather the children around the train set and show them two train cars with different quantities of bricks loaded onto them (e.g., one with six green bricks and the number six brick, and another with four light green bricks and the number four brick). • Place the cars side by side and ask the children to compare. - Which train car has the most bricks? - Which train car has the least bricks? • Explain that sometimes people want to know how many items there are altogether and that we can figure this out by adding. • Encourage the children to count the total number of bricks on both train cars. Point out that the two quantities (i.e., six and four) added together make ten, and so we can call six and four best friends of ten! ## Tip If making sums of ten is too difficult, use smaller numbers. ## Construct • Explain that several sets of numbers are “best friends of ten.” • Tell the children they will be working with a partner to create “best friends of ten” train cars. • Ask each pair of children to find two number bricks that add u p to ten. You may use the in-box card as an example. • Ask each child to place a number brick on a train car and find the corresponding number of bricks in the same color. If they need help, show them the inspiration photos, which they can duplicate. • Once the children have finished building, have them count the bricks on each train car. Then ask them to count their pair’s bricks altogether--which make ten! ## Contemplate • Facilitate a discussion about “best friends of ten.” - How many sets of “best friends of ten” there are? - What other numbers add up to ten? • Write out the sets as children say them. If time allows, encourage the children to build each set of “best friends” and share with the group. ## Continue • Challenge children to create “best friends of twelve” train cars. • The combinations which can be built using the set include: - Two and ten - Three and nine - Four and eight - Five and seven • Look at the in-box cards and use the ideas shown on the cards to help the children find the bricks needed to add up to twelve. ## Did you notice? Observing the following skills can help you monitor whether the children are developing the necessary competencies in math. • Counting using number names, and beginning to recognize the number of objects in a set • Comparing two or more objects • Exploring simple operations, such as adding ## Unterstützung für Lehrkräfte Children will: • Count and recognize the number of objects in a set • Match numbers to quantities • Compare quantities For up to 4 children The Mathematics guidelines from the National Association for the Education of Young Children (NAEYC) and HeadStart have been used to develop the Math Train lessons. Please refer to the learning grid for an overview of the learning values referenced throughout this Teacher Guide. The learning goals listed at the end of each lesson can be used to determine whether or not each child is developing the relevant early math skills. These bullet points target specific skills or pieces of information that are practiced or presented during each lesson.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Geometry (FL B.E.S.T.)>Unit 2 Lesson 3: Rotations # Rotating shapes about the origin by multiples of 90° Learn how to draw the image of a given shape under a given rotation about the origin by any multiple of 90°. ## Introduction In this article we will practice the art of rotating shapes. Mathematically speaking, we will learn how to draw the image of a given shape under a given rotation. This article focuses on rotations by multiples of ${90}^{\circ }$, both positive (counterclockwise) and negative (clockwise). ## Part 1: Rotating points by ${90}^{\circ }$‍ , ${180}^{\circ }$‍ , and $-{90}^{\circ }$‍ ### Let's study an example problem We want to find the image ${A}^{\prime }$ of the point $A\left(3,4\right)$ under a rotation by ${90}^{\circ }$ about the origin. Let's start by visualizing the problem. Positive rotations are counterclockwise, so our rotation will look something like this: Cool, we estimated ${A}^{\prime }$ visually. But now we need to find exact coordinates. There are two ways to do this. ### Solution method 1: The visual approach We can imagine a rectangle that has one vertex at the origin and the opposite vertex at $A$. A rotation by ${90}^{\circ }$ is like tipping the rectangle on its side: Now we see that the image of $A\left(3,4\right)$ under the rotation is ${A}^{\prime }\left(-4,3\right)$. Notice it's easier to rotate the points that lie on the axes, and these help us find the image of $A$: Point$\left(3,0\right)$$\left(0,4\right)$$\left(3,4\right)$ Image$\left(0,3\right)$$\left(-4,0\right)$$\left(-4,3\right)$ ### Solution method 2: The algebraic approach Let's take a closer look at points $A$ and ${A}^{\prime }$: Point$x$-coordinate$y$-coordinate $A$$3$$4$ ${A}^{\prime }$$-4$$3$ Notice an interesting phenomenon: The $x$-coordinate of $A$ became the $y$-coordinate of ${A}^{\prime }$, and the opposite of the $y$-coordinate of $A$ became the $x$-coordinate of ${A}^{\prime }$. We can represent this mathematically as follows: ${R}_{\left(0,0\right),{90}^{\circ }}\left(x,y\right)=\left(-y,x\right)$ It turns out that this is true for any point, not just our $A$. Here are a few more examples: Furthermore, it turns out that rotations by ${180}^{\circ }$ or $-{90}^{\circ }$ follow similar patterns: ${R}_{\left(0,0\right),{180}^{\circ }}\left(x,y\right)=\left(-x,-y\right)$ ${R}_{\left(0,0\right),-{90}^{\circ }}\left(x,y\right)=\left(y,-x\right)$ We can use these to rotate any point we want by plugging its coordinates in the appropriate equation. #### Problem 1 Draw the image of $B\left(-7,-3\right)$ under the rotation ${R}_{\left(0,0\right),-{90}^{\circ }}$. #### Problem 2 Draw the image of $C\left(5,-6\right)$ under the rotation ${R}_{\left(0,0\right),{180}^{\circ }}$. ### Graphical method vs. algebraic method In general, everyone is free to choose which of the two methods to use. Different strokes for different folks! The algebraic method takes less work and less time, but you need to remember those patterns. The graphical method is always at your disposal, but it might take you longer to solve. ## Part 2: Extending to any multiple of ${90}^{\circ }$‍ ### Let's study an example problem We want to find the image ${D}^{\prime }$ of the point $D\left(-5,4\right)$ under a rotation by ${270}^{\circ }$ about the origin. ### Solution Since rotating by ${270}^{\circ }$ is the same as rotating by ${90}^{\circ }$ three times, we can solve this graphically by performing three consecutive ${90}^{\circ }$ rotations: But wait! We could just rotate by $-{90}^{\circ }$ instead of ${270}^{\circ }$. These rotations are equivalent. Check it out: For the same reason, we can also use the pattern ${R}_{\left(0,0\right),-{90}^{\circ }}\left(x,y\right)=\left(y,-x\right)$: ${R}_{\left(0,0\right),{270}^{\circ }}\left(-5,4\right)=\left(4,5\right)$ ### Let's study one more example problem We want to find the image of $\left(-9,-7\right)$ under a rotation by ${810}^{\circ }$ about the origin. ### Solution A rotation by ${810}^{\circ }$ is the same as two consecutive rotations by ${360}^{\circ }$ followed by a rotation by ${90}^{\circ }$ (because $810=2\cdot 360+90$). A rotation by ${360}^{\circ }$ maps every point onto itself. In other words, it doesn't change anything. So a rotation by ${810}^{\circ }$ is the same as a rotation by ${90}^{\circ }$. Therefore, we can simply use the pattern ${R}_{\left(0,0\right),{90}^{\circ }}\left(x,y\right)=\left(-y,x\right)$: ${R}_{\left(0,0\right),{810}^{\circ }}\left(-9,-7\right)=\left(7,-9\right)$ #### Problem 1 Draw the image of $E\left(8,6\right)$ under the rotation ${R}_{\left(0,0\right),-{270}^{\circ }}$. #### Problem 2 Which rotation is equivalent to the rotation ${R}_{\left(0,0\right),-{990}^{\circ }}$? ## Part 3: Rotating polygons ### Let's study an example problem Consider quadrilateral $DEFG$ drawn below. Let's draw its image, ${D}^{\prime }{E}^{\prime }{F}^{\prime }{G}^{\prime }$, under the rotation ${R}_{\left(0,0\right),{270}^{\circ }}$. ### Solution Similar to translations, when we rotate a polygon, all we need is to perform the rotation on all of the vertices, and then we can connect the images of the vertices to find the image of the polygon. Draw the image of $\mathrm{△}HIJ$ below, under the rotation ${R}_{\left(0,0\right),{90}^{\circ }}$. ## Want to join the conversation? • im confused i dont get this Rotating by 90 degrees: If you have a point on (2, 1) and rotate it by 90 degrees, it will end up at (-1, 2) When you rotate by 90 degrees, you take your original X and Y, swap them, and make Y negative. So from 0 degrees you take (x, y), swap them, and make y negative (-y, x) and then you have made a 90 degree rotation. What if we rotate another 90 degrees? Same thing. Our point is at (-1, 2) so when we rotate it 90 degrees, it will be at (-2, -1) X and Y swaps, and Y becomes negative. What about 90 degrees again? Same thing! But remember that a negative and a negative gives a positive so when we swap X and Y, and make Y negative, Y actually becomes positive. Our point is as (-2, -1) so when we rotate it 90 degrees, it will be at (1, -2) Another 90 degrees will bring us back where we started. From (1, -2) to (2, 1) Rotating by -90 degrees: If you understand everything so far, then rotating by -90 degrees should be no issue for you. We do the same thing, except X becomes a negative instead of Y. So from (x, y) to (y, -x) Rotating by 180 degrees: If you have a point on (2, 1) and rotate it by 180 degrees, it will end up at (-2, -1) When you rotate by 180 degrees, you take your original x and y, and make them negative. So from 0 degrees you take (x, y) and make them negative (-x, -y) and then you've made a 180 degree rotation. Remember! A negative and a negative gives a positive! So if we rotate another 180 degrees we go from (-2, -1) to (2, 1) And if we have another point like (-3, 2) and rotate it 180 degrees, it will end up on (3, -2) • i am actually at the end of my rope this is my downfall • real this makes me pluck out my eyebrows • I made a summary of the formulas if anyone dont want to use the visualize method. R=rotateR(0,0)90°/-270°[counterclock wise/clockwise](x,y)➔(-y,x)R(0,0)180°/-180°[counterclock wise/clock wise](x,y)➔(-x,-y)R(0,0)-90°/270°[clock wise/counterclock wise](x,y)➔(y,-x)R(0,0)360°/-360°(x,y)=(x,y) In all honesty, there are only two that need to be memorized. R 90°/-270°and R -90°/270°since depends on the direction it rotates the x and y value sometimes turns into negative, but 100% of the time the x and y swaps. The rest of two you just need to know how it works, R 180°/-180 is just reflected through origin so you dont have to swap x and y, all you have to do is add - sign to them. And R360°/-360° is just rotating back to where it was, so nothing is changed. • Anyone have any tips for visualization? I am having a really hard time seeing a triangle and where the point should go in my head. Any suggestions are appreciated very much! Thank you, Clarebugg • you can try thinking of it as a mountain the bottom is the 2 points that stretch out and the top is the peak. there are many different ways to think about it. you are problem-solving by trying to visualize. • For the first problem, why does the solution say a rotation of 90 degrees when its asking for -270 • A rotation of 90 degrees is the same thing as -270 degrees. They are called coterminal angles. A full way around a circle is 360 degrees, right? So you can find an angle by adding 360. -270 + 360 = 90 Hope this helps! • -90 and 270: (y, -x) reverse, neg x 90 and -270 (-y,x) reverse, neg y (-)180 (-x,-y) easiest to remember - just negify everything • i think i understand and then i don't understand.... • Amazing. Just Amazing

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